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Gabarito completo anton_calculo_8ed_caps_01_08

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Gabarito completo anton_calculo_8ed_caps_01_08Document Transcript

• 1CHAPTER 1FunctionsEXERCISE SET 1.11. (a) −2.9, −2.0, 2.35, 2.9 (b) none (c) y = 0(d) −1.75 ≤ x ≤ 2.15 (e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.22. (a) x = −1, 4 (b) none (c) y = −1(d) x = 0, 3, 5 (e) ymax = 9 at x = 6; ymin = −2 at x = 03. (a) yes (b) yes(c) no (vertical line test fails) (d) no (vertical line test fails)4. (a) The natural domain of f is x = −1, and for g it is the set of all x. Whenever x = −1,f(x) = g(x), but they have diﬀerent domains.(b) The domain of f is the set of all x ≥ 0; the domain of g is the same.5. (a) around 1943 (b) 1960; 4200(c) no; you need the year’s population (d) war; marketing techniques(e) news of health risk; social pressure, antismoking campaigns, increased taxation6. (a) around 1983 (b) 1966(c) the former (d) no, it appears to be levelling out7. (a) 1999, \$34,400 (b) 1985, \$37,000(c) second year; graph has a larger (negative) slope8. (a) In thousands, approximately43.2 − 37.86=5.46per yr, or \$900/yr(b) The median income during 1993 increased from \$37.8K to \$38K (K for ’kilodollars’; all ﬁguresapproximate). During 1996 it increased from \$40K to \$42K, and during 1999 it decreasedslightly from \$43.2K to \$43.1K. Thus the average rate of change measured on January 1 was(40 - 37.8)/3 for the ﬁrst three-yr period and (43.2 - 40)/3 for the second-year period, andhence the median income as measured on January 1 increased more rapidly in the secondthree-year period. Measured on December 31, however, the numbers are (42 - 38)/3 and (43.1- 42)/3, and the former is the greater number. Thus the answer to the question depends onwhere in the year the median income is measured.(c) 19939. (a) f(0) = 3(0)2−2 = −2; f(2) = 3(2)2−2 = 10; f(−2) = 3(−2)2−2 = 10; f(3) = 3(3)2−2 = 25;f(√2) = 3(√2)2− 2 = 4; f(3t) = 3(3t)2− 2 = 27t2− 2(b) f(0) = 2(0) = 0; f(2) = 2(2) = 4; f(−2) = 2(−2) = −4; f(3) = 2(3) = 6; f(√2) = 2√2;f(3t) = 1/3t for t > 1 and f(3t) = 6t for t ≤ 1.10. (a) g(3) =3 + 13 − 1= 2; g(−1) =−1 + 1−1 − 1= 0; g(π) =π + 1π − 1; g(−1.1) =−1.1 + 1−1.1 − 1=−0.1−2.1=121;g(t2− 1) =t2− 1 + 1t2 − 1 − 1=t2t2 − 2(b) g(3) =√3 + 1 = 2; g(−1) = 3; g(π) =√π + 1; g(−1.1) = 3; g(t2− 1) = 3 if t2< 2 andg(t2− 1) =√t2 − 1 + 1 = |t| if t2≥ 2.
• 2 Chapter 111. (a) x = 3 (b) x ≤ −√3 or x ≥√3(c) x2− 2x + 5 = 0 has no real solutions so x2− 2x + 5 is always positive or always negative. Ifx = 0, then x2− 2x + 5 = 5 > 0; domain: (−∞, +∞).(d) x = 0 (e) sin x = 1, so x = (2n + 12 )π, n = 0, ±1, ±2, . . .12. (a) x = −75(b) x − 3x2must be nonnegative; y = x − 3x2is a parabola that crosses the x-axis at x = 0, 13and opens downward, thus 0 ≤ x ≤ 13(c)x2− 4x − 4> 0, so x2− 4 > 0 and x − 4 > 0, thus x > 4; or x2− 4 < 0 and x − 4 < 0, thus−2 < x < 2(d) x = −1 (e) cos x ≤ 1 < 2, 2 − cos x > 0, all x13. (a) x ≤ 3 (b) −2 ≤ x ≤ 2 (c) x ≥ 0 (d) all x (e) all x14. (a) x ≥ 23 (b) −32 ≤ x ≤ 32 (c) x ≥ 0 (d) x = 0 (e) x ≥ 015. (a) Breaks could be caused by war, pestilence, ﬂood, earthquakes, for example.(b) C decreases for eight hours, takes a jump upwards, and then repeats.16. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperaturechange.(b) No; the number is always an integer, so the changes are in movements (jumps) of at leastone unit.17.th 18. Tt19. (a) x = 2, 4 (b) none (c) x ≤ 2; 4 ≤ x (d) ymin = −1; no maximum value20. (a) x = 9 (b) none (c) x ≥ 25 (d) ymin = 1; no maximum value21. The cosine of θ is (L − h)/L (side adjacent over hypotenuse), so h = L(1 − cos θ).22. The sine of θ/2 is (L/2)/10 (side opposite over hypotenuse), so that L = 20 sin(θ/2).23. (a) If x < 0, then |x| = −x so f(x) = −x + 3x + 1 = 2x + 1. If x ≥ 0, then |x| = x sof(x) = x + 3x + 1 = 4x + 1;f(x) =2x + 1, x < 04x + 1, x ≥ 0(b) If x < 0, then |x| = −x and |x − 1| = 1 − x so g(x) = −x + 1 − x = 1 − 2x. If 0 ≤ x < 1, then|x| = x and |x − 1| = 1 − x so g(x) = x + 1 − x = 1. If x ≥ 1, then |x| = x and |x − 1| = x − 1so g(x) = x + x − 1 = 2x − 1;g(x) =1 − 2x, x < 01, 0 ≤ x < 12x − 1, x ≥ 1
• Exercise Set 1.1 324. (a) If x < 5/2, then |2x − 5| = 5 − 2x so f(x) = 3 + (5 − 2x) = 8 − 2x. If x ≥ 5/2, then|2x − 5| = 2x − 5 so f(x) = 3 + (2x − 5) = 2x − 2;f(x) =8 − 2x, x < 5/22x − 2, x ≥ 5/2(b) If x < −1, then |x − 2| = 2 − x and |x + 1| = −x − 1 so g(x) = 3(2 − x) − (−x − 1) = 7 − 2x.If −1 ≤ x < 2, then |x − 2| = 2 − x and |x + 1| = x + 1 so g(x) = 3(2 − x) − (x + 1) = 5 − 4x.If x ≥ 2, then |x − 2| = x − 2 and |x + 1| = x + 1 so g(x) = 3(x − 2) − (x + 1) = 2x − 7;g(x) =7 − 2x, x < −15 − 4x, −1 ≤ x < 22x − 7, x ≥ 225. (a) V = (8 − 2x)(15 − 2x)x(b) 0 ≤ x ≤ 4(c) 0 ≤ V ≤ 91(d) As x increases, V increases and then decreases; themaximum value could be approximated by zooming inon the graph.10000 426. (a) V = (6 − 2x)2x(b) 0 < x < 3(c) 0 < V < 16(d) A x increases, V increases and then decreases; themaximum value occurs somewhere on 0 < x < 3,and can be approximated by zooming with a graphingcalculator.2000 327. (a) The side adjacent to the buildinghas length x, so L = x + 2y.(b) A = xy = 1000, so L = x + 2000/x.(c) all x = 01208020 80(d) L ≈ 89.44 ft28. (a) x = 3000 tan θ(b) θ = nπ + π/2 for any integer n,−∞ < n < ∞(c) 3000 ft600000 629. (a) V = 500 = πr2h so h =500πr2. ThenC = (0.02)(2)πr2+ (0.01)2πrh = 0.04πr2+ 0.02πr500πr2= 0.04πr2+10r; Cmin ≈ 4.39 cents at r ≈ 3.4 cm,h ≈ 13.8 cm741.5 6
• 4 Chapter 1(b) C = (0.02)(2)(2r)2+ (0.01)2πrh = 0.16r2+10r. Since0.04π < 0.16, the top and bottom now get more weight.Since they cost more, we diminish their sizes in thesolution, and the cans become taller.741.5 5.5(c) r ≈ 3.1 cm, h ≈ 16.0 cm, C ≈ 4.76 cents30. (a) The length of a track with straightaways of length L and semicircles of radius r isP = (2)L + (2)(πr) ft. Let L = 360 and r = 80 to get P = 720 + 160π = 1222.65 ft.Since this is less than 1320 ft (a quarter-mile), a solution is possible.(b) P = 2L + 2πr = 1320 and 2r = 2x + 160, soL = 12 (1320 − 2πr) = 12 (1320 − 2π(80 + x))= 660 − 80π − πx.45000 100(c) The shortest straightaway is L = 360, so x = 15.49 ft.(d) The longest straightaway occurs when x = 0,so L = 660 − 80π = 408.67 ft.31. (i) x = 1, −2 causes division by zero (ii) g(x) = x + 1, all x32. (i) x = 0 causes division by zero (ii) g(x) =√x + 1 for x ≥ 033. (a) 25◦F (b) 13◦F (c) 5◦F34. If v = 48 then −60 = WCT ≈ 1.4157T − 30.6763; thus T ≈ −21◦F when WCT = −60.35. As in the previous exercise, WCT≈ 1.4157T − 30.6763; thus T ≈ 15◦F when WCT = −10.36. The WCT is given by two formulae, but the ﬁrst doesn’t work with the data. Hence−5 = WCT = −27.2v0.16+ 48.17 and v ≈ 66◦F.EXERCISE SET 1.21. (e) seems best,though only (a) is bad.–0.50.5y–1 1x2. (e) seems best, thoughonly (a) is bad and (b)is not good.–0.50.5y–1 1x3. (b) and (c) are good;(a) is very bad.121314y–1 1x
• Exercise Set 1.2 54. (b) and (c) are good;(a) is very bad.–14–13–12–11y–2 –1 0 1 2x5. [−3, 3] × [0, 5]2yx6. [−4, 2] × [0, 3]–2yx7. (a) window too narrow, too short(b) window wide enough, but too short(c) good window, good spacing–400–200y–5 5 10 20x(d) window too narrow, too short(e) window too narrow, too short8. (a) window too narrow(b) window too short(c) good window, good tick spacing–250–10050y–16 –8 –4 4x(d) window too narrow, too short(e) shows one local minimum only,window too narrow, too short9. [−5, 14] × [−60, 40]–602040y–5 5 10x10. [6, 12] × [−100, 100]–50100y10 12x11. [−0.1, 0.1] × [−3, 3]–223y–0.1x0.112. [−1000, 1000] × [−13, 13]–5510y–1000x100013. [−250, 1050] × [−1500000, 600000]–500000y–1000 1000x
• 6 Chapter 114. [−3, 20] × [−3500, 3000]–2000–100010005 10 15xy15. [−2, 2] × [−20, 20]–20–101020yx–2 –1 1 216. [1.6, 2] × [0, 2]0.512y1.6 1.7 1.8 2x17. depends on graphing utility–2246y–4 2 4x18. depends on graphing utility–2246y–4 –2 2 4x19. (a) f(x) =√16 − x213–2 2 4xy(b) f(x) = −√16 − x2–3–1xy–4 –2 2 4(c)–22xy–2 2(d)141 4yx(e) No; the vertical linetest fails.20. (a) y = ±3 1 − x2/4–33–2 2xy(b) y = ±√x2 + 1–4–224–4 –2 2 4xy
• Exercise Set 1.2 721. (a)–1 1xy1(b)1x12y(c)–1 1x–1y (d)2cc1yx(e)c1yxC–1(f)1x1–1y–122.1xy–1 123. The portions of the graph of y = f(x) which lie below the x-axis are reﬂected over the x-axis togive the graph of y = |f(x)|.24. Erase the portion of the graph of y = f(x) which lies in the left-half plane and replace it with thereﬂection over the y-axis of the portion in the right-half plane (symmetry over the y-axis) and youobtain the graph of y = f(|x|).25. (a) for example, let a = 1.1ayx(b)123y1 2 3x26. They are identical.xy27.51015y–1 1 2 3x
• 8 Chapter 128. This graph is very complex. We show three views, small (near the origin), medium and large:(a) yx–1–1 2(b)–50y x10(c)1000y40x29. (a)1y–3 –1 1 3x(b)–0.50.51–3 –1 1 3xy(c)0.51.5y–1 1 2 3x(d)0.51–2 –1 1xy30. y12x31. (a) stretches or shrinks thegraph in the y-direction;reﬂects it over the x-axisif c changes sign–224c = 2c = 1c = –1.5–1 1xy(b) As c increases, the lowerpart of the parabolamoves down and to theleft; as c decreases, themotion is down and tothe right.28–1 2xc = 2c = 1c = –1.5y(c) The graph rises or fallsin the y-direction withchanges in c.4268–2 1 2xc = 2c = 0.5c = –1yAs c increases, the lower part of the parabola moves down and to the left; as c decreases, themotion is down and to the right
• Exercise Set 1.3 932. (a) y–221 2x(b) x-intercepts at x = 0, a, b. Assume a < b and let a approach b. The two branches of thecurve come together. If a moves past b then a and b switch roles.–3–1131 2 3xa = 0.5b = 1.5y–3–1131 2 3xa = 1b = 1.5y–3–1132 3xa = 1.5b = 1.6y33. The curve oscillates betweenthe lines y = x and y = −x withincreasing rapidity as |x| increases.–30–101030y–30 –10 10 30x34. The curve oscillates between the linesy = +1 and y = −1, inﬁnitely manytimes in any neighborhood of x = 0.y–14–2xEXERCISE SET 1.31. (a)–101y–1 1 2x(b)21y1 2 3x(c) 1y–1 1 2x(d)2y–4 –2 2x
• 10 Chapter 12. (a) xy1–2(b)x1y1(c)x1 3y1(d)x1y1–13. (a)–11yx–2 –1 1 2(b)–11y1x(c)–11 y–1 1 2 3x(d)–11y–1 1 2 3x4.–11 y–1 1 2 3x5. Translate left 1 unit, stretch vertically bya factor of 2, reﬂect over x-axis, translatedown 3 units.–60–20–6 –2 2 6xy6. Translate right 3 units,compress vertically bya factor of 12 , andtranslate up 2 units.yx247. y = (x + 3)2− 9;translate left 3 unitsand down 9 units.–51015–8 –4 –2 2xy8. y = (x + 3)2− 19;translate left 3 unitsand down 19 units.yx–5–4
• Exercise Set 1.3 119. y = −(x − 1)2+ 2;translate right 1 unit,reﬂect over x-axis,translate up 2 units.–6–22–2 –1 1 3xy10. y = 12 [(x − 1)2+ 2];translate right 1 unitand up 2 units,compress verticallyby a factor of 12 .yx12111. Translate left 1 unit,reﬂect over x-axis,translate up 3 units.122 8xy12. Translate right 4 unitsand up 1 unit.yx144 1013. Compress verticallyby a factor of 12 ,translate up 1 unit.2y1 2 3x14. Stretch vertically bya factor of√3 andreﬂect over x-axis.y x–1215. Translate right 3 units.–1010y4 6x16. Translate right 1 unitand reﬂect over x-axis.yx–42–2 417. Translate left 1 unit,reﬂect over x-axis,translate up 2 units.–6–26y–3 1 2x18. y = 1 − 1/x;reﬂect over x-axis,translate up 1 unit.yx–55219. Translate left 2 unitsand down 2 units.–2–4 –2xy
• 12 Chapter 120. Translate right 3 units, reﬂectover x-axis, translate up 1 unit.yx–11521. Stretch vertically by a factor of 2,translate right 1/2 unit and up 1 unit.yx24222. y = |x − 2|; translate right 2 units.yx122 423. Stretch vertically by a factor of 2,reﬂect over x-axis, translate up 1 unit.–113–2 2xy24. Translate right 2 unitsand down 3 units.y x–2225. Translate left 1 unitand up 2 units.13y–3 –1 1x26. Translate right 2 units,reﬂect over x-axis.yx–11427. (a) 2y–1 1x(b) y =0 if x ≤ 02x if 0 < x28. yx–5229. (f + g)(x) = 3√x − 1, x ≥ 1; (f − g)(x) =√x − 1, x ≥ 1; (fg)(x) = 2x − 2, x ≥ 1;(f/g)(x) = 2, x > 130. (f + g)(x) = (2x2+ 1)/[x(x2+ 1)], all x = 0; (f − g)(x) = −1/[x(x2+ 1)], all x = 0; (fg)(x) =1/(x2+ 1), all x = 0; (f/g)(x) = x2/(x2+ 1), all x = 031. (a) 3 (b) 9 (c) 2 (d) 232. (a) π − 1 (b) 0 (c) −π2+ 3π − 1 (d) 1
• Exercise Set 1.3 1333. (a) t4+ 1 (b) t2+ 4t + 5 (c) x2+ 4x + 5 (d)1x2+ 1(e) x2+ 2xh + h2+ 1 (f) x2+ 1 (g) x + 1 (h) 9x2+ 134. (a)√5s + 2 (b)√x + 2 (c) 3√5x (d) 1/√x(e) 4√x (f) 0 (g) 1/ 4√x (h) |x − 1|35. (f ◦ g)(x) = 1 − x, x ≤ 1; (g ◦ f)(x) =√1 − x2, |x| ≤ 136. (f ◦ g)(x) =√x2 + 3 − 3, |x| ≥√6; (g ◦ f)(x) =√x, x ≥ 337. (f ◦ g)(x) =11 − 2x, x =12, 1; (g ◦ f)(x) = −12x−12, x = 0, 138. (f ◦ g)(x) =xx2 + 1, x = 0; (g ◦ f)(x) =1x+ x, x = 039. x−6+ 1 40.xx + 141. (a) g(x) =√x, h(x) = x + 2 (b) g(x) = |x|, h(x) = x2− 3x + 542. (a) g(x) = x + 1, h(x) = x2(b) g(x) = 1/x, h(x) = x − 343. (a) g(x) = x2, h(x) = sin x (b) g(x) = 3/x, h(x) = 5 + cos x44. (a) g(x) = 3 sin x, h(x) = x2(b) g(x) = 3x2+ 4x, h(x) = sin x45. (a) f(x) = x3, g(x) = 1 + sin x, h(x) = x2(b) f(x) =√x, g(x) = 1 − x, h(x) = 3√x46. (a) f(x) = 1/x, g(x) = 1 − x, h(x) = x2(b) f(x) = |x|, g(x) = 5 + x, h(x) = 2x47. yx–4–22–2 248. {−2, −1, 0, 1, 2, 3}49. Note thatf(g(−x)) = f(−g(x)) = f(g(x)),so f(g(x)) is even.f (g(x))xy1–3 –1–1–3150. Note that g(f(−x)) = g(f(x)),so g(f(x)) is even.xy–1–213–113–3g( f(x))
• 14 Chapter 151. f(g(x)) = 0 when g(x) = ±2, so x = ±1.4; g(f(x)) = 0 when f(x) = 0, so x = ±2.52. f(g(x)) = 0 at x = −1 and g(f(x)) = 0 at x = −153.3w2− 5 − (3x2− 5)w − x=3(w − x)(w + x)w − x= 3w + 3x3(x + h)2− 5 − (3x2− 5)h=6xh + 3h2h= 6x + 3h;54.w2+ 6w − (x2+ 6x)w − x= w + x + 6(x + h)2+ 6(x + h) − (x2+ 6x)h=2xh + h2+ 6hh= 2x + h + 6;55.1/w − 1/xw − x=x − wwx(w − x)= −1xw;1/(x + h) − 1/xh=x − (x + h)xh(x + h)=−1x(x + h)56.1/w2− 1/x2w − x=x2− w2x2w2(w − x)= −x + wx2w2;1/(x + h)2− 1/x2h=x2− (x + h)2x2h(x + h)2= −2x + hx2(x + h)257. neither; odd; even58. (a) x −3 −2 −1 0 1 2 3f(x) 1 −5 −1 0 −1 −5 1(b) x −3 −2 −1 0 1 2 3f(x) 1 5 −1 0 1 −5 −159. (a)xy (b)xy(c)xy60. (a)xy (b)xy61. (a) even (b) odd (c) odd (d) neither62. (a) the origin (b) the x-axis (c) the y-axis (d) none
• Exercise Set 1.3 1563. (a) f(−x) = (−x)2= x2= f(x), even (b) f(−x) = (−x)3= −x3= −f(x), odd(c) f(−x) = | − x| = |x| = f(x), even (d) f(−x) = −x + 1, neither(e) f(−x) =(−x)5− (−x)1 + (−x)2= −x5− x1 + x2= −f(x), odd(f) f(−x) = 2 = f(x), even64. (a) x-axis, because x = 5(−y)2+ 9 gives x = 5y2+ 9(b) x-axis, y-axis, and origin, because x2− 2(−y)2= 3, (−x)2− 2y2= 3, and(−x)2− 2(−y)2= 3 all give x2− 2y2= 3(c) origin, because (−x)(−y) = 5 gives xy = 565. (a) y-axis, because (−x)4= 2y3+ y gives x4= 2y3+ y(b) origin, because (−y) =(−x)3 + (−x)2gives y =x3 + x2(c) x-axis, y-axis, and origin because (−y)2= |x| − 5, y2= | − x| − 5, and (−y)2= | − x| − 5 allgive y2= |x| − 566. (a) x-axis(b) origin, both axes(c) origin67. (a) y-axis(b) none(c) origin, both axes68. 3–3–4 469. 2–2–3 370. (a) Whether we replace x with −x, y with −y, or both, we obtain the same equation, so byTheorem 1.3.3 the graph is symmetric about the x-axis, the y-axis and the origin.(b) y = (1 − x2/3)3/2(c) For quadrant II, the same; for III and IV use y = −(1 − x2/3)3/2. (For graphing it may behelpful to use the tricks that precede Exercise 29 in Section 1.2.)71.25y1 2x72. yx2273. (a)1yCxO c o(b)2yOxC c o
• 16 Chapter 174. (a)x1y1–1(b)x1y1–1 ‚2–‚2 ‚3(c)x1y–13(d)xy1c/2C75. Yes, e.g. f(x) = xkand g(x) = xnwhere k and n are integers.76. If x ≥ 0 then |x| = x and f(x) = g(x). If x < 0 then f(x) = |x|p/qif p is even and f(x) = −|x|p/qif p is odd; in both cases f(x) agrees with g(x).EXERCISE SET 1.41. (a) y = 3x + b (b) y = 3x + 6 (c)–1010y–2 2xy = 3x + 6y = 3x + 2y = 3x – 42. Since the slopes are negative reciprocals, y = −13 x + b.3. (a) y = mx + 2(b) m = tan φ = tan 135◦= −1, so y = −x + 2 (c)1345–2 1 2xym = –1 m = 1m = 1.54. (a) y = mx (b) y = m(x − 1)(c) y = −2 + m(x − 1) (d) 2x + 4y = C5. Let the line be tangent to the circle at the point (x0, y0) where x20 + y20 = 9. The slope of thetangent line is the negative reciprocal of y0/x0 (why?), so m = −x0/y0 and y = −(x0/y0)x + b.Substituting the point (x0, y0) as well as y0 = ± 9 − x20 we get y = ±9 − x0x9 − x20.
• Exercise Set 1.4 176. Solve the simultaneous equations to get the point (−2, 1/3) of intersection. Then y = 13 +m(x+2).7. The x-intercept is x = 10 so that with depreciation at 10% per year the ﬁnal value is always zero,and hence y = m(x − 10). The y-intercept is the original value.y2 6 10x8. A line through (6, −1) has the form y + 1 = m(x − 6). The intercepts are x = 6 + 1/m andy = −6m − 1. Set −(6 + 1/m)(6m + 1) = 3, or 36m2+ 15m + 1 = (12m + 1)(3m + 1) = 0 withroots m = −1/12, −1/3; thus y + 1 = −(1/3)(x − 6) and y + 1 = −(1/12)(x − 6).9. (a) The slope is −1.–3–25–1 1 2xy(b) The y-intercept is y = −1.–424–1 1xy(c) They pass through the point (−4, 2).yx–226–6 –4(d) The x-intercept is x = 1.–3–1131 2xy10. (a) horizontal linesxy(b) The y-intercept is y = −1/2.xy– 222
• 18 Chapter 1(c) The x-intercept is x = −1/2.xy11(d) They pass through (−1, 1).xy–211(–1, 1)11. (a) VI (b) IV (c) III (d) V (e) I (f) II12. In all cases k must be positive, or negative values would appear in the chart. Only kx−3decreases,so that must be f(x). Next, kx2grows faster than kx3/2, so that would be g(x), which grows fasterthan h(x) (to see this, consider ratios of successive values of the functions). Finally, experimenta-tion (a spreadsheet is handy) for values of k yields (approximately) f(x) = 10x−3, g(x) = x2/2,h(x) = 2x1.5.13. (a)–30–101030–2 1 2xy–40–10–2 1 2xy(b)–224y–42 4x610y–4 2 4x(c)–212–1 2 3xy1y1 2 3x
• Exercise Set 1.4 1914. (a)xy240–40xy240–280(b)xy24–2–4xy2–2–2–4(c)xy–1–24–3–2 2 4–2xy15. (a)–10–5510y–2 2x(b)–246y–2 1 2x(c)–10–5510y–2 2x
• 20 Chapter 116. (a)xy23(b)xy22(c)xy2217. (a)26y–3 –1 1x(b)–80–202080y1 2 3 4 5x(c)–40–10y–3x(d)–40–202040y21 3 4 5x18. (a)xy11(b)–2–6 2–2–112xy(c)xy24–2–4(d)xy–24–619. (a)–10.51.5y–3 –2 1x(b)–1–0.50.51y1 3x
• Exercise Set 1.4 21(c)–201030y–1 1 3x(d)–201030y–2 1 2x20. (a)–1010y1 3 4x(b)48y–5 3x(c)–106106y–1 1x(d)510y–2 2x21. y = x2+ 2x = (x + 1)2− 1510y–2 2x22. (a)1y–2 2x(b)1y–2 2x23. (a) N·m (b) k = 20 N·m(c) V (L) 0.25 0.5 1.0 1.5 2.0P (N/m2) 80 × 10340 × 10320 × 10313.3 × 10310 × 103(d)102030P(N/m2)10 20V(m3)
• 22 Chapter 124. If the side of the square base is x and the height of the container is y then V = x2y = 100; minimizeA = 2x2+ 4xy = 2x2+ 400/x. A graphing utility with a zoom feature suggests that the solutionis a cube of side 10013 cm.25. (a) F = k/x2so 0.0005 = k/(0.3)2and k = 0.000045 N·m2.(b) F = 0.000005 N(c)t5 10F10–5(d) When they approach one another, the force becomes inﬁnite; when they get far apart it tendsto zero.26. (a) 2000 = C/(4000)2, so C = 3.2 × 1010lb·mi2(b) W = C/50002= (3.2 × 1010)/(25 × 106) = 1280 lb.(c)500015000W2000 8000x(d) No, but W is very small when x is large.27. (a) II; y = 1, x = −1, 2 (b) I; y = 0, x = −2, 3(c) IV; y = 2 (d) III; y = 0, x = −228. The denominator has roots x = ±1, so x2− 1 is the denominator. To determine k use the point(0, −1) to get k = 1, y = 1/(x2− 1).29. (a) y = 3 sin(x/2)(b) y = 4 cos 2x(c) y = −5 sin 4x30. (a) y = 1 + cos πx(b) y = 1 + 2 sin x(c) y = −5 cos 4x31. (a) y = sin(x + π/2)(b) y = 3 + 3 sin(2x/9)(c) y = 1 + 2 sin(2(x − π/4))32. V = 120√2 sin(120πt)33. (a) 3, π/2yx–2136(b) 2, 2–222 4xy(c) 1, 4πyx1232c 4c 6c
• Exercise Set 1.5 2334. (a) 4, πyx–4–223 9 f l(b) 1/2, 2π/3yx–0.20.48c(c) 4, 6πyx–4–224i15c221c235. (a) A sin(ωt + θ) = A sin(ωt) cos θ + A cos(ωt) sin θ = A1 sin(ωt) + A2 cos(ωt)(b) A1 = A cos θ, A2 = A sin θ, so A = A21 + A22 and θ = tan−1(A2/A1).(c) A = 5√13/2, θ = tan−1 12√3;x =5√132sin 2πt + tan−1 12√310–10^ 636. three; x = 0, x = ±1.89553–3–3 3EXERCISE SET 1.51. (a) f(g(x)) = 4(x/4) = x, g(f(x)) = (4x)/4 = x, f and g are inverse functions(b) f(g(x)) = 3(3x − 1) + 1 = 9x − 2 = x so f and g are not inverse functions(c) f(g(x)) = 3(x3 + 2) − 2 = x, g(f(x)) = (x − 2) + 2 = x, f and g are inverse functions(d) f(g(x)) = (x1/4)4= x, g(f(x)) = (x4)1/4= |x| = x, f and g are not inverse functions2. (a) They are inverse functions.2–2–2 2(b) The graphs are not reﬂections ofeach other about the line y = x.2–2–2 2
• 24 Chapter 1(c) They are inverse functions providedthe domain of g is restricted to [0, +∞)500 5(d) They are inverse functions providedthe domain of f(x) is restrictedto [0, +∞)200 23. (a) yes (b) yes (c) no (d) yes (e) no (f) no4. (a) no, the horizontal line test fails6–2–3 3(b) yes, by the horizontal line test10–10–1 35. (a) yes; all outputs (the elements of row two) are distinct(b) no; f(1) = f(6)6. (a) Since the point (0, 0) lies on the graph, no other point on the line x = 0 can lie on the graph,by the vertical line test. Thus the hour hand cannot point straight up or straight down. Thusnoon, midnight, 6AM and 6PM are impossible. To show other times are possible, suppose(a, b) lies on the graph with a = 0. Then the function y = Ax1/3passes through (0, 0) and(a, b) provided A = b/a1/3.(b) same as (a)(c) Then the minute hand cannot point to 6 or 12, so in addition to (a), times of the form 1:00,1:30, 2:00, 2:30, , 12:30 are also impossible.7. (a) f has an inverse because the graph passes the horizontal line test. To compute f−1(2) startat 2 on the y-axis and go to the curve and then down, so f−1(2) = 8; similarly, f−1(−1) = −1and f−1(0) = 0.(b) domain of f−1is [−2, 2], range is [−8, 8] (c)–2 1 2–8–448yx8. (a) the horizontal line test fails(b) −3 < x ≤ −1; −1 ≤ x ≤ 2; and 2 ≤ x < 4.9. y = f−1(x), x = f(y) = 7y − 6, y =17(x + 6) = f−1(x)
• Exercise Set 1.5 2510. y = f−1(x), x = f(y) =y + 1y − 1, xy − x = y + 1, (x − 1)y = x + 1, y =x + 1x − 1= f−1(x)11. y = f−1(x), x = f(y) = 3y3− 5, y = 3(x + 5)/3 = f−1(x)12. y = f−1(x), x = f(y) = 5√4y + 2, y =14(x5− 2) = f−1(x)13. y = f−1(x), x = f(y) = 3√2y − 1, y = (x3+ 1)/2 = f−1(x)14. y = f−1(x), x = f(y) =5y2 + 1, y =5 − xx= f−1(x)15. y = f−1(x), x = f(y) = 3/y2, y = − 3/x = f−1(x)16. y = f−1(x), x = f(y) =2y, y ≤ 0y2, y > 0, y = f−1(x) =x/2, x ≤ 0√x, x > 017. y = f−1(x), x = f(y) =5/2 − y, y < 21/y, y ≥ 2, y = f−1(x) =5/2 − x, x > 1/21/x, 0 < x ≤ 1/218. y = p−1(x), x = p(y) = y3− 3y2+ 3y − 1 = (y − 1)3, y = x1/3+ 1 = p−1(x)19. y = f−1(x), x = f(y) = (y + 2)4for y ≥ 0, y = f−1(x) = x1/4− 2 for x ≥ 1620. y = f−1(x), x = f(y) =√y + 3 for y ≥ −3, y = f−1(x) = x2− 3 for x ≥ 021. y = f−1(x), x = f(y) = −√3 − 2y for y ≤ 3/2, y = f−1(x) = (3 − x2)/2 for x ≤ 022. y = f−1(x), x = f(y) = 3y2+ 5y − 2 for y ≥ 0, 3y2+ 5y − 2 − x = 0 for y ≥ 0,y = f−1(x) = (−5 +√12x + 49)/6 for x ≥ −223. y = f−1(x), x = f(y) = y − 5y2for y ≥ 1, 5y2− y + x = 0 for y ≥ 1,y = f−1(x) = (1 +√1 − 20x)/10 for x ≤ −424. (a) C =59(F − 32)(b) how many degrees Celsius given the Fahrenheit temperature(c) C = −273.15◦C is equivalent to F = −459.67◦F, so the domain is F ≥ −459.67, the rangeis C ≥ −273.1525. (a) y = f(x) = (6.214 × 10−4)x (b) x = f−1(y) =1046.214y(c) how many meters in y miles26. (a) f(g(x)) = f(√x)= (√x)2= x, x > 1;g(f(x)) = g(x2)=√x2 = x, x > 1(b)xyy = f(x)y = g(x)(c) no, because f(g(x)) = x for every x in the domain of g is not satisﬁed(the domain of g is x ≥ 0)
• 26 Chapter 127. (a) f(f(x)) =3 −3 − x1 − x1 −3 − x1 − x=3 − 3x − 3 + x1 − x − 3 + x= x so f = f−1(b) symmetric about the line y = x28. y = f−1(x), x = f(y) = ay2+ by + c, ay2+ by + c − x = 0, use the quadratic formula to gety =−b ± b2 − 4a(c − x)2a;(a) f−1(x) =−b + b2 − 4a(c − x)2a(b) f−1(x) =−b − b2 − 4a(c − x)2a29. if f−1(x) = 1, then x = f(1) = 2(1)3+ 5(1) + 3 = 1030. if f−1(x) = 2, then x = f(2) = (2)3/[(2)2+ 1] = 8/531. f(f(x)) = x thus f = f−1so the graph is symmetric about y = x.32. (a) Suppose x1 = x2 where x1 and x2 are in the domain of g and g(x1), g(x2) are in the domain off then g(x1) = g(x2) because g is one-to-one so f(g(x1)) = f(g(x2)) because f is one-to-onethus f ◦ g is one-to-one because (f ◦ g)(x1) = (f ◦ g)(x2) if x1 = x2.(b) f, g, and f ◦ g all have inverses because they are all one-to-one. Let h = (f ◦ g)−1then(f ◦ g)(h(x)) = f[g(h(x))] = x, apply f−1to both sides to get g(h(x)) = f−1(x), then applyg−1to get h(x) = g−1(f−1(x)) = (g−1◦ f−1)(x), so h = g−1◦ f−1.33.xy34. Suppose that g and h are both inverses of f then f(g(x)) = x, h[f(g(x))] = h(x), buth[f(g(x))] = g(x) because h is an inverse of f so g(x) = h(x).␪34535. tan θ = 4/3, 0 < θ < π/2; use the triangle shown toget sin θ = 4/5, cos θ = 3/5, cot θ = 3/4, sec θ = 5/3,csc θ = 5/4␪2.612.436. sec θ = 2.6, 0 < θ < π/2; use the triangle shown to getsin θ = 2.4/2.6 = 12/13, cos θ = 1/2.6 = 5/13,tan θ = 2.4 = 12/5, cot θ = 5/12, csc θ = 13/1237. (a) 0 ≤ x ≤ π (b) −1 ≤ x ≤ 1(c) −π/2 < x < π/2 (d) −∞ < x < +∞
• Exercise Set 1.5 2738. Let θ = sin−1(−3/4) thensin θ = −3/4, −π/2 < θ < 0and (see ﬁgure) sec θ = 4/√7␪–34√739. Let θ = cos−1(3/5),sin 2θ = 2 sin θ cos θ = 2(4/5)(3/5) = 24/25␪34540. (a) sin(cos−1x) =√1 − x21x√1 – x2cos–1 x(b) tan(cos−1x) =√1 − x2x1xcos–1 x√1 – x2(c) csc(tan−1x) =√1 + x2x1xtan–1 x√1 + x2(d) sin(tan−1x) =x√1 + x21xtan–1 x1 + x241. (a) cos(tan−1x) =1√1 + x21xtan–1 x1 + x2(b) tan(cos−1x) =√1 − x2xx1cos–1 x1 – x2(c) sin(sec−1x) =√x2 − 1x1xsec–1 xx2 – 1(d) cot(sec−1x) =1√x2 − 11xx2 – 1sec–1 x
• 28 Chapter 142. (a) x −1.00 −0.80 −0.6 −0.40 −0.20 0.00 0.20 0.40 0.60 0.80 1.00sin−1x −1.57 −0.93 −0.64 −0.41 −0.20 0.00 0.20 0.41 0.64 0.93 1.57cos−1x 3.14 2.50 2.21 1.98 1.77 1.57 1.37 1.16 0.93 0.64 0.00(b) yx11(c) yx–11230.5 143. (a)–0.5 0.5xyc/2–c/2(b)xyc/2–c/244. (a) x = π − sin−1(0.37) ≈ 2.7626 rad (b) θ = 180◦+ sin−1(0.61) ≈ 217.6◦45. (a) x = π + cos−1(0.85) ≈ 3.6964 rad (b) θ = − cos−1(0.23) ≈ −76.7◦46. (a) x = tan−1(3.16) − π ≈ −1.8773 (b) θ = 180◦− tan−1(0.45) ≈ 155.8◦47. (a) sin−10.9 > 1, so it is not in the domain of sin−1x(b) −1 ≤ sin−1x ≤ 1 is necessary, or −0.841471 ≤ x ≤ 0.84147148. sin 2θ = gR/v2= (9.8)(18)/(14)2= 0.9, 2θ = sin−1(0.9) or 2θ = 180◦− sin−1(0.9) soθ = 12 sin−1(0.9) ≈ 32◦or θ = 90◦− 12 sin−1(0.9) ≈ 58◦. The ball will have a lowerparabolic trajectory for θ = 32◦and hence will result in the shorter time of ﬂight.49. (a) yx–10 10c/2cyxc/25(b) The domain of cot−1x is (−∞, +∞), the range is (0, π); the domain of csc−1x is (−∞, −1]∪[1, +∞), the range is [−π/2, 0) ∪ (0, π/2].50. (a) y = cot−1x; if x > 0 then 0 < y < π/2 and x = cot y, tan y = 1/x, y = tan−1(1/x);if x < 0 then π/2 < y < π and x = cot y = cot(y − π), tan(y − π) = 1/x, y = π + tan−1 1x(b) y = sec−1x, x = sec y, cos y = 1/x, y = cos−1(1/x)(c) y = csc−1x, x = csc y, sin y = 1/x, y = sin−1(1/x)51. (a) 55.0◦(b) 33.6◦(c) 25.8◦52. (b) θ = sin−1 RR + h= sin−1 637816, 378≈ 23◦
• Exercise Set 1.5 2953. (a) If γ = 90◦, then sin γ = 1, 1 − sin2φ sin2γ = 1 − sin2φ = cos φ,D = tan φ tan λ = (tan 23.45◦)(tan 65◦) ≈ 0.93023374 so h ≈ 21.1 hours.(b) If γ = 270◦, then sin γ = −1, D = − tan φ tan λ ≈ −0.93023374 so h ≈ 2.9 hours.54. 42= 22+ 32− 2(2)(3) cos θ, cos θ = −1/4, θ = cos−1(−1/4) ≈ 104◦55. y = 0 when x2= 6000v2/g, x = 10v 60/g = 1000√30 for v = 400 and g = 32;tan θ = 3000/x = 3/√30, θ = tan−1(3/√30) ≈ 29◦.a␪␣␤bx56. θ = α − β, cot α =xa + band cot β =xbsoθ = cot−1 xa + b− cot−1 xb57. (a) Let θ = sin−1(−x) then sin θ = −x, −π/2 ≤ θ ≤ π/2. But sin(−θ) = − sin θ and−π/2 ≤ −θ ≤ π/2 so sin(−θ) = −(−x) = x, −θ = sin−1x, θ = − sin−1x.(b) proof is similar to that in Part (a)58. (a) Let θ = cos−1(−x) then cos θ = −x, 0 ≤ θ ≤ π. But cos(π − θ) = − cos θ and0 ≤ π − θ ≤ π so cos(π − θ) = x, π − θ = cos−1x, θ = π − cos−1x(b) Let θ = sec−1(−x) for x ≥ 1; then sec θ = −x and π/2 < θ ≤ π. So 0 ≤ π − θ < π/2 andπ − θ = sec−1sec(π − θ) = sec−1(− sec θ) = sec−1x, or sec−1(−x) = π − sec−1x.1 xsin–1 x√1 – x259. (a) sin−1x = tan−1 x√1 − x2(see ﬁgure)(b) sin−1x + cos−1x = π/2; cos−1x = π/2 − sin−1x = π/2 − tan−1 x√1 − x260. tan(α + β) =tan α + tan β1 − tan α tan β,tan(tan−1x + tan−1y) =tan(tan−1x) + tan(tan−1y)1 − tan(tan−1x) tan(tan−1y)=x + y1 − xyso tan−1x + tan−1y = tan−1 x + y1 − xy61. (a) tan−1 12+ tan−1 13= tan−1 1/2 + 1/31 − (1/2) (1/3)= tan−11 = π/4(b) 2 tan−1 13= tan−1 13+ tan−1 13= tan−1 1/3 + 1/31 − (1/3) (1/3)= tan−1 34,2 tan−1 13+ tan−1 17= tan−1 34+ tan−1 17= tan−1 3/4 + 1/71 − (3/4) (1/7)= tan−11 = π/462. sin(sec−1x) = sin(cos−1(1/x)) = 1 −1x2=√x2 − 1|x|
• 30 Chapter 1EXERCISE SET 1.61. (a) −4 (b) 4 (c) 1/42. (a) 1/16 (b) 8 (c) 1/33. (a) 2.9690 (b) 0.03414. (a) 1.8882 (b) 0.93815. (a) log2 16 = log2(24) = 4 (b) log2132= log2(2−5) = −5(c) log4 4 = 1 (d) log9 3 = log9(91/2) = 1/26. (a) log10(0.001) = log10(10−3) = −3 (b) log10(104) = 4(c) ln(e3) = 3 (d) ln(√e) = ln(e1/2) = 1/27. (a) 1.3655 (b) −0.30118. (a) −0.5229 (b) 1.14479. (a) 2 ln a +12ln b +12ln c = 2r + s/2 + t/2 (b) ln b − 3 ln a − ln c = s − 3r − t10. (a)13ln c − ln a − ln b = t/3 − r − s (b)12(ln a + 3 ln b − 2 ln c) = r/2 + 3s/2 − t11. (a) 1 + log x +12log(x − 3) (b) 2 ln |x| + 3 ln sin x −12ln(x2+ 1)12. (a)13log(x + 2) − log cos 5x (b)12ln(x2+ 1) −12ln(x3+ 5)13. log24(16)3= log(256/3) 14. log√x − log(sin32x) + log 100 = log100√xsin32x15. ln3√x(x + 1)2cos x16. 1 + x = 103= 1000, x = 99917.√x = 10−1= 0.1, x = 0.01 18. x2= e4, x = ±e219. 1/x = e−2, x = e220. x = 7 21. 2x = 8, x = 422. log10 x3= 30, x3= 1030, x = 101023. log10 x = 5, x = 10524. ln 4x − ln x6= ln 2, ln4x5= ln 2,4x5= 2, x5= 2, x = 5√225. ln 2x2= ln 3, 2x2= 3, x2= 3/2, x = 3/2 (we discard − 3/2 because it does not satisfy theoriginal equation)26. ln 3x= ln 2, x ln 3 = ln 2, x =ln 2ln 327. ln 5−2x= ln 3, −2x ln 5 = ln 3, x = −ln 32 ln 528. e−2x= 5/3, −2x = ln(5/3), x = −12 ln(5/3)
• Exercise Set 1.6 3129. e3x= 7/2, 3x = ln(7/2), x =13ln(7/2)30. ex(1 − 2x) = 0 so ex= 0 (impossible) or 1 − 2x = 0, x = 1/231. e−x(x + 2) = 0 so e−x= 0 (impossible) or x + 2 = 0, x = −232. e2x− ex− 6 = (ex− 3)(ex+ 2) = 0 so ex= −2 (impossible) or ex= 3, x = ln 333. e−2x− 3e−x+ 2 = (e−x− 2)(e−x− 1) = 0 so e−x= 2, x = − ln 2 or e−x= 1, x = 034. (a) yx–222 6(b) yx2–2 235. (a) yx246–2 4(b) yx–42–4 236. (a) yx–10–1(b) yx–1337. log2 7.35 = (log 7.35)/(log 2) = (ln 7.35)/(ln 2) ≈ 2.8777;log5 0.6 = (log 0.6)/(log 5) = (ln 0.6)/(ln 5) ≈ −0.317438. 10–50 239. 2–30 340. (a) Let X = logb x and Y = loga x. Then bX= x and aY= x so aY= bX, or aY/X= b, whichmeans loga b = Y/X. Substituting for Y and X yieldsloga xlogb x= loga b, logb x =loga xloga b.(b) Let x = a to get logb a = (loga a)/(loga b) = 1/(loga b) so (loga b)(logb a) = 1.(log2 81)(log3 32) = (log2[34])(log3[25]) = (4 log2 3)(5 log3 2) = 20(log2 3)(log3 2) = 20
• 32 Chapter 141. x = y ≈ 1.4710, x = y ≈ 7.8571801 842. Since the units are billions, one trillion is 1,000 units. Solve 1000 = 0.051517(1.1306727)xfor x bytaking common logarithms, resulting in 3 = log 0.051517 + x log 1.1306727, which yields x ≈ 80.4,so the debt ﬁrst reached one trillion dollars around 1980.43. (a) no, the curve passes through the origin (b) y = 2x/4(c) y = 2−x(d) y = (√5)x50–1 244. (a) As x → +∞ the function grows very slowly, but it is always increasing and tends to +∞. Asx → 1+the function tends to −∞.(b)1 2–55xy45. log(1/2) < 0 so 3 log(1/2) < 2 log(1/2)46. Let x = logb a and y = logb c, so a = bxand c = by.First, ac = bxby= bx+yor equivalently, logb(ac) = x + y = logb a + logb c.Secondly, a/c = bx/by= bx−yor equivalently, logb(a/c) = x − y = logb a − logb c.Next, ar= (bx)r= brxor equivalently, logb ar= rx = r logb a.Finally, 1/c = 1/by= b−yor equivalently, logb(1/c) = −y = − logb c.47. 75e−t/125= 15, t = −125 ln(1/5) = 125 ln 5 ≈ 201 days.48. (a) If t = 0, then Q = 12 grams(b) Q = 12e−0.055(4)= 12e−0.22≈ 9.63 grams(c) 12e−0.055t= 6, e−0.055t= 0.5, t = −(ln 0.5)/(0.055) ≈ 12.6 hours
• Exercise Set 1.7 3349. (a) 7.4; basic (b) 4.2; acidic (c) 6.4; acidic (d) 5.9; acidic50. (a) log[H+] = −2.44, [H+] = 10−2.44≈ 3.6 × 10−3mol/L(b) log[H+] = −8.06, [H+] = 10−8.06≈ 8.7 × 10−9mol/L51. (a) 140 dB; damage (b) 120 dB; damage(c) 80 dB; no damage (d) 75 dB; no damage52. Suppose that I1 = 3I2 and β1 = 10 log10 I1/I0, β2 = 10 log10 I2/I0. ThenI1/I0 = 3I2/I0, log10 I1/I0 = log10 3I2/I0 = log10 3 + log10 I2/I0, β1 = 10 log10 3 + β2,β1 − β2 = 10 log10 3 ≈ 4.8 decibels.53. Let IA and IB be the intensities of the automobile and blender, respectively. Thenlog10 IA/I0 = 7 and log10 IB/I0 = 9.3, IA = 107I0 and IB = 109.3I0, so IB/IA = 102.3≈ 200.54. The decibel level of the nth echo is 120(2/3)n;120(2/3)n< 10 if (2/3)n< 1/12, n <log(1/12)log(2/3)=log 12log 1.5≈ 6.13 so 6 echoes can be heard.55. (a) log E = 4.4 + 1.5(8.2) = 16.7, E = 1016.7≈ 5 × 1016J(b) Let M1 and M2 be the magnitudes of earthquakes with energies of E and 10E,respectively. Then 1.5(M2 − M1) = log(10E) − log E = log 10 = 1,M2 − M1 = 1/1.5 = 2/3 ≈ 0.67.56. Let E1 and E2 be the energies of earthquakes with magnitudes M and M + 1, respectively. Thenlog E2 − log E1 = log(E2/E1) = 1.5, E2/E1 = 101.5≈ 31.6.EXERCISE SET 1.71. The sum of the squares for the residuals for line I is approximately12+ 12+ 12+ 02+ 22+ 12+ 12+ 12= 10, and the same for line II is approximately02+ (0.4)2+ (1.2)2+ 02+ (2.2)2+ (0.6)2+ (0.2)2+ 02= 6.84; line II is the regression line.2. (a) The data appear to be periodic, so a trigonometric model may be appropriate.(b) The data appear to lie near a parabola, so a quadratic model may be appropriate.(c) The data appear to lie near a line, so a linear model may be appropriate.(d) The data appear to be randomly distributed, so no model seems to be appropriate.3. Least squares line S = 0.6414t − 1054.517, correlation coeﬃcient 0.57541989 1992 1995 1998 2001 2004220230240tS4. (a) p = 0.0154T + 4.19 (b) 3.42 atm5. (a) Least squares line p = 0.0146T + 3.98, correlation coeﬃcient 0.9999(b) p = 3.25 atm (c) T = −272◦C
• 34 Chapter 16. (a) p = 0.0203 T + 5.54, r = 0.9999 (b) T = −273(c) 1.05 p = 0.0203(T)(1.1) + 5.54 and p = 0.0203T + 5.54, subtract to get.05p = 0.1(0.0203T), p = 2(0.0203T). But p = 0.0203T + 5.54,equate the right hand sides, get T = 5.54/0.0203 ≈ 273◦C7. (a) R = 0.00723T + 1.55 (b) T = −214◦C8. (a) y = 0.0236x + 4.79 (b) T = −203◦C9. (a) S = 0.50179w − 0.00643 (b) S = 8, w = 16 lb10. (a) S = 0.756w − 0.0133(b) 2S = 0.756(w + 5) − 0.0133, subtract one equation from the other to get S = 5(0.756) = 3.7811. (a) Let h denote the height in inches and y the number of rebounds per minute. Theny = 0.0174h − 1.0549, r = 0.7095(b)65 750.10.20.3hy (c) The least squares line is a fair model for these data,since the correlation coeﬃcient is 0.7095.12. (a) Let h denote the height in inches and y the weight in pounds.Then y = 5.45h − 218; r = 0.8074(b)70 80175200225250hy (c) 223 lb13. (a) (0 + b − 0)2+ (1 + b − 0)2+ (1 + b − 1)2= 3b2+ 2b + 1(b) The function of Part (a) is a parabola in the variable b, and has its minimum when b = −13 ,hence the desired line is y = x − 13 .14. Let the points be (A, B) and (A, C). Let the regression line pass through the point (A, b) and haveslope m.The line y = m(x − A) + b is such a line.The sum of the squares of the residues is (b − B)2+ (b − C)2which is easily seen to be minimizedat b = (B +C)/2, which means the line passes through the point midway between the two originalgiven data points.Since the slope of the line is arbitary, and since the residual errors are independent of the slope,it follows that there are inﬁnitely many such lines of regression, and they all pass through themidpoint.
• Exercise Set 1.7 3515. Let the points be (A, B), (A, C) and (A , B ) and let the line of regression have the formy = m(x − A) + b. Then the sum of the squares of the residues is(B − b)2+ (C − b)2+ (m(x − A ) + b − B )2.Choose m and b so that the line passes through the point (A , B ), which makes the third termzero, and passes through the midpoint between (A, B) and (A, C), which minimizes the sum(B − b)2+ (C − b)2(see Exercise 15). These two conditions together minimize the sum of thesquares of the residuals, and they determine the slope and one point, and thus they determine theline of regression.16. Let the four vertices of the trapezoid be (A, B), (A, B ), (A , C) and (A , C ). From Exercise 14 itfollows that the sum of the squares of residuals at the two points (A, B) and (A, B ) is minimized ifthe line of regression passes through the midpoint between (A, B) and (A, B ). Similarly the sumof the sum of the squares of the residuals at the two points (A , B ) and (A , C ) is minimized byhaving the line of regression pass through the midpoint between these two points. Altogether wehave determined two points through which the line of regression must pass, and hence determinedthe line.17. (a) H ≈ 20000/110 ≈ 182 km/s/Mly(b) One light year is 9.408 × 1012km andt =dv=1H=120km/s/Mly=9.408 × 1018km20km/s= 4.704 × 1017s = 1.492 × 1010years.(c) The Universe would be even older.18. The population is modeled by P = 0.0045833t2− 16.378t + 14635; in the year 2000 the popula-tion would be P(2000) = 212, 200, 000. This is far short; in 1990 the population of the US wasapproximately 250,000,000.19. As in Example 4, a possible model is of the form T = D + A sin B t −CB. Since the longestday has 993 minutes and the shortest has 706, take 2A = 993 − 706 = 287 or A = 143.5. Themidpoint between the longest and shortest days is 849.5 minutes, so there is a vertical shift ofD = 849.5. The period is about 365.25 days, so 2π/B = 365.25 or B = π/183. Note that the sinefunction takes the value −1 when t −CB= −91.8125, and T is a minimum at about t = 0. Thusthe phase shiftCB≈ 91.5. Hence T = 849.5+143.5 sinπ183t −π2is a model for the temperature.20. As in Example 4, a possible model for the fraction f of illumination is of the formf = D + A sin B t −CB. Since the greatest fraction of illumination is 1 and the least is 0,2A = 1, A = 1/2. The midpoint of the fraction of illumination is 1/2, so there is a vertical shiftof D = 1/2. The period is approximately 30 days, so 2π/B = 30 or B = π/15. The phase shift isapproximately 49/2, so C/B = 49/2, and f = 1/2 + 1/2 sinπ15t −49221. t = 0.445√d2.300 2522. (a) t = 0.373 r1.5(b) 238,000 km(c) 1.89 days
• 36 Chapter 1EXERCISE SET 1.81. (a) x + 1 = t = y − 1, y = x + 2246y2 4t = 0t = 1t = 2t = 3t = 4t = 5x(c) t 0 1 2 3 4 5x −1 0 1 2 3 4y 1 2 3 4 5 62. (a) x2+ y2= 11y–1 1xt = 1t = 0.5t = 0.75 t = 0.25t = 0(c) t 0 0.2500 0.50 0.7500 1x 1 0.7071 0.00 −0.7071 −1y 0 0.7071 1.00 0.7071 03. t = (x + 4)/3; y = 2x + 10–8 612xy4. t = x + 3; y = 3x + 2, −3 ≤ x ≤ 0xy(0, 2)(–3, –7)5. cos t = x/2, sin t = y/5;x2/4 + y2/25 = 1–5 5–55xy6. t = x2; y = 2x2+ 4,x ≥ 0148xy7. cos t = (x − 3)/2,sin t = (y − 2)/4;(x − 3)2/4 + (y − 2)2/16 = 17–26xy8. sec2t − tan2t = 1;x2− y2= 1, x ≤ −1and y ≥ 0–1xy9. cos 2t = 1 − 2 sin2t;x = 1 − 2y2,−1 ≤ y ≤ 1–1 1–11xy10. t = (x − 3)/4;y = (x − 3)2− 9xy(3, –9)
• Exercise Set 1.8 3711. x/2 + y/3 = 1, 0 ≤ x ≤ 2, 0 ≤ y ≤ 3xy2312. y = x − 1, x ≥ 1, y ≥ 011xy13. x = 5 cos t, y = −5 sin t, 0 ≤ t ≤ 2π5–5–7.5 7.514. x = cos t, y = sin t, π ≤ t ≤ 3π/21–1–1 115. x = 2, y = t2–10 316. x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ 2π3–3–2 217. x = t2, y = t, −1 ≤ t ≤ 11–10 118. x = 1 + 4 cos t, y = −3 + 4 sin t, 0 ≤ t ≤ 2π2–8–7 819. (a) 140–35 8(b) t 0 1 2 3 4 5x 0 5.5 8 4.5 −8 −32.5y 1 1.5 3 5.5 9 13.5(c) x = 0 when t = 0, 2√3.(d) for 0 < t < 2√2(e) at t = 2
• 38 Chapter 120. (a) 50–2 14(b) y is always ≥ 1 since cos t ≤ 1(c) greater than 5, since cos t ≥ −121. (a) 3–50 20(b) oO–1 122. (a) 1.7–1.7–2.3 2.3(b)–10 10^623. (a) Eliminate t to getx − x0x1 − x0=y − y0y1 − y0(b) Set t = 0 to get (x0, y0); t = 1 for (x1, y1).(c) x = 1 + t, y = −2 + 6t (d) x = 2 − t, y = 4 − 6t24. (a) x = −3 − 2t, y = −4 + 5t, 0 ≤ t ≤ 1 (b) x = at, y = b(1 − t), 0 ≤ t ≤ 125. (a) |R−P|2= (x−x0)2+(y−y0)2= t2[(x1−x0)2+(y1−y0)2] and |Q−P|2= (x1−x0)2+(y1−y0)2,so r = |R − P| = |Q − P|t = qt.(b) t = 1/2 (c) t = 3/426. x = 2 + t, y = −1 + 2t(a) (5/2, 0) (b) (9/4, −1/2) (c) (11/4, 1/2)27. (a) IV, because x always increases whereas y oscillates.(b) II, because (x/2)2+ (y/3)2= 1, an ellipse.(c) V, because x2+ y2= t2increases in magnitude while x and y keep changing sign.(d) VI; examine the cases t < −1 and t > −1 and you see the curve lies in the ﬁrst, second andfourth quadrants only.(e) III because y > 0.(f) I; since x and y are bounded, the answer must be I or II; but as t runs, say, from 0 to π, xgoes directly from 2 to −2, but y goes from 0 to 1 to 0 to −1 and back to 0, which describesI but not II.
• Exercise Set 1.8 3928. (a) from left to right (b) counterclockwise (c) counterclockwise(d) As t travels from −∞ to −1, the curve goes from (near) the origin in the third quadrantand travels up and left. As t travels from −1 to +∞ the curve comes from way down inthe second quadrant, hits the origin at t = 0, and then makes the loop clockwise and ﬁnallyapproaches the origin again as t → +∞.(e) from left to right(f) Starting, say, at (1, 0), the curve goes up into the ﬁrst quadrant, loops back through theorigin and into the third quadrant, and then continues the ﬁgure-eight.29. The two branches corresponding to −1 ≤ t ≤ 0 and 0 ≤ t ≤ 1 coincide.30. (a) Eliminatet − t0t1 − t0to obtainy − y0x − x0=y1 − y0x1 − x0.(b) from (x0, y0) to (x1, y1)(c) x = 3 − 2(t − 1), y = −1 + 5(t − 1)5–20 531. (a)x − ba=y − dc(b)123y1 2 3x32. (a) If a = 0 the line segment is vertical; if c = 0 it is horizontal.(b) The curve degenerates to the point (b, d).33. 6–2–2 634. 10–5–5 1035. 300 336. 600 637.12y0.5 1x38. x = 1/2 − 4t, y = 1/2 for 0 ≤ t ≤ 1/4x = −1/2, y = 1/2 − 4(t − 1/4) for 1/4 ≤ t ≤ 1/2x = −1/2 + 4(t − 1/2), y = −1/2 for 1/2 ≤ t ≤ 3/4x = 1/2, y = −1/2 + 4(t − 3/4) for 3/4 ≤ t ≤ 1
• 40 Chapter 139. (a) x = 4 cos t, y = 3 sin t (b) x = −1 + 4 cos t, y = 2 + 3 sin t(c) 3–3–4 45–1–5 340. (a) t = x/(v0 cos α),so y = x tan α − gx2/(2v20 cos2α).(b)200060001000040000 80000xy41. (a) From Exercise 40, x = 400√2t,y = 400√2t − 4.9t2.(b) 16,326.53 m(c) 65,306.12 m42. (a) 15–15–25 25(b) 15–15–25 25(c) 15–15–25 25a = 3, b = 215–15–25 25a = 2, b = 315–15–25 25a = 2, b = 7
• Exercise Set 1.8 4143. Assume that a = 0 and b = 0; eliminate the parameter to get (x − h)2/a2+ (y − k)2/b2= 1. If|a| = |b| the curve is a circle with center (h, k) and radius |a|; if |a| = |b| the curve is an ellipsewith center (h, k) and major axis parallel to the x-axis when |a| > |b|, or major axis parallel to they-axis when |a| < |b|.(a) ellipses with a ﬁxed center and varying axes of symmetry(b) (assume a = 0 and b = 0) ellipses with varying center and ﬁxed axes of symmetry(c) circles of radius 1 with centers on the line y = x − 144. Refer to the diagram to get bθ = aφ, θ = aφ/b butθ − α = φ + π/2 so α = θ − φ − π/2 = (a/b − 1)φ − π/2x = (a − b) cos φ − b sin α= (a − b) cos φ + b cosa − bbφ,y = (a − b) sin φ − b cos α= (a − b) sin φ − b sina − bbφ.xy␪␣␾␾a␪ba − b45. (a)–aa–a ayx(b) Use b = a/4 in the equations of Exercise 44 to getx = 34 a cos φ + 14 a cos 3φ, y = 34 a sin φ − 14 a sin 3φ;but trigonometric identities yield cos 3φ = 4 cos3φ − 3 cos φ, sin 3φ = 3 sin φ − 4 sin3φ,so x = a cos3φ, y = a sin3φ.(c) x2/3+ y2/3= a2/3(cos2φ + sin2φ) = a2/346. (a)1–11xya = 1/23–3–113xya = 35–5–115xya = 5(b) (x − a)2+ y2= (2a cos2t − a)2+ (2a cos t sin t)2= 4a2cos4t − 4a2cos2t + a2+ 4a2cos2t sin2t= 4a2cos4t − 4a2cos2t + a2+ 4a2cos2t(1 − cos2t) = a2,a circle about (a, 0) of radius a
• 42 Chapter 1REVIEW EXERCISES CHAPTER 11. yx–15–5 52. (a) f(−2) = 2, g(3) = 2 (b) x = −3, 3 (c) x < −2, x > 3(d) the domain is −5 ≤ x ≤ 5 and the range is −5 ≤ y ≤ 4(e) the domain is −4 ≤ x ≤ 4.1, the range is −3 ≤ y ≤ 5(f) f(x) = 0 at x = −3, 5; g(x) = 0 at x = −3, 23.4050700 2 4 6tT4. Assume that the paint is applied in a thin veneer of uniform thickness, so that the quantity of paintto be used is proportional to the area covered. If P is the amount of paint to be used, P = kπr2.The constant k depends on physical factors, such as the thickness of the paint, absorption of thewood, etc.5. (a) If the side has length x and height h, then V = 8 = x2h, so h = 8/x2. Then the costC = 5x2+ 2(4)(xh) = 5x2+ 64/x.(b) The domain of C is (0, +∞) because x can be very large (just take h very small).6. (a) Suppose the radius of the uncoated ball is r and that of the coated ball is r + h. Then theplastic has volume equal to the diﬀerence of the volumes, i.e.V = 43 π(r+h)3− 43 πr3= 43 πh[3r2+3rh+h2] in3. But r = 3 and hence V = 43 πh[27+9h+h2].(b) 0 < h < ∞7. (a) The base has sides (10 − 2x)/2 and 6 − 2x, and the height is x, so V = (6 − 2x)(5 − x)x ft3.(b) From the picture we see that x < 5 and 2x < 6, so 0 < x < 3.(c) 3.57 ft ×3.79 ft ×1.21 ft8. (a) d = (x − 1)2 + 1/x2;(b) −∞ < x < 0, 0 < x < +∞(c) d ≈ 0.82 at x ≈ 1.380.811.21.620.5 1 2 3yx9.–21y–2 1 2x
• Review Exercises Chapter 1 4310. (a) On the interval [−20, 30]the curve seems tame,–10 20 30200000300000xy(b) but seen close up on the interval[−1/1000, +1/1000] we see that thereis some wiggling near the origin.–0.0005 0.001xy–0.00002–0.0000111. x −4 −3 −2 −1 0 1 2 3 4f(x) 0 −1 2 1 3 −2 −3 4 −4g(x) 3 2 1 −3 −1 −4 4 −2 0(f ◦ g)(x) 4 −3 −2 −1 1 0 −4 2 3(g ◦ f)(x) −1 −3 4 −4 −2 1 2 0 312. f ◦ g(x) = −1/|x| with domain x = 0, and g ◦ f(x) is nowhere deﬁned, with domain ∅.13. f(g(x)) = (3x+2)2+1, g(f(x)) = 3(x2+1)+2, so 9x2+12x+5 = 3x2+5, 6x2+12x = 0, x = 0, −214. (a) (3 − x)/x(b) no; the deﬁnition of f(g(x)) requires g(x) to be deﬁned, so x = 1, and f(g(x)) requiresg(x) = −1, so we must have g(x) = −1, i.e. x = 0; whereas h(x) only requires x = 015. When f(g(h(x))) is deﬁned, we require g(h(x)) = 1 and h(x) = 0. The ﬁrst requirement isequivalent to x = ±1, the second is equivalent to x = ± (2). For all other x, f · g · h = 1/(2 − x2).16. g(x) = x2+ 2x17. (a) even × odd = odd (b) a square is even(c) even + odd is neither (d) odd × odd = even18. (a) y = |x − 1|, y = |(−x) − 1| = |x + 1|,y = 2|x + 1|, y = 2|x + 1| − 3,y = −2|x + 1| + 3(b) yx–13–3 –1 219. (a) The circle of radius 1 centered at (a, a2); therefore, the family of all circles of radius 1 withcenters on the parabola y = x2.(b) All parabolas which open up, have latus rectum equal to 1 and vertex on the line y = x/2.
• 44 Chapter 120. Let y = ax2+ bx + c. Then 4a + 2b + c = 0, 64a + 8b + c = 18, 64a − 8b + c = 18, from which b = 0and 60a = 18, or ﬁnally y = 310 x2− 65 .21. (a)–202060y100 300t(b) when2π365(t − 101) =3π2, or t = 374.75, which is the same date as t = 9.75, so during thenight of January 10th-11th(c) from t = 0 to t = 70.58 and from t = 313.92 to t = 365 (the same date as t = 0) , for a totalof about 122 days22. Let y = A + B sin(at + b). Since the maximum and minimum values of y are 35 and 5, A + B = 35and A−B = 5, A = 20, B = 15. The period is 12 hours, so 12a = 2π and a = π/6. The maximumoccurs at t = 2, so 1 = sin(2a + b) = sin(π/3 + b), π/3 + b = π/2, b = π/2 − π/3 = π/6 andy = 20 + 15 sin(πt/6 + π/6).23. When x = 0 the value of the green curve is higher than that of the blue curve, therefore the bluecurve is given by y = 1 + 2 sin x.The points A, B, C, D are the points of intersection of the two curves, i.e. where1+2 sin x = 2 sin(x/2)+2 cos(x/2). Let sin(x/2) = p, cos(x/2) = q. Then 2 sin x = 4 sin(x/2) cos(x/2)(basic trigonometric identity, so the equation which yields the points of intersection becomes1+4pq = 2p+2q, 4pq−2p−2q+1 = 0, (2p−1)(2q−1) = 0; thus whenever either sin(x/2) = 1/2 orcos(x/2) = 1/2, i.e. when x/2 = π/6, 5π/6, ±π/3. Thus A has coordinates (−2π/3, 1−√3), B hascoordinates (π/3, 1+√3), C has coordinates (2π/3, 1+√3), and D has coordinates (5π/3, 1−√3).24. (a) R = R0 is the R-intercept, R0k is the slope, and T = −1/k is the T-intercept(b) −1/k = −273, or k = 1/273(c) 1.1 = R0(1 + 20/273), or R0 = 1.025(d) T = 126.55◦C25. (a) f(g(x)) = x for all x in the domain of g, and g(f(x)) = x for all x in the domain of f.(b) They are reﬂections of each other through the line y = x.(c) The domain of one is the range of the other and vice versa.(d) The equation y = f(x) can always be solved for x as a function of y. Functions with noinverses include y = x2, y = sin x.26. (a) For sin x, −π/2 ≤ x ≤ π/2; for cos x, 0 ≤ x ≤ π; for tan x, −π/2 < x < π/2; for sec x,0 ≤ x < π/2 or π/2 < x ≤ π.
• Review Exercises Chapter 1 45(b) yx–11c/2y = sin–1 xy = sin xyx–1cy = cos–1 xy = cos xyx–22c/2–c/2y = tan–1 xy = tanxyx–12c/2y = sec–1 x y = sec xy = sec–1 xy = sec x27. (a) x = f(y) = 8y3− 1; y = f−1(x) =x + 181/3=12(x + 1)1/3(b) f(x) = (x − 1)2; f does not have an inverse because f is not one-to-one, for examplef(0) = f(2) = 1.(c) x = f(y) = (ey)2+ 1; y = f−1(x) = ln√x − 1 = 12 ln(x − 1)(d) x = f(y) =y + 2y − 1; y = f−1(x) =x + 2x − 1(e) x = f(y) = sin1 − 2yy; y =12 + sin−1x(f) x =11 + 3 tan−1y; y = tan1 − x3x28. It is necessary and suﬃcient that the graph of f pass the horizontal line test. Suppose to the con-trary thatah + bch + d=ak + bck + dfor h = k. Then achk+bck+adh+bd = achk+adk+bch+bd, bc(h−k) =ad(h − k). It follows from h = k that ad − bc = 0. These steps are reversible, hence f−1exists ifand only if ad − bc = 0, and if so, thenx =ay + bcy + d, xcy + xd = ay + b, y(cx − a) = b − xd, y =b − xdcx − a= f−1(x)29. Draw equilateral triangles of sides 5, 12, 13, and 3, 4, 5. Then sin[cos−1(4/5)] = 3/5,sin[cos−1(5/13)] = 12/13, cos[sin−1(4/5)] = 3/5, cos[sin−1(5/13)] = 12/13(a) cos[cos−1(4/5) + sin−1(5/13)] = cos(cos−1(4/5)) cos(sin−1(5/13))− sin(cos−1(4/5)) sin(sin−1(5/13))=451213−35513=3365.(b) sin[sin−1(4/5) + cos−1(5/13)] = sin(sin−1(4/5)) cos(cos−1(5/13))+ cos(sin−1(4/5)) sin(cos−1(5/13))=45513+351213=5665.
• 46 Chapter 130. (a) yxc/21(b) yxc/21(c) yx–c/25(d) yxc/2131. y = 5 ft = 60 in, so 60 = log x, x = 1060in ≈ 1.58 × 1055mi.32. y = 100 mi = 12 × 5280 × 100 in, so x = log y = log 12 + log 5280 + log 100 ≈ 6.8018 in33. 3 ln e2x(ex)3+ 2 exp(ln 1) = 3 ln e2x+ 3 ln(ex)3+ 2 · 1 = 3(2x) + (3 · 3)x + 2 = 15x + 234. Y = ln(Cekt) = ln C + ln ekt= ln C + kt, a line with slope k and Y -intercept ln C35. (a) yx–224(b) The curve y = e−x/2sin 2x has x−intercepts at x = −π/2, 0, π/2, π, 3π/2. It intersects thecurve y = e−x/2at x = π/4, 5π/4 and it intersects the curve y = −e−x/2at x = −π/4, 3π/4.36. (a)520v1 2 3 4 5t(b) limt→∞(1 − e−1.3t) = 1,and thus as t → 0, v → 24.61 ft/s.(c) For large t the velocity approaches c = 24.61.(d) No; but it comes very close (arbitrarily close).(e) 3.009 s37. (a)100200N10 30 50t(b) N = 80 when t = 9.35 yrs(c) 220 sheep
• Review Exercises Chapter 1 4738. (a) The potato is done in the interval 27.65 < t < 32.71.(b) 91.54 min. The oven temperature is always 400·F, so the diﬀerence between the oventemperature and the potato temperature is D = 400 − T. Initially D = 325, so solveD = 75 + 325/2 = 237.5 for t, t ≈ 22.76.39. (a) The function ln x − x0.2is negative at x = 1 and positive at x = 4, so it is reasonable toexpect it to be zero somewhere in between. (This will be established later in this book.)(b) x = 3.654 and 3.32105 × 105–1 1 3 5–5–3–11xy40. (a) If xk= exthen k ln x = x, orln xx=1k. The steps are reversible.(b) By zooming it is seen that the maximum value of y isapproximately 0.368 (actually, 1/e), so there are two distinctsolutions of xk= exwhenever k > 1/0.368 ≈ 2.717.(c) x ≈ 1.15541. (a) 1.90 1.92 1.94 1.96 1.98 2.00 2.02 2.04 2.06 2.08 2.103.4161 3.4639 3.5100 3.5543 3.5967 3.6372 3.6756 3.7119 3.7459 3.7775 3.8068(b) y = 1.9589x − 0.2910(c) y − 3.6372 = 1.9589(x − 2), or y = 1.9589x − 0.2806(d) As one zooms in on the point (2, f(2))the two curves seem to converge to one line.3.83.41.9 2.242. (a) −0.10 −0.08 −0.06 −0.04 −0.02 0.00 0.02 0.04 0.06 0.08 0.100.9950 0.9968 0.9982 0.9992 0.9998 1.0000 0.9998 0.9992 0.9982 0.9968 0.9950(b) y = 13.669x2+ 2.865x + 0.733(c) y = −12 x2+ 1(d) As one zooms in on the point (0, f(0)) thetwo curves seem to converge to one curve. 1.2–0.8–1.7 1.7
• 48 Chapter 143. The data are periodic, so it is reasonable that a trigonometric function might approximate them.A possible model is of the form T = D + A sin B t −CB. Since the highest level is 1.032meters and the lowest is 0.045, take 2A = 1.032 − 0.042 = 0.990 or A = 0.495. The midpointbetween the lowest and highest levels is 0.537 meters, so there is a vertical shift of D = 0.537.The period is about 12 hours, so 2π/B = 12 or B = π/6. The phase shiftCB≈ 6.5. HenceT = 0.537 + 0.495 sinπ6(t − 6.5) is a model for the temperature.10 20 30 40 50 60 700.20.40.60.81tT44. (a) y = (415/458)x + 9508/1603 ≈ 0.906x + 5.931(b) 85.6745. x(t) =√2 cos t, y(t) = −√2 sin t, 0 ≤ t ≤ 3π/246. x = f(1 − t), y = g(1 − t) 47.–2–112y–1 1 2x48. (a) The three functions x2, tan(x) and ln(x) are all increasing on the indicated interval, and thusso is f(x) and it has an inverse.(b) The asymptotes for f(x) are x = 0, x = π/2. The asymptotes for f−1(x) are y = 0, y = π/2.3#x = 63xyy = xy = f(x)x = f(y)
• 49CHAPTER 2Limits and ContinuityEXERCISE SET 2.11. (a) 0 (b) 0 (c) 0 (d) 32. (a) +∞ (b) +∞ (c) +∞ (d) undef3. (a) −∞ (b) −∞ (c) −∞ (d) 14. (a) 1 (b) −∞ (c) does not exist (d) −25. for all x0 = −4 6. for all x0 = −6, 313. (a) 2 1.5 1.1 1.01 1.001 0 0.5 0.9 0.99 0.9990.1429 0.2105 0.3021 0.3300 0.3330 1.0000 0.5714 0.3690 0.3367 0.3337100 2The limit is 1/3.(b) 2 1.5 1.1 1.01 1.001 1.00010.4286 1.0526 6.344 66.33 666.3 6666.35001 2The limit is +∞.(c) 0 0.5 0.9 0.99 0.999 0.9999−1 −1.7143 −7.0111 −67.001 −667.0 −6667.00–500 1 The limit is −∞.
• 50 Chapter 214. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.250.5359 0.5132 0.5001 0.5000 0.5000 0.4999 0.4881 0.47210.60–0.25 0.25The limit is 1/2.(b) 0.25 0.1 0.001 0.00018.4721 20.488 2000.5 2000110000 0.25The limit is +∞.(c) −0.25 −0.1 −0.001 −0.0001−7.4641 −19.487 −1999.5 −200000–100–0.25 0 The limit is −∞.15. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.252.7266 2.9552 3.0000 3.0000 3.0000 3.0000 2.9552 2.726632–0.25 0.25The limit is 3.
• Exercise Set 2.1 51(b) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.0011 1.7552 6.2161 54.87 541.1 −0.1415 −4.536 −53.19 −539.560–60–1.5 0The limit does not exist.16. (a) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.0011.5574 1.0926 1.0033 1.0000 1.0000 1.0926 1.0033 1.0000 1.00001.51–1.5 0The limit is 1.(b) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.251.9794 2.4132 2.5000 2.5000 2.5000 2.5000 2.4132 1.97942.52–0.25 0.25The limit is 5/2.17. msec =x2− 1x + 1= x − 1 which gets close to −2 as x gets close to −1, thus y − 1 = −2(x + 1) ory = −2x − 118. msec =x2x= x which gets close to 0 as x gets close to 0 (doh!), thus y = 019. msec =x4− 1x − 1= x3+ x2+ x + 1 which gets close to 4 as x gets close to 1, thus y − 1 = 4(x − 1)or y = 4x − 320. msec =x4− 1x + 1= x3−x2+x−1 which gets close to −4 as x gets close to −1, thus y−1 = −4(x+1)or y = −4x − 3
• 52 Chapter 221. (a) The limit appears to be 3.3.52.5–1 1(b) The limit appears to be 3.3.52.5–0.001 0.001(c) The limit does not exist.3.52.5–0.000001 0.00000122. (a) 0.01 0.001 0.0001 0.000011.666 0.1667 0.1667 0.17(b) 0.000001 0.0000001 0.00000001 0.000000001 0.00000000010 0 0 0 0(c) it can be misleading23. (a) The plot over the interval [−a, a] becomes subject to catastrophic subtraction if a is smallenough (the size depending on the machine).(c) It does not.24. (a) the mass of the object while at rest(b) the limiting mass as the velocity approaches the speed of light; the mass is unbounded25. (a) The length of the rod while at rest(b) The limit is zero. The length of the rod approaches zeroEXERCISE SET 2.21. (a) −6 (b) 13 (c) −8 (d) 16 (e) 2 (f) −1/2(g) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.(h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.2. (a) 0(b) The limit doesn’t exist because lim f doesn’t exist and lim g does.(c) 0 (d) 3 (e) 0(f) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.(g) The limit doesn’t exist because f(x) is not deﬁned for 0 ≤ x < 2.(h) 1
• Exercise Set 2.3 533. 6 4. 27 5. 3/4 6. -3 7. 4 8. 129. -4/5 10. 0 11. −3 12. 1 13. 3/2 14. 4/315. +∞ 16. −∞ 17. does not exist 18. +∞19. −∞ 20. does not exist 21. +∞ 22. −∞23. does not exist 24. −∞ 25. +∞ 26. does not exist27. +∞ 28. +∞ 29. 6 30. 431. (a) 2(b) 2(c) 232. (a) -2(b) 0(c) does not exist33. (a) 3(b) yx4134. (a) −6(b) F(x) = x − 335. (a) Theorem 2.2.2(a) doesn’t apply; moreover one cannot add/subtract inﬁnities.(b) limx→0+1x−1x2= limx→0+x − 1x2= −∞36. limx→0−1x+1x2= limx→0−x + 1x2= +∞ 37. limx→0xx√x + 4 + 2=1438. limx→0x2x√x + 4 + 2= 039. The left and/or right limits could be plus or minus inﬁnity; or the limit could exist, or equal anypreassigned real number. For example, let q(x) = x − x0 and let p(x) = a(x − x0)nwhere n takeson the values 0, 1, 2.40. (a) If x < 0 then f(x) =ax + bx − ax + bx2x= b, so the limit is b.(b) Similarly if x > 0 then f(x) = a, so the limit is a.(c) Since the left limit is a and the right limit is b, the limit can only exist if a = b, in which casef(x) = a for all x = 0 and the limit is a.EXERCISE SET 2.31. (a) −∞(b) +∞2. (a) 2(b) 03. (a) 0(b) −14. (a) does not exist(b) 0
• 54 Chapter 25. (a) −12 (b) 21 (c) −15 (d) 25(e) 2 (f) −3/5 (g) 0(h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.6. (a) 20 (b) 0 (c) +∞ (d) −∞(e) (−42)1/3(f) −6/7 (g) 7 (h) −7/127. −∞ 8. +∞ 9. +∞ 10. +∞ 11. 3/212. 5/2 13. 0 14. 0 15. 0 16. 5/317. −51/3/2 18. 33/2 19. −√5 20.√5 21. 1/√622. −1/√6 23.√3 24.√3 25. −∞ 26. +∞27. −1/7 28. 4/729. It appears that limt→+∞n(t) = +∞, and limt→+∞e(t) = c.30. (a) It is the initial temperature of the potato (400◦F).(b) It is the ambient temperature, i.e. the temperature of the room.31. (a) +∞ (b) −5 32. (a) 0 (b) −633. limx→+∞( x2 + 3 − x)√x2 + 3 + x√x2 + 3 + x= limx→+∞3√x2 + 3 + x= 034. limx→+∞( x2 − 3x − x)√x2 − 3x + x√x2 − 3x + x= limx→+∞−3x√x2 − 3x + x= −3/235. limx→+∞x2 + ax − x√x2 + ax + x√x2 + ax + x= limx→+∞ax√x2 + ax + x= a/236. limx→+∞x2 + ax − x2 + bx√x2 + ax +√x2 + bx√x2 + ax +√x2 + bx= limx→+∞(a − b)x√x2 + ax +√x2 + bx=a − b237. limx→+∞p(x) = (−1)n∞ and limx→−∞p(x) = +∞38. If m > n the limits are both zero. If m = n the limits are both 1. If n > m the limits are(−1)n+m∞ and +∞, respectively.39. If m > n the limits are both zero. If m = n the limits are both equal to am, the leading coeﬃcientof p. If n > m the limits are ±∞ where the sign depends on the sign of am and whether n is evenor odd.40. (a) p(x) = q(x) = x (b) p(x) = x, q(x) = x2(c) p(x) = x2, q(x) = x (d) p(x) = x + 3, q(x) = x41. If m > n the limit is 0. If m = n the limit is −3. If m < n and n − m is odd, then the limit is+∞; if m < n and n − m is even, then the limit is −∞.42. If m > n the limit is zero. If m = n the limit is cm/dm. If n > m the limit is +∞ if cmdm > 0and −∞ if cmdm < 0.
• Exercise Set 2.3 5543. +∞ 44. 0 45. +∞ 46. 047. 1 48. −1 49. 1 50. −151. −∞ 52. −∞ 53. −∞ 54. +∞55. 1 56. −1 57. +∞ 58. −∞59. (a)4 8 12 16 204080120160200tv(b) limt→∞v = 190 1 − limt→∞e−0.168t= 190, so the asymptote is v = c = 190 ft/sec.(c) Due to air resistance (and other factors) this is the maximum speed that a sky diver canattain.60. (a) 50371.7/(151.3 + 181.626) ≈ 151.3 million(b)2000 2050 2100 2150200250300350tP(c) limt→∞p(t) =50371.7151.33 + 181.626 limt→∞ e−0.031636(t−1950)=50371.7151.33≈ 333 million(d) The population becomes stable at this number.61. limx→+∞f(x) = limx→−∞f(x) = L62. (a) Make the substitution t = 1/x to see that they are equal.(b) Make the substitution t = 1/x to see that they are equal.63.x + 1x= 1 +1x, so limx→+∞(x + 1)xxx= e from Figure 1.6.6.64. If x is large enough, then 1 +1x> 0, and so|x + 1|x|x|x= 1 +1xx, hencelimx→−∞|x + 1|x|x|x= limx→−∞1 +1xx= e by Figure 1.6.6.65. Set t = −x, then get limt→−∞1 +1tt= e by Figure 1.6.6.
• 56 Chapter 266. Set t = −x then limt→+∞1 +1tt= e67. Same as limx→+∞1 −1xx=1eby Exercise 65.68. Same as limt→+∞|t − 1|−t|t|−t=1eby Exercise 64.69. limx→+∞x +2x3x≥ limx→+∞x3xwhich is clearly +∞.70. If x ≤ −1 then2x≥ −2, consequently |x|+2x≥ |x|−2. But |x|−2 gets arbitrarily large, and hence|x| −2x3xgets arbitrarily small, since the exponent is negative. Thus limx→−∞|x| +2x3x= 0.71. Set t = 1/x, then limt→−∞1 +1tt= e 72. Set t = 1/x then limt→+∞1 +1tt= e73. Set t = −1/x, then limt→+∞1 +1t−t=1e74. Set x = −1/t, then limt→−∞1 +1t−t=1e.75. Set t = −1/(2x), then limt→+∞1 −1t−6t= e676. Set t = 1/(2x), then limt→+∞1 +1t6t= e677. Set t = 1/(2x), then limt→−∞1 −1t6t=1e678. Set t = 1/(2x), then limt→+∞1 −1t6t=1e679. f(x) = x + 2 +2x − 2,so limx→±∞(f(x) − (x + 2)) = 0and f(x) is asymptotic to y = x + 2.–12 –6 3 9 15–15–9–33915xx = 2yy = x + 280. f(x) = x2− 1 + 3/x,so limx→±∞[f(x) − (x2− 1) = 0]and f(x) is asymptotic to y = x2− 1.–4 –2 2 4–2135xyy = x2 – 1
• Exercise Set 2.4 5781. f(x) = −x2+ 1 + 2/(x − 3)so limx→±∞[f(x) − (−x2+ 1)] = 0and f(x) is asymptotic to y = −x2+ 1.–4 –2 2 4–12–6612xx = 3yy = –x2 + 182. f(x) = x3+32(x − 1)−32(x + 1)so limx→±∞[f(x) − x3] = 0and f(x) is asymptotic to y = x3.–2 2–15515xyy = x3x = 1x = –183. f(x) − sin x = 0 and f(x) is asymptotic to y = sin x.–4 2 8–435xyy = sin xx = 184. Note that the function is not deﬁned for −1 < x ≤ 1. Forx outside this interval we have f(x) = x2 +2x − 1whichsuggests that limx→±∞[f(x)−|x| ] = 0 (this can be checked witha CAS) and hence f(x) is asymptotic to y = |x| . –4 –2 2 4–4–224xyy = |x|x = 1EXERCISE SET 2.41. (a) |f(x) − f(0)| = |x + 2 − 2| = |x| < 0.1 if and only if |x| < 0.1(b) |f(x) − f(3)| = |(4x − 5) − 7| = 4|x − 3| < 0.1 if and only if |x − 3| < (0.1)/4 = 0.0025(c) |f(x)−f(4)| = |x2−16| < if |x−4| < δ. We get f(x) = 16+ = 16.001 at x = 4.000124998,which corresponds to δ = 0.000124998; and f(x) = 16 − = 15.999 at x = 3.999874998, forwhich δ = 0.000125002. Use the smaller δ: thus |f(x) − 16| < provided |x − 4| < 0.000125(to six decimals).2. (a) |f(x) − f(0)| = |2x + 3 − 3| = 2|x| < 0.1 if and only if |x| < 0.05(b) |f(x) − f(0)| = |2x + 3 − 3| = 2|x| < 0.01 if and only if |x| < 0.005(c) |f(x) − f(0)| = |2x + 3 − 3| = 2|x| < 0.0012 if and only if |x| < 0.00063. (a) x1 = (1.95)2= 3.8025, x2 = (2.05)2= 4.2025(b) δ = min ( |4 − 3.8025|, |4 − 4.2025| ) = 0.1975
• 58 Chapter 24. (a) x1 = 1/(1.1) = 0.909090 . . . , x2 = 1/(0.9) = 1.111111 . . .(b) δ = min( |1 − 0.909090|, |1 − 1.111111| ) = 0.0909090 . . .5. |(x3− 4x + 5) − 2| < 0.05, −0.05 < (x3− 4x + 5) − 2 < 0.05,1.95 < x3− 4x + 5 < 2.05; x3− 4x + 5 = 1.95 atx = 1.0616, x3− 4x + 5 = 2.05 at x = 0.9558; δ =min (1.0616 − 1, 1 − 0.9558) = 0.04422.21.90.9 1.16.√5x + 1 = 3.5 at x = 2.25,√5x + 1 = 4.5 at x = 3.85, soδ = min(3 − 2.25, 3.85 − 3) = 0.75502 47. With the TRACE feature of a calculator we discover that (to ﬁve decimal places) (0.87000, 1.80274)and (1.13000, 2.19301) belong to the graph. Set x0 = 0.87 and x1 = 1.13. Since the graph of f(x)rises from left to right, we see that if x0 < x < x1 then 1.80274 < f(x) < 2.19301, and therefore1.8 < f(x) < 2.2. So we can take δ = 0.13.8. From a calculator plot we conjecture that limx→0f(x) = 2. Using the TRACE feature we see thatthe points (±0.2, 1.94709) belong to the graph. Thus if −0.2 < x < 0.2 then 1.95 < f(x) ≤ 2 andhence |f(x) − L| < 0.05 < 0.1 = .9. |2x − 8| = 2|x − 4| < 0.1 if |x − 4| < 0.05, δ = 0.0510. |5x − 2 − 13| = 5|x − 3| < 0.01 if |x − 3| < 0.002, δ = 0.00211.x2− 9x − 3− 6 = |x + 3 − 6| = |x − 3| < 0.05 if |x − 3| < 0.05, δ = 0.0512.4x2− 12x + 1+ 2 = |2x − 1 + 2| = |2x + 1| < 0.05 if x +12< 0.025, δ = 0.02513. On the interval [1, 3] we have |x2+ x + 2| ≤ 14, so |x3− 8| = |x − 2||x2+ x + 2| ≤ 14|x − 2| < 0.001provided |x − 2| < 0.001 ·114; but 0.00005 <0.00114, so for convenience we take δ = 0.00005 (thereis no need to choose an ’optimal’ δ).14. Since√x > 0, |√x − 2| =|x − 4||√x + 2|<|x − 4|2< 0.001 if |x − 4| < 0.002, δ = 0.00215. if δ ≤ 1 then |x| > 3, so1x−15=|x − 5|5|x|≤|x − 5|15< 0.05 if |x − 5| < 0.75, δ = 3/416. |x − 0| = |x| < 0.05 if |x| < 0.05, δ = 0.05
• Exercise Set 2.4 5917. (a) limx→4f(x) = 3(b) |10f(x) − 30| = 10|f(x) − 3| < 0.005 provided |f(x) − 3| < 0.0005, which is true for|x − 3| < 0.0001, δ = 0.000118. (a) limx→3f(x) = 7; limx→3g(x) = 5(b) |7f(x) − 21| < 0.03 is equivalent to |f(x) − 7| < 0.01, so let = 0.01 in condition (i): thenwhen |x − 3| < δ = 0.012= 0.0001, it follows that |f(x) − 7| < 0.01, or |3f(x) − 21| < 0.03.19. It suﬃces to have |10f(x)+2x−38| ≤ |10f(x)−30|+2|x−4| < 0.01, by the triangle inequality. Toensure |10f(x) − 30| < 0.005 use Exercise 17 (with = 0.0005) to get δ = 0.0001. Then |x − 4| < δyields |10f(x)+2x−38| ≤ 10|f(x)−3|+2|x−4| ≤ (10)˙0.0005+(2)˙0.0001 ≤ 0.005+0.0002 < 0.0120. Let δ = 0.0009. By the triangle inequality |3f(x) + g(x) − 26| ≤ 3|f(x) − 7| + |g(x) − 5| ≤3 ·√0.0009 + 0.0072 = 0.03 + 0.0072 < 0.06.21. |3x − 15| = 3|x − 5| < if |x − 5| < 13 , δ = 1322. |7x + 5 + 2| = 7|x + 1| < if |x + 1| < 17 , δ = 1723.2x2+ xx− 1 = |2x| < if |x| < 12 , δ = 1224.x2− 9x + 3− (−6) = |x + 3| < if |x + 3| < , δ =25. |f(x) − 3| = |x + 2 − 3| = |x − 1| < if 0 < |x − 1| < , δ =26. |9 − 2x − 5| = 2|x − 2| < if 0 < |x − 2| < 12 , δ = 1227. (a) |(3x2+ 2x − 20 − 300| = |3x2+ 2x − 320| = |(3x + 32)(x − 10)| = |3x + 32| · |x − 10|(b) If |x − 10| < 1 then |3x + 32| < 65, since clearly x < 11(c) δ = min(1, /65); |3x + 32| · |x − 10| < 65 · |x − 10| < 65 · /65 =28. (a)283x + 1− 4 =28 − 12x − 43x + 1=−12x + 243x + 1=123x + 1· |x − 2|(b) If |x−2| < 4 then −2 < x < 6, so x can be very close to −1/3, hence123x + 1is not bounded.(c) If |x − 2| < 1 then 1 < x < 3 and 3x + 1 > 4, so123x + 1<124= 3(d) δ = min(1, /3);123x + 1· |x − 2| < 3 · |x − 2| < 3 · /3 =29. if δ < 1 then |2x2− 2| = 2|x − 1||x + 1| < 6|x − 1| < if |x − 1| < 16 , δ = min(1, 16 )30. If δ < 1 then |x2+ x − 12| = |x + 4| · |x − 3| < 5|x − 3| < if |x − 3| < /5, δ = min(1, 15 )31. If δ < 12 and |x − (−2)| < δ then −52 < x < −32 , x + 1 < −12 , |x + 1| > 12 ; then1x + 1− (−1) =|x + 2||x + 1|< 2|x + 2| < if |x + 2| <12, δ = min12,1232. If δ < 14 then2x + 3x− 8 =|6x − 3||x|<6|x − 12 |14= 24|x −12| < if |x − 12 | < /24, δ = min(14 , 24 )
• 60 Chapter 233. |√x − 2| = (√x − 2)√x + 2√x + 2=x − 4√x + 2<12|x − 4| < if |x − 4| < 2 , δ = 234. If x < 2 then |f(x) − 5| = |9 − 2x − 5| = 2|x − 2| < if |x − 2| < 12 , δ1 = 12 . If x > 2 then|f(x) − 5| = |3x − 1 − 5| = 3|x − 2| < if |x − 2| < 13 , δ2 = 13 Now let δ = min(δ1, δ2) then forany x with |x − 2| < δ, |f(x) − 5| <35. (a) |f(x) − L| =1x2< 0.1 if x >√10, N =√10(b) |f(x) − L| =xx + 1− 1 =1x + 1< 0.01 if x + 1 > 100, N = 99(c) |f(x) − L| =1x3<11000if |x| > 10, x < −10, N = −10(d) |f(x) − L| =xx + 1− 1 =1x + 1< 0.01 if |x + 1| > 100, −x − 1 > 100, x < −101,N = −10136. (a)1x3< 0.1, x > 101/3, N = 101/3(b)1x3< 0.01, x > 1001/3, N = 1001/3(c)1x3< 0.001, x > 10, N = 1037. (a)x211 + x21= 1 − , x1 = −1 −;x221 + x22= 1 − , x2 =1 −(b) N =1 −(c) N = −1 −38. (a) x1 = −1/ 3; x2 = 1/ 3(b) N = 1/ 3(c) N = −1/ 339.1x2< 0.01 if |x| > 10, N = 1040.1x + 2< 0.005 if |x + 2| > 200, x > 198, N = 19841.xx + 1− 1 =1x + 1< 0.001 if |x + 1| > 1000, x > 999, N = 99942.4x − 12x + 5− 2 =112x + 5< 0.1 if |2x + 5| > 110, 2x > 105, N = 52.543.1x + 2− 0 < 0.005 if |x + 2| > 200, −x − 2 > 200, x < −202, N = −20244.1x2< 0.01 if |x| > 10, −x > 10, x < −10, N = −1045.4x − 12x + 5− 2 =112x + 5< 0.1 if |2x+5| > 110, −2x−5 > 110, 2x < −115, x < −57.5, N = −57.546.xx + 1− 1 =1x + 1< 0.001 if |x + 1| > 1000, −x − 1 > 1000, x < −1001, N = −1001
• Exercise Set 2.4 6147.1x2< if |x| >1√ , N =1√48.1x + 2< if |x + 2| >1, x + 2 >1, x >1− 2, N =1− 249.4x − 12x + 5− 2 =112x + 5< if |2x + 5| >11, −2x − 5 >11, 2x < −11− 5, x < −112−52,N = −52−11250.xx + 1− 1 =1x + 1< if |x + 1| >1, −x − 1 >1, x < −1 −1, N = −1 −151.2√x√x − 1− 2 =2√x − 1< if√x − 1 >2,√x > 1 +2, x > 1 +22, N > 1 +2252. 2x< if x < log2 , N < log253. (a)1x2> 100 if |x| <110(b)1|x − 1|> 1000 if |x − 1| <11000(c)−1(x − 3)2< −1000 if |x − 3| <110√10(d) −1x4< −10000 if x4<110000, |x| <11054. (a)1(x − 1)2> 10 if and only if |x − 1| <1√10(b)1(x − 1)2> 1000 if and only if |x − 1| <110√10(c)1(x − 1)2> 100000 if and only if |x − 1| <1100√1055. if M > 0 then1(x − 3)2> M, 0 < (x − 3)2<1M, 0 < |x − 3| <1√M, δ =1√M56. if M < 0 then−1(x − 3)2< M, 0 < (x − 3)2< −1M, 0 < |x − 3| <1√−M, δ =1√−M57. if M > 0 then1|x|> M, 0 < |x| <1M, δ =1M58. if M > 0 then1|x − 1|> M, 0 < |x − 1| <1M, δ =1M59. if M < 0 then −1x4< M, 0 < x4< −1M, |x| <1(−M)1/4, δ =1(−M)1/460. if M > 0 then1x4> M, 0 < x4<1M, x <1M1/4, δ =1M1/461. if x > 2 then |x + 1 − 3| = |x − 2| = x − 2 < if 2 < x < 2 + , δ =62. if x < 1 then |3x + 2 − 5| = |3x − 3| = 3|x − 1| = 3(1 − x) < if 1 − x < 13 , 1 − 13 < x < 1, δ = 13
• 62 Chapter 263. if x > 4 then√x − 4 < if x − 4 < 2, 4 < x < 4 + 2, δ = 264. if x < 0 then√−x < if −x < 2, − 2< x < 0, δ = 265. if x > 2 then |f(x) − 2| = |x − 2| = x − 2 < if 2 < x < 2 + , δ =66. if x < 2 then |f(x) − 6| = |3x − 6| = 3|x − 2| = 3(2 − x) < if 2 − x < 13 , 2 − 13 < x < 2, δ = 1367. (a) if M < 0 and x > 1 then11 − x< M, x − 1 < −1M, 1 < x < 1 −1M, δ = −1M(b) if M > 0 and x < 1 then11 − x> M, 1 − x <1M, 1 −1M< x < 1, δ =1M68. (a) if M > 0 and x > 0 then1x> M, x <1M, 0 < x <1M, δ =1M(b) if M < 0 and x < 0 then1x< M, −x < −1M,1M< x < 0, δ = −1M69. (a) Given any M > 0 there corresponds N > 0 such that if x > N then f(x) > M, x + 1 > M,x > M − 1, N = M − 1.(b) Given any M < 0 there corresponds N < 0 such that if x < N then f(x) < M, x + 1 < M,x < M − 1, N = M − 1.70. (a) Given any M > 0 there corresponds N > 0 such that if x > N then f(x) > M, x2− 3 > M,x >√M + 3, N =√M + 3.(b) Given any M < 0 there corresponds N < 0 such that if x < N then f(x) < M, x3+ 5 < M,x < (M − 5)1/3, N = (M − 5)1/3.71. if δ ≤ 2 then |x − 3| < 2, −2 < x − 3 < 2, 1 < x < 5, and |x2− 9| = |x + 3||x − 3| < 8|x − 3| < if|x − 3| < 18 , δ = min 2, 1872. (a) We don’t care about the value of f at x = a, because the limit is only concerned with valuesof x near a. The condition that f be deﬁned for all x (except possibly x = a) is necessary,because if some points were excluded then the limit may not exist; for example, let f(x) = xif 1/x is not an integer and f(1/n) = 6. Then limx→0f(x) does not exist but it would if thepoints 1/n were excluded.(b) if x < 0 then√x is not deﬁned (c) yes; if δ ≤ 0.01 then x > 0, so√x is deﬁned73. (a) 0.4 amperes (b) [0.3947, 0.4054] (c)37.5 + δ,37.5 − δ(d) 0.0187 (e) It becomes inﬁnite.EXERCISE SET 2.51. (a) no, x = 2 (b) no, x = 2 (c) no, x = 2 (d) yes(e) yes (f) yes2. (a) no, x = 2 (b) no, x = 2 (c) no, x = 2 (d) yes(e) no, x = 2 (f) yes3. (a) no, x = 1, 3 (b) yes (c) no, x = 1 (d) yes(e) no, x = 3 (f) yes
• Exercise Set 2.5 634. (a) no, x = 3 (b) yes (c) yes (d) yes(e) no, x = 3 (f) yes5. (a) 3 (b) 3 6. −2/57. (a) yx3(b) yx11 3(c) yx–111(d) yx2 38. f(x) = 1/x, g(x) =0 if x = 0sin1xif x = 09. (a) Ct1\$42(b) One second could cost you one dollar.10. (a) no; disasters (war, ﬂood, famine, pestilence, for example) can cause discontinuities(b) continuous(c) not usually continuous; see Exercise 9(d) continuous11. none 12. none 13. none 14. x = −2, 2 15. x = 0, −1/216. none 17. x = −1, 0, 1 18. x = −4, 0 19. none 20. x = −1, 021. none; f(x) = 2x + 3 is continuous on x < 4 and f(x) = 7 +16xis continuous on 4 < x;limx→4−f(x) = limx→4+f(x) = f(4) = 11 so f is continuous at x = 422. limx→1f(x) does not exist so f is discontinuous at x = 123. (a) f is continuous for x < 1, and for x > 1; limx→1−f(x) = 5, limx→1+f(x) = k, so if k = 5 then f iscontinuous for all x(b) f is continuous for x < 2, and for x > 2; limx→2−f(x) = 4k, limx→2+f(x) = 4+k, so if 4k = 4+k,k = 4/3 then f is continuous for all x
• 64 Chapter 224. (a) f is continuous for x < 3, and for x > 3; limx→3−f(x) = k/9, limx→3+f(x) = 0, so if k = 0 then fis continuous for all x(b) f is continuous for x < 0, and for x > 0; limx→0−f(x) doesn’t exist unless k = 0, and if so thenlimx→0−f(x) = +∞; limx→0+f(x) = 9, so no value of k25. f is continuous for x < −1, −1 < x < 2 and x > 2; limx→−1−f(x) = 4, limx→−1+f(x) = k, so k = 4 isrequired. Next, limx→2−f(x) = 3m + k = 3m + 4, limx→2+f(x) = 9, so 3m + 4 = 9, m = 5/3 and f iscontinuous everywhere if k = 4, m = 5/326. (a) no, f is not deﬁned at x = 2 (b) no, f is not deﬁned for x ≤ 2(c) yes (d) no, f is not deﬁned for x ≤ 227. (a) yxc(b) yxc28. (a) f(c) = limx→cf(x)(b) limx→1f(x) = 2 limx→1g(x) = 1yx11–1yx11(c) Deﬁne f(1) = 2 and redeﬁne g(1) = 1.29. (a) x = 0, limx→0−f(x) = −1 = +1 = limx→0+f(x) so the discontinuity is not removable(b) x = −3; deﬁne f(−3) = −3 = limx→−3f(x), then the discontinuity is removable(c) f is undeﬁned at x = ±2; at x = 2, limx→2f(x) = 1, so deﬁne f(2) = 1 and f becomescontinuous there; at x = −2, limx→−2does not exist, so the discontinuity is not removable30. (a) f is not deﬁned at x = 2; limx→2f(x) = limx→2x + 2x2 + 2x + 4=13, so deﬁne f(2) =13and fbecomes continuous there(b) limx→2−f(x) = 1 = 4 = limx→2+f(x), so f has a nonremovable discontinuity at x = 2(c) limx→1f(x) = 8 = f(1), so f has a removable discontinuity at x = 1
• Exercise Set 2.5 6531. (a) discontinuity at x = 1/2, not removable;at x = −3, removableyx–555(b) 2x2+ 5x − 3 = (2x − 1)(x + 3)32. (a) there appears to be one discontinuitynear x = −1.524–4–3 3(b) one discontinuity at x ≈ −1.5233. For x > 0, f(x) = x3/5= (x3)1/5is the composition (Theorem 2.5.6) of the two continuousfunctions g(x) = x3and h(x) = x1/5and is thus continuous. For x < 0, f(x) = f(−x) which is thecomposition of the continuous functions f(x) (for positive x) and the continuous function y = −x.Hence f(−x) is continuous for all x > 0. At x = 0, f(0) = limx→0f(x) = 0.34. x4+ 7x2+ 1 ≥ 1 > 0, thus f(x) is the composition of the polynomial x4+ 7x2+ 1, the square root√x, and the function 1/x and is therefore continuous by Theorem 2.5.6.35. (a) Let f(x) = k for x = c and f(c) = 0; g(x) = l for x = c and g(c) = 0. If k = −l then f + g iscontinuous; otherwise it’s not.(b) f(x) = k for x = c, f(c) = 1; g(x) = l = 0 for x = c, g(c) = 1. If kl = 1, then fg iscontinuous; otherwise it’s not.36. A rational function is the quotient f(x)/g(x) of two polynomials f(x) and g(x). By Theorem 2.5.2f and g are continuous everywhere; by Theorem 2.5.3 f/g is continuous except when g(x) = 0.37. Since f and g are continuous at x = c we know that limx→cf(x) = f(c) and limx→cg(x) = g(c). In thefollowing we use Theorem 2.2.2.(a) f(c) + g(c) = limx→cf(x) + limx→cg(x) = limx→c(f(x) + g(x)) so f + g is continuous at x = c.(b) same as (a) except the + sign becomes a − sign(c) f(c) · g(c) = limx→cf(x) limx→cg(x) = limx→cf(x) · g(x) so f · g is continuous at x = c38. h(x) = f(x) − g(x) satisﬁes h(a) > 0, h(b) < 0. Use the Intermediate Value Theorem or Theorem2.5.8.39. Of course such a function must be discontinuous. Let f(x) = 1 on 0 ≤ x < 1, and f(x) = −1 on1 ≤ x ≤ 2.
• 66 Chapter 240. (a) (i) no (ii) yes (b) (i) no (ii) no (c) (i) no (ii) no41. The cone has volume πr2h/3. The function V (r) = πr2h (for variable r and ﬁxed h) gives thevolume of a right circular cylinder of height h and radius r, and satisﬁes V (0) < πr2h/3 < V (r).By the Intermediate Value Theorem there is a value c between 0 and r such that V (c) = πr2h/3,so the cylinder of radius c (and height h) has volume equal to that of the cone.42. A square whose diagonal has length r has area f(r) = r2/2. Note thatf(r) = r2/2 < πr2/2 < 2r2= f(2r). By the Intermediate Value Theorem there must be a valuec between r and 2r such that f(c) = πr2/2, i.e. a square of diagonal c whose area is πr2/2.43. If f(x) = x3+ x2− 2x then f(−1) = 2, f(1) = 0. Use the Intermediate Value Theorem.44. Since limx→−∞p(x) = −∞ and limx→+∞p(x) = +∞ (or vice versa, if the leading coeﬃcient of p isnegative), it follows that for M = −1 there corresponds N1 < 0, and for M = 1 there is N2 > 0,such that p(x) < −1 for x < N1 and p(x) > 1 for x > N2. Choose x1 < N1 and x2 > N2 and useTheorem 2.5.8 on the interval [x1, x2] to ﬁnd a solution of p(x) = 0.45. For the negative root, use intervals on the x-axis as follows: [−2, −1]; since f(−1.3) < 0 andf(−1.2) > 0, the midpoint x = −1.25 of [−1.3, −1.2] is the required approximation of the root.For the positive root use the interval [0, 1]; since f(0.7) < 0 and f(0.8) > 0, the midpoint x = 0.75of [0.7, 0.8] is the required approximation.46. x = −1.22 and x = 0.72.10–5–2 –11–10.7 0.847. For the negative root, use intervals on the x-axis as follows: [−2, −1]; since f(−1.7) < 0 andf(−1.6) > 0, use the interval [−1.7, −1.6]. Since f(−1.61) < 0 and f(−1.60) > 0 the midpointx = −1.605 of [−1.61, −1.60] is the required approximation of the root. For the positive root usethe interval [1, 2]; since f(1.3) > 0 and f(1.4) < 0, use the interval [1.3, 1.4]. Since f(1.37) > 0and f(1.38) < 0, the midpoint x = 1.375 of [1.37, 1.38] is the required approximation.48. x = −1.603 and x = 1.3799.1–2–1.7 –1.61–0.51.3 1.449. x = 2.24
• Exercise Set 2.6 6750. Set f(x) =ax − 1+bx − 3. Since limx→1+f(x) = +∞ and limx→3−f(x) = −∞ there exist x1 > 1 andx2 < 3 (with x2 > x1) such that f(x) > 1 for 1 < x < x1 and f(x) < −1 for x2 < x < 3. Choosex3 in (1, x1) and x4 in (x2, 3) and apply Theorem 2.5.8 on [x3, x4].51. The uncoated sphere has volume 4π(x − 1)3/3 and the coated sphere has volume 4πx3/3. If thevolume of the uncoated sphere and of the coating itself are the same, then the coated sphere hastwice the volume of the uncoated sphere. Thus 2(4π(x−1)3/3) = 4πx3/3, or x3−6x2+6x−2 = 0,with the solution x = 4.847 cm.52. Let g(t) denote the altitude of the monk at time t measured in hours from noon of day one, andlet f(t) denote the altitude of the monk at time t measured in hours from noon of day two. Theng(0) < f(0) and g(12) > f(12). Use Exercise 38.53. We must show limx→cf(x) = f(c). Let > 0; then there exists δ > 0 such that if |x − c| < δ then|f(x) − f(c)| < . But this certainly satisﬁes Deﬁnition 2.4.1.54. (a) yx0.410.2 0.8(b) Let g(x) = x − f(x). Then g(1) ≥ 0 and g(0) ≤ 0; by the Intermediate Value Theorem thereis a solution c in [0, 1] of g(c) = 0.EXERCISE SET 2.61. none 2. x = π 3. nπ, n = 0, ±1, ±2, . . .4. x = nπ + π/2, n = 0, ±1, ±2, . . . 5. x = nπ, n = 0, ±1, ±2, . . .6. none 7. 2nπ + π/6, 2nπ + 5π/6, n = 0, ±1, ±2, . . .8. x = nπ + π/2, n = 0, ±1, ±2, . . . 9. [−1, 1]10. (−∞, −1] ∪ [1, ∞) 11. (0, 3) ∪ (3, +∞)12. (−∞, 0) ∪ (0, +∞), and if f is deﬁned to be e at x = 0, then continuous for all x13. (−∞, −1] ∪ [1, ∞) 14. (−3, 0) ∪ (0, ∞)15. (a) sin x, x3+ 7x + 1 (b) |x|, sin x (c) x3, cos x, x + 1(d)√x, 3 + x, sin x, 2x (e) sin x, sin x (f) x5− 2x3+ 1, cos x16. (a) Use Theorem 2.5.6. (b) g(x) = cos x, g(x) =1x2 + 1, g(x) = x2+ 117. cos limx→+∞1x= cos 0 = 1 18. sin limx→+∞πx2 − 3x= sin −π3= −√3219. sin−1limx→+∞x1 − 2x= sin−1−12= −π6
• 68 Chapter 220. ln limx→+∞x + 1x= ln(1) = 021. 3 limθ→0sin 3θ3θ= 3 22.12limh→0sin hh=1223. limθ→0+1θlimθ→0+sin θθ= +∞ 24. limθ→0sin θ limθ→0sin θθ= 025.tan 7xsin 3x=73 cos 7xsin 7x7x3xsin 3xso limx→0tan 7xsin 3x=73(1)(1)(1) =7326.sin 6xsin 8x=68sin 6x6x8xsin 8x, so limx→0sin 6xsin 8x=68=3427.15limx→0+√x limx→0+sin xx= 0 28.13limx→0sin xx2=1329. limx→0x limx→0sin x2x2= 030.sin h1 − cos h=sin h1 − cos h1 + cos h1 + cos h=sin h(1 + cos h)1 − cos2 h=1 + cos hsin h; no limit31.t21 − cos2 t=tsin t2, so limt→0t21 − cos2 t= 132. cos(12 π − x) = sin(12 π) sin x = sin x, so limx→0xcos 12 π − x= 133.θ21 − cos θ1 + cos θ1 + cos θ=θ2(1 + cos θ)1 − cos2 θ=θsin θ2(1 + cos θ) so limθ→0θ21 − cos θ= (1)22 = 234.1 − cos 3hcos2 5h − 11 + cos 3h1 + cos 3h=sin23h− sin25h11 + cos 3h, solimx→01 − cos 3hcos2 5h − 1= limx→0sin23h− sin25h11 + cos 3h= −35212= −95035. limx→0+sin1x= limt→+∞sin t; limit does not exist36. limx→0x − 3 limx→0sin xx= −337.2 − cos 3x − cos 4xx=1 − cos 3xx+1 − cos 4xx. Note that1 − cos 3xx=1 − cos 3xx1 + cos 3x1 + cos 3x=sin23xx(1 + cos 3x)=sin 3xxsin 3x1 + cos 3x. Thuslimx→02 − cos 3x − cos 4xx= limx→0sin 3xxsin 3x1 + cos 3x+ limx→0sin 4xxsin 4x1 + cos 4x= 0 + 0 = 0
• Exercise Set 2.6 6938.tan 3x2+ sin25xx2=3cos 3x2sin 3x23x2+ 52 sin25x(5x)2, solimit = limx→03cos 3x2limx→0sin 3x23x2+ 25 limx→0sin 5x5x2= 3 + 25 = 2839. a/b 40. k2s41. 5.1 5.01 5.001 5.0001 5.00001 4.9 4.99 4.999 4.9999 4.999990.098845 0.099898 0.99990 0.099999 0.100000 0.10084 0.10010 0.10001 0.10000 0.10000The limit is 0.1.42. 2.1 2.01 2.001 2.0001 2.00001 1.9 1.99 1.999 1.9999 1.999990.484559 0.498720 0.499875 0.499987 0.499999 0.509409 0.501220 0.500125 0.500012 0.500001The limit is 0.5.43. −1.9 −1.99 −1.999 −1.9999 −1.99999 −2.1 −2.01 −2.001 −2.0001 −2.00001−0.898785 −0.989984 −0.999000 −0.999900 −0.999990 −1.097783 −1.009983 −1.001000 −1.000100 −1.000010The limit is −1.44. −0.9 −0.99 −0.999 −0.9999 −0.99999 −1.1 −1.01 −1.001 −1.0001 −1.000010.405086 0.340050 0.334001 0.333400 0.333340 0.271536 0.326717 0.332667 0.333267 0.333327The limit is 1/3.45. Since limx→0sin(1/x) does not exist, no conclusions can be drawn.46. k = f(0) = limx→0sin 3xx= 3 limx→0sin 3x3x= 3, so k = 347. limx→0−f(x) = k limx→0sin kxkx cos kx= k, limx→0+f(x) = 2k2, so k = 2k2, k =1248. No; sin x/|x| has unequal one-sided limits.49. (a) limt→0+sin tt= 1 (b) limt→0−1 − cos tt= 0 (Theorem 2.6.4)(c) sin(π − t) = sin t, so limx→ππ − xsin x= limt→0tsin t= 150. cosπ2− t = sin t, so limx→2cos(π/x)x − 2= limt→0(π − 2t) sin t4t= limt→0π − 2t4limt→0sin tt=π451. t = x − 1; sin(πx) = sin(πt + π) = − sin πt; and limx→1sin(πx)x − 1= − limt→0sin πtt= −π52. t = x − π/4; tan x − 1 =2 sin tcos t − sin t; limx→π/4tan x − 1x − π/4= limt→02 sin tt(cos t − sin t)= 253. t = x − π/4,cos x − sin xx − π/4= −√2 sin tt; limx→π/4cos x − sin xx − π/4= −√2 limt→0sin tt= −√2
• 70 Chapter 254. limx→0h(x) = L = h(0) so h is continuous at x = 0.Apply the Theorem to h ◦ g to obtain on the one hand h(g(0)) = L, and on the otherh(g(x)) =f(g(x))g(x), x = 0L, x = 0Since f(g(x)) = x and g = f−1this shows that limt→0xf−1(x)= L55. limx→0xsin−1x= limx→0sin xx= 156. tan(tan−1x) = x, so limx→0tan−1xx= limx→0xtan x= (limx→0cos x) limx→0xsin x= 157. 5 limx→0sin−1x5x= 5 limx→05xsin 5x= 558. limx→11x + 1limx→1sin−1(x − 1)x − 1=12limx→1x − 1sin(x − 1)=1259. 3 limx→0e3x− 13x= 360. With y = ln(1 + x), eln(1+x)= 1 + x, x = ey− 1, so limx→0ln(1 + x)x= limx→0xex − 1= 161. (limx→0x) limx→0ex2− 1x2= 0 · 1 = 0 62. 5 limx→0ln(1 + 5x)5x= 5 · 1 = 5 (Exercise 60)63. −|x| ≤ x cos50πx≤ |x| 64. −x2≤ x2sin50π3√x≤ x265. limx→0f(x) = 1 by the Squeezing Theorem–101–1 1xyy = cos xy = 1 – x2y = f (x)66. limx→+∞f(x) = 0 by the Squeezing Theoremyx–1467. Let g(x) = −1xand h(x) =1x; thus limx→+∞sin xx= 0 by the Squeezing Theorem.68. yxyx
• Exercise Set 2.6 7169. (a) sin x = sin t where x is measured in degrees, t is measured in radians and t =πx180. Thuslimx→0sin xx= limt→0sin t(180t/π)=π180.70. cos x = cos t where x is measured in degrees, t in radians, and t =πx180. Thuslimx→01 − cos xx= limt→01 − cos t(180t/π)= 0.71. (a) sin 10◦= 0.17365 (b) sin 10◦= sinπ18≈π18= 0.1745372. (a) cos θ = cos 2α = 1 − 2 sin2(θ/2)≈ 1 − 2(θ/2)2= 1 − 12 θ2(b) cos 10◦= 0.98481(c) cos 10◦= 1 −12π182≈ 0.9847773. (a) 0.08749 (b) tan 5◦≈π36= 0.0872774. (a) h = 52.55 ft(b) Since α is small, tan α◦≈πα180is a good approximation.(c) h ≈ 52.36 ft75. (a) Let f(x) = x − cos x; f(0) = −1, f (π/2) = π/2. By the IVT there must be a solution off(x) = 0.(b) yx00.511.5y = cos xc/2y = x(c) 0.73976. (a) f(x) = x + sin x − 1; f(0) = −1, f (π/6) = π/6 − 1/2 > 0. By the IVT there must be asolution of f(x) = 0 in the interval.(b) yx00.5c/6y = xy = 1 – sin x(c) x = 0.511
• 72 Chapter 277. (a) Gravity is stronger at the poles than at the equator.30 60 909.809.829.84fg(b) Let g(φ) be the given function. Then g(38) < 9.8 and g(39) > 9.8, so by the IntermediateValue Theorem there is a value c between 38 and 39 for which g(c) = 9.8 exactly.78. (a) does not exist (b) the limit is zero(c) For part (a) consider the fact that given any δ > 0 there are inﬁnitely many rational numbersx satisfying |x| < δ and there are inﬁnitely many irrational numbers satisfying the samecondition. Thus if the limit were to exist, it could not be zero because of the rational numbers,and it could not be 1 because of the irrational numbers, and it could not be anything elsebecause of all the numbers. Hence the limit cannot exist. For part (b) use the SqueezingTheorem with +x and −x as the ‘squeezers’.REVIEW EXERCISES CHAPTER 21. (a) 1 (b) no limit (c) no limit(d) 1 (e) 3 (f) 0(g) 0 (h) 2 (i) 1/22. (a) 0.222 . . . , 0.24390, 0.24938, 0.24994, 0.24999, 0.25000; for x = 2, f(x) =1x + 2,so the limit is 1/4.(b) 1.15782, 4.22793, 4.00213, 4.00002, 4.00000, 4.00000; to prove,usetan 4xx=sin 4xx cos 4x=4cos 4xsin 4x4x, the limit is 4.3. (a) x 1 0.1 0.01 0.001 0.0001 0.00001 0.000001f(x) 1.000 0.443 0.409 0.406 0.406 0.405 0.405(b) yx0.5–1 14. x 3.1 3.01 3.001 3.0001 3.00001 3.000001f(x) 5.74 5.56 5.547 5.545 5.5452 5.54518A CAS yields 5.545177445
• Review Exercises Chapter 2 735. 1 6. For x = 1,x3− x2x − 1= x2, so limx→−1x3− x2x − 1= 17. If x = −3 then3x + 9x2 + 4x + 3=3x + 1with limit −328. −∞ 9.253=32310.√x2 + 4 − 2x2√x2 + 4 + 2√x2 + 4 + 2=x2x2(√x2 + 4 + 2)=1√x2 + 4 + 2, solimx→0√x2 + 4 − 2x2= limx→01√x2 + 4 + 2=1411. (a) y = 0 (b) none (c) y = 212. (a)√5, no limit,√10,√10, no limit, +∞, no limit(b) −1, +1, −1, −1, no limit, −1, +113. 1 14. 2 15. 3 − k16. limθ→0tan1 − cos θθ= tan limθ→01 − cos θθ= tan limθ→01 − cos2θθ(1 + cos θ)= tan 0 = 017. +∞ 18. ln(2 sin θ cos θ) − ln tan θ = ln 2 + 2 ln cos θ so the limit is ln 2.19. 1 +3x−x= 1 +3xx/3 (−3)so the limit is e−320. 1 +axbx= 1 +axx/a (ab)so the limit is eab21. \$2,001.60, \$2,009.66, \$2,013.62, \$2013.7523. (a) f(x) = 2x/(x − 1) (b) yx101024. Given any window of height 2 centered at the point x = a, y = L there exists a width 2δ suchthat the window of width 2δ and height 2 contains all points of the graph of the function for x inthat interval.25. (a) limx→2f(x) = 5 (b) 0.004526. δ ≈ 0.07747 (use a graphing utility)
• 74 Chapter 227. (a) |4x − 7 − 1| < 0.01, 4|x − 2| < 0.01, |x − 2| < 0.0025, let δ = 0.0025(b)4x2− 92x − 3− 6 < 0.05, |2x + 3 − 6| < 0.05, |x − 1.5| < 0.025, take δ = 0.025(c) |x2− 16| < 0.001; if δ < 1 then |x + 4| < 9 if |x − 4| < 1; then |x2− 16| = |x − 4||x + 4| ≤9|x − 4| < 0.001 provided |x − 4| < 0.001/9, take δ = 0.0001, then |x2− 16| < 9|x − 4| <9(0.0001) = 0.0009 < 0.00128. (a) Given > 0 then |4x − 7 − 1| < provided |x − 2| < /4, take δ = /4(b) Given > 0 the inequality4x2− 92x − 3− 6 < holds if |2x + 3 − 6| < , |x − 1.5| < /2, takeδ = /229. Let = f(x0)/2 > 0; then there corresponds δ > 0 such that if |x−x0| < δ then |f(x)−f(x0)| < ,− < f(x) − f(x0) < , f(x) > f(x0) − = f(x0)/2 > 0 for x0 − δ < x < x0 + δ.30. (a) x 1.1 1.01 1.001 1.0001 1.00001 1.000001f(x) 0.49 0.54 0.540 0.5403 0.54030 0.54030(b) cos 131. (a) f is not deﬁned at x = ±1, continuous elsewhere(b) none(c) f is not deﬁned at x = 0, −332. (a) continuous everywhere except x = ±3(b) deﬁned and continuous for x ≤ −1, x ≥ 1(c) continuous for x > 033. For x < 2 f is a polynomial and is continuous; for x > 2 f is a polynomial and is continuous. Atx = 2, f(2) = −13 = 13 = limx→2+f(x) so f is not continuous there.35. f(x) = −1 for a ≤ x <a + b2and f(x) = 1 fora + b2≤ x ≤ b36. If, on the contrary, f(x0) < 0 for some x0 in [0, 1], then by the Intermediate Value Theorem wewould have a solution of f(x) = 0 in [0, x0], contrary to the hypothesis.37. f(−6) = 185, f(0) = −1, f(2) = 65; apply Theorem 2.4.8 twice, once on [−6, 0] and once on [0, 2]
• 75CHAPTER 3The DerivativeEXERCISE SET 3.11. (a) mtan = (50 − 10)/(15 − 5)= 40/10= 4 m/s(b)t (s)410 20v (m/s)2. (a) mtan ≈ (90 − 0)/(10 − 2)= 90/8= 11.25 m/s(b) mtan ≈ (140 − 0)/(10 − 4)= 140/6≈ 23.33 m/s3. (a) mtan = (600 − 0)/(20 − 2.2)= 600/17.8≈ 33.71 m/s(b) mtan ≈ (820 − 600)/(20 − 16)= 220/4= 55 m/sThe speed is increasing with time.4. (a) (10 − 10)/(3 − 0) = 0 cm/s(b) t = 0, t = 2, and t = 4.2 (horizontal tangent line)(c) maximum: t = 1 (slope > 0) minimum: t = 3 (slope < 0)(d) (3 − 18)/(4 − 2) = −7.5 cm/s (slope of estimated tangent line to curve at t = 3)5. From the ﬁgure:tst0 t1 t2(a) The particle is moving faster at time t0 because the slope of the tangent to the curve at t0 isgreater than that at t2.(b) The initial velocity is 0 because the slope of a horizontal line is 0.(c) The particle is speeding up because the slope increases as t increases from t0 to t1.(d) The particle is slowing down because the slope decreases as t increases from t1 to t2.6.tst0 t17. It is a straight line with slope equal to the velocity.8. (a) decreasing (slope of tangent line decreases with increasing time)(b) increasing (slope of tangent line increases with increasing time)(c) increasing (slope of tangent line increases with increasing time)(d) decreasing (slope of tangent line decreases with increasing time)
• 76 Chapter 39. (a) msec =f(1) − f(0)1 − 0=21= 2(b) mtan = limx1→0f(x1) − f(0)x1 − 0= limx1→02x21 − 0x1 − 0= limx1→02x1 = 0(c) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x02x21 − 2x20x1 − x0= limx1→x0(2x1 + 2x0)= 4x0(d)–2 –1 1 2–1123xySecantTangent10. (a) msec =f(2) − f(1)2 − 1=23− 131= 7(b) mtan = limx1→1f(x1) − f(1)x1 − 1= limx1→1x31 − 1x1 − 1= limx1→1(x1 − 1)(x21 + x1 + 1)x1 − 1= limx1→1(x21 + x1 + 1) = 3(c) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x0x31 − x30x1 − x0= limx1→x0(x21 + x1x0 + x20)= 3x20(d)xySecantTangent5911. (a) msec =f(3) − f(2)3 − 2=1/3 − 1/21= −16(b) mtan = limx1→2f(x1) − f(2)x1 − 2= limx1→21/x1 − 1/2x1 − 2= limx1→22 − x12x1(x1 − 2)= limx1→2−12x1= −14(c) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x01/x1 − 1/x0x1 − x0= limx1→x0x0 − x1x0x1(x1 − x0)= limx1→x0−1x0x1= −1x20(d)xySecantTangent14
• Exercise Set 3.1 7712. (a) msec =f(2) − f(1)2 − 1=1/4 − 11= −34(b) mtan = limx1→1f(x1) − f(1)x1 − 1= limx1→11/x21 − 1x1 − 1= limx1→11 − x21x21(x1 − 1)= limx1→1−(x1 + 1)x21= −2(c) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x01/x21 − 1/x20x1 − x0= limx1→x0x20 − x21x20x21(x1 − x0)= limx1→x0−(x1 + x0)x20x21= −2x30(d)xyTangent Secant1213. (a) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x0(x21 − 1) − (x20 − 1)x1 − x0= limx1→x0(x21 − x20)x1 − x0= limx1→x0(x1 + x0) = 2x0(b) mtan = 2(−1) = −214. (a) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x0(x21 + 3x1 + 2) − (x20 + 3x0 + 2)x1 − x0= limx1→x0(x21 − x20) + 3(x1 − x0)x1 − x0= limx1→x0(x1 + x0 + 3) = 2x0 + 3(b) mtan = 2(2) + 3 = 715. (a) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x0√x1 −√x0x1 − x0= limx1→x01√x1 +√x0=12√x0(b) mtan =12√1=1216. (a) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x01/√x1 − 1/√x0x1 − x0= limx1→x0√x0 −√x1√x0√x1 (x1 − x0)= limx1→x0−1√x0√x1 (√x1 +√x0 )= −12x3/20(b) mtan = −12(4)3/2= −11617. (a) 72◦F at about 4:30 P.M. (b) about (67 − 43)/6 = 4◦F/h(c) decreasing most rapidly at about 9 P.M.; rate of change of temperature is about −7◦F/h(slope of estimated tangent line to curve at 9 P.M.)
• 78 Chapter 318. For V = 10 the slope of the tangent line is about −0.25 atm/L, for V = 25 the slope isabout −0.04 atm/L.19. (a) during the ﬁrst year after birth(b) about 6 cm/year (slope of estimated tangent line at age 5)(c) the growth rate is greatest at about age 14; about 10 cm/year(d)t (yrs)Growth rate(cm/year)5 10 15 201020304020. (a) The rock will hit the ground when 16t2= 576, t2= 36, t = 6 s (only t ≥ 0 is meaningful)(b) vave =16(6)2− 16(0)26 − 0= 96 ft/s(c) vave =16(3)2− 16(0)23 − 0= 48 ft/s(d) vinst = limt1→616t21 − 16(6)2t1 − 6= limt1→616(t21 − 36)t1 − 6= limt1→616(t1 + 6) = 192 ft/s21. (a) (40)3/√10 = 20,238.6 ft (b) vave = 20,238.6/40 = 505.96 ft/s(c) Solve s = t3/√10 = 135, t ≈ 7.53 so vave = 135/7.53 = 17.93 ft/s.(d) vinst = limt1→40t31/√10 − (40)3/√10t1 − 40= limt1→40(t31 − 403)(t1 − 40)√10= limt1→401√10(t21 + 40t1 + 1600) = 1517.89 ft/s22. (a) vave =4.5(12)2− 4.5(0)212 − 0= 54 ft/s(b) vinst = limt1→64.5t21 − 4.5(6)2t1 − 6= limt1→64.5(t21 − 36)t1 − 6= limt1→64.5(t1 + 6)(t1 − 6)t1 − 6= limt1→64.5(t1 + 6) = 54 ft/s23. (a) vave =6(4)4− 6(2)44 − 2= 720 ft/min(b) vinst = limt1→26t41 − 6(2)4t1 − 2= limt1→26(t41 − 16)t1 − 2= limt1→26(t21 + 4)(t21 − 4)t1 − 2= limt1→26(t21 + 4)(t1 + 2) = 192 ft/min
• Exercise Set 3.2 79EXERCISE SET 3.21. f (1) = 2, f (3) = 0, f (5) = −2, f (6) = −12. f (4) < f (0) < f (2) < 0 < f (−3)3. (b) m = f (2) = 3 (c) the same, f (2) = 34. m =2 − (−1)1 − (−1)=32, y − 2 = m(x − 1), y =32x +125.xy–16.xy7. y − (−1) = 5(x − 3), y = 5x − 16 8. y − 3 = −4(x + 2), y = −4x − 59. f (1) = limh→0f(1 + h) − f(1)h= limh→02(1 + h)2− 2 · 12h= limh→04h + 2h2h= 4;y − 2 = 4(x − 1), y = 4x − 210. f (−1) = limh→0f((−1) + h) − f(−1)h= limh→01/(h − 1)2− 1h= limh→01 − (h − 1)2h(h − 1)2= limh→02h − h2h(h − 1)2= limh→02 − h(h − 1)2= 2;y − 1 = 2(x + 1), y = 2x + 311. f (0) = limh→0f(0 + h) − f(0)h= limh→0h3− 03h= limh→0h2= 0;f (0) = 0 so y − 0 = (0)(x − 0), y = 012. f (−1) = limh→0f(−1 + h) − f(−1)h= limh→02(−1 + h)3+ 1 − (2(−1)3+ 1)h= limh→0(6 − 6h + 2h2) = 6;f(−1) = 2(−1)3+ 1 = −1, f (−1) = 6 so y + 1 = 6(x + 1), y = 6x + 513. f (8) = limh→0f(8 + h) − f(8)h= limh→0√9 + h − 3h= limh→0√9 + h − 3h√9 + h + 3√9 + h + 3= limh→0hh(√9 + h + 3)=16;f(8) =√8 + 1 = 3, f (8) =16so y − 3 =16(x − 8), y =16x +53
• 80 Chapter 314. f (4) = limh→0f(4 + h) − f(4)h= limh→02(4 + h) + 1 −√2 · 4 + 1h√9 + 2h + 3√9 + 2h + 3= limh→02hh(√9 + 2h + 3)= limh→02√9 + 2h + 3=13f(4) = 2(4) + 1 =√9 = 3, f (4) = 1/3 so y − 3 =13(x − 4), y =13x +5315. f (x) = lim∆x→01x + ∆x−1x∆x= lim∆x→0x − (x + ∆x)x(x + ∆x)∆x= lim∆x→0−∆xx∆x(x + ∆x)= lim∆x→0−1x(x + ∆x)= −1x216. f (x) = lim∆x→01(x + ∆x) + 1−1x + 1∆x= lim∆x→0(x + 1) − (x + ∆x + 1)(x + 1)(x + ∆x + 1)∆x= lim∆x→0x + 1 − x − ∆x − 1∆x(x + 1)(x + ∆x + 1)= lim∆x→0−∆x∆x(x + 1)(x + ∆x + 1)= lim∆x→0−1(x + 1)(x + ∆x + 1)= −1(x + 1)217. f (x) = lim∆x→0(x + ∆x)2− (x + ∆x) − (x2− x)∆x= lim∆x→02x∆x + ∆x2− ∆x∆x= lim∆x→0(2x − 1 + ∆x) = 2x − 118. f (x) = lim∆x→0(x + ∆x)4− x4∆x= lim∆x→04x3∆x + 6x2(∆x)2+ 4x(∆x)3+ (∆x)4∆x= lim∆x→0(4x3+ 6x2∆x + 4x(∆x)2+ (∆x)3) = 4x319. f (x) = lim∆x→01√x + ∆x−1√x∆x= lim∆x→0√x −√x + ∆x∆x√x√x + ∆x= lim∆x→0x − (x + ∆x)∆x√x√x + ∆x(√x +√x + ∆x)= lim∆x→0−1√x√x + ∆x(√x +√x + ∆x)= −12x3/220. f (x) = lim∆x→01√x + ∆x − 1−1√x − 1∆x= lim∆x→0√x − 1 −√x + ∆x − 1∆x√x − 1√x + ∆x − 1√x − 1 +√x + ∆x − 1√x − 1 +√x + ∆x − 1= lim∆x→0−∆x∆x√x − 1√x + ∆x − 1(√x − 1 +√x + ∆x − 1)= lim∆x→0−1√x − 1√x + ∆x − 1(√x − 1 +√x + ∆x − 1)= −12(x − 1)3/2
• Exercise Set 3.2 8121. f (t) = limh→0f(t + h) − f(t)h= limh→0[4(t + h)2+ (t + h)] − [4t2+ t]h= limh→04t2+ 8th + 4h2+ t + h − 4t2− th= limh→08th + 4h2+ hh= limh→0(8t + 4h + 1) = 8t + 122.dVdr= limh→043π(r + h)3−43πr3h= limh→043π(r3+ 3r2h + 3rh2+ h3− r3)h= limh→043π(3r2+ 3rh + h2) = 4πr223. (a) D (b) F (c) B (d) C (e) A (f) E24. The point (√2/2,√2/2) lies on the line y = x.xm = –1y125. (a)xy (b)xy–1(c)xy1 226. (a)xy (b)xy (c)xy27. (a) f(x) =√x and a = 1 (b) f(x) = x2and a = 328. (a) f(x) = cos x and a = π (b) f(x) = x7and a = 129.dydx= limh→0(1 − (x + h)2) − (1 − x2)h= limh→0−2xh − h2h= limh→0−2(x + h) = −2x,anddydx x=1= −2
• 82 Chapter 330.dydx= limh→0x + 2 + hx + h−x + 2xh= limh→0x(x + 2 + h) − (x + 2)(x + h)hx(x + h)= limh→0−2x(x + h)=−2x2,and thusdydx x=−2= −1231. y = −2x + 15–2 2–332. 1.500 2.533. (b) h 0.5 0.1 0.01 0.001 0.0001 0.00001(f(1 + h) − f(1))/h 1.6569 1.4355 1.3911 1.3868 1.3863 1.386334. (b) h 0.5 0.1 0.01 0.001 0.0001 0.00001(f(1 + h) − f(1))/h 0.50489 0.67060 0.70356 0.70675 0.70707 0.7071035. (a) dollars/ft(b) As you go deeper the price per foot may increase dramatically, so f (x) is roughly the priceper additional foot.(c) If each additional foot costs extra money (this is to be expected) then f (x) remains positive.(d) From the approximation 1000 = f (300) ≈f(301) − f(300)301 − 300we see that f(301) ≈ f(300) + 1000, so the extra foot will cost around \$1000.36. (a) gallons/dollar(b) The increase in the amount of paint that would be sold for one extra dollar.(c) It should be negative since an increase in the price of paint would decrease the amount ofpaint sold.(d) From −100 = f (10) ≈f(11) − f(10)11 − 10we see that f(11) ≈ f(10) − 100, so an increase of onedollar would decrease the amount of paint sold by around 100 gallons.37. (a) F ≈ 200 lb, dF/dθ ≈ 50 lb/rad (b) µ = (dF/dθ)/F ≈ 50/200 = 0.2538. The derivative at time t = 100 of the velocity with respect to time is equal to the slope of thetangent line, which is approximately m ≈10272 − 0120 − 40= 128.4 ft/s2. Thus the mass is approximatelyM(100) ≈Tdv/dt=7680982128.4≈ 59820.73 lb/ft/s2.39. (a) T ≈ 115◦F, dT/dt ≈ −3.35◦F/min(b) k = (dT/dt)/(T − T0) ≈ (−3.35)/(115 − 75) = −0.084
• Exercise Set 3.2 8341. limx→0f(x) = limx→03√x = 0 = f(0), so f is continuous at x = 0.limh→0f(0 + h) − f(0)h= limh→03√h − 0h= limh→01h2/3= +∞, sof (0) does not exist.xy2–2242. limx→2f(x) = limx→2(x − 2)2/3= 0 = f(2) so f is continuous atx = 2. limh→0f(2 + h) − f(2)h= limh→0h2/3− 0h= limh→01h1/3which does not exist so f (2) does not exist.xy2543. limx→1−f(x) = limx→1+f(x) = f(1), so f is continuous at x = 1.limh→0−f(1 + h) − f(1)h= limh→0−[(1 + h)2+ 1] − 2h= limh→0−(2 + h) = 2;limh→0+f(1 + h) − f(1)h= limh→0+2(1 + h) − 2h= limh→0+2 = 2, so f (1) = 2.xy3–3544. limx→1−f(x) = limx→1+f(x) = f(1) so f is continuous at x = 1.limh→0−f(1 + h) − f(1)h= limh→0−[(1 + h)2+ 2] − 3h= limh→0−(2 + h) = 2;limh→0+f(1 + h) − f(1)h= limh→0+[(1 + h) + 2] − 3h= limh→0+1 = 1,so f (1) does not exist.xy3–3545. Since −|x| ≤ x sin(1/x) ≤ |x| it follows by the Squeezing Theorem(Theorem 2.6.3) that limx→0x sin(1/x) = 0. The derivative cannotexist: considerf(x) − f(0)x= sin(1/x). This function oscillatesbetween −1 and +1 and does not tend to zero as x tends to zero.xy46. For continuity, compare with ±x2to establish that the limit is zero. The diﬀerential quotient isx sin(1/x) and (see Exercise 45) this has a limit of zero at the origin.
• 84 Chapter 347. Let = |f (x0)/2|. Then there exists δ > 0 such that if 0 < |x − x0| < δ, thenf(x) − f(x0)x − x0− f (x0) < . Since f (x0) > 0 and = f (x0)/2 it follows thatf(x) − f(x0)x − x0> > 0. If x = x1 < x0 then f(x1) < f(x0) and if x = x2 > x0 then f(x2) > f(x0).48. g (x1) = limh→0g(x1 + h) − g(x1)h= limh→0f(m(x1 + h) + b) − f(mx1 + b)h= m limh→0f(x0 + mh) − f(x0)mh= mf (x0)49. (a) Let = |m|/2. Since m = 0, > 0. Since f(0) = f (0) = 0 we know there exists δ > 0such thatf(0 + h) − f(0)h< whenever |h| < δ. It follows that |f(h)| < 12 |hm| for |h| < δ.Replace h with x to get the result.(b) For |x| < δ, |f(x)| < 12 |mx|. Moreover |mx| = |mx − f(x) + f(x)| ≤ |f(x) − mx| + |f(x)|,which yields |f(x) − mx| ≥ |mx| − |f(x)| > 12 |mx| > |f(x)|, i.e. |f(x) − mx| > |f(x)|.(c) If any straight line y = mx + b is to approximate the curve y = f(x) for small values of x,then b = 0 since f(0) = 0. The inequality |f(x) − mx| > |f(x)| can also be interpreted as|f(x) − mx| > |f(x) − 0|, i.e. the line y = 0 is a better approximation than is y = mx.50. If m = f (x0) then let = 12 |f (x0) − m|. Then there exists δ > 0 such that if |x − x0| < δ thenf(x) − f(x0)x − x0− f (x0) <12|f (x0) − m|. Then |mx| ≤ |f(x) − mx| + |f(x)|,so |f(x) − mx| >|mx| − |f(x)| ≥ 12 |mx| > |f(x)|, i.e. |f(x) − mx| > |f(x)|, which shows that the straight line y = 0is a better approximating straight line than any other choice y = mx + b.EXERCISE SET 3.31. 28x62. −36x113. 24x7+ 2 4. 2x35. 0 6.√2 7. −13(7x6+ 2) 8.25x9. −3x−4− 7x−810.12√x−1x211. 24x−9+ 1/√x 12. −42x−7−52√x13. 12x(3x2+ 1) 14. f(x) = x10+ 4x6+ 4x2, f (x) = 10x9+ 24x5+ 8x15. 3ax2+ 2bx + c 16.1a2x +1b17. y = 10x − 3, y (1) = 718. y =12√x−2x2, y (1) = −3/2 19. 2t − 1 20.13−13t221. dy/dx = 1 + 2x + 3x2+ 4x3+ 5x4, dy/dx|x=1 = 1522.dydx=−3x4−2x3−1x2+ 1 + 2x + 3x2,dydx x=1= 0
• Exercise Set 3.3 8523. y = (1 − x2)(1 + x2)(1 + x4) = (1 − x4)(1 + x4) = 1 − x8dydx= −8x7, dy/dx|x=1 = −824. dy/dx = 24x23+ 24x11+ 24x7+ 24x5, dy/dx|x=1 = 9625. f (1) ≈f(1.01) − f(1)0.01=−0.999699 − (−1)0.01= 0.0301, and by diﬀerentiation, f (1) = 3(1)2−3 = 026. f (1) ≈f(1.01) − f(1)0.01≈0.980296 − 10.01≈ −1.9704, and by diﬀerentiation, f (1) = −2/13= −22.10.9 1.11.927. From the graph, f (1) ≈2.0088 − 2.01111.0981 − 0.9000≈ −0.0116,and by diﬀerentiation, f (1) = 03.30 1.1028. f(x) =√x + 2x, f (1) ≈3.2488 − 2.74871.1 − 0.9≈ 2.5006,f (x) =12√x+ 2, f (1) =5229. 32t 30. 2π 31. 3πr232. −2α−2+ 133. (a)dVdr= 4πr2(b)dVdr r=5= 4π(5)2= 100π34.ddλλλ0 + λ62 − λ0=12 − λ0ddλ(λλ0 + λ6) =12 − λ0(λ0 + 6λ5) =λ0 + 6λ52 − λ035. y − 2 = 5(x + 3), y = 5x + 17 36. y + 2 = −(x − 2), y = −x37. (a) dy/dx = 21x2− 10x + 1, d2y/dx2= 42x − 10(b) dy/dx = 24x − 2, d2y/dx2= 24(c) dy/dx = −1/x2, d2y/dx2= 2/x3(d) y = 35x5− 16x3− 3x, dy/dx = 175x4− 48x2− 3, d2y/dx2= 700x3− 96x38. (a) y = 28x6− 15x2+ 2, y = 168x5− 30x(b) y = 3, y = 0(c) y =25x2, y = −45x3(d) y = 2x4+ 3x3− 10x − 15, y = 8x3+ 9x2− 10, y = 24x2+ 18x
• 86 Chapter 339. (a) y = −5x−6+ 5x4, y = 30x−7+ 20x3, y = −210x−8+ 60x2(b) y = x−1, y = −x−2, y = 2x−3, y = −6x−4(c) y = 3ax2+ b, y = 6ax, y = 6a40. (a) dy/dx = 10x − 4, d2y/dx2= 10, d3y/dx3= 0(b) dy/dx = −6x−3− 4x−2+ 1, d2y/dx2= 18x−4+ 8x−3, d3y/dx3= −72x−5− 24x−4(c) dy/dx = 4ax3+ 2bx, d2y/dx2= 12ax2+ 2b, d3y/dx3= 24ax41. (a) f (x) = 6x, f (x) = 6, f (x) = 0, f (2) = 0(b)dydx= 30x4− 8x,d2ydx2= 120x3− 8,d2ydx2x=1= 112(c)ddxx−3= −3x−4,d2dx2x−3= 12x−5,d3dx3x−3= −60x−6,d4dx4x−3= 360x−7,d4dx4x−3x=1= 36042. (a) y = 16x3+ 6x2, y = 48x2+ 12x, y = 96x + 12, y (0) = 12(b) y = 6x−4,dydx= −24x−5,d2ydx2= 120x−6,d3ydx3= −720x−7,d4ydx4= 5040x−8,d4ydx4x=1= 504043. y = 3x2+ 3, y = 6x, and y = 6 soy + xy − 2y = 6 + x(6x) − 2(3x2+ 3) = 6 + 6x2− 6x2− 6 = 044. y = x−1, y = −x−2, y = 2x−3sox3y + x2y − xy = x3(2x−3) + x2(−x−2) − x(x−1) = 2 − 1 − 1 = 045. The graph has a horizontal tangent at points wheredydx= 0,butdydx= x2− 3x + 2 = (x − 1)(x − 2) = 0 if x = 1, 2. Thecorresponding values of y are 5/6 and 2/3 so the tangentline is horizontal at (1, 5/6) and (2, 2/3).1.500 346. Find where f (x) = 0; f (x) = 1 − 9/x2= 0, x2= 9, x = ±3. 14–7 7–14
• Exercise Set 3.3 8747. The y-intercept is −2 so the point (0, −2) is on the graph; −2 = a(0)2+ b(0) + c, c = −2. Thex-intercept is 1 so the point (1,0) is on the graph; 0 = a + b − 2. The slope is dy/dx = 2ax + b; atx = 0 the slope is b so b = −1, thus a = 3. The function is y = 3x2− x − 2.48. Let P(x0, y0) be the point where y = x2+k is tangent to y = 2x. The slope of the curve isdydx= 2xand the slope of the line is 2 thus at P, 2x0 = 2 so x0 = 1. But P is on the line, so y0 = 2x0 = 2.Because P is also on the curve we get y0 = x20 + k so k = y0 − x20 = 2 − (1)2= 1.49. The points (−1, 1) and (2, 4) are on the secant line so its slope is (4 − 1)/(2 + 1) = 1. The slopeof the tangent line to y = x2is y = 2x so 2x = 1, x = 1/2.50. The points (1, 1) and (4, 2) are on the secant line so its slope is 1/3. The slope of the tangent lineto y =√x is y = 1/(2√x) so 1/(2√x) = 1/3, 2√x = 3, x = 9/4.51. y = −2x, so at any point (x0, y0) on y = 1 − x2the tangent line is y − y0 = −2x0(x − x0), ory = −2x0x + x20 + 1. The point (2, 0) is to be on the line, so 0 = −4x0 + x20 + 1, x20 − 4x0 + 1 = 0.Use the quadratic formula to get x0 =4 ±√16 − 42= 2 ±√3.52. Let P1(x1, ax21) and P2(x2, ax22) be the points of tangency. y = 2ax so the tangent lines at P1 andP2 are y − ax21 = 2ax1(x − x1) and y − ax22 = 2ax2(x − x2). Solve for x to get x = 12 (x1 + x2)which is the x-coordinate of a point on the vertical line halfway between P1 and P2.53. y = 3ax2+ b; the tangent line at x = x0 is y − y0 = (3ax20 + b)(x − x0) where y0 = ax30 + bx0.Solve with y = ax3+ bx to get(ax3+ bx) − (ax30 + bx0) = (3ax20 + b)(x − x0)ax3+ bx − ax30 − bx0 = 3ax20x − 3ax30 + bx − bx0x3− 3x20x + 2x30 = 0(x − x0)(x2+ xx0 − 2x20) = 0(x − x0)2(x + 2x0) = 0, so x = −2x0.54. Let (x0, y0) be the point of tangency. Note that y0 = 1/x0. Since y = −1/x2, the tangent linehas the form y − y0 = (−1/x20)(x − x0), or y −1x0= −1x20x +1x0or y = −1x20x +2x0, withintercepts at 0,2x0and (2x0, 0). The distance from the y-intercept to the point of tangencyis (0 − x0)2 + (y0 − 2y0)2, and the distance from the x-intercept to the point of tangency is(x0 − 2x0)2 + y20 so that they are equal (and equal the distance from the point of tangency tothe origin).55. y = −1x2; the tangent line at x = x0 is y − y0 = −1x20(x − x0), or y = −xx20+2x0. The tangent linecrosses the x-axis at 2x0, the y-axis at 2/x0, so that the area of the triangle is12(2/x0)(2x0) = 2.56. f (x) = 3ax2+ 2bx + c; there is a horizontal tangent where f (x) = 0. Use the quadratic formulaon 3ax2+ 2bx + c = 0 to get x = (−b ±√b2 − 3ac)/(3a) which gives two real solutions, one realsolution, or none if(a) b2− 3ac > 0 (b) b2− 3ac = 0 (c) b2− 3ac < 057. F = GmMr−2,dFdr= −2GmMr−3= −2GmMr3
• 88 Chapter 358. dR/dT = 0.04124 − 3.558 × 10−5T which decreases as T increases from 0 to 700. When T = 0,dR/dT = 0.04124 Ω/◦C; when T = 700, dR/dT = 0.01633 Ω/◦C. The resistance is most sensitiveto temperature changes at T = 0◦C, least sensitive at T = 700◦C.59. f (x) = 1 + 1/x2> 0 for all x = 06–6 6–660. f (x) = 3x2− 3 = 0 when x = ±1;increasing for −∞ < x < −1and 1 < x < +∞–2 –1 1 2–22xy(1, –2)(–1, 2)61. f is continuous at 1 because limx→1−f(x) = limx→1+f(x) = f(1); also limx→1−f (x) = limx→1−(2x + 1) = 3and limx→1+f (x) = limx→1+3 = 3 so f is diﬀerentiable at 1.62. f is not continuous at x = 9 because limx→9−f(x) = −63 and limx→9+f(x) = 36.f cannot be diﬀerentiable at x = 9, for if it were, then f would also be continuous, which it is not.63. f is continuous at 1 because limx→1−f(x) = limx→1+f(x) = f(1), also limx→1−f (x) = limx→1−2x = 2 andlimx→1+f (x) = limx→1+12√x=12so f is not diﬀerentiable at 1.64. f is continuous at 1/2 because limx→1/2−f(x) = limx→1/2+f(x) = f(1/2), alsolimx→1/2−f (x) = limx→1/2−3x2= 3/4 and limx→1/2+f (x) = limx→1/2+3x/2 = 3/4 so f (1/2) = 3/4, andf is diﬀerentiable at x = 1/2.65. (a) f(x) = 3x − 2 if x ≥ 2/3, f(x) = −3x + 2 if x < 2/3 so f is diﬀerentiable everywhereexcept perhaps at 2/3. f is continuous at 2/3, also limx→2/3−f (x) = limx→2/3−(−3) = −3 andlimx→2/3+f (x) = limx→2/3+(3) = 3 so f is not diﬀerentiable at x = 2/3.(b) f(x) = x2− 4 if |x| ≥ 2, f(x) = −x2+ 4 if |x| < 2 so f is diﬀerentiable everywhereexcept perhaps at ±2. f is continuous at −2 and 2, also limx→2−f (x) = limx→2−(−2x) = −4and limx→2+f (x) = limx→2+(2x) = 4 so f is not diﬀerentiable at x = 2. Similarly, f is notdiﬀerentiable at x = −2.66. (a) f (x) = −(1)x−2, f (x) = (2 · 1)x−3, f (x) = −(3 · 2 · 1)x−4f(n)(x) = (−1)n n(n − 1)(n − 2) · · · 1xn+1(b) f (x) = −2x−3, f (x) = (3 · 2)x−4, f (x) = −(4 · 3 · 2)x−5f(n)(x) = (−1)n (n + 1)(n)(n − 1) · · · 2xn+2
• Exercise Set 3.4 8967. (a)d2dx2[cf(x)] =ddxddx[cf(x)] =ddxcddx[f(x)] = cddxddx[f(x)] = cd2dx2[f(x)]d2dx2[f(x) + g(x)] =ddxddx[f(x) + g(x)] =ddxddx[f(x)] +ddx[g(x)]=d2dx2[f(x)] +d2dx2[g(x)](b) yes, by repeated application of the procedure illustrated in Part (a)68. limh→0f (2 + h) − f (2)h= f (2); f (x) = 8x7− 2, f (x) = 56x6, so f (2) = 56(26) = 3584.69. (a) f (x) = nxn−1, f (x) = n(n − 1)xn−2, f (x) = n(n − 1)(n − 2)xn−3, . . .,f(n)(x) = n(n − 1)(n − 2) · · · 1(b) from Part (a), f(k)(x) = k(k − 1)(k − 2) · · · 1 so f(k+1)(x) = 0 thus f(n)(x) = 0 if n > k(c) from Parts (a) and (b), f(n)(x) = ann(n − 1)(n − 2) · · · 170. (a) If a function is diﬀerentiable at a point then it is continuous at that point, thus f is continuouson (a, b) and consequently so is f.(b) f and all its derivatives up to f(n−1)(x) are continuous on (a, b)71. Let g(x) = xn, f(x) = (mx + b)n. Use Exercise 48 in Section 3.2, but with f and g permuted. Ifx0 = mx1 + b then Exercise 48 says that f is diﬀerentiable at x1 and f (x1) = mg (x0). Sinceg (x0) = nxn−10 , the result follows.72. f (x) = 2 · 2(2x + 3) = 8x + 12 73. f (x) = 3 · 3(3x − 1)2= 81x2− 54x + 974. f (x) = 1 · (−1)(x − 1)−2= −1/(x − 1)275. f (x) = 2 · 3 · (−2)(2x + 1)−3= −12/(2x + 1)376. f(x) =x + 1 − 1x + 1= 1 −1x + 1, and f (x) = −1(x + 1)−2= −1/(x + 1)277. f(x) =2x2+ 4x + 2 + 1(x + 1)2= 2 +1(x + 1)2, so f (x) = −2(x + 1)−3= −2/(x + 1)3EXERCISE SET 3.41. (a) f(x) = 2x2+ x − 1, f (x) = 4x + 1(b) f (x) = (x + 1) · (2) + (2x − 1) · (1) = 4x + 12. (a) f(x) = 3x4+ 5x2− 2, f (x) = 12x3+ 10x(b) f (x) = (3x2− 1) · (2x) + (x2+ 2) · (6x) = 12x3+ 10x3. (a) f(x) = x4− 1, f (x) = 4x3(b) f (x) = (x2+ 1) · (2x) + (x2− 1) · (2x) = 4x34. (a) f(x) = x3+ 1, f (x) = 3x2(b) f (x) = (x + 1)(2x − 1) + (x2− x + 1) · (1) = 3x2
• 90 Chapter 35. f (x) = (3x2+ 6)ddx2x −14+ 2x −14ddx(3x2+ 6) = (3x2+ 6)(2) + 2x −14(6x)= 18x2−32x + 126. f (x) = (2 − x − 3x3)ddx(7 + x5) + (7 + x5)ddx(2 − x − 3x3)= (2 − x − 3x3)(5x4) + (7 + x5)(−1 − 9x2)= −24x7− 6x5+ 10x4− 63x2− 77. f (x) = (x3+ 7x2− 8)ddx(2x−3+ x−4) + (2x−3+ x−4)ddx(x3+ 7x2− 8)= (x3+ 7x2− 8)(−6x−4− 4x−5) + (2x−3+ x−4)(3x2+ 14x)= −15x−2− 14x−3+ 48x−4+ 32x−58. f (x) = (x−1+ x−2)ddx(3x3+ 27) + (3x3+ 27)ddx(x−1+ x−2)= (x−1+ x−2)(9x2) + (3x3+ 27)(−x−2− 2x−3) = 3 + 6x − 27x−2− 54x−39. f (x) = x2+ 2x + 4 + (2x + 2)(x − 2) = 3x210. f(x) = x4− x2, f (x) = 4x3− 2x11.dydx=(5x − 3)ddx(1) − (1)ddx(5x − 3)(5x − 3)2= −5(5x − 3)2; y (1) = −5/412.dydx=(√x + 2)ddx(3) − 3ddx(√x + 2)(√x + 2)2= −3/(2√x(√x + 2)2); y (1) = −3/18 = −1/613.dydx=(x + 3)ddx(2x − 1) − (2x − 1)ddx(x + 3)(x + 3)2=(x + 3)(2) − (2x − 1)(1)(x + 3)2=7(x + 3)2;dydx x=1=71614.dydx=(x2− 5)ddx(4x + 1) − (4x + 1)ddx(x2− 5)(x2 − 5)2=(x2− 5)(4) − (4x + 1)(2x)(x2 − 5)2= −4x2+ 2x + 20(x2 − 5)2;dydx x=1=13815.dydx=3x + 2xddxx−5+ 1 + x−5+ 1ddx3x + 2x=3x + 2x−5x−6+ x−5+ 1x(3) − (3x + 2)(1)x2=3x + 2x−5x−6+ x−5+ 1 −2x2;dydx x=1= 5(−5) + 2(−2) = −29
• Exercise Set 3.4 9116.dydx= (2x7− x2)ddxx − 1x + 1+x − 1x + 1ddx(2x7− x2)= (2x7− x2)(x + 1)(1) − (x − 1)(1)(x + 1)2+x − 1x + 1(14x6− 2x)= (2x7− x2) ·2(x + 1)2+x − 1x + 1(14x6− 2x);dydx x=1= (2 − 1)24+ 0(14 − 2) =1217. f (1) = 0 18. f (1) = 119. (a) g (x) =√xf (x) +12√xf(x), g (4) = (2)(−5) +14(3) = −37/4(b) g (x) =xf (x) − f(x)x2, g (4) =(4)(−5) − 316= −23/1620. (a) g (x) = 6x − 5f (x), g (3) = 6(3) − 5(4) = −2(b) g (x) =2f(x) − (2x + 1)f (x)f2(x), g (3) =2(−2) − 7(4)(−2)2= −821. (a) F (x) = 5f (x) + 2g (x), F (2) = 5(4) + 2(−5) = 10(b) F (x) = f (x) − 3g (x), F (2) = 4 − 3(−5) = 19(c) F (x) = f(x)g (x) + g(x)f (x), F (2) = (−1)(−5) + (1)(4) = 9(d) F (x) = [g(x)f (x) − f(x)g (x)]/g2(x), F (2) = [(1)(4) − (−1)(−5)]/(1)2= −122. (a) F (x) = 6f (x) − 5g (x), F (π) = 6(−1) − 5(2) = −16(b) F (x) = f(x) + g(x) + x(f (x) + g (x)), F (π) = 10 − 3 + π(−1 + 2) = 7 + π(c) F (x) = 2f(x)g (x) + 2f (x)g(x) = 2(20) + 2(3) = 46(d) F (x) =(4 + g(x))f (x) − f(x)g (x)(4 + g(x))2=(4 − 3)(−1) − 10(2)(4 − 3)2= −2123.dydx=2x(x + 2) − (x2− 1)(x + 2)2,dydx= 0 if x2+ 4x + 1 = 0. By the quadratic formula,x =−4 ±√16 − 42= −2 ±√3. The tangent line is horizontal at x = −2 ±√3.24.dydx=2x(x − 1) − (x2+ 1)(x − 1)2=x2− 2x − 1(x − 1)2. The tangent line is horizontal when it has slope 0, i.e.x2− 2x − 1 = 0 which, by the quadratic formula, has solutions x =2 ±√4 + 42= 1 ±√2, thetangent line is horizontal when x = 1 ±√2.25. The tangent line is parallel to the line y = x when it has slope 1.dydx=2x(x + 1) − (x2+ 1)(x + 1)2=x2+ 2x − 1(x + 1)2= 1 if x2+2x−1 = (x+1)2, which reduces to −1 = +1,impossible. Thus the tangent line is never parallel to the line y = x.
• 92 Chapter 326. The tangent line is perpendicular to the line y = x when the tangent line has slope −1.y =x + 2 + 1x + 2= 1 +1x + 2, hencedydx= −1(x + 2)2= −1 when (x + 2)2= 1, x2+ 4x + 3 = 0,(x + 1)(x + 3) = 0, x = −1, −3. Thus the tangent line is perpendicular to the line y = x at thepoints (−1, 2), (−3, 0).27. Fix x0. The slope of the tangent line to the curve y =1x + 4at the point (x0, 1/(x0 + 4)) isgiven bydydx=−1(x + 4)2x=x0=−1(x0 + 4)2. The tangent line to the curve at (x0, y0) thus has theequation y − y0 =−1(x0 + 4)2(x − x0), and this line passes through the origin if its constant termy0 − x0−1(x0 + 4)2is zero. Then1x0 + 4=−x0(x0 + 4)2, x0 + 4 = −x0, x0 = −2.28. y =2x + 5x + 2=2x + 4 + 1x + 2= 2 +1x + 2, and hencedydx=−1(x + 2)2, thus the tangent line at thepoint (x0, y0) is given by y − y0 =−1(x0 + 2)2(x − x0) , where y0 = 2 +1x0 + 2.If this line is to pass through (0, 2), then2 − y0 =−1(x0 + 2)2(−x0),−1x0 + 2=x0(x0 + 2)2, −x0 − 2 = x0, x0 = −129. (b) They intersect when1x=12 − x, x = 2 − x, x = 1, y = 1. The ﬁrst curve has derivativey = −1x2, so the slope when x = 1 is m = −1. Second curve has derivative y =1(2 − x)2sothe slope when x = 1 is m = 1. Since the two slopes are negative reciprocals of each other,the tangent lines are perpendicular at the point (1, 1).30. The curves intersect when a/(x − 1) = x2− 2x + 1, or (x − 1)3= a, x = 1 + a1/3. They areperpendicular when their slopes are negative reciprocals of each other, i.e.−a(x − 1)2(2x − 2) = −1,which has the solution x = 2a + 1. Solve x = 1 + a1/3= 2a + 1, 2a2/3= 1, a = 2−3/2. Thus thecurves intersect and are perpendicular at the point (2a + 1, 1/2) provided a = 2−3/2.31. F (x) = xf (x) + f(x), F (x) = xf (x) + f (x) + f (x) = xf (x) + 2f (x)32. (a) F (x) = xf (x) + 3f (x)(b) Assume that Fn)(x) = xf(n)(x) + nf(n−1)(x) for some n (for instance n = 3, as in part (a)).Then F(n+1)(x) = xf(n+1)(x) + (1 + n)f(n)(x) = xf(n+1)(x) + (n + 1)f(n)(x), which is aninductive proof.33. (f · g · h) = [(f · g) · h] = (f · g)h + h(f · g) = (f · g)h + h[fg + f g] = fgh + fg h + f gh34. (f1f2 · · · fn) = (f1f2 · · · fn) + (f1f2 · · · fn) + · · · + (f1f2 · · · fn)
• Exercise Set 3.5 9335. (a) 2(1 + x−1)(x−3+ 7) + (2x + 1)(−x−2)(x−3+ 7) + (2x + 1)(1 + x−1)(−3x−4)(b) (x7+ 2x − 3)3= (x7+ 2x − 3)(x7+ 2x − 3)(x7+ 2x − 3) soddx(x7+ 2x − 3)3= (7x6+ 2)(x7+ 2x − 3)(x7+ 2x − 3)+(x7+ 2x − 3)(7x6+ 2)(x7+ 2x − 3)+(x7+ 2x − 3)(x7+ 2x − 3)(7x6+ 2)= 3(7x6+ 2)(x7+ 2x − 3)236. (a) −5x−6(x2+ 2x)(4 − 3x)(2x9+ 1) + x−5(2x + 2)(4 − 3x)(2x9+ 1)+x−5(x2+ 2x)(−3)(2x9+ 1) + x−5(x2+ 2x)(4 − 3x)(18x8)(b) (x2+ 1)50= (x2+ 1)(x2+ 1) · · · (x2+ 1), where (x2+ 1) occurs 50 times soddx(x2+ 1)50= [(2x)(x2+ 1) · · · (x2+ 1)] + [(x2+ 1)(2x) · · · (x2+ 1)]+ · · · + [(x2+ 1)(x2+ 1) · · · (2x)]= 2x(x2+ 1)49+ 2x(x2+ 1)49+ · · · + 2x(x2+ 1)49= 100x(x2+ 1)49because 2x(x2+ 1)49occurs 50 times.37. By the product rule, g (x) is the sum of n terms, each containing n factors of the formf (x)f(x)f(x) . . . f(x); the function f(x) occurs n − 1 times, and f (x) occurs once. Each of thesen terms is equal to f (x)(f(x))n−1, and so g (x) = n(f(x))n−1f (x).38. g (x) = 10(x2− 1)9(2x) = 20x(x2− 1)939. f(x) =1xnso f (x) =xn· (0) − 1 · (nxn−1)x2n= −nxn+140. f(x) = g(x)h(x), f (x) = g (x)h(x) + g(x)h (x), solve for h : h (x) =f (x) − g (x)h(x)g(x), buth = f/g soddxf(x)g(x)=f (x)g(x)−f(x)g (x)g(x)2=f (x)g(x) − f(x)g (x)g(x)2EXERCISE SET 3.51. f (x) = −4 sin x + 2 cos x 2. f (x) =−10x3+ cos x3. f (x) = 4x2sin x − 8x cos x 4. f (x) = 4 sin x cos x5. f (x) =sin x(5 + sin x) − cos x(5 − cos x)(5 + sin x)2=1 + 5(sin x − cos x)(5 + sin x)26. f (x) =(x2+ sin x) cos x − sin x(2x + cos x)(x2 + sin x)2=x2cos x − 2x sin x(x2 + sin x)27. f (x) = sec x tan x −√2 sec2x8. f (x) = (x2+ 1) sec x tan x + (sec x)(2x) = (x2+ 1) sec x tan x + 2x sec x
• 94 Chapter 39. f (x) = −4 csc x cot x + csc2x 10. f (x) = − sin x − csc x + x csc x cot x11. f (x) = sec x(sec2x) + (tan x)(sec x tan x) = sec3x + sec x tan2x12. f (x) = (csc x)(− csc2x) + (cot x)(− csc x cot x) = − csc3x − csc x cot2x13. f (x) =(1 + csc x)(− csc2x) − cot x(0 − csc x cot x)(1 + csc x)2=csc x(− csc x − csc2x + cot2x)(1 + csc x)2but1 + cot2x = csc2x (identity) thus cot2x − csc2x = −1 sof (x) =csc x(− csc x − 1)(1 + csc x)2= −csc x1 + csc x14. f (x) =(1 + tan x)(sec x tan x) − (sec x)(sec2x)(1 + tan x)2=sec x tan x + sec x tan2x − sec3x(1 + tan x)2=sec x(tan x + tan2x − sec2x)(1 + tan x)2=sec x(tan x − 1)(1 + tan x)215. f(x) = sin2x + cos2x = 1 (identity) so f (x) = 016. f (x) = 2 sec x tan x sec x − 2 tan x sec2x =2 sin xcos3 x− 2sin xcos3 x= 0OR f(x) = sec2x − tan2x = 1 (identity), f (x) = 017. f(x) =tan x1 + x tan x(because sin x sec x = (sin x)(1/ cos x) = tan x),f (x) =(1 + x tan x)(sec2x) − tan x[x(sec2x) + (tan x)(1)](1 + x tan x)2=sec2x − tan2x(1 + x tan x)2=1(1 + x tan x)2(because sec2x − tan2x = 1)18. f(x) =(x2+ 1) cot x3 − cot x(because cos x csc x = (cos x)(1/ sin x) = cot x),f (x) =(3 − cot x)[2x cot x − (x2+ 1) csc2x] − (x2+ 1) cot x csc2x(3 − cot x)2=6x cot x − 2x cot2x − 3(x2+ 1) csc2x(3 − cot x)219. dy/dx = −x sin x + cos x, d2y/dx2= −x cos x − sin x − sin x = −x cos x − 2 sin x20. dy/dx = − csc x cot x, d2y/dx2= −[(csc x)(− csc2x) + (cot x)(− csc x cot x)] = csc3x + csc x cot2x21. dy/dx = x(cos x) + (sin x)(1) − 3(− sin x) = x cos x + 4 sin x,d2y/dx2= x(− sin x) + (cos x)(1) + 4 cos x = −x sin x + 5 cos x22. dy/dx = x2(− sin x) + (cos x)(2x) + 4 cos x = −x2sin x + 2x cos x + 4 cos x,d2y/dx2= −[x2(cos x) + (sin x)(2x)] + 2[x(− sin x) + cos x] − 4 sin x = (2 − x2) cos x − 4(x + 1) sin x23. dy/dx = (sin x)(− sin x) + (cos x)(cos x) = cos2x − sin2x,d2y/dx2= (cos x)(− sin x) + (cos x)(− sin x) − [(sin x)(cos x) + (sin x)(cos x)] = −4 sin x cos x
• Exercise Set 3.5 9524. dy/dx = sec2x; d2y/dx2= 2 sec2x tan x25. Let f(x) = tan x, then f (x) = sec2x.(a) f(0) = 0 and f (0) = 1 so y − 0 = (1)(x − 0), y = x.(b) fπ4= 1 and fπ4= 2 so y − 1 = 2 x −π4, y = 2x −π2+ 1.(c) f −π4= −1 and f −π4= 2 so y + 1 = 2 x +π4, y = 2x +π2− 1.26. Let f(x) = sin x, then f (x) = cos x.(a) f(0) = 0 and f (0) = 1 so y − 0 = (1)(x − 0), y = x(b) f(π) = 0 and f (π) = −1 so y − 0 = (−1)(x − π), y = −x + π(c) fπ4=1√2and fπ4=1√2so y −1√2=1√2x −π4, y =1√2x −π4√2+1√227. (a) If y = x sin x then y = sin x + x cos x and y = 2 cos x − x sin x so y + y = 2 cos x.(b) If y = x sin x then y = sin x+x cos x and y = 2 cos x−x sin x so y +y = 2 cos x; diﬀerentiatetwice more to get y(4)+ y = −2 cos x.28. (a) If y = cos x then y = − sin x and y = − cos x so y + y = (− cos x) + (cos x) = 0;if y = sin x then y = cos x and y = − sin x so y + y = (− sin x) + (sin x) = 0.(b) y = A cos x − B sin x, y = −A sin x − B cos x soy + y = (−A sin x − B cos x) + (A sin x + B cos x) = 0.29. (a) f (x) = cos x = 0 at x = ±π/2, ±3π/2.(b) f (x) = 1 − sin x = 0 at x = −3π/2, π/2.(c) f (x) = sec2x ≥ 1 always, so no horizontal tangent line.(d) f (x) = sec x tan x = 0 when sin x = 0, x = ±2π, ±π, 030. (a) 0.5-0.50 2c(b) y = sin x cos x = (1/2) sin 2x and y = cos 2x. So y = 0 when 2x = (2n + 1)π/2 forn = 0, 1, 2, 3 or x = π/4, 3π/4, 5π/4, 7π/431. x = 10 sin θ, dx/dθ = 10 cos θ; if θ = 60◦, thendx/dθ = 10(1/2) = 5 ft/rad = π/36 ft/deg ≈ 0.087 ft/deg32. s = 3800 csc θ, ds/dθ = −3800 csc θ cot θ; if θ = 30◦, thends/dθ = −3800(2)(√3) = −7600√3 ft/rad = −380√3π/9 ft/deg ≈ −230 ft/deg33. D = 50 tan θ, dD/dθ = 50 sec2θ; if θ = 45◦, thendD/dθ = 50(√2)2= 100 m/rad = 5π/9 m/deg ≈ 1.75 m/deg34. (a) From the right triangle shown, sin θ = r/(r + h) so r + h = r csc θ, h = r(csc θ − 1).(b) dh/dθ = −r csc θ cot θ; if θ = 30◦, thendh/dθ = −6378(2)(√3) ≈ −22, 094 km/rad ≈ −386 km/deg
• 96 Chapter 335. (a)d4dx4sin x = sin x, sod4kdx4ksin x = sin x;d87dx87sin x =d3dx3d4·21dx4·21sin x =d3dx3sin x = − cos x(b)d100dx100cos x =d4kdx4kcos x = cos x36.ddx[x sin x] = x cos x + sin xd2dx2[x sin x] = −x sin x + 2 cos xd3dx3[x sin x] = −x cos x − 3 sin xd4dx4[x sin x] = x sin x − 4 cos xBy mathematical induction one can showd4kdx4k[x sin x] = x sin x − (4k) cos x;d4k+1dx4k+1[x sin x] = x cos x + (4k + 1) sin x;d4k+2dx4k+2[x sin x] = −x sin x + (4k + 2) cos x;d4k+3dx4k+3[x sin x] = −x cos x − (4k + 3) sin x;Since 17 = 4 · 4 + 1,d17dx17[x sin x] = x cos x + 17 sin x37. f (x) = − sin x, f (x) = − cos x, f (x) = sin x, and f(4)(x) = cos x with higher order derivativesrepeating this pattern, so f(n)(x) = sin x for n = 3, 7, 11, . . .38. f(x) = sin x, f (x) = cos x, f (x) = − sin x, f(4)(x) = − cos x, f(5)(x) = sin x, and the right-handsides continue with a period of 4, so that f(n)(x) = sin x when n = 4k for some k > 0.39. (a) all x (b) all x(c) x = π/2 + nπ, n = 0, ±1, ±2, . . . (d) x = nπ, n = 0, ±1, ±2, . . .(e) x = π/2 + nπ, n = 0, ±1, ±2, . . . (f) x = nπ, n = 0, ±1, ±2, . . .(g) x = (2n + 1)π, n = 0, ±1, ±2, . . . (h) x = nπ/2, n = 0, ±1, ±2, . . .(i) all x40. (a)ddx[cos x] = limh→0cos(x + h) − cos xh= limh→0cos x cos h − sin x sin h − cos xh= limh→0cos xcos h − 1h− sin xsin hh= (cos x)(0) − (sin x)(1) = − sin x(b)ddx[cot x] =ddxcos xsin x=sin x(− sin x) − cos x(cos x)sin2x=− sin2x − cos2xsin2x=−1sin2x= − csc2x(c)ddx[sec x] =ddx1cos x=0 · cos x − (1)(− sin x)cos2 x=sin xcos2 x= sec x tan x(d)ddx[csc x] =ddx1sin x=(sin x)(0) − (1)(cos x)sin2x= −cos xsin2x= − csc x cot x
• Exercise Set 3.5 9741.ddxsin x = limw→xsin w − sin xw − x= limw→x2 sin w−x2 cos w+x2w − x= limw→xsin w−x2w−x2cosw + x2= 1 · cos x = cos x42.ddx[cos x] = limw→xcos w − cos xw − x= limw→x−2 sin(w−x2 ) sin(w+x2 )w − x= − limw→xsinw + x2limw→xsin(w−x2 )w−x2= − sin x43. (a) limh→0tan hh= limh→0sin hcos hh= limh→0sin hhcos h=11= 1(b)ddx[tan x] = limh→0tan(x + h) − tan xh= limh→0tan x + tan h1 − tan x tan h− tan xh= limh→0tan x + tan h − tan x + tan2x tan hh(1 − tan x tan h)= limh→0tan h(1 + tan2x)h(1 − tan x tan h)= limh→0tan h sec2xh(1 − tan x tan h)= sec2x limh→0tan hh1 − tan x tan h= sec2xlimh→0tan hhlimh→0(1 − tan x tan h)= sec2x44. limx→0tan(x + y) − tan yx= limh→0tan(y + h) − tan yh=ddy(tan y) = sec2y45.ddx[cos kx] = limh→0cos k(x + h) − cos kxh= k limkh→0cos(kx + kh) − cos kxkh= −k sin kx46. The position function is given by s = −4 cos πt, so by Exercise 45 the velocity is the derivativedsdt= 4π sin πt. The position function shows that the top of the mass moves from a low point ofs = −4π to a high point of s = 4π, and passes through the origin when t = π/2, 3π/2, . . .. On theﬁrst pass through the origin the velocity is v = 4π sin(π/2) = 4π.47. Let t be the radian measure, then h =180πt and cos h = cos t, sin h = sin t. Thenlimh→0cos h − 1h= limt→0cos t − 1180t/π=π180limt→0cos t − 1t= 0limh→0sin hh= limt→0sin t180t/π=π180limt→0sin tt=π180(a)ddx[sin x] = sin x limh→0cos h − 1h+ cos x limh→0sin hh= (sin x)(0) + (cos x)(π/180) =π180cos x(b)ddx[cos x] = limh→0cos(x + h) + cos xh= limh→0cos(x) cos h − sin x sin h − cos xh= cos x limcos h − 1h− sin x × limsin hh= 0 − 1 · sin x = − sin x
• 98 Chapter 3EXERCISE SET 3.61. (f ◦ g) (x) = f (g(x))g (x) so (f ◦ g) (0) = f (g(0))g (0) = f (0)(3) = (2)(3) = 62. (f ◦ g) (2) = f (g(2))g (2) = 5(−3) = −153. (a) (f ◦g)(x) = f(g(x)) = (2x−3)5and (f ◦g) (x) = f (g(x))g (x) = 5(2x−3)4(2) = 10(2x−3)4(b) (g ◦ f)(x) = g(f(x)) = 2x5− 3 and (g ◦ f) (x) = g (f(x))f (x) = 2(5x4) = 10x44. (a) (f ◦ g)(x) = 5√4 + cos x and (f ◦ g) (x) = f (g(x))g (x) =52√4 + cos x(− sin x)(b) (g ◦ f)(x) = 4 + cos(5√x) and (g ◦ f) (x) = g (f(x))f (x) = − sin(5√x)52√x5. (a) F (x) = f (g(x))g (x) = f (g(3))g (3) = −1(7) = −7(b) G (x) = g (f(x))f (x) = g (f(3))f (3) = 4(−2) = −86. (a) F (x) = f (g(x))g (x), F (−1) = f (g(−1))g (−1) = f (2)(−3) = (4)(−3) = −12(b) G (x) = g (f(x))f (x), G (−1) = g (f(−1))f (−1) = −5(3) = −157. f (x) = 37(x3+ 2x)36 ddx(x3+ 2x) = 37(x3+ 2x)36(3x2+ 2)8. f (x) = 6(3x2+ 2x − 1)5 ddx(3x2+ 2x − 1) = 6(3x2+ 2x − 1)5(6x + 2) = 12(3x2+ 2x − 1)5(3x + 1)9. f (x) = −2 x3−7x−3ddxx3−7x= −2 x3−7x−33x2+7x210. f(x) = (x5− x + 1)−9,f (x) = −9(x5− x + 1)−10 ddx(x5− x + 1) = −9(x5− x + 1)−10(5x4− 1) = −9(5x4− 1)(x5 − x + 1)1011. f(x) = 4(3x2− 2x + 1)−3,f (x) = −12(3x2− 2x + 1)−4 ddx(3x2− 2x + 1) = −12(3x2− 2x + 1)−4(6x − 2) =24(1 − 3x)(3x2 − 2x + 1)412. f (x) =12√x3 − 2x + 5ddx(x3− 2x + 5) =3x2− 22√x3 − 2x + 513. f (x) =12 4 + 3√xddx(4 + 3√x) =34√x 4 + 3√x14. f (x) =12√xddx(√x) =12x1/412√x=14x3/415. f (x) = cos(1/x2)ddx(1/x2) = −2x3cos(1/x2)16. f (x) = sec2√xddx√x = sec2√x12√x17. f (x) = 20 cos4xddx(cos x) = 20 cos4x(− sin x) = −20 cos4x sin x
• Exercise Set 3.6 9918. f (x) = 4 + 20(sin3x)ddx(sin x) = 4 + 20 sin3x cos x19. f (x) = 2 cos(3√x)ddx[cos(3√x)] = −2 cos(3√x) sin(3√x)ddx(3√x) = −3 cos(3√x) sin(3√x)√x20. f (x) = 4 tan3(x3)ddx[tan(x3)] = 4 tan3(x3) sec2(x3)ddx(x3) = 12x2tan3(x3) sec2(x3)21. f (x) = 4 sec(x7)ddx[sec(x7)] = 4 sec(x7) sec(x7) tan(x7)ddx(x7) = 28x6sec2(x7) tan(x7)22. f (x) = 3 cos2 xx + 1ddxcosxx + 1= 3 cos2 xx + 1− sinxx + 1(x + 1)(1) − x(1)(x + 1)2= −3(x + 1)2cos2 xx + 1sinxx + 123. f (x) =12 cos(5x)ddx[cos(5x)] = −5 sin(5x)2 cos(5x)24. f (x) =12 3x − sin2(4x)ddx[3x − sin2(4x)] =3 − 8 sin(4x) cos(4x)2 3x − sin2(4x)25. f (x) = −3 x + csc(x3+ 3)−4 ddxx + csc(x3+ 3)= −3 x + csc(x3+ 3)−41 − csc(x3+ 3) cot(x3+ 3)ddx(x3+ 3)= −3 x + csc(x3+ 3)−41 − 3x2csc(x3+ 3) cot(x3+ 3)26. f (x) = −4 x4− sec(4x2− 2)−5 ddxx4− sec(4x2− 2)= −4 x4− sec(4x2− 2)−54x3− sec(4x2− 2) tan(4x2− 2)ddx(4x2− 2)= −16x x4− sec(4x2− 2)−5x2− 2 sec(4x2− 2) tan(4x2− 2)27.dydx= x3(2 sin 5x)ddx(sin 5x) + 3x2sin25x = 10x3sin 5x cos 5x + 3x2sin25x28.dydx=√x 3 tan2(√x) sec2(√x)12√x+12√xtan3(√x) =32tan2(√x) sec2(√x) +12√xtan3(√x)29.dydx= x5sec1xtan1xddx1x+ sec1x(5x4)= x5sec1xtan1x−1x2+ 5x4sec1x= −x3sec1xtan1x+ 5x4sec1x
• 100 Chapter 330.dydx=sec(3x + 1) cos x − 3 sin x sec(3x + 1) tan(3x + 1)sec2(3x + 1)= cos x cos(3x + 1) − 3 sin x sin(3x + 1)31.dydx= − sin(cos x)ddx(cos x) = − sin(cos x)(− sin x) = sin(cos x) sin x32.dydx= cos(tan 3x)ddx(tan 3x) = 3 sec23x cos(tan 3x)33.dydx= 3 cos2(sin 2x)ddx[cos(sin 2x)] = 3 cos2(sin 2x)[− sin(sin 2x)]ddx(sin 2x)= −6 cos2(sin 2x) sin(sin 2x) cos 2x34.dydx=(1 − cot x2)(−2x csc x2cot x2) − (1 + csc x2)(2x csc2x2)(1 − cot x2)2= −2x csc x2 1 + cot x2+ csc x2(1 − cot x2)235.dydx= (5x + 8)7 ddx(1 −√x)6+ (1 −√x)6 ddx(5x + 8)7= 6(5x + 8)7(1 −√x)5 −12√x+ 7 · 5(1 −√x)6(5x + 8)6=−3√x(5x + 8)7(1 −√x)5+ 35(1 −√x)6(5x + 8)636.dydx= (x2+ x)5 ddxsin8x + (sin8x)ddx(x2+ x)5= 8(x2+ x)5sin7x cos x + 5(sin8x)(x2+ x)4(2x + 1)37.dydx= 3x − 52x + 12ddxx − 52x + 1= 3x − 52x + 12·11(2x + 1)2=33(x − 5)2(2x + 1)438.dydx= 171 + x21 − x216ddx1 + x21 − x2= 171 + x21 − x216(1 − x2)(2x) − (1 + x2)(−2x)(1 − x2)2= 171 + x21 − x2164x(1 − x2)2=68x(1 + x2)16(1 − x2)1839.dydx=(4x2− 1)8(3)(2x + 3)2(2) − (2x + 3)3(8)(4x2− 1)7(8x)(4x2 − 1)16=2(2x + 3)2(4x2− 1)7[3(4x2− 1) − 32x(2x + 3)](4x2 − 1)16= −2(2x + 3)2(52x2+ 96x + 3)(4x2 − 1)940.dydx= 12[1 + sin3(x5)]11 ddx[1 + sin3(x5)]= 12[1 + sin3(x5)]113 sin2(x5)ddxsin(x5) = 180x4[1 + sin3(x5)]11sin2(x5) cos(x5)41.dydx= 5 x sin 2x + tan4(x7)4 ddxx sin 2x tan4(x7)= 5 x sin 2x + tan4(x7)4x cos 2xddx(2x) + sin 2x + 4 tan3(x7)ddxtan(x7)= 5 x sin 2x + tan4(x7)42x cos 2x + sin 2x + 28x6tan3(x7) sec2(x7)
• Exercise Set 3.6 10142.dydx= 4 tan32 +(7 − x)√3x2 + 5x3 + sin xsec22 +(7 − x)√3x2 + 5x3 + sin x× −√3x2 + 5x3 + sin x+ 3(7 − x)x√3x2 + 5 (x3 + sin x)−(7 − x)√3x2 + 5 (3x2+ cos x)(x3 + sin x)243.dydx= cos 3x − 3x sin 3x; if x = π thendydx= −1 and y = −π, so the equation of the tangent lineis y + π = −(x − π), y = −x44.dydx= 3x2cos(1 + x3); if x = −3 then y = − sin 26,dydx= −27 cos 26, so the equation of the tangentline is y + sin 26 = −27(cos 26)(x + 3), y = −27(cos 26)x − 81 cos 26 − sin 2645.dydx= −3 sec3(π/2 − x) tan(π/2 − x); if x = −π/2 thendydx= 0, y = −1 so the equation of thetangent line is y + 1 = 0, y = −146.dydx= 3 x −1x21 +1x2; if x = 2 then y =278,dydx= 39454=13516so the equation of thetangent line is y − 27/8 = (135/16)(x − 2), y =13516x −27247.dydx= sec2(4x2)ddx(4x2) = 8x sec2(4x2),dydx x=√π= 8√π sec2(4π) = 8√π. When x =√π,y = tan(4π) = 0, so the equation of the tangent line is y = 8√π(x −√π) = 8√πx − 8π.48.dydx= 12 cot3xddxcot x = −12 cot3x csc2x,dydx x=π/4= −24. When x = π/4, y = 3, so theequation of the tangent line is y − 3 = −24(x − π/4), or y = −24x + 3 + 6π.49.dydx= 2x 5 − x2 +x22√5 − x2(−2x),dydx x=1= 4−1/2 = 7/2. When x = 1, y = 2, so the equationof the tangent line is y − 2 = (7/2)(x − 1), or y =72x −32.50.dydx=1√1 − x2−x2(1 − x2)3/2(−2x),dydx x=0= 1. When x = 0, y = 0, so the equation of thetangent line is y = x.51.dydx= x(− sin(5x))ddx(5x) + cos(5x) − 2 sin xddx(sin x)= −5x sin(5x) + cos(5x) − 2 sin x cos x = −5x sin(5x) + cos(5x) − sin(2x),d2ydx2= −5x cos(5x)ddx(5x) − 5 sin(5x) − sin(5x)ddx(5x) − cos(2x)ddx(2x)= −25x cos(5x) − 10 sin(5x) − 2 cos(2x)52.dydx= cos(3x2)ddx(3x2) = 6x cos(3x2),d2ydx2= 6x(− sin(3x2))ddx(3x2) + 6 cos(3x2) = −36x2sin(3x2) + 6 cos(3x2)
• 102 Chapter 353.dydx=(1 − x) + (1 + x)(1 − x)2=2(1 − x)2= 2(1 − x)−2andd2ydx2= −2(2)(−1)(1 − x)−3= 4(1 − x)−354.dydx= x sec2 1xddx1x+ tan1x= −1xsec2 1x+ tan1x,d2ydx2= −2xsec1xddxsec1x+1x2sec2 1x+ sec2 1xddx1x=2x3sec2 1xtan1x55. y = cot3(π − θ) = − cot3θ so dy/dx = 3 cot2θ csc2θ56. 6au + bcu + d5ad − bc(cu + d)257.ddω[a cos2πω + b sin2πω] = −2πa cos πω sin πω + 2πb sin πω cos πω= π(b − a)(2 sin πω cos πω) = π(b − a) sin 2πω58. 2 csc2(π/3 − y) cot(π/3 − y)59. (a) 2–2 2–2(c) f (x) = x−x√4 − x2+ 4 − x2 =4 − 2x2√4 − x22–2 2–6(d) f(1) =√3 and f (1) =2√3so the tangent line has the equation y −√3 =2√3(x − 1).300 260. (a) 0.50^ 6(c) f (x) = 2x cos(x2) cos x − sin x sin(x2)1.2–1.2^ 6
• Exercise Set 3.6 103(d) f(1) = sin 1 cos 1 and f (1) = 2 cos21 − sin21, so the tangent line has the equationy − sin 1 cos 1 = (2 cos21 − sin21)(x − 1).0.80^ 661. (a) dy/dt = −Aω sin ωt, d2y/dt2= −Aω2cos ωt = −ω2y(b) one complete oscillation occurs when ωt increases over an interval of length 2π, or if t increasesover an interval of length 2π/ω(c) f = 1/T(d) amplitude = 0.6 cm, T = 2π/15 s/oscillation, f = 15/(2π) oscillations/s62. dy/dt = 3A cos 3t, d2y/dt2= −9A sin 3t, so −9A sin 3t + 2A sin 3t = 4 sin 3t,−7A sin 3t = 4 sin 3t, −7A = 4, A = −4/763. f(x) =43 x + 5, x < 0−52 x + 5, x > 0ddxx + f(x) =72]√21x+45, x < 0− 32√−6x+20, x > 0=7√624when x = −164. 2 sin(π/6) = 1, so we can assume f(x) = −52 x + 5. Thus for suﬃciently small values of |x − π/6|we haveddx[f(2 sin x)]x=π/6= f (2 sin x)ddx2 sin xx=π/6= −522 cos xx=π/6= −522√32= −52√365. (a) p ≈ 10 lb/in2, dp/dh ≈ −2 lb/in2/mi(b)dpdt=dpdhdhdt≈ (−2)(0.3) = −0.6 lb/in2/s66. (a) F =45cos θ + 0.3 sin θ,dFdθ= −45(− sin θ + 0.3 cos θ)(cos θ + 0.3 sin θ)2;if θ = 30◦, then dF/dθ ≈ 10.5 lb/rad ≈ 0.18 lb/deg(b)dFdt=dFdθdθdt≈ (0.18)(−0.5) = −0.09 lb/s67. With u = sin x,ddx(| sin x|) =ddx(|u|) =ddu(|u|)dudx=ddu(|u|) cos x =cos x, u > 0− cos x, u < 0=cos x, sin x > 0− cos x, sin x < 0=cos x, 0 < x < π− cos x, −π < x < 068.ddx(cos x) =ddx[sin(π/2 − x)] = − cos(π/2 − x) = − sin x
• 104 Chapter 369. (a) For x = 0, |f(x)| ≤ |x|, and limx→0|x| = 0, so by the Squeezing Theorem, limx→0f(x) = 0.(b) If f (0) were to exist, then the limitf(x) − f(0)x − 0= sin(1/x) would have to exist, but itdoesn’t.(c) For x = 0, f (x) = x cos1x−1x2+ sin1x= −1xcos1x+ sin1x(d) limx→0f(x) − f(0)x − 0= limx→0sin(1/x), which does not exist, thus f (0) does not exist.70. (a) −x2≤ x2sin(1/x) ≤ x2, so by the Squeezing Theorem limx→0f(x) = 0.(b) f (0) = limx→0f(x) − f(0)x − 0= limx→0x sin(1/x) = 0 by Exercise 69, Part a.(c) For x = 0, f (x) = 2x sin(1/x) + x2cos(1/x)(−1/x2) = 2x sin(1/x) − cos(1/x)(d) If f (x) were continuous at x = 0 then so would cos(1/x) = f (x) − 2x sin(1/x) be, since2x sin(1/x) is continuous there. But cos(1/x) oscillates at x = 0.71. (a) g (x) = 3[f(x)]2f (x), g (2) = 3[f(2)]2f (2) = 3(1)2(7) = 21(b) h (x) = f (x3)(3x2), h (2) = f (8)(12) = (−3)(12) = −3672. F (x) = f (g(x))g (x) = 3(x2 − 1) + 4(2x) = 2x√3x2 + 173. F (x) = f (g(x))g (x) = f (√3x − 1)32√3x − 1=√3x − 1(3x − 1) + 132√3x − 1=12x74.ddx[f(x2)] = f (x2)(2x), thus f (x2)(2x) = x2so f (x2) = x/2 if x = 075.ddx[f(3x)] = f (3x)ddx(3x) = 3f (3x) = 6x, so f (3x) = 2x. Let u = 3x to get f (u) =23u;ddx[f(x)] = f (x) =23x.76. (a) If f(−x) = f(x), thenddx[f(−x)] =ddx[f(x)], f (−x)(−1) = f (x), f (−x) = −f (x) sof is odd.(b) If f(−x) = −f(x), thenddx[f(−x)] = −ddx[f(x)], f (−x)(−1) = −f (x), f (−x) = f (x) sof is even.77. For an even function, the graph is symmetric about the y-axis; the slope of the tangent line at(a, f(a)) is the negative of the slope of the tangent line at (−a, f(−a)). For an odd function, thegraph is symmetric about the origin; the slope of the tangent line at (a, f(a)) is the same as theslope of the tangent line at (−a, f(−a)).yxf(x)f(x)yxf(x)f(x)
• Exercise Set 3.7 10578.dydx=dydududvdvdwdwdx79.ddx[f(g(h(x)))] =ddx[f(g(u))], u = h(x)=ddu[f(g(u))]dudx= f (g(u))g (u)dudx= f (g(h(x)))g (h(x))h (x)EXERCISE SET 3.71.dydt= 3dxdt(a)dydt= 3(2) = 6 (b) −1 = 3dxdt,dxdt= −132.dxdt+ 4dydt= 0(a) 1 + 4dydt= 0 sodydt= −14when x = 2. (b)dxdt+ 4(4) = 0 sodxdt= −16 when x = 3.3. 8xdxdt+ 18ydydt= 0(a) 812√2· 3 + 1813√2dydt= 0,dydt= −2 (b) 813dxdt− 18√59· 8 = 0,dxdt= 6√54. 2xdxdt+ 2ydydt= 2dxdt+ 4dydt(a) 2 · 3(−5) + 2 · 1dydt= 2(−5) + 4dydt,dydt= −10(b) 2(1 +√2)dxdt+ 2(2 +√3) · 6 = 2dxdt+ 4 · 6,dxdt= −12√32√2= −3√3√25. (b) A = x2(c)dAdt= 2xdxdt(d) FinddAdt x=3given thatdxdt x=3= 2. From Part (c),dAdt x=3= 2(3)(2) = 12 ft2/min.6. (b) A = πr2(c)dAdt= 2πrdrdt(d) FinddAdt r=5given thatdrdt r=5= 2. From Part (c),dAdt r=5= 2π(5)(2) = 20π cm2/s.7. (a) V = πr2h, sodVdt= π r2 dhdt+ 2rhdrdt.(b) FinddVdt h=6,r=10given thatdhdt h=6,r=10= 1 anddrdt h=6,r=10= −1. From Part (a),dVdt h=6,r=10= π[102(1) + 2(10)(6)(−1)] = −20π in3/s; the volume is decreasing.
• Exercise Set 3.7 10715. FinddVdt r=9given thatdrdt= −15. From V =43πr3we getdVdt= 4πr2 drdtsodVdt r=9= 4π(9)2(−15) = −4860π. Air must be removed at the rate of 4860π cm3/min.16. Let x and y be the distances shown in the diagram.We want to ﬁnddydt y=8given thatdxdt= 5. Fromx2+y2= 172we get 2xdxdt+2ydydt= 0, sodydt= −xydxdt.When y = 8, x2+82= 172, x2= 289−64 = 225, x = 15sodydt y=8= −158(5) = −758ft/s; the top of the ladderis moving down the wall at a rate of 75/8 ft/s.17xy17. Finddxdt y=5given thatdydt= −2. From x2+ y2= 132we get 2xdxdt+ 2ydydt= 0 sodxdt= −yxdydt. Usex2+ y2= 169 to ﬁnd that x = 12 when y = 5 sodxdt y=5= −512(−2) =56ft/s.13xy18. Let θ be the acute angle, and x the distance of the bottom of the plank from the wall. Finddθdt x=2given thatdxdt x=2= −12ft/s. The variables θ and x are related by the equation cos θ =x10so− sin θdθdt=110dxdt,dθdt= −110 sin θdxdt. When x = 2, the top of the plank is√102 − 22 =√96 ftabove the ground so sin θ =√96/10 anddθdt x=2= −1√96−12=12√96≈ 0.051 rad/s.19. Let x denote the distance from ﬁrst base and y thedistance from home plate. Then x2+ 602= y2and2xdxdt= 2ydydt. When x = 50 then y = 10√61 sodydt=xydxdt=5010√61(25) =125√61ft/s.60ftyxHomeFirst
• 108 Chapter 320. Finddxdt x=4given thatdydt x=4= 2000. Fromx2+ 52= y2we get 2xdxdt= 2ydydtsodxdt=yxdydt.Use x2+ 25 = y2to ﬁnd that y =√41 when x = 4 sodxdt x=4=√414(2000) = 500√41 mi/h.Rocketyx5 miRadarstation21. Finddydt x=4000given thatdxdt x=4000= 880. Fromy2= x2+ 30002we get 2ydydt= 2xdxdtsodydt=xydxdt.If x = 4000, then y = 5000 sodydt x=4000=40005000(880) = 704 ft/s.RocketCamera3000 ftyx22. Finddxdt φ=π/4given thatdφdt φ=π/4= 0.2. But x = 3000 tan φ sodxdt= 3000(sec2φ)dφdt,dxdt φ=π/4= 3000 sec2 π4(0.2) = 1200 ft/s.23. (a) If x denotes the altitude, then r − x = 3960, the radius of the Earth. θ = 0 at perigee, sor = 4995/1.12 ≈ 4460; the altitude is x = 4460 − 3960 = 500 miles. θ = π at apogee, sor = 4995/0.88 ≈ 5676; the altitude is x = 5676 − 3960 = 1716 miles.(b) If θ = 120◦, then r = 4995/0.94 ≈ 5314; the altitude is 5314 − 3960 = 1354 miles. The rateof change of the altitude is given bydxdt=drdt=drdθdθdt=4995(0.12 sin θ)(1 + 0.12 cos θ)2dθdt.Use θ = 120◦and dθ/dt = 2.7◦/min = (2.7)(π/180) rad/min to get dr/dt ≈ 27.7 mi/min.24. (a) Let x be the horizontal distance shown in the ﬁgure. Then x = 4000 cot θ anddxdt= −4000 csc2θdθdt, sodθdt= −sin2θ4000dxdt. Use θ = 30◦anddx/dt = 300 mi/h = 300(5280/3600) ft/s = 440 ft/s to getdθ/dt = −0.0275 rad/s ≈ −1.6◦/s; θ is decreasing at the rate of 1.6◦/s.(b) Let y be the distance between the observation point and the aircraft. Then y = 4000 csc θso dy/dt = −4000(csc θ cot θ)(dθ/dt). Use θ = 30◦and dθ/dt = −0.0275 rad/s to getdy/dt ≈ 381 ft/s.
• Exercise Set 3.7 10925. Finddhdt h=16given thatdVdt= 20. The volume ofwater in the tank at a depth h is V =13πr2h. Usesimilar triangles (see ﬁgure) to getrh=1024so r =512hthus V =13π512h2h =25432πh3,dVdt=25144πh2 dhdt;dhdt=14425πh2dVdt,dhdt h=16=14425π(16)2(20) =920πft/min.hr241026. Finddhdt h=6given thatdVdt= 8. V =13πr2h,but r =12h so V =13πh22h =112πh3,dVdt=14πh2 dhdt,dhdt=4πh2dVdt,dhdt h=6=4π(6)2(8) =89πft/min.hr27. FinddVdt h=10given thatdhdt= 5. V =13πr2h, butr =12h so V =13πh22h =112πh3,dVdt=14πh2 dhdt,dVdt h=10=14π(10)2(5) = 125π ft3/min.hrhr28. Let r and h be as shown in the ﬁgure. If C is the circumferenceof the base, then we want to ﬁnddCdt h=8given thatdVdt= 10.It is given that r =12h, thus C = 2πr = πh sodCdt= πdhdt(1)Use V =13πr2h =112πh3to getdVdt=14πh2 dhdt, sodhdt=4πh2dVdt(2)Substitution of (2) into (1) givesdCdt=4h2dVdtsodCdt h=8=464(10) =58ft/min.
• 110 Chapter 329. With s and h as shown in the ﬁgure, we want to ﬁnddhdtgiven thatdsdt= 500. From the ﬁgure, h = s sin 30◦=12ssodhdt=12dsdt=12(500) = 250 mi/h.shGround30°30. Finddxdt y=125given thatdydt= −20. From x2+ 102= y2we get 2xdxdt= 2ydydtsodxdt=yxdydt. Use x2+ 100 = y2to ﬁnd that x =√15, 525 = 15√69 when y = 125 sodxdt y=125=12515√69(−20) = −5003√69.The boat is approaching the dock at the rate of5003√69ft/min.yx BoatPulley1031. Finddydtgiven thatdxdt y=125= −12. From x2+ 102= y2we get 2xdxdt= 2ydydtsodydt=xydxdt. Use x2+ 100 = y2to ﬁnd that x = 15, 525 = 15√69 when y = 125 sodydt=15√69125(−12) = −36√6925. The rope must be pulledat the rate of36√6925ft/min.yxBoatPulley1032. (a) Let x and y be as shown in the ﬁgure. It is required toﬁnddxdt, given thatdydt= −3. By similar triangles,x6=x + y18, 18x = 6x + 6y, 12x = 6y, x =12y, sodxdt=12dydt=12(−3) = −32ft/s. 618ManShadowLightyx(b) The tip of the shadow is z = x + y feet from the street light, thus the rate at which it ismoving is given bydzdt=dxdt+dydt. In part (a) we found thatdxdt= −32whendydt= −3 sodzdt= (−3/2) + (−3) = −9/2 ft/s; the tip of the shadow is moving at the rate of 9/2 ft/stoward the street light.
• Exercise Set 3.7 11133. Finddxdt θ=π/4given thatdθdt=2π10=π5rad/s.Then x = 4 tan θ (see ﬁgure) sodxdt= 4 sec2θdθdt,dxdt θ=π/4= 4 sec2 π4π5= 8π/5 km/s.x4Shipθ34. If x, y, and z are as shown in the ﬁgure, then we wantdzdt x=2,y=4given thatdxdt= −600 anddydt x=2,y=4= −1200. Butz2= x2+ y2so 2zdzdt= 2xdxdt+ 2ydydt,dzdt=1zxdxdt+ ydydt.When x = 2 and y = 4, z2= 22+ 42= 20, z =√20 = 2√5 sodzdt x=2,y=4=12√5[2(−600) + 4(−1200)] = −3000√5= −600√5 mi/h;the distance between missile and aircraft is decreasing at the rate of 600√5 mi/h.AircraftPMissilexyz35. We wish to ﬁnddzdt x=2,y=4givendxdt= −600 anddydt x=2,y=4= −1200 (see ﬁgure). From the law of cosines,z2= x2+ y2− 2xy cos 120◦= x2+ y2− 2xy(−1/2)= x2+y2+xy, so 2zdzdt= 2xdxdt+2ydydt+xdydt+ydxdt,dzdt=12z(2x + y)dxdt+ (2y + x)dydt.AircraftPMissilexyz120ºWhen x = 2 and y = 4, z2= 22+ 42+ (2)(4) = 28, so z =√28 = 2√7, thusdzdt x=2,y=4=12(2√7)[(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = −4200√7= −600√7 mi/h;the distance between missile and aircraft is decreasing at the rate of 600√7 mi/h.36. (a) Let P be the point on the helicopter’s path that lies directly above the car’s path. Let x,y, and z be the distances shown in the ﬁrst ﬁgure. Finddzdt x=2,y=0given thatdxdt= −75 anddydt= 100. In order to ﬁnd an equation relating x, y, and z, ﬁrst draw the line segmentthat joins the point P to the car, as shown in the second ﬁgure. Because triangle OPC is aright triangle, it follows that PC has length x2 + (1/2)2; but triangle HPC is also a righttriangle so z2= x2 + (1/2)22+ y2= x2+ y2+ 1/4 and 2zdzdt= 2xdxdt+ 2ydydt+ 0,
• 112 Chapter 3dzdt=1zxdxdt+ ydydt. Now, when x = 2 and y = 0, z2= (2)2+ (0)2+ 1/4 = 17/4,z =√17/2 sodzdt x=2,y=0=1(√17/2)[2(−75) + 0(100)] = −300/√17 mi/hWest EastNorthCarHelicopterxzyPmi12 CHxOzyPmi12(b) decreasing, becausedzdt< 0.37. (a) We wantdydt x=1,y=2given thatdxdt x=1,y=2= 6. For convenience, ﬁrst rewrite the equation asxy3=85+85y2then 3xy2 dydt+ y3 dxdt=165ydydt,dydt=y3165y − 3xy2dxdtsodydt x=1,y=2=23165(2) − 3(1)22(6) = −60/7 units/s.(b) falling, becausedydt< 038. Finddxdt (2,5)given thatdydt (2,5)= 2. Square and rearrange to get x3= y2− 17so 3x2 dxdt= 2ydydt,dxdt=2y3x2dydt,dxdt (2,5)=56(2) =53units/s.39. The coordinates of P are (x, 2x), so the distance between P and the point (3, 0) isD = (x − 3)2 + (2x − 0)2 =√5x2 − 6x + 9. FinddDdt x=3given thatdxdt x=3= −2.dDdt=5x − 3√5x2 − 6x + 9dxdt, sodDdt x=3=12√36(−2) = −4 units/s.40. (a) Let D be the distance between P and (2, 0). FinddDdt x=3given thatdxdt x=3= 4.D = (x − 2)2 + y2 = (x − 2)2 + x =√x2 − 3x + 4 sodDdt=2x − 32√x2 − 3x + 4;dDdt x=3=32√4=34units/s.
• Exercise Set 3.7 113(b) Let θ be the angle of inclination. Finddθdt x=3given thatdxdt x=3= 4.tan θ =yx − 2=√xx − 2so sec2θdθdt= −x + 22√x(x − 2)2dxdt,dθdt= − cos2θx + 22√x(x − 2)2dxdt.When x = 3, D = 2 so cos θ =12anddθdt x=3= −1452√3(4) = −52√3rad/s.41. Solvedxdt= 3dydtgiven y = x/(x2+ 1). Then y(x2+ 1) = x. Diﬀerentiating with respect to x,(x2+ 1)dydx+ y(2x) = 1. Butdydx=dy/dtdx/dt=13so (x2+ 1)13+ 2xy = 1, x2+ 1 + 6xy = 3,x2+1+6x2/(x2+1) = 3, (x2+1)2+6x2−3x2−3 = 0, x4+5x2−3 = 0. By the binomial theoremapplied to x2we obtain x2= (−5 ±√25 + 12)/2. The minus sign is spurious since x2cannot benegative, so x2= (√37 − 5)/2, x ≈ ±0.7357861545, y = ±0.4773550654.42. 32xdxdt+ 18ydydt= 0; ifdydt=dxdt= 0, then (32x + 18y)dxdt= 0, 32x + 18y = 0, y = −169x so16x2+ 925681x2= 144,4009x2= 144, x2=8125, x = ±95. If x =95, then y = −16995= −165.Similarly, if x = −95, then y =165. The points are95, −165and −95,165.43. FinddSdt s=10given thatdsdt s=10= −2. From1s+1S=16we get −1s2dsdt−1S2dSdt= 0, sodSdt= −S2s2dsdt. If s = 10, then110+1S=16which gives S = 15. SodSdt s=10= −225100(−2) = 4.5cm/s. The image is moving away from the lens.44. Suppose that the reservoir has height H and that the radius at the top is R. At any instant of timelet h and r be the corresponding dimensions of the cone of water (see ﬁgure). We want to showthatdhdtis constant and independent of H and R, given thatdVdt= −kA where V is the volumeof water, A is the area of a circle of radius r, and k is a positive constant. The volume of a cone ofradius r and height h is V =13πr2h. By similar trianglesrh=RH, r =RHh thus V =13πRH2h3sodVdt= πRH2h2 dhdt(1)But it is given thatdVdt= −kA or, becauseA = πr2= πRH2h2,dVdt= −kπRH2h2,which when substituted into equation (1) gives−kπRH2h2= πRH2h2 dhdt,dhdt= −k.hrHR
• 114 Chapter 345. Let r be the radius, V the volume, and A the surface area of a sphere. Show thatdrdtis a constantgiven thatdVdt= −kA, where k is a positive constant. Because V =43πr3,dVdt= 4πr2 drdt(1)But it is given thatdVdt= −kA or, because A = 4πr2,dVdt= −4πr2k which when substituted intoequation (1) gives −4πr2k = 4πr2 drdt,drdt= −k.46. Let x be the distance between the tips of the minute and hour hands, and α and β the anglesshown in the ﬁgure. Because the minute hand makes one revolution in 60 minutes,dαdt=2π60= π/30 rad/min; the hour hand makes one revolution in 12 hours (720 minutes), thusdβdt=2π720= π/360 rad/min. We want to ﬁnddxdt α=2π,β=3π/2given thatdαdt= π/30 anddβdt= π/360.Using the law of cosines on the triangle shown in the ﬁgure,x2= 32+ 42− 2(3)(4) cos(α − β) = 25 − 24 cos(α − β), so2xdxdt= 0 + 24 sin(α − β)dαdt−dβdt,dxdt=12xdαdt−dβdtsin(α − β). When α = 2π and β = 3π/2,x2= 25 − 24 cos(2π − 3π/2) = 25, x = 5; sodxdt α=2π,β=3π/2=125(π/30 − π/360) sin(2π − 3π/2) =11π150in/min.34␣x␤47. Extend sides of cup to complete the cone and let V0be the volume of the portion added, then (see ﬁgure)V =13πr2h − V0 whererh=412=13so r =13h andV =13πh32h − V0 =127πh3− V0,dVdt=19πh2 dhdt,dhdt=9πh2dVdt,dhdt h=9=9π(9)2(20) =209πcm/s.r42h66EXERCISE SET 3.81. (a) f(x) ≈ f(1) + f (1)(x − 1) = 1 + 3(x − 1)(b) f(1 + ∆x) ≈ f(1) + f (1)∆x = 1 + 3∆x(c) From Part (a), (1.02)3≈ 1 + 3(0.02) = 1.06. From Part (b), (1.02)3≈ 1 + 3(0.02) = 1.06.2. (a) f(x) ≈ f(2) + f (2)(x − 2) = 1/2 + (−1/22)(x − 2) = (1/2) − (1/4)(x − 2)(b) f(2 + ∆x) ≈ f(2) + f (2)∆x = 1/2 − (1/4)∆x(c) From Part (a), 1/2.05 ≈ 0.5 − 0.25(0.05) = 0.4875, and from Part (b),1/2.05 ≈ 0.5 − 0.25(0.05) = 0.4875.
• Exercise Set 3.8 1153. (a) f(x) ≈ f(x0) + f (x0)(x − x0) = 1 + (1/(2√1)(x − 0) = 1 + (1/2)x, so with x0 = 0 andx = −0.1, we have√0.9 = f(−0.1) ≈ 1 + (1/2)(−0.1) = 1 − 0.05 = 0.95. With x = 0.1 wehave√1.1 = f(0.1) ≈ 1 + (1/2)(0.1) = 1.05.(b) yxdy0.1–0.1∆y ∆ydy4. (a) f(x) ≈ f(x0)+f (x0)(x−x0) = 1/2− 1/(2 · 43/2) (x−4) = 1/2−(x−4)/16, so with x0 = 4and x = 3.9 we have 1/√3.9 = f(3.9) ≈ 0.5 − (−0.1)/16 = 0.50625. If x0 = 4 and x = 4.1then 1/√4.1 = f(4.1) ≈ 0.5 − (0.1)/16 = 0.49375(b)xy43.9 4.1∆y∆ydydy5. f(x) = (1+x)15and x0 = 0. Thus (1+x)15≈ f(x0)+f (x0)(x−x0) = 1+15(1)14(x−0) = 1+15x.6. f(x) =1√1 − xand x0 = 0, so1√1 − x≈ f(x0)+f (x0)(x−x0) = 1+12(1 − 0)3/2(x−0) = 1+x/27. tan x ≈ tan(0) + sec2(0)(x − 0) = x 8.11 + x≈ 1 +−1(1 + 0)2(x − 0) = 1 − x9. x4≈ (1)4+ 4(1)3(x − 1). Set ∆x = x − 1; then x = ∆x + 1 and (1 + ∆x)4= 1 + 4∆x.10.√x ≈√1 +12√1(x − 1), and x = 1 + ∆x, so√1 + ∆x ≈ 1 + ∆x/211.12 + x≈12 + 1−1(2 + 1)2(x − 1), and 2 + x = 3 + ∆x, so13 + ∆x≈13−19∆x12. (4 + x)3≈ (4 + 1)3+ 3(4 + 1)2(x − 1) so, with 4 + x = 5 + ∆x we get (5 + ∆x)3≈ 125 + 75∆x13. f(x) =√x + 3 and x0 = 0, so√x + 3 ≈√3 +12√3(x − 0) =√3 +12√3x, andf(x) −√3 +12√3x < 0.1 if |x| < 1.692.0–0.1–2 2|f(x) – ( 3 + x)|12 3
• 116 Chapter 314. f(x) =1√9 − xso1√9 − x≈1√9+12(9 − 0)3/2(x − 0) =13+154x,and f(x) −13+154x < 0.1 if |x| < 5.51140.060–6 6|f (x) – ( +13154x)|15. tan 2x ≈ tan 0 + (sec20)(2x − 0) = 2x,and | tan 2x − 2x| < 0.1 if |x| < 0.31580.060–0.8 0.8|f(x) – 2x|16.1(1 + 2x)5≈1(1 + 2 · 0)5+−5(2)(1 + 2 · 0)6(x − 0) = 1 − 10x,and |f(x) − (1 − 10x)| < 0.1 if |x| < 0.03720.120–0.04 0.04| f(x) – (1 – 10x)|17. (a) The local linear approximation sin x ≈ x gives sin 1◦= sin(π/180) ≈ π/180 = 0.0174533 anda calculator gives sin 1◦= 0.0174524. The relative error | sin(π/180)−(π/180)|/(sin π/180) =0.000051 is very small, so for such a small value of x the approximation is very good.(b) Use x0 = 45◦(this assumes you know, or can approximate,√2/2).(c) 44◦=44π180radians, and 45◦=45π180=π4radians. With x =44π180and x0 =π4we obtainsin 44◦= sin44π180≈ sinπ4+ cosπ444π180−π4=√22+√22−π180= 0.694765. With acalculator, sin 44◦= 0.694658.18. (a) tan x ≈ tan 0 + sec20(x − 0) = x, so tan 2◦= tan(2π/180) ≈ 2π/180 = 0.034907, and with acalculator tan 2◦= 0.034921(b) use x0 = π/3 because we know tan 60◦= tan(π/3) =√3(c) with x0 =π3=60π180and x =61π180we havetan 61◦= tan61π180≈ tanπ3+ sec2 π361π180−π3=√3 + 4π180= 1.8019,and with a calculator tan 61◦= 1.8040
• Exercise Set 3.8 11719. f(x) = x4, f (x) = 4x3, x0 = 3, ∆x = 0.02; (3.02)4≈ 34+ (108)(0.02) = 81 + 2.16 = 83.1620. f(x) = x3, f (x) = 3x2, x0 = 2, ∆x = −0.03; (1.97)3≈ 23+ (12)(−0.03) = 8 − 0.36 = 7.6421. f(x) =√x, f (x) =12√x, x0 = 64, ∆x = 1;√65 ≈√64 +116(1) = 8 +116= 8.062522. f(x) =√x, f (x) =12√x, x0 = 25, ∆x = −1;√24 ≈√25 +110(−1) = 5 − 0.1 = 4.923. f(x) =√x, f (x) =12√x, x0 = 81, ∆x = −0.1;√80.9 ≈√81 +118(−0.1) ≈ 8.994424. f(x) =√x, f (x) =12√x, x0 = 36, ∆x = 0.03;√36.03 ≈√36 +112(0.03) = 6 + 0.0025 = 6.002525. f(x) = sin x, f (x) = cos x, x0 = 0, ∆x = 0.1; sin 0.1 ≈ sin 0 + (cos 0)(0.1) = 0.126. f(x) = tan x, f (x) = sec2x, x0 = 0, ∆x = 0.2; tan 0.2 ≈ tan 0 + (sec20)(0.2) = 0.227. f(x) = cos x, f (x) = − sin x, x0 = π/6, ∆x = π/180;cos 31◦≈ cos 30◦+ −12π180=√32−π360≈ 0.857328. (a) Let f(x) = (1 + x)kand x0 = 0. Then (1 + x)k≈ 1k+ k(1)k−1(x − 0) = 1 + kx. Set k = 37and x = 0.001 to obtain (1.001)37≈ 1.037.(b) With a calculator (1.001)37= 1.03767.(c) The approximation is (1.1)37≈ 1 + 37(0.1) = 4.7, and the calculator value is 34.004. Theerror is due to the relative largeness of f (1)∆x = 37(0.1) = 3.7.29. (a) dy = f (x)dx = 2xdx = 4(1) = 4 and∆y = (x + ∆x)2− x2= (2 + 1)2− 22= 5(b)xydy∆y42 330. (a) dy = 3x2dx = 3(1)2(1) = 3 and∆y = (x + ∆x)3− x3= (1 + 1)3− 13= 7(b)1yx81 2(1,1)dx∆ydy
• 118 Chapter 331. (a) dy = (−1/x2)dx = (−1)(−0.5) = 0.5 and∆y = 1/(x + ∆x) − 1/x= 1/(1 − 0.5) − 1/1 = 2 − 1 = 1(b) yx120.5 1dy=0.5y∆ =132. (a) dy = (1/2√x)dx = (1/(2 · 3))(−1) = −1/6 ≈ −0.167 and∆y =√x + ∆x −√x = 9 + (−1) −√9 =√8 − 3 ≈ −0.172(b) yx38 9dy = –0.167∆y = –0.17133. dy = 3x2dx;∆y = (x + ∆x)3− x3= x3+ 3x2∆x + 3x(∆x)2+ (∆x)3− x3= 3x2∆x + 3x(∆x)2+ (∆x)334. dy = 8dx; ∆y = [8(x + ∆x) − 4] − [8x − 4] = 8∆x35. dy = (2x − 2)dx;∆y = [(x + ∆x)2− 2(x + ∆x) + 1] − [x2− 2x + 1]= x2+ 2x ∆x + (∆x)2− 2x − 2∆x + 1 − x2+ 2x − 1 = 2x ∆x + (∆x)2− 2∆x36. dy = cos x dx; ∆y = sin(x + ∆x) − sin x37. (a) dy = (12x2− 14x)dx(b) dy = x d(cos x) + cos x dx = x(− sin x)dx + cos xdx = (−x sin x + cos x)dx38. (a) dy = (−1/x2)dx (b) dy = 5 sec2x dx39. (a) dy =√1 − x −x2√1 − xdx =2 − 3x2√1 − xdx(b) dy = −17(1 + x)−18dx40. (a) dy =(x3− 1)d(1) − (1)d(x3− 1)(x3 − 1)2=(x3− 1)(0) − (1)3x2dx(x3 − 1)2= −3x2(x3 − 1)2dx(b) dy =(2 − x)(−3x2)dx − (1 − x3)(−1)dx(2 − x)2=2x3− 6x2+ 1(2 − x)2dx41. dy =32√3x − 2dx, x = 2, dx = 0.03; ∆y ≈ dy =34(0.03) = 0.022542. dy =x√x2 + 8dx, x = 1, dx = −0.03; ∆y ≈ dy = (1/3)(−0.03) = −0.01
• Exercise Set 3.8 11943. dy =1 − x2(x2 + 1)2dx, x = 2, dx = −0.04; ∆y ≈ dy = −325(−0.04) = 0.004844. dy =4x√8x + 1+√8x + 1 dx, x = 3, dx = 0.05; ∆y ≈ dy = (37/5)(0.05) = 0.3745. (a) A = x2where x is the length of a side; dA = 2x dx = 2(10)(±0.1) = ±2 ft2.(b) relative error in x is ≈dxx=±0.110= ±0.01 so percentage error in x is ≈ ±1%; relative errorin A is ≈dAA=2x dxx2= 2dxx= 2(±0.01) = ±0.02 so percentage error in A is ≈ ±2%46. (a) V = x3where x is the length of a side; dV = 3x2dx = 3(25)2(±1) = ±1875 cm3.(b) relative error in x is ≈dxx=±125= ±0.04 so percentage error in x is ≈ ±4%; relative errorin V is ≈dVV=3x2dxx3= 3dxx= 3(±0.04) = ±0.12 so percentage error in V is ≈ ±12%47. (a) x = 10 sin θ, y = 10 cos θ (see ﬁgure),dx = 10 cos θdθ = 10 cosπ6±π180= 10√32±π180≈ ±0.151 in,dy = −10(sin θ)dθ = −10 sinπ6±π180= −1012±π180≈ ±0.087 in10″xyθ(b) relative error in x is ≈dxx= (cot θ)dθ = cotπ6±π180=√3 ±π180≈ ±0.030so percentage error in x is ≈ ±3.0%;relative error in y is ≈dyy= − tan θdθ = − tanπ6±π180= −1√3±π180≈ ±0.010so percentage error in y is ≈ ±1.0%48. (a) x = 25 cot θ, y = 25 csc θ (see ﬁgure);dx = −25 csc2θdθ = −25 csc2 π3±π360= −2543±π360≈ ±0.291 cm,dy = −25 csc θ cot θdθ = −25 cscπ3cotπ3±π360= −252√31√3±π360≈ ±0.145 cm25 cmxyθ(b) relative error in x is ≈dxx= −csc2θcot θdθ = −4/31/√3±π360≈ ±0.020 so percentageerror in x is ≈ ±2.0%; relative error in y is ≈dyy= − cot θdθ = −1√3±π360≈ ±0.005so percentage error in y is ≈ ±0.5%
• 120 Chapter 349.dRR=(−2k/r3)dr(k/r2)= −2drr, butdrr≈ ±0.05 sodRR≈ −2(±0.05) = ±0.10; percentage error in Ris ≈ ±10%50. h = 12 sin θ thus dh = 12 cos θdθ so, with θ = 60◦= π/3 radians and dθ = −1◦= −π/180 radians,dh = 12 cos(π/3)(−π/180) = −π/30 ≈ −0.105 ft51. A =14(4)2sin 2θ = 4 sin 2θ thus dA = 8 cos 2θdθ so, with θ = 30◦= π/6 radians anddθ = ±15 = ±1/4◦= ±π/720 radians, dA = 8 cos(π/3)(±π/720) = ±π/180 ≈ ±0.017 cm252. A = x2where x is the length of a side;dAA=2x dxx2= 2dxx, butdxx≈ ±0.01 sodAA≈ 2(±0.01) = ±0.02; percentage error in A is ≈ ±2%53. V = x3where x is the length of a side;dVV=3x2dxx3= 3dxx, butdxx≈ ±0.02sodVV≈ 3(±0.02) = ±0.06; percentage error in V is ≈ ±6%.54.dVV=4πr2dr4πr3/3= 3drr, butdVV≈ ±0.03 so 3drr≈ ±0.03,drr≈ ±0.01; maximum permissiblepercentage error in r is ≈ ±1%.55. A =14πD2where D is the diameter of the circle;dAA=(πD/2)dDπD2/4= 2dDD, butdAA≈ ±0.01 so2dDD≈ ±0.01,dDD≈ ±0.005; maximum permissible percentage error in D is ≈ ±0.5%.56. V = x3where x is the length of a side; approximate ∆V by dV if x = 1 and dx = ∆x = 0.02,dV = 3x2dx = 3(1)2(0.02) = 0.06 in3.57. V = volume of cylindrical rod = πr2h = πr2(15) = 15πr2; approximate ∆V by dV if r = 2.5 anddr = ∆r = 0.001. dV = 30πr dr = 30π(2.5)(0.001) ≈ 235.62 cm3.58. P =2π√g√L, dP =2π√g12√LdL =π√g√LdL,dPP=12dLLso the relative error in P ≈12therelative error in L. Thus the percentage error in P is ≈12the percentage error in L.59. (a) α = ∆L/(L∆T) = 0.006/(40 × 10) = 1.5 × 10−5/◦C(b) ∆L = 2.3 × 10−5(180)(25) ≈ 0.1 cm, so the pole is about 180.1 cm long.60. ∆V = 7.5 × 10−4(4000)(−20) = −60 gallons; the truck delivers 4000 − 60 = 3940 gallons.
• Review Exercises Chapter 3 121REVIEW EXERCISES CHAPTER 32. (a) msec =f(4) − f(3)4 − 3=(4)2/2 − (3)2/21=72(b) mtan = limx1→3f(x1) − f(3)x1 − 3= limx1→3x21/2 − 9/2x1 − 3= limx1→3x21 − 92(x1 − 3)= limx1→3(x1 + 3)(x1 − 3)2(x1 − 3)= limx1→3x1 + 32= 3(c) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x0x21/2 − x20/2x1 − x0= limx1→x0x21 − x202(x1 − x0)= limx1→x0x1 + x02= x0(d)xyTangentSecant5103. (a) mtan = limx1→x0f(x1) − f(x0)x1 − x0= limx1→x0(x21 + 1) − (x20 + 1)x1 − x0= limx1→x0x21 − x20x1 − x0= limx1→x0(x1 + x0) = 2x0(b) mtan = 2(2) = 44. To average 60 mi/h one would have to complete the trip in two hours. At 50 mi/h, 100 miles arecompleted after two hours. Thus time is up, and the speed for the remaining 20 miles would haveto be inﬁnite.5. vinst = limh→03(h + 1)2.5+ 580h − 310h= 58 +110ddx3x2.5x=1= 58 +110(2.5)(3)(1)1.5= 58.75 ft/s6. 164 ft/s250001 207. (a) vave =[3(3)2+ 3] − [3(1)2+ 1]3 − 1= 13 mi/h(b) vinst = limt1→1(3t21 + t1) − 4t1 − 1= limt1→1(3t1 + 4)(t1 − 1)t1 − 1= limt1→1(3t1 + 4) = 7 mi/h
• 122 Chapter 39. (a)dydx= limh→09 − 4(x + h) −√9 − 4xh= limh→09 − 4(x + h) − (9 − 4x)h( 9 − 4(x + h) +√9 − 4x)= limh→0−4hh( 9 − 4(x + h) +√9 − 4x)=−42√9 − 4x=−2√9 − 4x(b)dydx= limh→0x + hx + h + 1−xx + 1h= limh→0(x + h)(x + 1) − x(x + h + 1)h(x + h + 1)(x + 1)= limh→0hh(x + h + 1)(x + 1)=1(x + 1)210. f(x) is continuous and diﬀerentiable at any x = 1, so we consider x = 1.(a) limx→1−(x2− 1) = limx→1+k(x − 1) = 0 = f(1), so any value of k gives continuity at x = 1.(b) limx→1−f (x) = limx→1−2x = 2, and limx→1+f (x) = limx→1+k = k, so only if k = 2 is f(x) diﬀeren-tiable at x = 1.11. (a) x = −2, −1, 1, 3(b) (−∞, −2), (−1, 1), (3, +∞)(c) (−2, −1), (1, 3)(d) g (x) = f (x) sin x + 2f (x) cos x − f(x) sin x; g (0) = 2f (0) cos 0 = 2(2)(1) = 412. yx113. (a) The slope of the tangent line ≈10 − 2.22050 − 1950= 0.078 billion, or in 2050 the world populationwas increasing at the rate of about 78 million per year.(b)dN/dtN≈0.0786= 0.013 = 1.3 %/year14. When x4− x − 1 > 0, f(x) = x4− 2x − 1; whenx4− x − 1 < 0, f(x) = −x4+ 1, and f is diﬀeren-tiable in both cases. The roots of x4− x − 1 = 0 arex1 = −0.724492, x2 = 1.220744. So x4− x − 1 > 0 on(−∞, x1) and (x2, +∞), and x4− x − 1 < 0 on (x1, x2).Then limx→x−1f (x) = limx→x−1(4x3− 2) = 4x31 − 2 andlimx→x+1f (x) = limx→x+1−4x3= −4x31 which is not equal to4x31 − 2, so f is not diﬀerentiable at x = x1; similarly fis not diﬀerentiable at x = x2.1.5–1.5–1.5 215. f (x) = 2x sin x + x2cos x 16. f (x) =1 − 2√x sin 2x2√x
• Review Exercises Chapter 3 12317. f (x) =6x2+ 8x − 17(3x + 2)218. f (x) =(1 + x2) sec2x − 2x tan x(1 + x2)219. (a)dWdt= 200(t − 15); at t = 5,dWdt= −2000; the water is running out at the rate of 2000gal/min.(b)W(5) − W(0)5 − 0=10000 − 225005= −2500; the average rate of ﬂow out is 2500 gal/min.20. (a)43− 234 − 2=562= 28 (b) (dV/d )| =5 = 3 2=5= 3(5)2= 7521. (a) f (x) = 2x, f (1.8) = 3.6(b) f (x) = (x2− 4x)/(x − 2)2, f (3.5) ≈ −0.777777822. (a) f (x) = 3x2− 2x, f (2.3) = 11.27(b) f (x) = (1 − x2)/(x2+ 1)2, f (−0.5) = 0.4823. f is continuous at x = 1 because it is diﬀerentiable there, thus limh→0f(1+h) = f(1) and so f(1) = 0because limh→0f(1 + h)hexists; f (1) = limh→0f(1 + h) − f(1)h= limh→0f(1 + h)h= 5.24. f (x) = limh→0f(x + h) − f(x)h= limh→0f(x)f(h) − f(x)h= limh→0f(x)[f(h) − 1]h= f(x) limh→0f(h) − f(0)h= f(x)f (0) = f(x)25. The equation of such a line has the form y = mx. The points (x0, y0) which lie on both the line andthe parabola and for which the slopes of both curves are equal satisfy y0 = mx0 = x30 −9x20 −16x0,so that m = x20 − 9x0 − 16. By diﬀerentiating, the slope is also given by m = 3x20 − 18x0 − 16.Equating, we have x20 −9x0 −16 = 3x20 −18x0 −16, or 2x20 −9x0 = 0. The root x0 = 0 correspondsto m = −16, y0 = 0 and the root x0 = 9/2 corresponds to m = −145/4, y0 = −1305/8. So the liney = −16x is tangent to the curve at the point (0, 0), and the line y = −145x/4 is tangent to thecurve at the point (9/2, −1305/8).26. The slope of the line x + 4y = 10 is m1 = −1/4, so we set the negative reciprocal4 = m2 =ddx(2x3− x2) = 6x2− 2x and obtain 6x2− 2x − 4 = 0 with rootsx =1 ±√1 + 246= 1, −2/3.27. The slope of the tangent line is the derivativey = 2xx= 12 (a+b)= a + b. The slope of the secant isa2− b2a − b= a + b, so they are equal.yxa ba+b2(a, a2)(b, b2)
• 124 Chapter 328. (a) f (1)g(1) + f(1)g (1) = 3(−2) + 1(−1) = −7(b)g(1)f (1) − f(1)g (1)g(1)2=−2(3) − 1(−1)(−2)2= −54(c)12 f(1)f (1) =12√1(3) =32(d) 0 (because f(1)g (1) is constant)29. (a) 8x7−32√x− 15x−4(b) 2 · 101(2x + 1)100(5x2− 7) + 10x(2x + 1)101(c) 2(x − 1)√3x + 1 +32√3x + 1(x − 1)2(d) 33x + 1x22x2(3) − (3x + 1)(2x)x4= −3(3x + 1)2(3x + 2)x730. (a) cos x − 6 cos2x sin x(b) (1 + sec x)(2x − sec2x) + (x2− tan x) sec x tan x(c) − csc2 csc 2xx3 + 5−2(x3+ 5) csc 2x cot 2x − 3x2csc 2x(x3 + 5)2(d) −2 + 3 sin2x cos x(2x + sin3x)231. Set f (x) = 0: f (x) = 6(2)(2x + 7)5(x − 2)5+ 5(2x + 7)6(x − 2)4= 0, so 2x + 7 = 0 or x − 2 = 0or, factoring out (2x + 7)5(x − 2)4, 12(x − 2) + 5(2x + 7) = 0. This reduces to x = −7/2, x = 2,or 22x + 11 = 0, so the tangent line is horizontal at x = −7/2, 2, −1/2.32. Set f (x) = 0: f (x) =4(x2+ 2x)(x − 3)3− (2x + 2)(x − 3)4(x2 + 2x)2, and a fraction can equal zeroonly if its numerator equals zero. So either x − 3 = 0 or, after factoring out (x − 3)3,4(x2+ 2x) − (2x + 2)(x − 3) = 0, 2x2+ 12x + 6 = 0, whose roots are (by the quadratic formula)x =−6 ±√36 − 4 · 32= −3 ±√6. So the tangent line is horizontal at x = 3, −3 ±√6.33. Let y = mx + b be a line tangent to y = x2+ 1 at the point (x0, x20 + 1) with slope m = 2x0.By inspection if x1 = −x0 then the same line is also tangent to the curve y = −x2−1 at (−x1, −y1),since y1 = −y0 = −x20 − 1 = −(x20 + 1) = −x21 − 1.Thus the tangent line passes through the points (x0, y0) and (x1, y1) = (−x0, −y0), so its slopem =y0 − y1x0 − x1=2y02x0=x20 + 1x0. But, from the above, m = 2x0; equate and getx20 + 1x0= 2x0, withsolution x0 = ±1. Thus the only possible such tangent lines are y = 2x and y = −2x.34. (a) Suppose y = mx+b is tangent to y = xn+n−1 at (x0, y0) and to y = −xn−n+1 at (x1, y1).Then m = nxn−10 = −nxn−11 , and hence x1 = −x0. Since n is even, y1 = −xn1 − n + 1 =−xn0 − n + 1 = −(xn0 + n − 1) = −y0. Thus the points (x0, y0) and (x1, y1) are symmetricwith respect to the origin and both lie on the tangent line and thus b = 0.The slope m is given by m = nxn−10 and by m = y0/x0 = (xn0 + n − 1)/x0, hence nxn0 =xn0 + n − 1, (n − 1)xn0 = n − 1, xn0 = 1. Since n is even, x0 = ±1. One easily checks thaty = nx is tangent to y = xn+ n − 1 at (1, n) and to y = −xn− n + 1 at (−1, −n).
• Review Exercises Chapter 3 125(b) Suppose there is such a common tangent line with slope m. The function y = xn+ n − 1 isalways increasing, so m ≥ 0. Moreover the function y = −xn− n + 1 is always decreasing, som ≤ 0. Thus the tangent line has slope 0, which only occurs on the curves for x = 0. Thiswould require the common tangent line to pass through (0, n − 1) and (0, −n + 1) and do sowith slope m = 0, which is impossible.35. The line y −x = 2 has slope m1 = 1 so we set m2 =ddx(3x−tan x) = 3−sec2x = 1, or sec2x = 2,sec x = ±√2 so x = nπ ± π/4 where n = 0, ±1, ±2, . . . .36. Solve 3x2− cos x = 0 to get x = ±0.535428.37. 3 = f(π/4) = (M + N)√2/2 and 1 = f (π/4) = (M − N)√2/2. Add these two equations to get4 =√2M, M = 23/2. Subtract to obtain 2 =√2N, N =√2. Thus f(x) = 2√2 sin x +√2 cos x.f3π4= −3 so tangent line is y − 1 = −3 x −3π4.38. f(x) = M tan x+N sec x, f (x) = M sec2x+N sec x tan x. At x = π/4, 2M+√2N, 0 = 2M+√2N.Add to get M = −2, subtract to get N =√2 + M/√2 = 2√2, f(x) = −2 tan x + 2√2 sec x.f (0) = −2 so tangent line is y − 2√2 = −2x.39. f (x) = 2xf(x), f(2) = 5(a) g(x) = f(sec x), g (x) = f (sec x) sec x tan x = 2 · 2f(2) · 2 ·√3 = 40√3.(b) h (x) = 4f(x)x − 13(x − 1)f (x) − f(x)(x − 1)2,h (2) = 4531f (2) − f(2)1= 4 · 53 2 · 2f(2) − f(2)1= 4 · 53· 3 · 5 = 750040. (a)dTdL=2√g12√L=1√gL(b) s/m(c) SincedTdL> 0 an increase in L gives an increase in T, which is the period. To speed up aclock, decrease the period; to decrease T, decrease L.(d)dTdg= −√Lg3/2< 0; a decrease in g will increase T and the clock runs slower(e)dTdg= 2√L−12g−3/2= −√Lg3/2(f) s3/m41. A = πr2anddrdt= −5, sodAdt=dAdrdrdt= 2πr(−5) = −500π, so the area is shrinking at a rate of500π m2/min.42. Finddθdt x=1y=1givendzdt= a anddydt= −b. From the ﬁguresin θ = y/z; when x = y = 1, z =√2. So θ = sin−1(y/z)anddθdt=11 − y2/z21zdydt−yz2dzdt= −b −a√2when x = y = 1.yxzθ
• 126 Chapter 343. (a) ∆x = 1.5 − 2 = −0.5; dy =−1(x − 1)2∆x =−1(2 − 1)2(−0.5) = 0.5; and∆y =1(1.5 − 1)−1(2 − 1)= 2 − 1 = 1.(b) ∆x = 0 − (−π/4) = π/4; dy = sec2(−π/4) (π/4) = π/2; and ∆y = tan 0 − tan(−π/4) = 1.(c) ∆x = 3 − 0 = 3; dy =−x√25 − x2=−025 − (0)2(3) = 0; and∆y =√25 − 32 −√25 − 02 = 4 − 5 = −1.44. cot 46◦= cot46π180; let x0 =π4and x =46π180. Thencot 46◦= cot x ≈ cotπ4− csc2 π4x −π4= 1 − 246π180−π4= 0.9651;with a calculator, cot 46◦= 0.9657.45. (a) h = 115 tan φ, dh = 115 sec2φ dφ; with φ = 51◦=51180π radians anddφ = ±0.5◦= ±0.5π180radians, h ± dh = 115(1.2349) ± 2.5340 = 142.0135 ± 2.5340, sothe height lies between 139.48 m and 144.55 m.(b) If |dh| ≤ 5 then |dφ| ≤5115cos2 51180π ≈ 0.017 radians, or |dφ| ≤ 0.98◦.
• 127CHAPTER 4Derivatives of Logarithmic, Exponential, andInverse Trigonometric FunctionsEXERCISE SET 4.11. y = (2x − 5)1/3; dy/dx =23(2x − 5)−2/32. dy/dx =132 + tan(x2)−2/3sec2(x2)(2x) =23x sec2(x2) 2 + tan(x2)−2/33. dy/dx =23x + 1x − 2−1/3x − 2 − (x + 1)(x − 2)2= −2(x + 1)1/3(x − 2)5/34. dy/dx =12x2+ 1x2 − 5−1/2ddxx2+ 1x2 − 5=12x2+ 1x2 − 5−1/2−12x(x2 − 5)2= −6x(x2 − 5)3/2√x2 + 15. dy/dx = x3−23(5x2+ 1)−5/3(10x) + 3x2(5x2+ 1)−2/3=13x2(5x2+ 1)−5/3(25x2+ 9)6. dy/dx = −3√2x − 1x2+1x23(2x − 1)2/3=−4x + 33x2(2x − 1)2/37. dy/dx =52[sin(3/x)]3/2[cos(3/x)](−3/x2) = −15[sin(3/x)]3/2cos(3/x)2x28. dy/dx = −12cos(x3)−3/2− sin(x3) (3x2) =32x2sin(x3) cos(x3)−3/29. (a) 1 + y + xdydx− 6x2= 0,dydx=6x2− y − 1x(b) y =2 + 2x3− xx=2x+ 2x2− 1,dydx= −2x2+ 4x(c) From Part (a),dydx= 6x −1x−1xy = 6x −1x−1x2x+ 2x2− 1 = 4x −2x210. (a)12y−1/2 dydx− cos x = 0 ordydx= 2√y cos x(b) y = (2 + sin x)2= 4 + 4 sin x + sin2x sodydx= 4 cos x + 2 sin x cos x(c) from Part (a),dydx= 2√y cos x = 2 cos x(2 + sin x) = 4 cos x + 2 sin x cos x11. 2x + 2ydydx= 0 sodydx= −xy12. 3x2+ 3y2 dydx= 3y2+ 6xydydx,dydx=3y2− 3x3y2 − 6xy=y2− x2y2 − 2xy13. x2 dydx+ 2xy + 3x(3y2)dydx+ 3y3− 1 = 0(x2+ 9xy2)dydx= 1 − 2xy − 3y3sodydx=1 − 2xy − 3y3x2 + 9xy2
• 128 Chapter 414. x3(2y)dydx+ 3x2y2− 5x2 dydx− 10xy + 1 = 0(2x3y − 5x2)dydx= 10xy − 3x2y2− 1 sodydx=10xy − 3x2y2− 12x3y − 5x215. −12x3/2−dydx2y3/2= 0,dydx= −y3/2x3/216. 2x =(x − y)(1 + dy/dx) − (x + y)(1 − dy/dx)(x − y)2,2x(x − y)2= −2y + 2xdydxsodydx=x(x − y)2+ yx17. cos(x2y2) x2(2y)dydx+ 2xy2= 1,dydx=1 − 2xy2cos(x2y2)2x2y cos(x2y2)18. − sin(xy2) y2+ 2xydydx=dydx,dydx= −y2sin(xy2)2xy sin(xy2) + 119. 3 tan2(xy2+ y) sec2(xy2+ y) 2xydydx+ y2+dydx= 1sodydx=1 − 3y2tan2(xy2+ y) sec2(xy2+ y)3(2xy + 1) tan2(xy2 + y) sec2(xy2 + y)20.(1 + sec y)[3xy2(dy/dx) + y3] − xy3(sec y tan y)(dy/dx)(1 + sec y)2= 4y3 dydx,multiply through by (1 + sec y)2and solve fordydxto getdydx=y(1 + sec y)4y(1 + sec y)2 − 3x(1 + sec y) + xy sec y tan y21. 4x − 6ydydx= 0,dydx=2x3y, 4 − 6dydx2− 6yd2ydx2= 0,d2ydx2= −3 dydx2− 23y=2(3y2− 2x2)9y3= −89y322.dydx= −x2y2,d2ydx2= −y2(2x) − x2(2ydy/dx)y4= −2xy2− 2x2y(−x2/y2)y4= −2x(y3+ x3)y5,but x3+ y3= 1 sod2ydx2= −2xy523.dydx= −yx,d2ydx2= −x(dy/dx) − y(1)x2= −x(−y/x) − yx2=2yx224. y + xdydx+ 2ydydx= 0,dydx= −yx + 2y, 2dydx+ xd2ydx2+ 2dydx2+ 2yd2ydx2= 0,d2ydx2=2y(x + y)(x + 2y)325.dydx= (1 + cos y)−1,d2ydx2= −(1 + cos y)−2(− sin y)dydx=sin y(1 + cos y)3
• Exercise Set 4.1 12926.dydx=cos y1 + x sin y,d2ydx2=(1 + x sin y)(− sin y)(dy/dx) − (cos y)[(x cos y)(dy/dx) + sin y](1 + x sin y)2= −2 sin y cos y + (x cos y)(2 sin2y + cos2y)(1 + x sin y)3,but x cos y = y, 2 sin y cos y = sin 2y, and sin2y + cos2y = 1 sod2ydx2= −sin 2y + y(sin2y + 1)(1 + x sin y)327. By implicit diﬀerentiation, 2x + 2y(dy/dx) = 0,dydx= −xy; at (1/2,√3/2),dydx= −√3/3; at(1/2, −√3/2),dydx= +√3/3. Directly, at the upper point y =√1 − x2,dydx=−x√1 − x2=−1/23/4= −1/√3 and at the lower point y = −√1 − x2,dydx=x√1 − x2= +1/√3.28. If y2− x + 1 = 0, then y =√x − 1 goes through the point (10, 3) so dy/dx = 1/(2√x − 1). Byimplicit diﬀerentiation dy/dx = 1/(2y). In both cases, dy/dx|(10,3) = 1/6. Similarly y = −√x − 1goes through (10, −3) so dy/dx = −1/(2√x − 1) = −1/6 which yields dy/dx = 1/(2y) = −1/6.29. 4x3+ 4y3 dydx= 0, sodydx= −x3y3= −1153/4≈ −0.1312.30. 3y2 dydx+ x2 dydx+ 2xy + 2x − 6ydydx= 0, sodydx= −2xy + 13y2 + x2 − 6y= 0 at x = 031. 4(x2+ y2) 2x + 2ydydx= 25 2x − 2ydydx,dydx=x[25 − 4(x2+ y2)]y[25 + 4(x2 + y2)]; at (3, 1)dydx= −9/1332.23x−1/3+ y−1/3 dydx= 0,dydx= −y1/3x1/3=√3 at (−1, 3√3)33. 4a3 dadt− 4t3= 6 a2+ 2atdadt, solve fordadtto getdadt=2t3+ 3a22a3 − 6at34.12u−1/2 dudv+12v−1/2= 0 sodudv= −√u√v35. 2a2ωdωdλ+ 2b2λ = 0 sodωdλ= −b2λa2ω36. 1 = (cos x)dxdysodxdy=1cos x= sec x
• 130 Chapter 437. (a)–4 4–22xy(b) Implicit diﬀerentiation of the equation of the curve yields (4y3+ 2y)dydx= 2x − 1 sodydx= 0only if x = 1/2 but y4+ y2≥ 0, so x = 1/2 is impossible.(c) x2− x − (y4+ y2) = 0, so by the Quadratic Formula x =1 ± 1 + 4y2 + 4y42= 1 + y2, −y2which gives the parabolas x = 1 + y2, x = −y2.38. (a) y–2021 2x(b) 2ydydx= (x − a)(x − b) + x(x − b) + x(x − a) = 3x2− 2(a + b)x + ab. Ifdydx= 0 then3x2− 2(a + b)x + ab = 0. By the Quadratic Formulax =2(a + b) ± 4(a + b)2 − 4 · 3ab6=13a + b ± (a2+ b2− ab)1/2.(c) y = ± x(x − a)(x − b). The square root is only deﬁned for nonnegative arguments, so it isnecessary that all three of the factors x, x − a, x − b be nonnegative, or that two of them benonpositive. If, for example, 0 < a < b then the function is deﬁned on the disjoint intervals0 < x < a and b < x < +∞, so there are two parts.39. (a) yx2–2 2–2(b) x ≈ ±1.1547(c) Implicit diﬀerentiation yields 2x − xdydx− y + 2ydydx= 0. Solve fordydx=y − 2x2y − x. Ifdydx= 0then y − 2x = 0 or y = 2x. Thus 4 = x2− xy + y2= x2− 2x2+ 4x2= 3x2, x = ±2√3.40. (a) See Exercise 39 (a)(b) Since the equation is symmetric in x and y, we obtain, as in Exercise 39, x ≈ ±1.1547.
• Exercise Set 4.1 131(c) Implicit diﬀerentiation yields 2x − xdydx− y + 2ydydx= 0. Solve fordxdy=2y − xy − 2x. Ifdxdy= 0then 2y − x = 0 or x = 2y. Thus 4 = 4y2− 2y2+ y2= 3y2, y = ±2√3, x = 2y = ±4√3.41. Solve the simultaneous equations y = x, x2−xy+y2= 4 to get x2−x2+x2= 4, x = ±2, y = x = ±2,so the points of intersection are (2, 2) and (−2, −2).From Exercise 39 part (c),dydx=y − 2x2y − x. When x = y = 2,dydx= −1; when x = y = −2,dydx= −1;the slopes are equal.42. Suppose a2− 2ab + b2= 4. Then (−a)2− 2(−a)(−b) + (−b)2= a2− 2ab + b2= 4 so if P(a, b) lieson C then so does Q(−a, −b).From Exercise 39 part (c),dydx=y − 2x2y − x. When x = a, y = b thendydx=b − 2a2b − a, and whenx = −a, y = −b, thendydx=b − 2a2b − a, so the slopes at P and Q are equal.43. The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; useimplicit diﬀerentiation to getdydx= −2xyx2 + 2ayso at (1,1), −21 + 2a= −43, 1 + 2a = 3/2, a = 1/4and hence b = 1 + 1/4 = 5/4.44. The slope of the line x + 2y − 2 = 0 is m1 = −1/2, so the line perpendicular has slope m = 2(negative reciprocal). The slope of the curve y3= 2x2can be obtained by implicit diﬀerentiation:3y2 dydx= 4x,dydx=4x3y2. Setdydx= 2;4x3y2= 2, x = (3/2)y2. Use this in the equation of the curve:y3= 2x2= 2((3/2)y2)2= (9/2)y4, y = 2/9, x =32292=227.45. (a)–3 –1 2–3–12xy (b) x ≈ 0.84(c) Use implicit diﬀerentiation to get dy/dx = (2y−3x2)/(3y2−2x), so dy/dx = 0 if y = (3/2)x2.Substitute this into x3− 2xy + y3= 0 to obtain 27x6− 16x3= 0, x3= 16/27, x = 24/3/3and hence y = 25/3/3.46. (a)–3 –1 2–3–12xy (b) Evidently the tangent line at the pointx = 1, y = 1 has slope −1.
• 132 Chapter 4(c) Use implicit diﬀerentiation to get dy/dx = (2y−3x2)/(3y2−2x), so dy/dx = −1 if 2y−3x2=−3y2+2x, 2(y−x)+3(y−x)(y+x) = 0. One solution is y = x; this together with x3+y3= 2xyyields x = y = 1. For these values dy/dx = −1, so that (1, 1) is a solution.To prove that there is no other solution, suppose y = x. From dy/dx = −1 it followsthat 2(y − x) + 3(y − x)(y + x) = 0. But y = x, so x + y = −2/3. Then x3+ y3=(x + y)(x2− xy + y2) = 2xy, so replacing x + y with −2/3 we get x2+ 2xy + y2= 0, or(x+y)2= 0, so y = −x. Substitute that into x3+y3= 2xy to obtain x3−x3= −2x2, x = 0.But at x = y = 0 the derivative is not deﬁned.47. (a) The curve is the circle (x − 2)2+ y2= 1 about the point (2, 0) of radius 1. One tangentline is tangent at a point P(x,y) in the ﬁrst quadrant. Let Q(2, 0) be the center of thecircle. Then OPQ is a right angle, with sides |PQ| = r = 1 and |OP| = x2 + y2. Bythe Pythagorean Theorem x2+ y2+ 12= 22. Substitute this into (x − 2)2+ y2= 1 toobtain 3 − 4x + 4 = 1, x = 3/2, y =√3 − x2 =√3/2. So the required tangent lines arey = ±(√3/3)x.(b) Let P(x0, y0) be a point where a line through the origin is tangent to the curvex2− 4x + y2+ 3 = 0. Implicit diﬀerentiation applied to the equation of the curve givesdy/dx = (2 − x)/y. At P the slope of the curve must equal the slope of the line so(2 − x0)/y0 = y0/x0, or y20 = 2x0 − x20. But x20 − 4x0 + y20 + 3 = 0 because (x0, y0) is on thecurve, and elimination of y20 in the latter two equations gives x20 − 4x0 + (2x0 − x20) + 3 = 0,x0 = 3/2 which when substituted into y20 = 2x0 − x20 yields y20 = 3/4, so y0 = ±√3/2. Theslopes of the lines are (±√3/2)/(3/2) = ±√3/3 and their equations are y = (√3/3)x andy = −(√3/3)x.48. Let P(x0, y0) be a point where a line through the origin is tangent to the curve2x2− 4x + y2+ 1 = 0. Implicit diﬀerentiation applied to the equation of the curve givesdy/dx = (2 − 2x)/y. At P the slope of the curve must equal the slope of the line so(2 − 2x0)/y0 = y0/x0, or y20 = 2x0(1 − x0). But 2x20 − 4x0 + y20 + 1 = 0 because (x0, y0) is on thecurve, and elimination of y20 in the latter two equations gives 2x0 = 4x0 − 1, x0 = 1/2 which whensubstituted into y20 = 2x0(1 − x0) yields y20 = 1/2, so y0 = ±√2/2. The slopes of the lines are(±√2/2)/(1/2) = ±√2 and their equations are y =√2x and y = −√2x.49. The linear equation axr−10 x + byr−10 y = c is the equation of a line . Implicit diﬀerentiation of theequation of the curve yields raxr−1+rbyr−1 dydx= 0,dydx= −axr−1byr−1. At the point (x0, y0) the slopeof the line must be −axr−10byr−10, which is the slope of . Moreover, the equation of is satisﬁed bythe point (x0, y0), so this point lies on . By the point-slope formula, must be the line tangentto the curve at (x0, y0).50. Implicit diﬀerentiation of the equation of the curve yields rxr−1+ryr−1 dydx= 0. At the point (1, 1)this becomes r + rdydx= 0,dydx= −1.51. By the chain rule,dydx=dydtdtdx. Use implicit diﬀerentiation on 2y3t + t3y = 1 to getdydt= −2y3+ 3t2y6ty2 + t3, butdtdx=1cos tsodydx= −2y3+ 3t2y(6ty2 + t3) cos t.52. (a) f (x) =43x1/3, f (x) =49x−2/3(b) f (x) =73x4/3, f (x) =289x1/3, f (x) =2827x−2/3(c) generalize parts (a) and (b) with k = (n − 1) + 1/3 = n − 2/3
• Exercise Set 4.2 13353. y = rxr−1, y = r(r − 1)xr−2so 3x2r(r − 1)xr−2+ 4x rxr−1− 2xr= 0,3r(r − 1)xr+ 4rxr− 2xr= 0, (3r2+ r − 2)xr= 0,3r2+ r − 2 = 0, (3r − 2)(r + 1) = 0; r = −1, 2/354. y = rxr−1, y = r(r − 1)xr−2so 16x2r(r − 1)xr−2+ 24x rxr−1+ xr= 0,16r(r − 1)xr+ 24rxr+ xr= 0, (16r2+ 8r + 1)xr= 0,16r2+ 8r + 1 = 0, (4r + 1)2= 0; r = −1/455. We shall ﬁnd when the curves intersect and check that the slopes are negative reciprocals. For theintersection solve the simultaneous equations x2+ (y − c)2= c2and (x − k)2+ y2= k2to obtaincy = kx =12(x2+ y2). Thus x2+ y2= cy + kx, or y2− cy = −x2+ kx, andy − cx= −x − ky.Diﬀerentiating the two families yields (black)dydx= −xy − c, and (gray)dydx= −x − ky. But it wasproven that these quantities are negative reciprocals of each other.56. Diﬀerentiating, we get the equations (black) xdydx+ y = 0 and (gray) 2x − 2ydydx= 0. The ﬁrstsays the (black) slope is = −yxand the second says the (gray) slope isxy, and these are negativereciprocals of each other.EXERCISE SET 4.21.15x(5) =1x2.1x/313=1x3.11 + x4.12 +√x12√x=12√x(2 +√x)5.1x2 − 1(2x) =2xx2 − 16.3x2− 14xx3 − 7x2 − 37.1x/(1 + x2)(1 + x2)(1) − x(2x)(1 + x2)2=1 − x2x(1 + x2)8.1(1 + x)/(1 − x)1 − x + 1 + x(1 − x)2=21 − x29.ddx(2 ln x) = 2ddxln x =2x10. 3 (ln x)2 1x11.12(ln x)−1/2 1x=12x√ln x12.1√x12√x=12x13. ln x + x1x= 1 + ln x14. x3 1x+ (3x2) ln x = x2(1 + 3 ln x) 15. 2x log2(3 − 2x) −2x2(3 − 2x) ln 216. log2(x2− 2x)3+ 3x log2(x2− 2x)2 2x − 2(x2 − 2x) ln 2
• 134 Chapter 417.2x(1 + log x) − x/(ln 10)(1 + log x)218. 1/[x(ln 10)(1 + log x)2]19.1ln x1x=1x ln x20.1ln(ln(x))1ln x1x21.1tan x(sec2x) = sec x csc x 22.1cos x(− sin x) = − tan x23. −1xsin(ln x) 24. 2 sin(ln x) cos(ln x)1x=sin(2 ln x)x=sin(ln x2)x25.1ln 10 sin2x(2 sin x cos x) = 2cot xln 1026.1(ln 10)(1 − sin2x)(−2 sin x cos x) = −2 sin x cos x(ln 10) cos2 x= −2 tan xln 1027.ddx3 ln(x − 1) + 4 ln(x2+ 1) =3x − 1+8xx2 + 1=11x2− 8x + 3(x − 1)(x2 + 1)28.ddx[2 ln cos x +12ln(1 + x4)] = −2 tan x +2x31 + x429.ddxln cos x −12ln(4 − 3x2) = − tan x +3x4 − 3x230.ddx12[ln(x − 1) − ln(x + 1)] =121x − 1−1x + 131. ln |y| = ln |x| +13ln |1 + x2|,dydx= x31 + x21x+2x3(1 + x2)32. ln |y| =15[ln |x − 1| − ln |x + 1|],dydx=155 x − 1x + 11x − 1−1x + 133. ln |y| =13ln |x2− 8| +12ln |x3+ 1| − ln |x6− 7x + 5|dydx=(x2− 8)1/3√x3 + 1x6 − 7x + 52x3(x2 − 8)+3x22(x3 + 1)−6x5− 7x6 − 7x + 534. ln |y| = ln | sin x| + ln | cos x| + 3 ln | tan x| −12ln |x|dydx=sin x cos x tan3x√xcot x − tan x +3 sec2xtan x−12x35. f (x) = exe−136. ln y = −√10 ln x,1ydydx= −√10x,dydx= −√10x1+√1037. (a) logx e =ln eln x=1ln x,ddx[logx e] = −1x(ln x)2(b) logx 2 =ln 2ln x,ddx[logx 2] = −ln 2x(ln x)2
• Exercise Set 4.2 13538. (a) From loga b =ln bln afor a, b > 0 it follows that log(1/x) e =ln eln(1/x)= −1ln x, henceddxlog(1/x) e =1x(ln x)2(b) log(ln x) e =ln eln(ln x)=1ln(ln x), soddxlog(ln x) e = −1(ln(ln x))21x ln x= −1x(ln x)(ln(ln x))239. f(x0) = f(e−1) = −1, f (x) =1x, f (x0) = e, y − (−1) = e(x − 1/e) = ex − 1, y = ex − 240. y0 = log 10 = 1, y = log x = (log e) ln x,dydx x=10= log e110,y − 1 =log e10(x − 10), y =log e10x + 1 − log e41. f(x0) = f(−e) = 1, f (x)x=−e= −1e,y − 1 = −1e(x + e), y = −1ex42. y − ln 2 = −12(x + 2), y = −12x + ln 2 − 143. Let the equation of the tangent line be y = mx and suppose that it meets the curve at (x0, y0).Then m =1x x=x0=1x0and y0 = mx0 = ln x0. So m =1x0=ln x0x0and ln x0 = 1, x0 = e, m =1eand the equation of the tangent line is y =1ex.44. Let y = mx + b be a line tangent to the curve at (x0, y0). Then b is the y-intercept and theslope of the tangent line is m =1x0. Moreover, at the point of tangency, mx0 + b = ln x0 or1x0x0 + b = ln x0, b = ln x0 − 1, as required.45. The area of the triangle PQR, given by |PQ||QR|/2 isrequired. |PQ| = w, and, by Exercise 44, |QR| = 1, soarea = w/2.2–21xyP (w, ln w)QRw46. Since y = 2 ln x, let y = 2z; then z = ln x and we apply the result of Exercise 45 to ﬁnd that thearea is, in the x-z plane, w/2. In the x-y plane, since y = 2z, the vertical dimension gets doubled,so the area is w.47. If x = 0 then y = ln e = 1, anddydx=1x + e. But ey= x + e, sodydx=1ey= e−y.48. When x = 0, y = − ln(e2) = −2. Next,dydx=1e2 − x. But ey= e− ln(e2−x)= (e2− x)−1, sodydx= ey.
• 136 Chapter 449. Let y = ln(x+a). Following Exercise 47 we getdydx=1x + a= e−y, and when x = 0, y = ln(a) = 0if a = 1, so let a = 1, then y = ln(x + 1).50. Let y = − ln(a − x), thendydx=1a − x. But ey=1a − x, sodydx= ey.If x = 0 then y = − ln(a) = − ln 2 provided a = 2, so y = − ln(2 − x).51. (a) f(x) = ln x; f (e2) = lim∆x→0ln(e2+ ∆x) − 2∆x=ddx(ln x)x=e2=1x x=e2= e−2(b) f(w) = ln w; f (1) = limh→0ln(1 + h) − ln 1h= limh→0ln(1 + h)h=1w w=1= 152. (a) Let f(x) = ln(cos x), then f(0) = ln(cos 0) = ln 1 = 0, so f (0) = limx→0f(x) − f(0)x=limx→0ln(cos x)x, and f (0) = − tan 0 = 0.(b) Let f(x) = x√2, then f(1) = 1, so f (1) = limh→0f(1 + h) − f(1)h= limh→0(1 + h)√2− 1h, andf (x) =√2x√2−1, f (1) =√2.53.ddx[logb x] = limh→0logb(x + h) − logb(x)h= limh→01hlogbx + hxTheorem 1.6.2(b)= limh→01hlogb 1 +hx= limv→01vxlogb(1 + v) Let v = h/x and note that v → 0 as h → 0=1xlimv→01vlogb(1 + v) h and v are variable, whereas x is constant=1xlimv→0logb(1 + v)1/vTheorem 1.6.2.(c)=1xlogb limv→0(1 + v)1/vTheorem 2.5.5=1xlogb e Formula 7 of Section 7.1EXERCISE SET 4.31. (a) f (x) = 5x4+ 3x2+ 1 ≥ 1 so f is one-to-one on −∞ < x < +∞.(b) f(1) = 3 so 1 = f−1(3);ddxf−1(x) =1f (f−1(x)), (f−1) (3) =1f (1)=192. (a) f (x) = 3x2+ 2ex; for −1 < x < 1, f (x) ≥ 2e−1= 2/e, and for |x| > 1, f (x) ≥ 3x2≥ 3, sof is increasing and one-to-one(b) f(0) = 2 so 0 = f−1(2);ddxf−1(x) =1f (f−1(x)), (f−1) (2) =1f (0)=12
• Exercise Set 4.3 1373. f−1(x) =2x− 3, so directlyddxf−1(x) = −2x2. Using Formula (1),f (x) =−2(x + 3)2, so1f (f−1(x))= −(1/2)(f−1(x) + 3)2,ddxf−1(x) = −(1/2)2x2= −2x24. f−1(x) =ex− 12, so directly,ddxf−1(x) =ex2. Next, f (x) =22x + 1, and using Formula (1),ddxf−1(x) =2f−1(x) + 12=ex25. (a) f (x) = 2x + 8; f < 0 on (−∞, −4) and f > 0 on (−4, +∞); not enough information. Byinspection, f(1) = 10 = f(−9), so not one-to-one(b) f (x) = 10x4+ 3x2+ 3 ≥ 3 > 0; f (x) is positive for all x, so f is one-to-one(c) f (x) = 2 + cos x ≥ 1 > 0 for all x, so f is one-to-one(d) f (x) = −(ln 2) 12x< 0 because ln 2 > 0, so f is one-to-one for all x.6. (a) f (x) = 3x2+ 6x = x(3x + 6) changes sign at x = −2, 0, so not enough information; byobservation (of the graph, and using some guesswork), f(−1 +√3) = −6 = f(−1 −√3), sof is not one-to-one.(b) f (x) = 5x4+ 24x2+ 2 ≥ 2 > 0; f is positive for all x, so f is one-to-one(c) f (x) =1(x + 1)2; f is one-to-one because:if x1 < x2 < −1 then f > 0 on [x1, x2], so f(x1) = f(x2)if −1 < x1 < x2 then f > 0 on [x1, x2], so f(x1) = f(x2)if x1 < −1 < x2 then f(x1) > 1 > f(x2) since f(x) > 1 on (−∞, −1) and f(x) < 1 on(−1, +∞)(d) Note that f(x) is only deﬁned for x > 0.ddxlogb x =1x ln b, which is always negative(0 < b < 1), so f is one-to-one.7. y = f−1(x), x = f(y) = 5y3+ y − 7,dxdy= 15y2+ 1,dydx=115y2 + 1;check: 1 = 15y2 dydx+dydx,dydx=115y2 + 18. y = f−1(x), x = f(y) = 1/y2,dxdy= −2y−3,dydx= −y3/2;check: 1 = −2y−3 dydx,dydx= −y3/29. y = f−1(x), x = f(y) = 2y5+ y3+ 1,dxdy= 10y4+ 3y2,dydx=110y4 + 3y2;check: 1 = 10y4 dydx+ 3y2 dydx,dydx=110y4 + 3y210. y = f−1(x), x = f(y) = 5y − sin 2y,dxdy= 5 − 2 cos 2y,dydx=15 − 2 cos 2y;check: 1 = (5 − 2 cos 2y)dydx,dydx=15 − 2 cos 2y
• 138 Chapter 411. 7e7x12. −10xe−5x213. x3ex+ 3x2ex= x2ex(x + 3) 14. −1x2e1/x15.dydx=(ex+ e−x)(ex+ e−x) − (ex− e−x)(ex− e−x)(ex + e−x)2=(e2x+ 2 + e−2x) − (e2x− 2 + e−2x)(ex + e−x)2= 4/(ex+ e−x)216. excos(ex)17. (x sec2x + tan x)ex tan x18.dydx=(ln x)ex− ex(1/x)(ln x)2=ex(x ln x − 1)x(ln x)219. (1 − 3e3x)e(x−e3x)20.152x2(1 + 5x3)−1/2exp( 1 + 5x3)21.(x − 1)e−x1 − xe−x=x − 1ex − x22.1cos(ex)[− sin(ex)]ex= −extan(ex)23. f (x) = 2xln 2; y = 2x, ln y = x ln 2,1yy = ln 2, y = y ln 2 = 2xln 224. f (x) = −3−xln 3; y = 3−x, ln y = −x ln 3,1yy = − ln 3, y = −y ln 3 = −3−xln 325. f (x) = πsin x(ln π) cos x;y = πsin x, ln y = (sin x) ln π,1yy = (ln π) cos x, y = πsin x(ln π) cos x26. f (x) = πx tan x(ln π)(x sec2x + tan x);y = πx tan x, ln y = (x tan x) ln π,1yy = (ln π)(x sec2x + tan x)y = πx tan x(ln π)(x sec2x + tan x)27. ln y = (ln x) ln(x3− 2x),1ydydx=3x2− 2x3 − 2xln x +1xln(x3− 2x),dydx= (x3− 2x)ln x 3x2− 2x3 − 2xln x +1xln(x3− 2x)28. ln y = (sin x) ln x,1ydydx=sin xx+ (cos x) ln x,dydx= xsin x sin xx+ (cos x) ln x29. ln y = (tan x) ln(ln x),1ydydx=1x ln xtan x + (sec2x) ln(ln x),dydx= (ln x)tan x tan xx ln x+ (sec2x) ln(ln x)
• Exercise Set 4.3 13930. ln y = (ln x) ln(x2+ 3),1ydydx=2xx2 + 3ln x +1xln(x2+ 3),dydx= (x2+ 3)ln x 2xx2 + 3ln x +1xln(x2+ 3)31. f (x) = exe−132. (a) because xxis not of the form axwhere a is constant(b) y = xx, ln y = x ln x,1yy = 1 + ln x, y = xx(1 + ln x)33.31 − (3x)2=3√1 − 9x234. −1/21 − x+122= −14 − (x + 1)235.11 − 1/x2(−1/x2) = −1|x|√x2 − 136.sin x√1 − cos2 x=sin x| sin x|=1, sin x > 0−1, sin x < 037.3x21 + (x3)2=3x21 + x638.5x4|x5| (x5)2 − 1=5|x|√x10 − 139. y = 1/ tan x = cot x, dy/dx = − csc2x40. y = (tan−1x)−1, dy/dx = −(tan−1x)−2 11 + x241.ex|x|√x2 − 1+ exsec−1x 42. −1(cos−1 x)√1 − x243. 0 44.3x2(sin−1x)2√1 − x2+ 2x(sin−1x)345. 0 46. −1/√e2x − 147. −11 + x12x−1/2= −12(1 + x)√x48. −12√cot−1x(1 + x2)49. (a) Let x = f(y) = cot y, 0 < y < π, −∞ < x < +∞. Then f is diﬀerentiable and one-to-oneand f (f−1(x)) = − csc2(cot−1x) = −x2− 1 = 0, andddx[cot−1x]x=0= limx→01f (f−1(x))= − limx→01x2 + 1= −1.(b) If x = 0 then, from Exercise 50(a) of Section 1.5,ddxcot−1x =ddxtan−1 1x= −1x211 + (1/x)2= −1x2 + 1. For x = 0, Part (a) shows the same;thus for −∞ < x < +∞,ddx[cot−1x] = −1x2 + 1.(c) For −∞ < u < +∞, by the chain rule it follows thatddx[cot−1u] = −1u2 + 1dudx.
• 140 Chapter 450. (a) By the chain rule,ddx[csc−1x] =ddxsin−1 1x= −1x211 − (1/x)2=−1|x|√x2 − 1(b) By the chain rule,ddx[csc−1u] =dudxddu[csc−1u] =−1|u|√u2 − 1dudx51. x3+ x tan−1y = ey, 3x2+x1 + y2y + tan−1y = eyy , y =(3x2+ tan−1y)(1 + y2)(1 + y2)ey − x52. sin−1(xy) = cos−1(x − y),11 − x2y2(xy + y) = −11 − (x − y)2(1 − y ),y =y 1 − (x − y)2 + 1 − x2y21 − x2y2 − x 1 − (x − y)253. (a) f(x) = x3− 3x2+ 2x = x(x − 1)(x − 2) so f(0) = f(1) = f(2) = 0 thus f is not one-to-one.(b) f (x) = 3x2− 6x + 2, f (x) = 0 when x =6 ±√36 − 246= 1 ±√3/3. f (x) > 0 (f isincreasing) if x < 1 −√3/3, f (x) < 0 (f is decreasing) if 1 −√3/3 < x < 1 +√3/3, so f(x)takes on values less than f(1 −√3/3) on both sides of 1 −√3/3 thus 1 −√3/3 is the largestvalue of k.54. (a) f(x) = x3(x − 2) so f(0) = f(2) = 0 thus f is not one to one.(b) f (x) = 4x3− 6x2= 4x2(x − 3/2), f (x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2]and increasing on [3/2, +∞) so 3/2 is the smallest value of k.55. (a) f (x) = 4x3+ 3x2= (4x + 3)x2= 0 only at x = 0. But on [0, 2], f has no sign change, so fis one-to-one.(b) F (x) = 2f (2g(x))g (x) so F (3) = 2f (2g(3))g (3). By inspection f(1) = 3, sog(3) = f−1(3) = 1 and g (3) = (f−1) (3) = 1/f (f−1(3)) = 1/f (1) = 1/7 becausef (x) = 4x3+ 3x2. Thus F (3) = 2f (2)(1/7) = 2(44)(1/7) = 88/7.F(3) = f(2g(3)) = f(2·1) = f(2) = 24, so the line tangent to F(x) at (3, 25) has the equationy − 25 = (88/7)(x − 3), y = (88/7)x − 89/7.56. (a) f (x) = −e4−x22 +1x2< 0 for all x > 0, so f is one-to-one.(b) By inspection, f(2) = 1/2, so 2 = f−1(1/2) = g(1/2). By inspection,f (2) = − 2 +14= −94, andF (1/2) = f ([g(x)]2)ddx[g(x)2]x=1/2= f ([g(x)]2)2g(x)g (x)x=1/2= f (22)2 · 21f (g(x)) x=1/2= 4f (4)f (2)= 4e−12(2 + 116 )(2 + 14 )=339e12=113e1257. (a) f (x) = kekx, f (x) = k2ekx, f (x) = k3ekx, . . . , f(n)(x) = knekx(b) g (x) = −ke−kx, g (x) = k2e−kx, g (x) = −k3e−kx, . . . , g(n)(x) = (−1)nkne−kx58.dydt= e−λt(ωA cos ωt − ωB sin ωt) + (−λ)e−λt(A sin ωt + B cos ωt)= e−λt[(ωA − λB) cos ωt − (ωB + λA) sin ωt]
• Exercise Set 4.3 14159. f (x) =1√2πσexp −12x − µσ2ddx−12x − µσ2=1√2πσexp −12x − µσ2−x − µσ1σ= −1√2πσ3(x − µ) exp −12x − µσ260. y = Aekt, dy/dt = kAekt= k(Aekt) = ky61. y = Ae2x+ Be−4x, y = 2Ae2x− 4Be−4x, y = 4Ae2x+ 16Be−4xsoy + 2y − 8y = (4Ae2x+ 16Be−4x) + 2(2Ae2x− 4Be−4x) − 8(Ae2x+ Be−4x) = 062. (a) y = −xe−x+ e−x= e−x(1 − x), xy = xe−x(1 − x) = y(1 − x)(b) y = −x2e−x2/2+ e−x2/2= e−x2/2(1 − x2), xy = xe−x2/2(1 − x2) = y(1 − x2)63.dydx= 100(−0.2)e−0.2x= −20y, k = −0.264. ln y = (5x + 1) ln 3 − (x/2) ln 4, soy /y = 5 ln 3 − (1/2) ln 4 = 5 ln 3 − ln 2, andy = (5 ln 3 − ln 2)y65. ln y = ln 60 − ln(5 + 7e−t),yy=7e−t5 + 7e−t=7e−t+ 5 − 55 + 7e−t= 1 −112y, sodydt= r 1 −yKy, with r = 1, K = 12.66. (a) 1200 9(b) P tends to 12 as t gets large; limt→+∞P(t) = limt→+∞605 + 7e−t=605 + 7 limt→+∞e−t =605= 12(c) the rate of population growth tends to zero3.200 967. limh→010h− 1h=ddx10xx=0=ddxex ln 10x=0= ln 10
• 142 Chapter 468. limh→0tan−1(1 + h) − π/4h=ddxtan−1xx=1=11 + x2x=1=1269. lim∆x→09[sin−1(√32 + ∆x)]2− π2∆x=ddx(3 sin−1x)2x=√3/2= 2(3 sin−1x)3√1 − x2x=√3/2= 2(3π3)31 − (3/4)= 12π70. lim∆x→0(2 + ∆x)(2+∆x)− 4∆x=ddxxxx=2=ddxex ln xx=2= (1 + ln x)ex ln xx=2= (1 + ln 2)22= 4(1 + ln 2)71. limw→23 sec−1w − πw − 2=ddx3 sec−1xx=2=3|2|√22 − 1=√3272. limw→14(tan−1w)w− πw − 1=ddx4(tan−1x)xx=1=ddx4ex ln tan−1xx=1= 4(tan−1x)xln tan−1x + x1/(1 + x2)tan−1x x=1= π ln(π/4) +124π== 2 + π ln(π/4)EXERCISE SET 4.41. (a) limx→2x2− 4x2 + 2x − 8= limx→2(x − 2)(x + 2)(x + 4)(x − 2)= limx→2x + 2x + 4=23(b) limx→+∞2x − 53x + 7=2 − limx→+∞5x3 + limx→+∞7x=232. (a)sin xtan x= sin xcos xsin x= cos x so limx→0sin xtan x= limx→0cos x = 1(b)x2− 1x3 − 1=(x − 1)(x + 1)(x − 1)(x2 + x + 1)=x + 1x2 + x + 1so limx→1x2− 1x3 − 1=233. Tf (x) = −2(x + 1), Tg(x) = −3(x + 1),limit = 2/34. Tf (x) = − x −π2, Tg(x) = − x −π2limit = 15. limx→0excos x= 1 6. limx→316x − 13= 1/57. limθ→0sec2θ1= 1 8. limt→0tet+ et−et= −19. limx→π+cos x1= −1 10. limx→0+cos x2x= +∞11. limx→+∞1/x1= 0 12. limx→+∞3e3x2x= limx→+∞9e3x2= +∞
• Exercise Set 4.4 14313. limx→0+− csc2x1/x= limx→0+−xsin2x= limx→0+−12 sin x cos x= −∞14. limx→0+−1/x(−1/x2)e1/x= limx→0+xe1/x= 015. limx→+∞100x99ex= limx→+∞(100)(99)x98ex= · · · = limx→+∞(100)(99)(98) · · · (1)ex= 016. limx→0+cos x/ sin xsec2 x/ tan x= limx→0+cos2x = 1 17. limx→02/√1 − 4x21= 218. limx→01 −11 + x23x2= limx→013(1 + x2)=1319. limx→+∞xe−x= limx→+∞xex= limx→+∞1ex= 020. limx→π(x − π) tan(x/2) = limx→πx − πcot(x/2)= limx→π1−(1/2) csc2(x/2)= −221. limx→+∞x sin(π/x) = limx→+∞sin(π/x)1/x= limx→+∞(−π/x2) cos(π/x)−1/x2= limx→+∞π cos(π/x) = π22. limx→0+tan x ln x = limx→0+ln xcot x= limx→0+1/x− csc2 x= limx→0+− sin2xx= limx→0+−2 sin x cos x1= 023. limx→(π/2)−sec 3x cos 5x = limx→(π/2)−cos 5xcos 3x= limx→(π/2)−−5 sin 5x−3 sin 3x=−5(+1)(−3)(−1)= −5324. limx→π(x − π) cot x = limx→πx − πtan x= limx→π1sec2 x= 125. y = (1 − 3/x)x, limx→+∞ln y = limx→+∞ln(1 − 3/x)1/x= limx→+∞−31 − 3/x= −3, limx→+∞y = e−326. y = (1 + 2x)−3/x, limx→0ln y = limx→0−3 ln(1 + 2x)x= limx→0−61 + 2x= −6, limx→0y = e−627. y = (ex+ x)1/x, limx→0ln y = limx→0ln(ex+ x)x= limx→0ex+ 1ex + x= 2, limx→0y = e228. y = (1 + a/x)bx, limx→+∞ln y = limx→+∞b ln(1 + a/x)1/x= limx→+∞ab1 + a/x= ab, limx→+∞y = eab29. y = (2 − x)tan(πx/2), limx→1ln y = limx→1ln(2 − x)cot(πx/2)= limx→12 sin2(πx/2)π(2 − x)= 2/π, limx→1y = e2/π30. y = [cos(2/x)]x2, limx→+∞ln y = limx→+∞ln cos(2/x)1/x2= limx→+∞(−2/x2)(− tan(2/x))−2/x3= limx→+∞− tan(2/x)1/x= limx→+∞(2/x2) sec2(2/x)−1/x2= −2, limx→+∞y = e−231. limx→01sin x−1x= limx→0x − sin xx sin x= limx→01 − cos xx cos x + sin x= limx→0sin x2 cos x − x sin x= 0
• 144 Chapter 432. limx→01 − cos 3xx2= limx→03 sin 3x2x= limx→092cos 3x =9233. limx→+∞(x2+ x) − x2√x2 + x + x= limx→+∞x√x2 + x + x= limx→+∞11 + 1/x + 1= 1/234. limx→0ex− 1 − xxex − x= limx→0ex− 1xex + ex − 1= limx→0exxex + 2ex= 1/235. limx→+∞[x − ln(x2+ 1)] = limx→+∞[ln ex− ln(x2+ 1)] = limx→+∞lnexx2 + 1,limx→+∞exx2 + 1= limx→+∞ex2x= limx→+∞ex2= +∞ so limx→+∞[x − ln(x2+ 1)] = +∞36. limx→+∞lnx1 + x= limx→+∞ln11/x + 1= ln(1) = 038. (a) limx→+∞ln xxn= limx→+∞1/xnxn−1= limx→+∞1nxn= 0(b) limx→+∞xnln x= limx→+∞nxn−11/x= limx→+∞nxn= +∞39. (a) L’Hˆopital’s Rule does not apply to the problem limx→13x2− 2x + 13x2 − 2xbecause it is not a00form.(b) limx→13x2− 2x + 13x2 − 2x= 240. L’Hˆopital’s Rule does not apply to the probleme3x2−12x+12x4 − 16, which is of the forme00, and fromwhich it follows that limx→2−and limx→2+exist, with values −∞ if x approaches 2 from the left and+∞ if from the right. The general limit limx→2does not exist.41. limx→+∞1/(x ln x)1/(2√x)= limx→+∞2√x ln x= 0 0.150100 1000042. y = xx, limx→0+ln y = limx→0+ln x1/x= limx→0+−x = 0, limx→0+y = 1 100 0.5
• Exercise Set 4.4 14525190 0.543. y = (sin x)3/ ln x,limx→0+ln y = limx→0+3 ln sin xln x= limx→0+(3 cos x)xsin x= 3,limx→0+y = e34.13.31.4 1.644. limx→π/2−4 sec2xsec x tan x= limx→π/2−4sin x= 40–160 345. ln x − ex= ln x −1e−x=e−xln x − 1e−x;limx→+∞e−xln x = limx→+∞ln xex= limx→+∞1/xex= 0 by L’Hˆopital’s Rule,so limx→+∞[ln x − ex] = limx→+∞e−xln x − 1e−x= −∞–0.6–1.20 1246. limx→+∞[ln ex− ln(1 + 2ex)] = limx→+∞lnex1 + 2ex= limx→+∞ln1e−x + 2= ln12;horizontal asymptote y = − ln 21.021100 1000047. y = (ln x)1/x,limx→+∞ln y = limx→+∞ln(ln x)x= limx→+∞1x ln x= 0;limx→+∞y = 1, y = 1 is the horizontal asymptote
• 146 Chapter 448. y =x + 1x + 2x, limx→+∞ln y = limx→+∞lnx + 1x + 21/x= limx→+∞−x2(x + 1)(x + 2)= −1;limx→+∞y = e−1is the horizontal asymptote100 5049. (a) 0 (b) +∞ (c) 0 (d) −∞ (e) +∞ (f) −∞50. (a) Type 00; y = x(ln a)/(1+ln x); limx→0+ln y = limx→0+(ln a) ln x1 + ln x= limx→0+(ln a)/x1/x= limx→0+ln a = ln a,limx→0+y = eln a= a(b) Type ∞0; same calculation as Part (a) with x → +∞(c) Type 1∞; y = (x + 1)(ln a)/x, limx→0ln y = limx→0(ln a) ln(x + 1)x= limx→0ln ax + 1= ln a,limx→0y = eln a= a51. limx→+∞1 + 2 cos 2x1does not exist, nor is it ±∞; limx→+∞x + sin 2xx= limx→+∞1 +sin 2xx= 152. limx→+∞2 − cos x3 + cos xdoes not exist, nor is it ±∞; limx→+∞2x − sin x3x + sin x= limx→+∞2 − (sin x)/x3 + (sin x)/x=2353. limx→+∞(2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; limx→+∞x(2 + sin 2x)x + 1= limx→+∞2 + sin 2x1 + 1/x,which does not exist because sin 2x oscillates between −1 and 1 as x → +∞54. limx→+∞1x+12cos x +sin x2xdoes not exist, nor is it ±∞;limx→+∞x(2 + sin x)x2 + 1= limx→+∞2 + sin xx + 1/x= 055. limR→0+V tL e−Rt/L1=V tL56. (a) limx→π/2(π/2 − x) tan x = limx→π/2π/2 − xcot x= limx→π/2−1− csc2 x= limx→π/2sin2x = 1(b) limx→π/21π/2 − x− tan x = limx→π/21π/2 − x−sin xcos x= limx→π/2cos x − (π/2 − x) sin x(π/2 − x) cos x= limx→π/2−(π/2 − x) cos x−(π/2 − x) sin x − cos x= limx→π/2(π/2 − x) sin x + cos x−(π/2 − x) cos x + 2 sin x= 0(c) 1/(π/2 − 1.57) ≈ 1255.765534, tan 1.57 ≈ 1255.765592;1/(π/2 − 1.57) − tan 1.57 ≈ 0.000058
• Review Exercises Chapter 4 14757. (b) limx→+∞x(k1/x− 1) = limt→0+kt− 1t= limt→0+(ln k)kt1= ln k(c) ln 0.3 = −1.20397, 1024 1024√0.3 − 1 = −1.20327;ln 2 = 0.69315, 1024 1024√2 − 1 = 0.6933858. If k = −1 then limx→0(k + cos x) = k + 1 = 0, so limx→0k + cos xx2= ±∞. Hence k = −1, and by therule limx→0−1 + cos xx2= limx→0− sin x2x= limx→0− 2cos x2= −22= −4 if = ±2√2.59. (a) No; sin(1/x) oscillates as x → 0. (b) 0.05–0.05–0.35 0.35(c) For the limit as x → 0+use the Squeezing Theorem together with the inequalities−x2≤ x2sin(1/x) ≤ x2. For x → 0−do the same; thus limx→0f(x) = 0.60. (a) Apply the rule to get limx→0− cos(1/x) + 2x sin(1/x)cos xwhich does not exist (nor is it ±∞).(b) Rewrite as limx→0xsin x[x sin(1/x)], but limx→0xsin x= limx→01cos x= 1 and limx→0x sin(1/x) = 0,thus limx→0xsin x[x sin(1/x)] = (1)(0) = 061. limx→0+sin(1/x)(sin x)/x, limx→0+sin xx= 1 but limx→0+sin(1/x) does not exist because sin(1/x) oscillates between−1 and 1 as x → +∞, so limx→0+x sin(1/x)sin xdoes not exist.62. Since f(0) = g(0) = 0, then for x = a,f(x)g(x)=(f(x) − f(0)/(x − a)(g(x) − g(0))/(x − a). Now take the limit:limx→af(x)g(x)= limx→a(f(x) − f(0)/(x − a)(g(x) − g(0))/(x − a)=f (a)g (a)REVIEW EXERCISES CHAPTER 41.14(6x − 5)3/4(6) =32(6x − 5)3/42.13(x2 + x)2/3(2x + 1) =2x + 13(x2 + x)2/33. dy/dx =32x − 1x + 21/2ddxx − 1x + 2=92(x + 2)2x − 1x + 21/24. dy/dx =x2 43(3 − 2x)1/3(−2) − (3 − 2x)4/3(2x)x4=2(3 − 2x)1/3(2x − 9)3x3
• 148 Chapter 45. (a) 3x2+ xdydx+ y − 2 = 0,dydx=2 − y − 3x2x(b) y = (1 + 2x − x3)/x = 1/x + 2 − x2, dy/dx = −1/x2− 2x(c)dydx=2 − (1/x + 2 − x2) − 3x2x= −1/x2− 2x6. (a) xy = x − y, xdydx+ y = 1 −dydx,dydx=1 − yx + 1(b) y(x + 1) = x, y =xx + 1, y =1(x + 1)2(c)dydx=1 − yx + 1=1 − xx+11 + x=1x2 + 17. −1y2dydx−1x2= 0 sodydx= −y2x28. 3x2− 3y2 dydx= 6(xdydx+ y), −(3y2+ 6x)dydx= 6y − 3x2sodydx=x2− 2yy2 + 2x9. xdydx+ y sec(xy) tan(xy) =dydx,dydx=y sec(xy) tan(xy)1 − x sec(xy) tan(xy)10. 2x =(1 + csc y)(− csc2y)(dy/dx) − (cot y)(− csc y cot y)(dy/dx)(1 + csc y)2,2x(1 + csc y)2= − csc y(csc y + csc2y − cot2y)dydx,but csc2y − cot2y = 1, sodydx= −2x(1 + csc y)csc y11.dydx=3x4y,d2ydx2=(4y)(3) − (3x)(4dy/dx)16y2=12y − 12x(3x/(4y))16y2=12y2− 9x216y3=−3(3x2− 4y2)16y3,but 3x2− 4y2= 7 sod2ydx2=−3(7)16y3= −2116y312.dydx=yy − x,d2ydx2=(y − x)(dy/dx) − y(dy/dx − 1)(y − x)2=(y − x)yy − x− yyy − x− 1(y − x)2=y2− 2xy(y − x)3but y2− 2xy = −3, sod2ydx2= −3(y − x)313.dydx= tan(πy/2) + x(π/2)dydxsec2(πy/2),dydx= 1 + (π/4)dydx(2),dydx=2π − 214. Let P(x0, y0) be the required point. The slope of the line 4x − 3y + 1 = 0 is 4/3 so the slope ofthe tangent to y2= 2x3at P must be −3/4. By implicit diﬀerentiation dy/dx = 3x2/y, so at P,3x20/y0 = −3/4, or y0 = −4x20. But y20 = 2x30 because P is on the curve y2= 2x3. Elimination ofy0 gives 16x40 = 2x30, x30(8x0 − 1) = 0, so x0 = 0 or 1/8. From y0 = −4x20 it follows that y0 = 0when x0 = 0, and y0 = −1/16 when x0 = 1/8. It does not follow, however, that (0, 0) is a solutionbecause dy/dx = 3x2/y (the slope of the curve as determined by implicit diﬀerentiation) is validonly if y = 0. Further analysis shows that the curve is tangent to the x-axis at (0, 0), so the point(1/8, −1/16) is the only solution.
• Review Exercises Chapter 4 14915. Substitute y = mx into x2+ xy + y2= 4 to get x2+ mx2+ m2x2= 4, which has distinct solutionsx = ±2/√m2 + m + 1. They are distinct because m2+ m + 1 = (m + 1/2)2+ 3/4 ≥ 3/4, som2+ m + 1 is never zero.Note that the points of intersection occur in pairs (x0, y0) and (−x0, y0). By implicit diﬀerentiation,the slope of the tangent line to the ellipse is given by dy/dx = −(2x + y)/(x + 2y). Since the slopeis unchanged if we replace (x, y) with (−x, −y), it follows that the slopes are equal at the two pointof intersection.Finally we must examine the special case x = 0 which cannot be written in the form y = mx. Ifx = 0 then y = ±2, and the formula for dy/dx gives dy/dx = −1/2, so the slopes are equal.16. Use implicit diﬀerentiation to get dy/dx = (y−3x2)/(3y2−x), so dy/dx = 0 if y = 3x2. Substitutethis into x3− xy + y3= 0 to obtain 27x6− 2x3= 0, x3= 2/27, x = 3√2/3 and hence y = 3√4/3.17. By implicit diﬀerentiation, 3x2− y − xy + 3y2y = 0, so y = (3x2− y)/(x − 3y2). This derivativeexists except when x = 3y2. Substituting this into the original equation x3− xy + y3= 0, one has27y6− 3y3+ y3= 0, y3(27y3− 2) = 0. The unique solution in the ﬁrst quadrant isy = 21/3/3, x = 3y2= 22/3/318. By implicit diﬀerentiation, dy/dx = k/(2y) so the slope of the tangent to y2= kx at (x0, y0) isk/(2y0) if y0 = 0. The tangent line in this case is y − y0 =k2y0(x − x0), or 2y0y − 2y20 = kx − kx0.But y20 = kx0 because (x0, y0) is on the curve y2= kx, so the equation of the tangent line becomes2y0y − 2kx0 = kx − kx0 which gives y0y = k(x + x0)/2. If y0 = 0, then x0 = 0; the graph ofy2= kx has a vertical tangent at (0, 0) so its equation is x = 0, but y0y = k(x + x0)/2 gives thesame result when x0 = y0 = 0.19. y = ln(x + 1) + 2 ln(x + 2) − 3 ln(x + 3) − 4 ln(x + 4), dy/dx =1x + 1+2x + 2−3x + 3−4x + 420. y =12ln x +13ln(x + 1) − ln sin x + ln cos x, sodydx=12x+13(x + 1)−cos xsin x−sin xcos x=5x + 36x(x + 1)− cot x − tan x21.12x(2) = 1/x 22. 2(ln x)1x=2 ln xx23.13x(ln x + 1)2/324. y = 13 ln(x + 1), y =13(x + 1)25. log10 ln x =ln ln xln 10, y =1(ln 10)(x ln x)26. y =1 + ln x/ ln 101 − ln x/ ln 10=ln 10 + ln xln 10 − ln x, y =(ln 10 − ln x)/x + (ln 10 + ln x)/x(ln 10 − ln x)2=2 ln 10x(ln 10 − ln x)227. y =32ln x +12ln(1 + x4), y =32x+2x3(1 + x4)28. y =12ln x + ln cos x − ln(1 + x2), y =12x−sin xcos x−2x1 + x2=1 − 3x22x(1 + x2)− tan x29. y = x2+ 1 so y = 2x.
• 150 Chapter 430. y = ln(1 + ex+ e2x)(1 − ex)(1 + ex + e2x)= − ln(1 − ex),dydx=ex1 − ex31. y = 2e√x+ 2xe√x ddx√x = 2e√x+√xe√x32. y =abe−x(1 + be−x)233. y =2π(1 + 4x2)34. y = e(sin−1x) ln 2, y =ln 2√1 − x22sin−1x35. ln y = exln x,yy= ex 1x+ ln x ,dydx= xexex 1x+ ln x = exxex−1+ xexln x36. ln y =ln(1 + x)x,yy=x/(1 + x) − ln(1 + x)x2=1x(1 + x)−ln(1 + x)x2,dydx=1x(1 + x)(1/x)−1−(1 + x)(1/x)x2ln(1 + x)37. y =2|2x + 1| (2x + 1)2 − 138. y =12√cos−1 x2ddxcos−1x2= −1√cos−1 x2x√1 − x439. ln y = 3 ln x −12ln(x2+ 1), y /y =3x−xx2 + 1, y =3x2√x2 + 1−x4(x2 + 1)3/240. ln y =13(ln(x2−1)−ln(x2+1)),yy=132xx2 − 1−2xx2 + 1=4x3(x4 − 1)so y =4x3(x4 − 1)3 x2 − 1x2 + 141. (b) yx2461 2 3 4(c)dydx=12−1xsodydx< 0 at x = 1 anddydx> 0 at x = e(d) The slope is a continuous function which goes from a negative value at x = 1 to a positivevalue at x = e; therefore it must take the value zero between, by the Intermediate ValueTheorem.(e)dydx= 0 when x = 242. β = 10 log I − 10 log I0,dβdI=10I ln 10(a)dβdI I=10I0=1I0 ln 10db/W/m2(b)dβdI I=100I0=110I0 ln 10db/W/m2(c)dβdI I=100I0=1100I0 ln 10db/W/m243. Solvedydt= 3dxdtgiven y = x ln x. Thendydt=dydxdxdt= (1 + ln x)dxdt, so 1 + ln x = 3, ln x = 2,x = e2.
• Review Exercises Chapter 4 15144. x = 2, y = 0; y = −2x/(5 − x2) = −4 at x = 2, so y − 0 = −4(x − 2) or y = −4x + 8yx2245. Set y = logb x and solve y = 1: y =1x ln b= 1 sox =1ln b. The curves intersect when (x, x) lies on thegraph of y = logb x, so x = logb x. From Formula (8),Section 1.6, logb x =ln xln bfrom which ln x = 1, x = e,ln b = 1/e, b = e1/e≈ 1.4447.46. (a) Find the point of intersection: f(x) =√x + k = ln x. Theslopes are equal, so m1 =1x= m2 =12√x,√x = 2, x = 4.Then ln 4 =√4 + k, k = ln 4 − 2.yx22(b) Since the slopes are equal m1 =k2√x= m2 =1x, sok√x = 2. At the point of intersection k√x = ln x, 2 = ln x,x = e2, k = 2/e.yx02547. Where f is diﬀerentiable and f = 0, g must be diﬀerentiable; this can be inferred from the graphs.In general, however, g need not be diﬀerentiable: consider f(x) = x3, g(x) = x1/3.48. (a) f (x) = −3/(x + 1)2. If x = f(y) = 3/(y + 1) then y = f−1(x) = (3/x) − 1, soddxf−1(x) = −3x2; and1f (f−1(x))= −(f−1(x) + 1)23= −(3/x)23= −3x2.(b) f(x) = ex/2, f (x) = 12 ex/2. If x = f(y) = ey/2then y = f−1(x) = 2 ln x, soddxf−1(x) =2x;and1f (f−1(x))= 2e−f−1(x)/2= 2e− ln x= 2x−1=2x49. Let P(x0, y0) be a point on y = e3xthen y0 = e3x0. dy/dx = 3e3xso mtan = 3e3x0at P and anequation of the tangent line at P is y − y0 = 3e3x0(x − x0), y − e3x0= 3e3x0(x − x0). If the linepasses through the origin then (0, 0) must satisfy the equation so −e3x0= −3x0e3x0which givesx0 = 1/3 and thus y0 = e. The point is (1/3, e).50. ln y = ln 5000 + 1.07x;dy/dxy= 1.07, ordydx= 1.07y51. ln y = 2x ln 3 + 7x ln 5;dy/dxy= 2 ln 3 + 7 ln 5, ordydx= (2 ln 3 + 7 ln 5)y52.dkdT= k0 exp −q(T − T0)2T0T−q2T2= −qk02T2exp −q(T − T0)2T0T53. y = aeaxsin bx + beaxcos bx and y = (a2− b2)eaxsin bx + 2abeaxcos bx, so y − 2ay + (a2+ b2)y= (a2− b2)eaxsin bx + 2abeaxcos bx − 2a(aeaxsin bx + beaxcos bx) + (a2+ b2)eaxsin bx = 0.
• 152 Chapter 454. sin(tan−1x) = x/√1 + x2 and cos(tan−1x) = 1/√1 + x2, and y =11 + x2, y =−2x(1 + x2)2, hencey + 2 sin y cos3y =−2x(1 + x2)2+ 2x√1 + x21(1 + x2)3/2= 0.55. (a) 100200 8(b) as t tends to +∞, the population tends to 19limt→+∞P(t) = limt→+∞955 − 4e−t/4=955 − 4 limt→+∞e−t/4=955= 19(c) the rate of population growth tends to zero 0–800 856. (a) y = (1 + x)π, limh→0(1 + h)π− 1h=ddx(1 + x)πx=0= π(1 + x)π−1x=0= π(b) Let y =1 − ln xln x. Then y(e) = 0, and limx→e1 − ln x(x − e) ln x=dydx x=e= −1/x(ln x)2= −1e57. In the case +∞ − (−∞) the limit is +∞; in the case −∞ − (+∞) the limit is −∞, becauselarge positive (negative) quantities are added to large positive (negative) quantities. The cases+∞ − (+∞) and −∞ − (−∞) are indeterminate; large numbers of opposite sign are subtracted,and more information about the sizes is needed.58. (a) when the limit takes the form 0/0 or ∞/∞(b) Not necessarily; only if limx→af(x) = 0. Consider g(x) = x; limx→0g(x) = 0. For f(x) choosecos x, x2, and |x|1/2. Then: limx→0cos xxdoes not exist, limx→0x2x= 0, and limx→0|x|1/2x2= +∞.59. limx→+∞(ex− x2) = limx→+∞x2(ex/x2− 1), but limx→+∞exx2= limx→+∞ex2x= limx→+∞ex2= +∞so limx→+∞(ex/x2− 1) = +∞ and thus limx→+∞x2(ex/x2− 1) = +∞60. limx→1ln xx4 − 1= limx→11/x4x3=14; limx→1ln xx4 − 1= limx→1ln xx4 − 1=1261. = limx→0(x2+ 2x)ex6 sin 3x cos 3x= limx→0(x2+ 2x)ex3 sin 6x= limx→0(x2+ 4x + 2)ex18 cos 6x=1962. limx→0axln a = ln a
• 153CHAPTER 5The Derivative in Graphing and ApplicationsEXERCISE SET 5.11. (a) f > 0 and f > 0yx(b) f > 0 and f < 0yx(c) f < 0 and f > 0yx(d) f < 0 and f < 0yx2. (a) yx(b) yx(c) yx(d) yx3. A: dy/dx < 0, d2y/dx2> 0B: dy/dx > 0, d2y/dx2< 0C: dy/dx < 0, d2y/dx2< 04. A: dy/dx < 0, d2y/dx2< 0B: dy/dx < 0, d2y/dx2> 0C: dy/dx > 0, d2y/dx2< 05. An inﬂection point occurs when f changes sign: at x = −1, 0, 1 and 2.6. (a) f(0) < f(1) since f > 0 on (0, 1). (b) f(1) > f(2) since f < 0 on (1, 2).(c) f (0) > 0 by inspection. (d) f (1) = 0 by inspection.(e) f (0) < 0 since f is decreasing there. (f) f (2) = 0 since f has a minimum there.7. (a) [4, 6] (b) [1, 4] and [6, 7] (c) (1, 2) and (3, 5)(d) (2, 3) and (5, 7) (e) x = 2, 3, 5
• 154 Chapter 58. (1, 2) (2, 3) (3, 4) (4, 5) (5, 6) (6, 7)f − − − + + −f + − + + − −9. (a) f is increasing on [1, 3] (b) f is decreasing on (−∞, 1], [3, +∞](c) f is concave up on (−∞, 2), (4, +∞) (d) f is concave down on (2, 4)(e) points of inﬂection at x = 2, 410. (a) f is increasing on (−∞, +∞) (b) f is nowhere decreasing(c) f is concave up on (−∞, 1), (3, +∞) (d) f is concave down on (1, 3)(e) f has points of inﬂection at x = 1, 311. f (x) = 2(x − 3/2)f (x) = 2(a) [3/2, +∞) (b) (−∞, 3/2](c) (−∞, +∞) (d) nowhere(e) none12. f (x) = −2(2 + x)f (x) = −2(a) (−∞, −2] (b) [−2, +∞)(c) nowhere (d) (−∞, +∞)(e) none13. f (x) = 6(2x + 1)2f (x) = 24(2x + 1)(a) (−∞, +∞) (b) nowhere(c) (−1/2, +∞) (d) (−∞, −1/2)(e) −1/214. f (x) = 3(4 − x2)f (x) = −6x(a) [−2, 2] (b) (−∞, −2], [2, +∞)(c) (−∞, 0) (d) (0, +∞)(e) 015. f (x) = 12x2(x − 1)f (x) = 36x(x − 2/3)(a) [1, +∞) (b) (−∞, 1](c) (−∞, 0), (2/3, +∞) (d) (0, 2/3)(e) 0, 2/316. f (x) = x(4x2− 15x + 18)f (x) = 6(x − 1)(2x − 3)(a) [0, +∞), (b) (−∞, 0](c) (−∞, 1), (3/2, +∞) (d) (1, 3/2)(e) 1, 3/217. f (x) = −3(x2− 3x + 1)(x2 − x + 1)3f (x) =6x(2x2− 8x + 5)(x2 − x + 1)4(a) [3−√52 , 3+√52 ] (b) (−∞, 3−√52 ], [3+√52 , +∞)(c) (0, 2 −√62 ), (2 +√62 , +∞) (d) (−∞, 0), (2 −√62 , 2 +√62 )(e) 0, 2 −√6/2, 2 +√6/218. f (x) = −x2− 2(x + 2)2f (x) =2x(x2− 6)(x + 2)3(a) (−∞, −√2), (√2, +∞) (b) (−√2,√2)(c) (−∞, −√6), (0,√6) (d) (−√6, 0), (√6, +∞)(e) none19. f (x) =2x + 13(x2 + x + 1)2/3f (x) = −2(x + 2)(x − 1)9(x2 + x + 1)5/3(a) [−1/2, +∞) (b) (−∞, −1/2](c) (−2, 1) (d) (−∞, −2), (1, +∞)(e) −2, 1
• Exercise Set 5.1 15520. f (x) =4(x − 1/4)3x2/3f (x) =4(x + 1/2)9x5/3(a) [1/4, +∞) (b) (−∞, 1/4](c) (−∞, −1/2), (0, +∞) (d) (−1/2, 0)(e) −1/2, 021. f (x) =4(x2/3− 1)3x1/3f (x) =4(x5/3+ x)9x7/3(a) [−1, 0], [1, +∞) (b) (−∞, −1], [0, 1](c) (−∞, 0), (0, +∞) (d) nowhere(e) none22. f (x) =23x−1/3− 1f (x) = −29x4/3(a) [−1, 0], [1, +∞) (b) (−∞, −1], [0, 1](c) (−∞, 0), (0, +∞) (d) nowhere(e) none23. f (x) = −xe−x2/2f (x) = (−1 + x2)e−x2/2(a) (−∞, 0] (b) [0, +∞)(c) (−∞, −1), (1, +∞) (d) (−1, 1)(e) −1, 124. f (x) = (2x2+ 1)ex2f (x) = 2x(2x2+ 3)ex2(a) (−∞, +∞) (b) none(c) (0, +∞) (d) (−∞, 0)(e) 025. f (x) =xx2 + 4f (x) = −x2− 4(x2 + 4)2(a) [0, +∞) (b) (−∞, 0](c) (−2, +2) (d) (−∞, −2), (2, +∞)(e) −2, +226. f (x) = x2(1 + 3 ln x)f (x) = x(5 + 6 ln x)(a) [e−1/3, +∞) (b) (0, e−1/3](c) (e−5/6, +∞) (d) (0, e−5/6)(e) e−5/627. f (x) =2x1 + (x2 − 1)2f (x) = −23x4− 2x2− 2[1 + (x2 − 1)2]2(a) [0 + ∞) (b) (−∞, 0](c) −√1+√7√3, −√1−√7√3,√1−√7√3,√1+√7√3(d) − ∞, −√1+√7√3, −√1−√7√3,√1−√7√3,√1+√7√3, +∞(e) four: ±1 ±√7328. f (x) =23x1/3√1 − x4/3f (x) =2 −1 + 3x4/39x4/3 1 − x4/3 3/2(a) [0, 1](b) [−1, 0](c) (−1, −3−3/4), (0, 3−3/4, 1)(d) (3−3/4, 0), (3−3/4(e) ±3−3/4
• 156 Chapter 529. f (x) = cos x + sin xf (x) = − sin x + cos x(a) [−π/4, 3π/4] (b) (−π, −π/4], [3π/4, π)(c) (−3π/4, π/4) (d) (−π, −3π/4), (π/4, π)(e) −3π/4, π/41.5–1.5^ 630. f (x) = (2 tan2x + 1) sec xf (x) = sec x tan x(6 tan2x + 5)(a) [−π/2, π/2] (b) nowhere(c) (0, π/2) (d) (−π/2, 0)(e) 010–10^ 631. f (x) = −12sec2(x/2)f (x) = −12tan(x/2) sec2(x/2))(a) nowhere (b) (−π, π)(c) (−π, 0) (d) (0, π)(e) 010–10C c32. f (x) = 2 − csc2xf (x) = 2 csc2x cot x = 2cos xsin3x(a) [π/4, 3π/4] (b) (0, π/4], [3π/4, π)(c) (0, π/2) (d) (π/2, π)(e) π/28–20 p33. f(x) = 1 + sin 2xf (x) = 2 cos 2xf (x) = −4 sin 2x(a) [−π, −3π/4], [−π/4, π/4], [3π/4, π](b) [−3π/4, −π/4], [π/4, 3π/4](c) (−π/2, 0), (π/2, π)(d) (−π, −π/2), (0, π/2)(e) −π/2, 0, π/220C c34. f (x) = 2 sin 4xf (x) = 8 cos 4x(a) (0, π/4], [π/2, 3π/4](b) [π/4, π/2], [3π/4, π](c) (0, π/8), (3π/8, 5π/8), (7π/8, π)(d) (π/8, 3π/8), (5π/8, 7π/8)(e) π/8, 3π/8, 5π/8, 7π/8100 p
• Exercise Set 5.1 15735. (a)24xy (b)24xy (c)24xy36. (a)24xy (b)24xy (c)24xy37. (a) g(x) has no zeros:There can be no zero of g(x) on the interval −∞ < x < 0 because if there were, sayg(x0) = 0 where x0 < 0, then g (x) would have to be positive between x = x0 and x = 0, sayg (x1) > 0 where x0 < x1 < 0. But then g (x) cannot be concave up on the interval (x1, 0),a contradiction.There can be no zero of g(x) on 0 < x < 4 because g(x) is concave up for 0 < x < 4 and thusthe graph of g(x), for 0 < x < 4, must lie above the line y = −23x + 2, which is the tangentline to the curve at (0, 2), and above the line y = 3(x − 4) + 3 = 3x − 9 also for 0 < x < 4(see ﬁgure). The ﬁrst condition says that g(x) could only be zero for x > 3 and the secondcondition says that g(x) could only be zero for x < 3, thus g(x) has no zeros for 0 < x < 4.Finally, if 4 < x < +∞, g(x) could only have a zero if g (x) were negative somewherefor x > 4, and since g (x) is decreasing there we would ultimately have g(x) < −10, acontradiction.(b) one, between 0 and 4(c) We must have limx→+∞g (x) = 0; if the limit were −5then g(x) would at some time cross the line g(x) =−10; if the limit were 5 then, since g is concave downfor x > 4 and g (4) = 3, g must decrease for x > 4and thus the limit would be ≤ 4.x144yslope 4slope –2/338. (a) f (x) = 3(x − a)2, f (x) = 6(x − a); inﬂection point is (a, 0)(b) f (x) = 4(x − a)3, f (x) = 12(x − a)2; no inﬂection points39. For n ≥ 2, f (x) = n(n − 1)(x − a)n−2; there is a sign change of f (point of inﬂection) at (a, 0)if and only if n is odd. For n = 1, y = x − a, so there is no point of inﬂection.40. If t is in the interval (a, b) and t < x0 then, because f is increasing,f(t) − f(x)t − x≥ 0. If x0 < tthenf(t) − f(x)t − x≥ 0. Thus f (x0) = limt→xf(t) − f(x)t − x≥ 0.
• 158 Chapter 541. f (x) = 1/3 − 1/[3(1 + x)2/3] so f is increasing on [0, +∞)thus if x > 0, then f(x) > f(0) = 0, 1 + x/3 − 3√1 + x > 0,3√1 + x < 1 + x/3.2.500 1042. f (x) = sec2x − 1 so f is increasing on [0, π/2)thus if 0 < x < π/2, then f(x) > f(0) = 0,tan x − x > 0, x < tan x.1000 643. x ≥ sin x on [0, +∞): let f(x) = x − sin x.Then f(0) = 0 and f (x) = 1 − cos x ≥ 0,so f(x) is increasing on [0, +∞).4–10 444. Let f(x) = 1 − x2/2 − cos x for x ≥ 0. Then f(0) = 0 and f (x) = −x + sin x. By Exercise 43,f (x) ≤ 0 for x ≥ 0, so f(x) ≤ 0 for all x ≥ 0, that is, cos x ≥ 1 − x2/2.45. (a) Let f(x) = x − ln(x + 1) for x ≥ 0. Then f(0) = 0 and f (x) = 1 − 1/(x + 1) ≥ 0 for x ≥ 0,so f is increasing for x ≥ 0 and thus ln(x + 1) ≤ x for x ≥ 0.(b) Let g(x) = x − 12 x2− ln(x + 1). Then g(0) = 0 and g (x) = 1 − x − 1/(x + 1) ≤ 0 for x ≥ 0since 1 − x2≤ 1. Thus g is decreasing and thus ln(x + 1) ≥ x − 12 x2for x ≥ 0.(c) 200 21.200 246. (a) Let h(x) = ex− 1 − x for x ≥ 0. Then h(0) = 0 and h (x) = ex− 1 ≥ 0 for x ≥ 0, so h(x) isincreasing.(b) Let h(x) = ex−1−x− 12 x2. Then h(0) = 0 and h (x) = ex−1−x. By Part (a), ex−1−x ≥ 0for x ≥ 0, so h(x) is increasing.
• Exercise Set 5.1 159(c) 600 2600 247. Points of inﬂection at x = −2, +2. Concave upon (−5, −2) and (2, 5); concave down on (−2, 2).Increasing on [−3.5829, 0.2513] and [3.3316, 5],and decreasing on [−5, −3.5829] and [0.2513, 3.3316].250–250–5 548. Points of inﬂection at x = ±1/√3. Concave upon [−5, −1/√3] and [1/√3, 5], and concave downon [−1/√3, 1/√3]. Increasing on [−5, 0]and decreasing on [0, 5].1–2–5 549. f (x) = 290x3− 81x2− 585x + 397(3x2 − 5x + 8)3. The denominator has complex roots, so is always positive;hence the x-coordinates of the points of inﬂection of f(x) are the roots of the numerator (if itchanges sign). A plot of the numerator over [−5, 5] shows roots lying in [−3, −2], [0, 1], and [2, 3].To six decimal places the roots are x = −2.464202, 0.662597, 2.701605.50. f (x) =2x5+ 5x3+ 14x2+ 30x − 7(x2 + 1)5/2. Points of inﬂection will occur when the numerator changessign, since the denominator is always positive. A plot of y = 2x5+5x3+14x2+30x−7 shows thatthere is only one root and it lies in [0, 1]. To six decimal place the point of inﬂection is located atx = 0.210970.51. f(x1)−f(x2) = x21 −x22 = (x1 +x2)(x1 −x2) < 0 if x1 < x2 for x1, x2 in [0, +∞), so f(x1) < f(x2)and f is thus increasing.52. f(x1) − f(x2) =1x1−1x2=x2 − x1x1x2> 0 if x1 < x2 for x1, x2 in (0, +∞), so f(x1) > f(x2) andthus f is decreasing.53. (a) If x1 < x2 where x1 and x2 are in I, then f(x1) < f(x2) and g(x1) < g(x2), sof(x1) + g(x1) < f(x2) + g(x2), (f + g)(x1) < (f + g)(x2). Thus f + g is increasing on I.(b) Case I: If f and g are ≥ 0 on I, and if x1 < x2 where x1 and x2 are in I, then0 < f(x1) < f(x2) and 0 < g(x1) < g(x2), so f(x1)g(x1) < f(x2)g(x2),(f · g)(x1) < (f · g)(x2). Thus f · g is increasing on I.Case II: If f and g are not necessarily positive on I then no conclusion can be drawn: forexample, f(x) = g(x) = x are both increasing on (−∞, 0), but (f · g)(x) = x2isdecreasing there.
• 160 Chapter 554. (a) f and g are increasing functions on the interval. By Exercise 53, f + g is increasing.(b) f and g are increasing functions on the interval. If f ≥ 0 and g ≥ 0 on the interval thenby Exercise 53, f · g is concave up on the interval.If the requirement of nonnegativity is removed, then the conclusion may not be true: letf(x) = g(x) = x2on (−∞, 0). Each is concave up,55. (a) f(x) = x, g(x) = 2x (b) f(x) = x, g(x) = x + 6 (c) f(x) = 2x, g(x) = x56. (a) f(x) = ex, g(x) = e2x(b) f(x) = g(x) = x (c) f(x) = e2x, g(x) = ex57. (a) f (x) = 6ax + 2b = 6a x +b3a, f (x) = 0 when x = −b3a. f changes its direction ofconcavity at x = −b3aso −b3ais an inﬂection point.(b) If f(x) = ax3+ bx2+ cx + d has three x-intercepts, then it has three roots, say x1, x2 andx3, so we can write f(x) = a(x − x1)(x − x2)(x − x3) = ax3+ bx2+ cx + d, from which itfollows that b = −a(x1 + x2 + x3). Thus −b3a=13(x1 + x2 + x3), which is the average.(c) f(x) = x(x2− 3x + 2) = x(x − 1)(x − 2) so the intercepts are 0, 1, and 2 and the average is1. f (x) = 6x − 6 = 6(x − 1) changes sign at x = 1.58. f (x) = 6x + 2b, so the point of inﬂection is at x = −b3. Thus an increase in b moves the point ofinﬂection to the left.59. (a) Let x1 < x2 belong to (a, b). If both belong to (a, c] or both belong to [c, b) then we havef(x1) < f(x2) by hypothesis. So assume x1 < c < x2. We know by hypothesis thatf(x1) < f(c), and f(c) < f(x2). We conclude that f(x1) < f(x2).(b) Use the same argument as in Part (a), but with inequalities reversed.60. By Theorem 5.1.2, f is increasing on any interval [(2n−1)π, 2(n+1)π] (n = 0, ±1, ±2, · · ·), becausef (x) = 1 + cos x > 0 on ((2n − 1)π, (2n + 1)π). By Exercise 59 (a) we can piece these intervalstogether to show that f(x) is increasing on (−∞, +∞).61. By Theorem 5.1.2, fis decreasing on any interval [(2nπ +π/2, 2(n+1)π +π/2] (n = 0, ±1, ±2, · · ·),because f (x) = − sin x + 1 < 0 on (2nπ + π/2, 2(n + 1)π + π/2). By Exercise 59 (b) we can piecethese intervals together to show that f(x) is decreasing on (−∞, +∞).62. By zooming on the graph of y (t), maximum increase is at x = −0.577 and maximum decrease isat x = 0.577.63.t12yinfl pt64.t12yno infl pts
• Exercise Set 5.2 16165. yxinfl pts123466. yxinfl pts123467. (a) y (t) =LAke−kt(1 + Ae−kt)2S, so y (0) =LAk(1 + A)2(b) The rate of growth increases to its maximum, which occurs when y is halfway between 0 andL, or when t =1kln A; it then decreases back towards zero.(c) From (2) one sees thatdydtis maximized when y lies half way between 0 and L, i.e. y = L/2.This follows since the right side of (2) is a parabola (with y as independent variable) withy-intercepts y = 0, L. The value y = L/2 corresponds to t =1kln A, from (4).68. Find t so that N (t) is maximum. The size of the population is increasing most rapidly whent = 8.4 years.69. t = 7.67 100000 1570. Since 0 < y < L the right-hand side of (3) of Example 9 can change sign only if the factor L − 2ychanges sign, which it does when y = L/2, at which point we haveL2=L1 + Ae−kt, 1 = Ae−kt,t =1kln A.EXERCISE SET 5.21. (a)f (x)xy (b)f (x)xy
• 162 Chapter 5(c)f (x)xy (d)f (x)xy2. (a) yx(b) yx(c) yx(d) yx3. (a) f (x) = 6x − 6 and f (x) = 6, with f (1) = 0. For the ﬁrst derivative test, f < 0 for x < 1and f > 0 for x > 1. For the second derivative test, f (1) > 0.(b) f (x) = 3x2− 3 and f (x) = 6x. f (x) = 0 at x = ±1. First derivative test: f > 0 forx < −1 and x > 1, and f < 0 for −1 < x < 1, so there is a relative maximum at x = −1,and a relative minimum at x = 1. Second derivative test: f < 0 at x = −1, a relativemaximum; and f > 0 at x = 1, a relative minimum.4. (a) f (x) = 2 sin x cos x = sin 2x (so f (0) = 0) and f (x) = 2 cos 2x. First derivative test: if x isnear 0 then f < 0 for x < 0 and f > 0 for x > 0, so a relative minimum at x = 0. Secondderivative test: f (0) = 2 > 0, so relative minimum at x = 0.(b) g (x) = 2 tan x sec2x (so g (0) = 0) and g (x) = 2 sec2x(sec2x + 2 tan2x). First derivativetest: g < 0 for x < 0 and g > 0 for x > 0, so a relative minimum at x = 0. Second derivativetest: g (0) = 2 > 0, relative minimum at x = 0.(c) Both functions are squares, and so are positive for values of x near zero; both functions arezero at x = 0, so that must be a relative minimum.5. (a) f (x) = 4(x − 1)3, g (x) = 3x2− 6x + 3 so f (1) = g (1) = 0.(b) f (x) = 12(x − 1)2, g (x) = 6x − 6, so f (1) = g (1) = 0, which yields no information.(c) f < 0 for x < 1 and f > 0 for x > 1, so there is a relative minimum at x = 1;g (x) = 3(x − 1)2> 0 on both sides of x = 1, so there is no relative extremum at x = 1.6. (a) f (x) = −5x4, g (x) = 12x3− 24x2so f (0) = g (0) = 0.(b) f (x) = −20x3, g (x) = 36x2− 48x, so f (0) = g (0) = 0, which yields no information.(c) f < 0 on both sides of x = 0, so there is no relative extremum there; g (x) = 12x2(x−2) < 0on both sides of x = 0 (for x near 0), so again there is no relative extremum there.7. f (x) = 16x3− 32x = 16x(x2− 2), so x = 0, ±√2 are stationary points.8. f (x) = 12x3+ 12 = 12(x + 1)(x2− x + 1), so x = −1 is the stationary point.9. f (x) =−x2− 2x + 3(x2 + 3)2, so x = −3, 1 are the stationary points.
• Exercise Set 5.2 16310. f (x) = −x(x3− 16)(x3 + 8)2, so stationary points at x = 0, 24/3.11. f (x) =2x3(x2 − 25)2/3; so x = 0 is the stationary point.12. f (x) =2x(4x − 3)3(x − 1)1/3, so x = 0, 3/4 are the stationary points.13. f(x) = | sin x| =sin x, sin x ≥ 0− sin x, sin x < 0so f (x) =cos x, sin x > 0− cos x, sin x < 0and f (x) does not existwhen x = nπ, n = 0, ±1, ±2, · · · (the points where sin x = 0) becauselimx→nπ−f (x) = limx→nπ+f (x) (see Theorem preceding Exercise 61, Section 3.3). Now f (x) = 0 when± cos x = 0 provided sin x = 0 so x = π/2 + nπ, n = 0, ±1, ±2, · · · are stationary points.14. When x > 0, f (x) = cos x, so x = (n + 12 )π, n = 0, 1, 2, . . . are stationary points.When x < 0, f (x) = − cos x, so x = (n + 12 )π, n = −1, −2, −3, . . . are stationary points.f is not diﬀerentiable at x = 0, so the latter is a critical point but not a stationary point.15. (a) none(b) x = 1 because f changes sign from + to − there(c) none because f = 0 (never changes sign)16. (a) x = 1 because f (x) changes sign from − to + there(b) x = 3 because f (x) changes sign from + to − there(c) x = 2 because f (x) changes sign there17. (a) x = 2 because f (x) changes sign from − to + there.(b) x = 0 because f (x) changes sign from + to − there.(c) x = 1, 3 because f (x) changes sign at these points.18. (a) x = 1 (b) x = 5 (c) x = −1, 0, 319. critical points x = 0, 51/3:f : 0051/30– –– –– ++ +–x = 0: neitherx = 51/3: relative minimum20. critical points x = −3/2, 0, 3/2: f : 00 00 3/2–3/2+ –+ –+– –– – + ++x = −3/2: relative minimum;x = 0: relative maximum;x = 3/2: relative minimum21. critical points x = 2/3: f : 02/3+ –+ –+ –x = 2/3: relative maximum;22. critical points x = ±√7: f : 00+ –+ –+ ++ +–– 7 7x = −√7: relative maximum;x =√7: relative minimum23. critical points x = 0: f : 00– +– +– +x = 0: relative minimum;
• 164 Chapter 524. critical points x = 0, ln 3: f : 00ln 30– –– –– ++ +–x = 0: neither;x = ln 3: relative minimum25. critical points x = −1, 1:f : 001–1– +– +– –– –+x = −1: relative minimum;x = 1: relative maximum26. critical points x = ln 2, ln 3: f : 00ln 3ln 2+ –+ –+ ++ +–x = ln 2: relative maximum;x = ln 3: relative minimum27. f (x) = 8 − 6x: critical point x = 4/3f (4/3) = −6 : f has a maximum of 19/3 at x = 4/328. f (x) = 4x3− 36x2: critical points at x = 0, 9f (0) = 0: Theorem 5.2.5 with m = 3: f has an inﬂection point at x = 0f (9) > 0: f has a minimum of −2187 at x = 929. f (x) = 2 cos 2x: critical points at x = π/4, 3π/4f (π/4) = −4: f has a maximum of 1 at x = π/4f (3π/4) = 4 : f has a minimum of 1 at x = 3π/430. f (x) = (x − 2)ex: critical point at x = 2f (2) = e2: f has a minimum of −e2at x = 231. f (x) = 4x3− 12x2+ 8x: critical points at x = 0, 1, 200 01 20+ –+ –+– –– – + ++relative minimum of 0 at x = 0relative maximum of 1 at x = 1relative minimum of 0 at x = 232. f (x) = 4x3− 36x2+ 96x − 64; critical points at x = 1, 4f (1) = 36: f has a relative minimum of −27 at x = 1f (4) = 0: Theorem 5.2.5 with m = 3: f has an inﬂection point at x = 433. f (x) = 5x4+ 8x3+ 3x2: critical points at x = −3/5, −1, 0f (−3/5) = 18/25 : f has a relative minimum of −108/3125 at x = −3/5f (−1) = −2 : f has a relative maximum of 0 at x = −1f (0) = 0: Theorem 5.2.5 with m = 3: f has an inﬂection point at x = 034. f (x) = 5x4+ 12x3+ 9x2+ 2x: critical points at x = −2/5, −1, 0f (−2/5) = −18/25 : f has a relative maximum of 108/3125 at x = −2/5f (−1) = 0: Theorem 5.2.5 with m = 3: f has an inﬂection point at x = −1f (0) = 2 : f has a relative maximum of 0 at x = 035. f (x) =2(x1/3+ 1)x1/3: critical point at x = −1, 0f (−1) = −23: f has a relative maximum of 1 at x = −1f does not exist at x = 0. By inspection it is a relative minimum of 0.36. f (x) =2x2/3+ 1x2/3: no critical point except x = 0; since f is an odd function, x = 0 is an inﬂectionpoint for f.
• Exercise Set 5.2 16537. f (x) = −5(x − 2)2; no extrema38. f (x) = −2x(x4− 16)(x4 + 16)2; critical points x = −2, 0, 2f (−2) = −18; f has a relative maximum of18at x = −2f (0) =18; f has a relative minimum of 0 at x = 0f (2) = −18; f has a relative maximum of18at x = 239. f (x) =2x2 + x2; critical point at x = 0f (0) = 1; f has a relative minimum of ln 2 at x = 040. f (x) =3x22 + x2; critical points at x = 0, −21/3: f :f (0) = 0 no relative extrema; f has no limit at x = −21/3000– +– +– ++ ++– 2341. f (x) = 2e2x− ex; critical point x = − ln 2f (− ln 2) = 1/2; relative minimum of −1/4 at x = − ln 242. f (x) = 2x(1 + x)e2x: critical point x = −1, 0f (−1) = −2/e2; relative maximum of 1/e2at x = −1f (0) = 2: relative minimum of 0 at x = 043. f (x) = ±(3 − 2x) : critical points at x = 3/2, 0, 3f (3/2) = −2, relative maximum of 9/4 at x = 3/2x = 0, 3 are not stationary points:by ﬁrst derivative test, x = 0 and x = 3 are each a minimum of 044. On each of the intervals (−∞, −1), (−1, +∞) the derivative is of the form y = ±13x2/3hence it isclear that the only critical points are possibly −1 or 0. Near x = 0, x = 0, y =13x2/3> 0 so yhas an inﬂection point at x = 0. At x = −1, y changes sign, thus the only extremum is a relativeminimum of 0 at x = −1.45.–2 1 3 5–6–42xy( , – )3225446.2 4 6 81020xy(4 – √17, 0)(4 + √17, 0)(4, 17)(0, 1)
• 166 Chapter 547.–802 4xy(–2, 49)(0, 5)(3, –76)( , – )12272( , 0)–7 – √574( , 0)–7 + √57448.–1 0.5 1.5–4–246xy135027,( )(0, 2)(2, 0)49.–1 1135yx( , )1 – √329 – 6√34( , )1 + √329 + 6√34(– , – √2 )1√254( , + √2 )1√25450.–1.5 0.5 1.5–4–213xy(–√5, 0)(–√3, –4) (√3, –4)(√5, 0)(1, 0)(0, 5)(–1, 0)51.–1 1–10.51.5xy(– , – )12271652.–0.4 0.6–0.5–0.3–0.1xy2916729,( )13127,( )49, 0( )(0, 0)53.10.20.4xy( , )√5516√5125( , )√1554√5125( , )–√155–4√5125( , )–√55–16√512554.–0.6 0.4 1.2–0.40.4xy–216√72401( , )–√217( , )√217216√72401(0, 0)(1, 0)(–1, 0)
• Exercise Set 5.2 16755. (a) limx→−∞y = −∞, limx→+∞y = +∞;curve crosses x-axis at x = 0, 1, −1yx–6–4–224–2 –1 1(b) limx→±∞y = +∞;curve never crosses x-axisyx0.2–1 1(c) limx→−∞y = −∞, limx→+∞y = +∞;curve crosses x-axis at x = −1yx–0.20.20.4–1 1(d) limx→±∞y = +∞;curve crosses x-axis at x = 0, 1yx0.4–1 156. (a) yxa byxa byxa b(b) yxa byxa b(c) yxa b57. f (x) = 2 cos 2x if sin 2x > 0,f (x) = −2 cos 2x if sin 2x < 0,f (x) does not exist when x = π/2, π, 3π/2;critical numbers x = π/4, 3π/4, 5π/4, 7π/4, π/2, π, 3π/2relative minimum of 0 at x = π/2, π, 3π/2;relative maximum of 1 at x = π/4, 3π/4, 5π/4, 7π/4100 o
• 168 Chapter 558. f (x) =√3 + 2 cos x;critical numbers x = 5π/6, 7π/6relative minimum of 7√3π/6 − 1 at x = 7π/6;relative maximum of 5√3π/6 + 1 at x = 5π/61200 o59. f (x) = − sin 2x;critical numbers x = π/2, π, 3π/2relative minimum of 0 at x = π/2, 3π/2;relative maximum of 1 at x = π100 o60. f (x) = (2 cos x − 1)/(2 − cos x)2;critical numbers x = π/3, 5π/3relative maximum of√3/3 at x = π/3,relative minimum of −√3/3 at x = 5π/30.80 o–0.861. f (x) = ln x + 1, f (x) = 1/x; f (1/e) = 0, f (1/e) > 0;relative minimum of −1/e at x = 1/e2.5–0.50 2.562. f (x) = −2ex− e−x(ex + e−x)2= 0 when x = 0.By the ﬁrst derivative test f (x) > 0 for x < 0and f (x) < 0 for x > 0; relative maximum of 1 at x = 010–2 263. f (x) = 2x(1 − x)e−2x= 0 at x = 0, 1.f (x) = (4x2− 8x + 2)e−2x;f (0) > 0 and f (1) < 0, so a relative minimum of 0 at x = 0and a relative maximum of 1/e2at x = 1.0.140–0.3 4
• Exercise Set 5.2 16964. f (x) = 10/x − 1 = 0 at x = 10; f (x) = −10/x2< 0;relative maximum of 10(ln(10) − 1) ≈ 13.03 at x = 1014–40 2065. Relative minima at x = −3.58, 3.33;relative maximum at x = 0.25250–250–5 566. Relative minimum at x = −0.84;relative maximum at x = 0.841.2–1.2-6 667. Relative maximum at x = −0.27,relative minimum at x = 0.22–4 22468xf(x)f"(x)y68. Relative maximum at x = −1.11,relative minimum at x = 0.47relative maximum at x = 2.04–5 –3 1 50.20.4yxf(x)f"(x)69. f (x) =4x3− sin 2x2√x4 + cos2 x,f (x) =6x2− cos 2x√x4 + cos2 x−(4x3− sin 2x)(4x3− sin 2x)4(x4 + cos2 x)3/2Relative minima at x ≈ ±0.62, relative maximum at x = 02–2–2 1xyf(x)f(x)70. Point of inﬂection at x = 0–0.1 0.1–0.20.10.2xyf(x)f"(x)
• 170 Chapter 571. (a) Let f(x) = x2+kx, then f (x) = 2x −kx2=2x3− kx2. f has a relative extremum when2x3− k = 0, so k = 2x3= 2(3)3= 54.(b) Let f(x) =xx2 + k, then f (x) =k − x2(x2 + k)2. f has a relative extremum when k − x2= 0, sok = x2= 32= 9.72. (a) relative minima at x ≈ ±0.6436,relative maximum at x = 0yx11.21.41.61.82–1.5 –0.5 0.5 1.5(b) x ≈ ±0.6436, 073. (a) (−2.2, 4), (2, 1.2), (4.2, 3)(b) f exists everywhere, so the critical numbers are when f = 0, i.e. when x = ±2 or r(x) = 0,so x ≈ −5.1, −2, 0.2, 2. At x = −5.1 f changes sign from − to +, so minimum; at x = −2f changes sign from + to −, so maximum; at x = 0.2 f doesn’t change sign, so neither; atx = 2 f changes sign from − to +, so minimum.Finally, f (1) = (12− 4)r (1) + 2r(1) ≈ −3(0.6) + 2(0.3) = −1.2.74. g (x) exists everywhere, so the critical points are the points when g (x) = 0, or r(x) = x, so r(x)crosses the line y = x. From the graph it appears that this happens precisely when x = 0.75. f (x) = 3ax2+ 2bx + c and f (x) has roots at x = 0, 1, so f (x) must be of the formf (x) = 3ax(x − 1); thus c = 0 and 2b = −3a, b = −3a/2. f (x) = 6ax + 2b = 6ax − 3a, sof (0) > 0 and f (1) < 0 provided a < 0. Finally f(0) = d, so d = 0; andf(1) = a + b + c + d = a + b = −a/2 so a = −2. Thus f(x) = −2x3+ 3x2.76. (a) one relative maximum, located at x = n 0.300 14(b) f (x) = cxn−1(−x + n)e−x= 0 at x = n.Since f (x) > 0 for x < n and f (x) < 0 forx > n it’s a maximum.77. (a) f (x) = −xf(x). Since f(x) is alwayspositive, f (x) = 0 at x = 0,f (x) > 0 for x < 0 andf (x) < 0 for x > 0,so x = 0 is a maximum.(b)µyx2c1 µ,( )2c1
• Exercise Set 5.3 17178. (a) Because h and g have relative maxima at x0, h(x) ≤ h(x0) for all x in I1 and g(x) ≤ g(x0)for all x in I2, where I1 and I2 are open intervals containing x0. If x is in both I1 and I2then both inequalities are true and by addition so is h(x)+g(x) ≤ h(x0)+g(x0) which showsthat h + g has a relative maximum at x0.(b) By counterexample; both h(x) = −x2and g(x) = −2x2have relative maxima at x = 0 buth(x) − g(x) = x2has a relative minimum at x = 0 so in general h − g does not necessarilyhave a relative maximum at x0.79. (a)xyx0( )f(x0) is not anextreme value.(b)xyx0( )f(x0) is a relativemaximum.(c)xyx0( )f(x0) is a relativeminimum.EXERCISE SET 5.31. Vertical asymptote x = 4horizontal asymptote y = −26 8–6–42xx = 4y = –2y2. Vertical asymptotes x = ±2horizontal asymptote y = 0–4 4–8–6–42468xx = 2x = –2y3. Vertical asymptotes x = ±2horizontal asymptote y = 0–44–4–224xx = 2x = –2y4. Vertical asymptotes x = ±2horizontal asymptote y = 1–4 4–6–248xyy = 1x = 2x = –2
• 172 Chapter 55. No vertical asymptoteshorizontal asymptote y = 1–40.751.25xyy = 1(– , )2√314 ( , )2√3146. No vertical asymptoteshorizontal asymptote y = 1–4 –2 2 40.8xyy = 1(1, 0)(–1, 0)(0, 1)(( ) )√ √18 – 3√333 – √33/313√18 – 3√33 – 1213( ,)( ,)(( ) )√ √18 + 3√333 + √33/313√18 – 3√33 – 12137. Vertical asymptote x = 1horizontal asymptote y = 12 4–224xyx = 1y = 1(– , – )1√21338. Vertical asymptote x = 0, −3horizontal asymptote y = 2–6 –4 –2 2–6–28xx = –3yy = 2(–2, )749. Vertical asymptote x = 0horizontal asymptote y = 3–5 548xy = 3y(6, )259(4, )114(2, 3)10. Vertical asymptote x = 1horizontal asymptote y = 33 661218xx = 1yy = 3(–2, )13(–1, 0)
• Exercise Set 5.3 17311. Vertical asymptote x = 1horizontal asymptote y = 9–10 102030xyy = 10x = 1( , 9)13(– , 0)13(–1, 1)12. Vertical asymptote x = 1horizontal asymptote y = 3–10 –6 –2 4 8 12691215xx = 1yy = 3(– , )5361172048(– , )737473250013. Vertical asymptote x = 1horizontal asymptote y = −12 4–4–2xx = 1y = –1y(–1, – )1214. Vertical asymptote x = 1horizontal asymptote y = 0–4 –2 4 6–4–235xx = 1y(–21/3, )22/33( , )(28 + 12√5)1/32(28 + 12√5)2/318 + 6√515. (a) horizontal asymptote y = 3 asx → ±∞, vertical asymptotes of x = ±2yx–5510–5 5(b) horizontal asymptote of y = 1 asx → ±∞, vertical asymptotes at x = ±1yx–1010–5 5
• 174 Chapter 5(c) horizontal asymptote of y = −1 asx → ±∞, vertical asymptotes at x = −2, 1yx–1010–5 5(d) horizontal asymptote of y = 1 asx → ±∞, vertical asymptote at x = −1, 2yx–1010–5 516. (a) yxa b(b) Symmetric about the line x =a + b2means fa + b2+ x = fa + b2− x for any x.Note thata + b2+ x − a = x −a − b2anda + b2+ x − b = x +a − b2, and the same equationsare true with x replaced by −x. Hencea + b2+ x − aa + b2+ x − b = x −a − b2x +a − b2= x2−a − b22The right hand side remains the same if we replace x with −x, and so the same is true ofthe left hand side, and the same is therefore true of the reciprocal of the left hand side. Butthe reciprocal of the left hand side is equal to fa + b2+ x . Since this quantity remainsunchanged if we replace x with −x, the condition of symmetry about the line x =a + b2hasbeen met.17. limx→±∞x2x − 3− (x + 3) = limx→±∞9x − 3= 010–510xx = 3y = x + 3y
• Exercise Set 5.3 17518.2 + 3x − x3x− (3 − x2) =2x→ 0 as x → ±∞–5 –3 –1 1 3 5–20–1051525xyy = 3 – x219. y = x2−1x=x3− 1x;y =2x3+ 1x2,y = 0 whenx = −3 12≈ −0.8;y =2(x3− 1)x3xy≈(–0.8, 1.9)(1, 0)20. y =x2− 2x= x −2xsoy = x is an oblique asymptote;y =x2+ 2x2,y = −4x3xyy = x4421. y =(x − 2)3x2= x − 6 +12x − 8x2soy = x − 6 is an oblique asymptote;y =(x − 2)2(x + 4)x3,y =24(x − 2)x4xy–10(–4, –13.5)y = x – 6(2, 0)1022. y = x −1x−1x2= x −1x+1x2soy = x is an oblique asymptote;y = 1 +1x2+1x3,y = −21x3− 61x4–6 –4 –2 2 4 646xyy = x(–3, – )259(–1, 1)
• 176 Chapter 523. y =x3− 4x − 8x + 2= x2− 2x −8x + 2soy = x2− 2x is a curvilinear asymptote as x → ±∞–4 41030xyy = x2 – 2x(–3, 23)(0, –4)x = –224. y =x5x2 + 1= x3− x +xx2 + 1soy = x3− x is a curvilinear asymptote–2 –1 1 2–10–5510xyy = x3 – x25. (a) VI (b) I (c) III (d) V (e) IV (f) II26. (a) When n is even the function is deﬁned only for x ≥ 0; as n increases the graph approachesthe line y = 1 for x > 0.yx(b) When n is odd the graph is symmetric with respect to the origin; as n increases the graphapproaches the line y = 1 for x > 0 and the line y = −1 for x < 0.yx
• Exercise Set 5.3 17727. y = 4x2 − 1y =4x√4x2 − 1y = −4(4x2 − 1)3/2soextrema when x = ±12,no inﬂection points–1 11234xy28. y =3x2 − 4;y =2x3(x2 − 4)2/3;y = −2(3x2+ 4)9(x2 − 4)5/3–2(–2, 0) (2, 0)(0, –2)2xy329. y = 2x + 3x2/3;y = 2 + 2x−1/3;y = −23x−4/3xy45(0, 0)30. y = 2x2− 3x4/3y = 4x − 4x1/3y = 4 −43x−2/3–4 –2 2 42468xy(–1, –1) (1, –1)31. y = x1/3(4 − x);y =4(1 − x)3x2/3;y = −4(x + 2)9x5/3–1010xy(1, 3)(–2, –6 )2332. y = 5x2/3+ x5/3y =5(x + 2)3x1/3y =10(x − 1)9x4/3–6 –4 –2 2 4–2468xy(1, 6)(–2, 3 × 22/3)
• 178 Chapter 533. y = x2/3− 2x1/3+ 4y =2(x1/3− 1)3x2/3y = −2(x1/3− 2)9x5/310 30–1026xy(–8, 12)(0, 4)(1, 3) (8, 4)34. y =8(√x − 1)x;y =4(2 −√x)x2;y =2(3√x − 8)x315xy(4, 2)649158,4( )35. y = x + sin x;y = 1 + cos x, y = 0 when x = π + 2nπ;y = − sin x; y = 0 when x = nπn = 0, ±1, ±2, . . .ccxy36. y = x − tan x;y = 1 − sec2x; y = 0 when x = 2nπy = −2 sec2x tan x = 0 when x = nπn = 0, ±1, ±2, . . .–5 –3 3 5–8–42610xx = –1.5 x = 1.5x = –4.5 x = 4.5y37. y =√3 cos x + sin x;y = −√3 sin x + cos x;y = 0 when x = π/6 + nπ;y = −√3 cos x − sin x;y = 0 when x = 2π/3 + nπo–22xy38. y = sin x + cos x;y = cos x − sin x;y = 0 when x = π/4 + nπ;y = − sin x − cos x;y = 0 when x = 3π/4 + nπ–o o–22xy
• Exercise Set 5.3 17939. y = sin2x − cos x;y = sin x(2 cos x + 1);y = 0 when x = ±π, 2π, 3π and whenx = −23π,23π,43π,83π;y = 4 cos2x + cos x − 2; y = 0 whenx ≈ ±2.57, ±0.94, 3.71, 5.35, 7.22, 8.86–4 6 10–11.5xy(5.35, 0.06)(–2.57, 1.13)(2.57, 1.13) (3.71, 1.13)(8.86, 1.13)(7.22, 0.06)(–0.94, 0.06) (0.94, 0.06)(o, –1)(å, 1)(C, 1) (c, 1)(8, )54(*, )5454(g, ) 54(w, )40. y =√tan x;y =sec2x2√tan x; so y > 0 always;y = sec2x3 tan2x − 14(tan x)3/2,y = 0 when x =π6246810xx = 6y341. (a) limx→+∞xex= +∞, limx→−∞xex= 0–1–3–51xy(–2, –0.27)(–1, –0.37)(b) y = xex;y = (x + 1)ex;y = (x + 2)exyx–0.80.21 2(1, )1e(2, )2e242. limx→+∞f(x) = 0, limx→−∞f(x) = −∞f (x) = (1 − x)e−x, f (x) = (x − 2)e−xcritical point at x = 1;relative maximum at x = 1point of inﬂection at x = 2horizontal asymptote y = 0 as x → +∞43. (a) limx→+∞x2e2x= 0, limx→−∞x2e2x= +∞1 2 30.3xy(0, 0)(0.29, 0.05)(1, 0.14)(1.71, 0.10)(b) y = x2/e2x= x2e−2x;y = 2x(1 − x)e−2x;y = 2(2x2− 4x + 1)e−2x;y = 0 if 2x2− 4x + 1 = 0, whenx =4 ±√16 − 84= 1 ±√2/2 ≈ 0.29, 1.71
• 180 Chapter 544. (a) limx→+∞x2e2x= +∞, limx→−∞x2e2x= 0.–3 –2 –1yx0.20.3(–1.71, 0.10)(–1, 0.14)(0, 0)(–0.29, 0.05)(b) y = x2e2x;y = 2x(x + 1)e2x;y = 2(2x2+ 4x + 1)e2x;y = 0 if 2x2+ 4x + 1 = 0, whenx =−4 ±√16 − 84= −1 ±√2/2 ≈ −0.29, −1.7145. (a) limx→±∞x2e−x2= 0(b)–1–3 1 30.1xy1e(–1, ) 1e(1, )(–1.51, 0.23) (1.51, 0.23)(–0.47, 0.18) (0.47, 0.18)y = x2e−x2;y = 2x(1 − x2)e−x2;y = 0 if x = 0, ±1;y = 2(1 − 5x2+ 2x4)e−x2y = 0 if 2x4− 5x2+ 1 = 0,x2=5 ±√174,x = ±12 5 +√17 ≈ ±1.51,x = ±12 5 −√17 ≈ ±0.4746. (a) limx→±∞f(x) = 1 yx0.41–10 –5 5 10(0, 0)(–√2/3, e–3/2) (√2/3, e–3/2)(b) f (x) = 2x−3e−1/x2so f (x) < 0 for x < 0 and f (x) > 0for x > 0. Set u = x2and use the given result to ﬁndlimx→0f (x) = 0, so (by the First Derivative Test) f(x)has a minimum at x = 0. f (x) = (−6x−4+ 4x−6)e−1/x2,so f(x) has points of inﬂection at x = ± 2/3.47. (a) limx→−∞f(x) = 0, limx→+∞f(x) = −∞–1 2 3 4–20–1010xx = 1y(2, –e2)(b) f (x) = −ex(x − 2)(x − 1)2sof (x) = 0 when x = 2f (x) = −ex(x2− 4x + 5)(x − 1)3sof (x) = 0 alwaysrelative maximum when x = 2, no point of inﬂectionasymptote x = 1
• Exercise Set 5.3 18148. (a) limx→−∞f(x) = 0, limx→+∞f(x) = +∞–2 –1 1123xy(–2/3, –2/32/3e2/3)(–1.48, 0.30) (0.15, 0.33)(b) f (x) =ex(3x + 2)3x1/3sof (x) = 0 when x = −23f (x) =ex(9x2+ 12x − 2)9x4/3sopoints of inﬂection when f (x) = 0 atx = −2 −√63, −2 +√63,relative maximum at −23, e−2/3−232/3absolute minimum at (0, 0)yx0.611.81 2 3 4(2, )(3.41, 1.04)(0.59, 0.52)(0, 0)4e49. limx→+∞f(x) = 0,limx→−∞f(x) = +∞f (x) = x(2 − x)e1−x, f (x) = (x2− 4x + 2)e1−xcritical points at x = 0, 2;relative minimum at x = 0,relative maximum at x = 2points of inﬂection at x = 2 ±√2horizontal asymptote y = 0 as x → +∞50. limx→+∞f(x) = +∞, limx→−∞f(x) = 0f (x) = x2(3 + x)ex−1, f (x) = x(x2+ 6x + 6)ex−1critical points at x = −3, 0;relative minimum at x = −3points of inﬂection at x = 0, −3 ±√3 ≈ 0, −4.7, −1.27horizontal asymptote y = 0 as x → −∞yx–0.40.40.8–4 –2(–4.7, –0.35)(0, 0)(–1.27, –0.21)(–3, –0.49)11xy(e–1, –e–1)51. (a) limx→0+y = limx→0+x ln x = limx→0+ln x1/x= limx→0+1/x−1/x2= 0;limx→+∞y = +∞(b) y = x ln x,y = 1 + ln x,y = 1/x,y = 0 when x = e−1
• 182 Chapter 552. (a) limx→0+y = limx→0+ln x1/x2= limx→0+1/x−2/x3= 0,limx→+∞y = +∞1–0.2–0.10.10.2xy(e–1/2, – e–1)12(e–3/2, – e–3)32(b) y = x2ln x, y = x(1 + 2 ln x),y = 3 + 2 ln x,y = 0 if x = e−1/2,y = 0 if x = e−3/2,limx→0+y = 053. (a) limx→0+x2ln x = 0 by the rule given, limx→+∞x2ln x = +∞ by inspection, and f(x) not deﬁnedfor x < 0yx38e312e3/2( ), –( )18e12, –12√e(b) y = x2ln 2x, y = 2x ln 2x + xy = 2 ln 2x + 3y = 0 if x = 1/(2√e),y = 0 if x = 1/(2e3/2)yx12–2 2(0, 0)(1, ln 2)(–1, ln 2)54. (a) limx→+∞f(x) = +∞; limx→0f(x) = 0(b) y = ln(x2+ 1), y = 2x/(x2+ 1)y = −2x2− 1(x2 + 1)2y = 0 if x = 0y = 0 if x = ±14 5–1135x(e3/2, )3e2(e–3/2, – )32ey55. (a) limx→+∞f(x) = +∞, limx→0+f(x) = 0(b) y = x2/3ln xy =2 ln x + 33x1/3y = 0 when ln x = −32, x = e−3/2y =−3 + 2 ln x9x4/3,y = 0 when ln x =32, x = e3/28 16 24 32 400.51xy(e15/4, )154e5/4(e3, )3e56. (a) limx→0+f(x) = −∞, limx→+∞f(x) = 0(b) y = x−1/3ln xy =3 − ln x3x4/3y = 0 when x = e3;y =4 ln x − 159x7/3y = 0 when x = e15/4
• Exercise Set 5.3 18357. (a) 0.4–0.2–0.5 3(b) y = (1 − bx)e−bx, y = b2(x − 2/b)e−bx;relative maximum at x = 1/b, y = 1/be;point of inﬂection at x = 2/b, y = 2/be2.Increasing b moves the relative maximumand the point of inﬂection to the left anddown, i.e. towards the origin.58. (a) 10–2 2(b) y = −2bxe−bx2,y = 2b(−1 + 2bx2)e−bx2;relative maximum at x = 0, y = 1; pointsof inﬂection at x = ± 1/2b, y = 1/√e.Increasing b moves the points of inﬂectiontowards the y-axis; the relative maximumdoesn’t move.59. (a) The oscillations of excos x about zero in-crease as x → +∞ so the limit does notexist, and limx→−∞excos x = 0.(b) yx–246–2 –1 1 2(0,1)(1.52, 0.22)(c) The curve y = eaxcos bx oscillates between y = eaxand y = −eax. The frequency ofoscillation increases when b increases.yx–55–1 2a = 1b = 1b = 2b = 3yx510–1 0.5 1b = 1a = 1a = 2a = 360. (a) 1000 6(b) y =n2− 2x2nxn−1e−x2/n,y =n4− n3− 4x2n2− 2x2n + 4x4n2xn−2e−x2/n,relative maximum when x =n√2, y =3e;point of inﬂection whenx =2n + 1 ±√8n + 14n.As n increases the maxima move up andto the right; the points of inﬂection moveto the right.61. (a) x = 1, 2.5, 4 and x = 3, the latter being a cusp(b) (−∞, 1], [2.5, 3)(c) relative maxima for x = 1, 3; relative minima for x = 2.5(d) x = 0.8, 1.9, 4
• 184 Chapter 562. (a) f (x) = −2h(x) + (1 − 2x)h (x), f (5) = −2h(5) − 9h (5). But from the graphh (5) ≈ −0.2 and f (5) = 0, so h(5) = −(9/2)h (5) ≈ 0.9(b) f (x) = −4h (x) + (1 − 2x)h (x), f (5) ≈ 0.8 − 9h (5) and since h (5) is clearly negative,f (5) > 0 and thus f has a minimum at x = 5.63. Let y be the length of the other side of the rectangle,then L = 2x+2y and xy = 400 so y = 400/x and henceL = 2x + 800/x. L = 2x is an oblique asymptoteL = 2x +800x=2(x2+ 400)x,L = 2 −800x2=2(x2− 400)x2,L =1600x3,L = 0 when x = 20, L = 8020100xL64. Let y be the height of the box, then S = x2+ 4xy andx2y = 500 so y = 500/x2and hence S = x2+ 2000/x.The graph approaches the curve S = x2asymptoticallyS = x2+2000x=x3+ 2000x,S = 2x −2000x2=2(x3− 1000)x2,S = 2 +4000x3=2(x3+ 2000)x3,S = 0 when x = 10, S = 300301000xS65. y = 0.1x4(6x − 5);critical numbers: x = 0, x = 5/6;relative minimum at x = 5/6,y ≈ −6.7 × 10−3–1 10.01xy66. y = 0.1x4(x + 1)(7x + 5);critical numbers: x = 0, x = −1, x = −5/7,relative maximum at x = −1, y = 0;relative minimum at x = −5/7, y ≈ −1.5 × 10−310.001 xy
• Exercise Set 5.4 185EXERCISE SET 5.41. relative maxima at x = 2, 6; absolute maximum at x = 6; relative and absolute minima at x = 0, 42. relative maximum at x = 3; absolute maximum at x = 7; relative minima at x = 1, 5; absoluteminima at x = 1, 53. (a) yx10(b) yx2 7(c) yx53 74. (a) yx(b) yx(c) yx–5 55. x = 1 is a point of discontinuity of f.6. Since f is monotonically increasing on (0, 1), one might expect a minimum at x = 0 and a maximumat x = 1. But both points are discontinuities of f.7. f (x) = 8x − 12, f (x) = 0 when x = 3/2; f(1) = 2, f(3/2) = 1, f(2) = 2 so the maximum valueis 2 at x = 1, 2 and the minimum value is 1 at x = 3/2.8. f (x) = 8 − 2x, f (x) = 0 when x = 4; f(0) = 0, f(4) = 16, f(6) = 12 so the maximum value is 16at x = 4 and the minimum value is 0 at x = 0.9. f (x) = 3(x − 2)2, f (x) = 0 when x = 2; f(1) = −1, f(2) = 0, f(4) = 8 so the minimum is −1 atx = 1 and the maximum is 8 at x = 4.
• 186 Chapter 510. f (x) = 6x2+ 6x − 12, f (x) = 0 when x = −2, 1; f(−3) = 9, f(−2) = 20, f(1) = −7, f(2) = 4, sothe minimum is −7 at x = 1 and the maximum is 20 at x = −2.11. f (x) = 3/(4x2+ 1)3/2, no critical points; f(−1) = −3/√5, f(1) = 3/√5 so the maximum value is3/√5 at x = 1 and the minimum value is −3/√5 at x = −1.12. f (x) =2(2x + 1)3(x2 + x)1/3, f (x) = 0 when x = −1/2 and f (x) does not exist when x = −1, 0;f(−2) = 22/3, f(−1) = 0, f(−1/2) = 4−2/3, f(0) = 0, f(3) = 122/3so the maximum value is 122/3at x = 3 and the minimum value is 0 at x = −1, 0.13. f (x) = 1 − 2 cos x, f (x) = 0 when x = π/3; then f(−π/4) = −π/4 +√2; f(π/3) = π/3 −√3; f(π/2) = π/2−2, so f has a minimum of π/3−√3 at x = π/3 and a maximum of −π/4+√2at x = −π/4.14. f (x) = cos x + sin x, f (x) = 0 for x in (0, π) when x = 3π/4; f(0) = −1, f(3π/4) =√2, f(π) = 1so the maximum value is√2 at x = 3π/4 and the minimum value is −1 at x = 0.15. f(x) = 1 + |9 − x2| =10 − x2, |x| ≤ 3−8 + x2, |x| > 3, f (x) =−2x, |x| < 32x, |x| > 3thus f (x) = 0 whenx = 0, f (x) does not exist for x in (−5, 1) when x = −3 because limx→−3−f (x) = limx→−3+f (x) (seeTheorem preceding Exercise 61, Section 3.3); f(−5) = 17, f(−3) = 1, f(0) = 10, f(1) = 9 so themaximum value is 17 at x = −5 and the minimum value is 1 at x = −3.16. f(x) = |6 − 4x| =6 − 4x, x ≤ 3/2−6 + 4x, x > 3/2, f (x) =−4, x < 3/24, x > 3/2, f (x) does not exist whenx = 3/2 thus 3/2 is the only critical point in (−3, 3); f(−3) = 18, f(3/2) = 0, f(3) = 6 so themaximum value is 18 at x = −3 and the minimum value is 0 at x = 3/2.17. f (x) = 2x − 1, f (x) = 0 when x = 1/2; f(1/2) = −9/4 and limx→±∞f(x) = +∞. Thus f has aminimum of −9/4 at x = 1/2 and no maximum.18. f (x) = −4(x + 1); critical point x = −1. Maximum value f(−1) = 5, no minimum.19. f (x) = 12x2(1 − x); critical points x = 0, 1. Maximum value f(1) = 1, no minimum becauselimx→+∞f(x) = −∞.20. f (x) = 4(x3+ 1); critical point x = −1. Minimum value f(−1) = −3, no maximum.21. No maximum or minimum because limx→+∞f(x) = +∞ and limx→−∞f(x) = −∞.22. No maximum or minimum because limx→+∞f(x) = +∞ and limx→−∞f(x) = −∞.23. limx→−1−f(x) = −∞, so there is no absolute minimum on the interval;f (x) = 0 at x ≈ −2.414213562, for which y ≈ −4.828427125. Also f(−5) = −13/2, so theabsolute maximum of f on the interval is y ≈ −4.828427125 taken at x ≈ −2.414213562.24. limx→−1+f(x) = −∞, so there is no absolute minimum on the interval.f (x) = 3/(x + 1)2> 0, so f is increasing on the interval (−1, 5] and the maximum must occur atthe endpoint x = 5 where f(5) = 1/2.
• Exercise Set 5.4 18725. limx→±∞= +∞ so there is no absolute maximum.f (x) = 4x(x − 2)(x − 1), f (x) = 0 when x = 0, 1, 2,and f(0) = 0, f(1) = 1, f(2) = 0 so f has an absoluteminimum of 0 at x = 0, 2.80–2 426. (x−1)2(x+2)2can never be less than zero because it isthe product of two squares; the minimum value is 0 forx = 1 or −2, no maximum because limx→+∞f(x) = +∞.150–3 227. f (x) =5(8 − x)3x1/3, f (x) = 0 when x = 8 and f (x)does not exist when x = 0; f(−1) = 21, f(0) = 0,f(8) = 48, f(20) = 0 so the maximum value is 48 atx = 8 and the minimum value is 0 at x = 0, 20.500–1 2028. f (x) = (2 − x2)/(x2+ 2)2, f (x) = 0 for x in theinterval (−1, 4) when x =√2; f(−1) = −1/3,f(√2) =√2/4, f(4) = 2/9 so the maximum value is√2/4 at x =√2 and the minimum value is −1/3 atx = −1.0.4–0.4–1 429. f (x) = −1/x2; no maximum or minimum becausethere are no critical points in (0, +∞).2500 10
• 188 Chapter 530. f (x) = −x(x − 2)(x2 − 2x + 2)2, and for 1 ≤ x < +∞, f (x) = 0when x = 2. Also limx→+∞f(x) = 2 and f(2) = 5/2 andf(1) = 2, hence f has an absolute minimum value of 2 atx = 1 and an absolute maximum value of 5/2 at x = 2.301 831. f (x) =1 − 2 cos xsin2x; f (x) = 0 on [π/4, 3π/4] only whenx = π/3. Then f(π/4) = 2√2 − 1, f(π/3) =√3 andf(3π/4) = 2√2 + 1, so f has an absolute maximum valueof 2√2 + 1 at x = 3π/4 and an absolute minimum value of√3 at x = π/3.303 932. f (x) = 2 sin x cos x−sin x = sin x(2 cos x−1), f (x) = 0 forx in (−π, π) when x = 0, ±π/3; f(−π) = −1, f(−π/3) =5/4, f(0) = 1, f(π/3) = 5/4, f(π) = −1 so the maximumvalue is 5/4 at x = ±π/3 and the minimum value is −1 atx = ±π.1.5–1.5C c0.201 433. f (x) = x2(3 − 2x)e−2x, f (x) = 0 for x in [1, 4] whenx = 3/2; if x = 1, 3/2, 4, then f(x) = e−2,278e−3, 64e−8;critical point at x = 3/2; absolute maximum of278e−3atx = 3/2, absolute minimum of 64e−8at x = 40.760.641 2.734. f (x) = (1 − ln 2x)/x2, f (x) = 0 on [1, e] for x = e/2;if x = 1, e/2, e then f(x) = ln 2, 2/e, (ln 2 + 1)/e;absolute minimum of1 + ln 2eat x = e,absolute maximum of 2/e at x = e/2
• Exercise Set 5.4 1893.0–2.50 435. f (x) = −3x2− 10x + 3x2 + 1, f (x) = 0 when x =13, 3.Then f(0) = 0, f13= 5 ln109− 1,f(3) = 5 ln 10 − 9, f(4) = 5 ln 17 − 12 and thus f has anabsolute minimum of 5(ln 10 − ln 9) − 1 at x = 1/3 and anabsolute maximum of 5 ln 10 − 9 at x = 3.20–20–2 236. f (x) = (x2+ 2x − 1)ex, f (x) = 0 at x = −2 +√2 andx = −1−√2 (discard), f(−1+√2) = (2−2√2)e(−1+√2)≈−1.25, absolute maximum at x = 2, f(2) = 3e2≈ 22.17,absolute minimum at x = −1 +√237. f (x) = −[cos(cos x)] sin x; f (x) = 0 if sin x = 0 or ifcos(cos x) = 0. If sin x = 0, then x = π is the criti-cal point in (0, 2π); cos(cos x) = 0 has no solutions be-cause −1 ≤ cos x ≤ 1. Thus f(0) = sin(1), f(π) =sin(−1) = − sin(1), and f(2π) = sin(1) so the maxi-mum value is sin(1) ≈ 0.84147 and the minimum valueis − sin(1) ≈ −0.84147.1–10 o38. f (x) = −[sin(sin x)] cos x; f (x) = 0 if cos x = 0 or ifsin(sin x) = 0. If cos x = 0, then x = π/2 is the criticalpoint in (0, π); sin(sin x) = 0 if sin x = 0, which gives nocritical points in (0, π). Thus f(0) = 1, f(π/2) = cos(1),and f(π) = 1 so the maximum value is 1 and the minimumvalue is cos(1) ≈ 0.54030.1.500 c39. f (x) =4, x < 12x − 5, x > 1so f (x) = 0 when x = 5/2, and f (x) does not exist when x = 1because limx→1−f (x) = limx→1+f (x) (see Theorem preceding Exercise 61, Section 3.3); f(1/2) = 0,f(1) = 2, f(5/2) = −1/4, f(7/2) = 3/4 so the maximum value is 2 and the minimum value is−1/4.40. f (x) = 2x + p which exists throughout the interval (0, 2) for all values of p so f (1) = 0 becausef(1) is an extreme value, thus 2 + p = 0, p = −2. f(1) = 3 so 12+ (−2)(1) + q = 3, q = 4 thusf(x) = x2− 2x + 4 and f(0) = 4, f(2) = 4 so f(1) is the minimum value.41. The period of f(x) is 2π, so check f(0) = 3, f(2π) = 3 and the critical points.f (x) = −2 sin x − 2 sin 2x = −2 sin x(1 + 2 cos x) = 0 on [0, 2π] at x = 0, π, 2π andx = 2π/3, 4π/3. Check f(π) = −1, f(2π/3) = −3/2, f(4π/3) = −3/2.Thus f has an absolute maximum on (−∞, +∞) of 3 at x = 2kπ, k = 0, ±1, ±2, . . .and an absolute minimum of −3/2 at x = 2kπ ± 2π/3, k = 0, ±1, ±2, . . ..
• 190 Chapter 542. cosx3has a period of 6π, and cosx2a period of 4π, so f(x) has a period of 12π. Consider theinterval [0, 12π]. f (x) = − sinx3− sinx2, f (x) = 0 when sinx3+ sinx2= 0 thus, by use ofthe trig identity sin a + sin b = 2 sina + b2cosa − b2, 2 sin5x12cos −x12= 0 so sin5x12= 0 orcosx12= 0. Solve sin5x12= 0 to get x = 12π/5, 24π/5, 36π/5, 48π/5 and then solve cosx12= 0to get x = 6π. The corresponding values of f(x) are −4.0450, 1.5450, 1.5450, −4.0450, 1, 5, 5 so themaximum value is 5 and the minimum value is −4.0450 (approximately).43. Let f(x) = x − sin x, then f (x) = 1 − cos x and so f (x) = 0 when cos x = 1 which has no solutionfor 0 < x < 2π thus the minimum value of f must occur at 0 or 2π. f(0) = 0, f(2π) = 2π so 0 isthe minimum value on [0, 2π] thus x − sin x ≥ 0, sin x ≤ x for all x in [0, 2π].44. Let h(x) = cos x − 1 + x2/2. Then h(0) = 0, and it is suﬃcient to show that h (x) ≥ 0 for0 < x < 2π. But h (x) = − sin x + x ≥ 0 by Exercise 43.45. Let m = slope at x, then m = f (x) = 3x2− 6x + 5, dm/dx = 6x − 6; critical point for m is x = 1,minimum value of m is f (1) = 246. (a) limx→0+f(x) = +∞, limx→(π/2)−f(x) = +∞, so f has no maximum value on the interval. Bytable 5.4.3 f must have a minimum value.(b) According to table 5.4.3, there is an absolute minimum value of f on (0, π/2). To ﬁnd theabsolute minimum value, we examine the critical points (Theorem 5.4.3).f (x) = sec x tan x − csc x cot x = 0 at x = π/4, where f(π/4) = 2√2, which must be theabsolute minimum value of f on the interval (0, π/2).47. limx→+∞f(x) = +∞, limx→8−f(x) = +∞, so there is no absolute maximum value of f for x > 8. Bytable 5.4.3 there must be a minimum. Since f (x) =2x(−520 + 192x − 24x2+ x3)(x − 8)3, we must solvea quartic equation to ﬁnd the critical points. But it is easy to see that x = 0 and x = 10 are realroots, and the other two are complex. Since x = 0 is not in the interval in question, we must havean absolute minimum of f on (8, +∞) of 125 at x = 10.48. (a)dCdt=Ka − bae−at− be−btsodCdt= 0 at t =ln(a/b)a − b. This is the only stationary point andC(0) = 0, limt→+∞C(t) = 0, C(t) > 0 for 0 < t < +∞, so it is an absolute maximum.(b) 0.700 1049. The slope of the line is −1, and the slope of the tangent to y = −x2is −2x so −2x = −1, x = 1/2.The line lies above the curve so the vertical distance is given by F(x) = 2 − x + x2; F(−1) = 4,F(1/2) = 7/4, F(3/2) = 11/4. The point (1/2, −1/4) is closest, the point (−1, −1) farthest.
• Exercise Set 5.5 19150. The slope of the line is 4/3; and the slope of the tangent to y = x3is 3x2so 3x2= 4/3, x2= 4/9,x = ±2/3. The line lies below the curve so the vertical distance is given by F(x) = x3− 4x/3 + 1;F(−1) = 4/3, F(−2/3) = 43/27, F(2/3) = 11/27, F(1) = 2/3. The closest point is (2/3, 8/27),the farthest is (−2/3, −8/27).51. The absolute extrema of y(t) can occur at the endpoints t = 0, 12 or when dy/dt = 2 sin t = 0, i.e.t = 0, 12, kπ, k = 1, 2, 3; the absolute maximum is y = 4 at t = π, 3π; the absolute minimum isy = 0 at t = 0, 2π.52. (a) The absolute extrema of y(t) can occur at the endpoints t = 0, 2π or whendy/dt = 2 cos 2t − 4 sin t cos t = 2 cos 2t − 2 sin 2t = 0, t = 0, 2π, π/8, 5π/8, 9π/8, 13π/8;the absolute maximum is y = 3.4142 at t = π/8, 9π/8; the absolute minimum is y = 0.5859at t = 5π/8, 13π/8.(b) The absolute extrema of x(t) occur at the endpoints t = 0, 2π or whendxdt= −2 sin t + 1(2 + sin t)2= 0, t = 7π/6, 11π/6. The absolute maximum is x = 0.5774 at t = 11π/6and the absolute minimum is x = −0.5774 at t = 7π/6.53. f (x) = 2ax + b; critical point is x = −b2af (x) = 2a > 0 so f −b2ais the minimum value of f, butf −b2a= a −b2a2+ b −b2a+ c =−b2+ 4ac4athus f(x) ≥ 0 if and only iff −b2a≥ 0,−b2+ 4ac4a≥ 0, −b2+ 4ac ≥ 0, b2− 4ac ≤ 054. Use the proof given in the text, replacing “maximum” by “minimum” and “largest” by “smallest”and reversing the order of all inequality symbols.EXERCISE SET 5.51. If y = x + 1/x for 1/2 ≤ x ≤ 3/2 then dy/dx = 1 − 1/x2= (x2− 1)/x2, dy/dx = 0 when x = 1. Ifx = 1/2, 1, 3/2 then y = 5/2, 2, 13/6 so(a) y is as small as possible when x = 1.(b) y is as large as possible when x = 1/2.2. Let x and y be nonnegative numbers and z the sum of their squares, then z = x2+ y2. Butx + y = 1, y = 1 − x so z = x2+ (1 − x)2= 2x2− 2x + 1 for 0 ≤ x ≤ 1. dz/dx = 4x − 2, dz/dx = 0when x = 1/2. If x = 0, 1/2, 1 then z = 1, 1/2, 1 so(a) z is as large as possible when one number is 0 and the other is 1.(b) z is as small as possible when both numbers are 1/2.3. A = xy where x + 2y = 1000 so y = 500 − x/2 andA = 500x − x2/2 for x in [0, 1000]; dA/dx = 500 − x,dA/dx = 0 when x = 500. If x = 0 or 1000 then A = 0,if x = 500 then A = 125, 000 so the area is maximumwhen x = 500 ft and y = 500 − 500/2 = 250 ft.xyStream
• 192 Chapter 54. The triangle in the ﬁgure is determined by the two sides oflength a and b and the angle θ. Suppose that a + b = 1000.Then h = a sin θ and area = Q = hb/2 = ab(sin θ)/2. Thisexpression is maximized by choosing θ = π/2 and thus Q =ab/2 = a(1000 − a)/2, which is maximized by a = 500. Thusthe maximal area is obtained by a right isosceles triangle, wherethe angle between two sides of 500 is the right angle.A BChabcθ5. Let x and y be the dimensions shown in the ﬁgure and A thearea, then A = xy subject to the cost condition 3(2x)+2(2y) =6000, or y = 1500 − 3x/2. Thus A = x(1500 − 3x/2) = 1500x −3x2/2 for x in [0, 1000]. dA/dx = 1500 − 3x, dA/dx = 0 whenx = 500. If x = 0 or 1000 then A = 0, if x = 500 thenA = 375, 000 so the area is greatest when x = 500 ft and (fromy = 1500 − 3x/2) when y = 750 ft.Heavy-dutyStandardxy6. Let x and y be the dimensions shown in the ﬁgure and A thearea of the rectangle, then A = xy and, by similar triangles,x/6 = (8 − y)/8, y = 8 − 4x/3 so A = x(8 − 4x/3) = 8x − 4x2/3for x in [0, 6]. dA/dx = 8 − 8x/3, dA/dx = 0 when x = 3. Ifx = 0, 3, 6 then A = 0, 12, 0 so the area is greatest when x = 3in and (from y = 8 − 4x/3) y = 4 in.8yx6107. Let x, y, and z be as shown in the ﬁgure and A the area of therectangle, then A = xy and, by similar triangles, z/10 = y/6,z = 5y/3; also x/10 = (8 − z)/8 = (8 − 5y/3)/8 thus y =24/5 − 12x/25 so A = x(24/5 − 12x/25) = 24x/5 − 12x2/25 forx in [0, 10]. dA/dx = 24/5 − 24x/25, dA/dx = 0 when x = 5. Ifx = 0, 5, 10 then A = 0, 12, 0 so the area is greatest when x = 5in. and y = 12/5 in.8zxy6108. A = (2x)y = 2xy where y = 16 − x2so A = 32x − 2x3for0 ≤ x ≤ 4; dA/dx = 32 − 6x2, dA/dx = 0 when x = 4/√3. Ifx = 0, 4/√3, 4 then A = 0, 256/(3√3), 0 so the area is largestwhen x = 4/√3 and y = 32/3. The dimensions of the rectanglewith largest area are 8/√3 by 32/3.–4 416xyyx9. A = xy where x2+ y2= 202= 400 so y =√400 − x2 andA = x√400 − x2 for 0 ≤ x ≤ 20;dA/dx = 2(200 − x2)/√400 − x2, dA/dx = 0 whenx =√200 = 10√2. If x = 0, 10√2, 20 then A = 0, 200, 0 so thearea is maximum when x = 10√2 and y =√400 − 200 = 10√2.xy10
• Exercise Set 5.5 19310. Let R denote the rectangle with corners (±8, ±10). Sup-pose the lower left corner of the square S is at the point(x0, y0) = (x0, −4x0). Let Q denote the desired region.It is clear that Q will be a rectangle, and that its left andbottom sides are subsets of the left and bottom sides of S. Theright and top edges of S will be subsets of the sides of R (if Sis big enough).From the drawing it is evident that the area of Q is (8−x0)(10+4x0) = −4x20 + 22x0 + 80. This function is maximized whenx0 = 11/4, for which the area of Q is 441/4.The maximum possible area for Q is 441/4, taken when x0 = 11/4.–6 2–6y = –4xy = –4x(x0, y0)RS11. Let x = length of each side that uses the \$1 per foot fencing,y = length of each side that uses the \$2 per foot fencing.The cost is C = (1)(2x) + (2)(2y) = 2x + 4y, but A = xy = 3200 thus y = 3200/x soC = 2x + 12800/x for x > 0,dC/dx = 2 − 12800/x2, dC/dx = 0 when x = 80, d2C/dx2> 0 soC is least when x = 80, y = 40.12. A = xy where 2x+2y = p so y = p/2−x and A = px/2−x2forx in [0, p/2]; dA/dx = p/2 − 2x, dA/dx = 0 when x = p/4. Ifx = 0 or p/2 then A = 0, if x = p/4 then A = p2/16 so the areais maximum when x = p/4 and y = p/2 − p/4 = p/4, which is asquare. xy13. Let x and y be the dimensions of a rectangle; the perimeter is p = 2x + 2y. But A = xy thusy = A/x so p = 2x + 2A/x for x > 0, dp/dx = 2 − 2A/x2= 2(x2− A)/x2, dp/dx = 0 whenx =√A, d2p/dx2= 4A/x3> 0 if x > 0 so p is a minimum when x =√A and y =√A and thusthe rectangle is a square.14. With x, y, r, and s as shown in the ﬁgure, the sum of theenclosed areas is A = πr2+s2where r =x2πand s =y4becausex is the circumference of the circle and y is the perimeter of thesquare, thus A =x24π+y216. But x + y = 12, so y = 12 − x andA =x24π+(12 − x)216=π + 416πx2−32x + 9 for 0 ≤ x ≤ 12.dAdx=π + 48πx −32,dAdx= 0 when x =12ππ + 4. If x = 0,12ππ + 4, 12then A = 9,36π + 4,36πso the sum of the enclosed areas isx y12r scut(a) a maximum when x = 12 in. (when all of the wire is used for the circle)(b) a minimum when x = 12π/(π + 4) in.
• 194 Chapter 515. (a)dNdt= 250(20 − t)e−t/20= 0 at t = 20, N(0) = 125,000, N(20) ≈ 161,788, andN(100) ≈ 128,369; the absolute maximum is N = 161788 at t = 20, the absoluteminimum is N = 125,000 at t = 0.(b) The absolute minimum ofdNdtoccurs whend2Ndt2= 12.5(t − 40)e−t/20= 0, t = 40.16. The area of the window is A = 2rh + πr2/2, the perimeteris p = 2r + 2h + πr thus h =12[p − (2 + π)r] soA = r[p − (2 + π)r] + πr2/2= pr − (2 + π/2)r2for 0 ≤ r ≤ p/(2 + π),dA/dr = p − (4 + π)r, dA/dr = 0 when r = p/(4 + π) andd2A/dr2< 0, so A is maximum when r = p/(4 + π). 2rrh17. Let the box have dimensions x, x, y, with y ≥ x. The constraint is 4x + y ≤ 108, and the volumeV = x2y. If we take y = 108 − 4x then V = x2(108 − 4x) and dV/dx = 12x(−x + 18) with rootsx = 0, 18. The maximum value of V occurs at x = 18, y = 36 with V = 11, 664 in2.18. Let the box have dimensions x, x, y with x ≥ y. The constraint is x+2(x+y) ≤ 108, and the volumeV = x2y. Take x = (108 − 2y)/3 = 36 − 2y/3, V = y(36 − 2y/3)2, dV/dy = (4/3)y2− 96y + 1296with roots y = 18, 54. Then d2V/dy2= (8/3)y − 96 is negative for y = 18, so by the secondderivative test, V has a maximum of 10, 368 in2at y = 18, x = 24.19. Let x be the length of each side of a square, then V = x(3 − 2x)(8 − 2x) = 4x3− 22x2+ 24xfor 0 ≤ x ≤ 3/2; dV/dx = 12x2− 44x + 24 = 4(3x − 2)(x − 3), dV/dx = 0 when x = 2/3 for0 < x < 3/2. If x = 0, 2/3, 3/2 then V = 0, 200/27, 0 so the maximum volume is 200/27 ft3.20. Let x = length of each edge of base, y = height. The cost isC = (cost of top and bottom) + (cost of sides) = (2)(2x2) + (3)(4xy) = 4x2+ 12xy, butV = x2y = 2250 thus y = 2250/x2so C = 4x2+ 27000/x for x > 0, dC/dx = 8x − 27000/x2,dC/dx = 0 when x = 3√3375 = 15, d2C/dx2> 0 so C is least when x = 15, y = 10.21. Let x = length of each edge of base, y = height, k = \$/cm2for the sides. The cost isC = (2k)(2x2) + (k)(4xy) = 4k(x2+ xy), but V = x2y = 2000 thus y = 2000/x2soC = 4k(x2+ 2000/x) for x > 0, dC/dx = 4k(2x − 2000/x2), dC/dx = 0 whenx = 3√1000 = 10, d2C/dx2> 0 so C is least when x = 10, y = 20.22. Let x and y be the dimensions shown in the ﬁgure and V the vol-ume, then V = x2y. The amount of material is to be 1000 ft2,thus (area of base) + (area of sides) = 1000, x2+ 4xy = 1000,y =1000 − x24xso V = x2 1000 − x24x=14(1000x − x3) for0 < x ≤ 10√10.dVdx=14(1000 − 3x2),dVdx= 0when x = 1000/3 = 10 10/3.If x = 0, 10 10/3, 10√10 then V = 0,5000310/3, 0;the volume is greatest for x = 10 10/3 ft and y = 5 10/3 ft.xxy
• Exercise Set 5.5 19523. Let x = height and width, y = length. The surface area is S = 2x2+ 3xy where x2y = V , soy = V/x2and S = 2x2+ 3V/x for x > 0; dS/dx = 4x − 3V/x2, dS/dx = 0 when x = 33V/4,d2S/dx2> 0 so S is minimum when x =3 3V4, y =433 3V4.24. Let r and h be the dimensions shown in the ﬁgure, thenthe volume of the inscribed cylinder is V = πr2h. Butr2+h22= R2thus r2= R2−h24so V = π R2−h24h = π R2h −h34for 0 ≤ h ≤ 2R.dVdh= π R2−34h2,dVdh= 0 h2hrRwhen h = 2R/√3. If h = 0, 2R/√3, 2Rthen V = 0,4π3√3R3, 0 so the volume is largestwhen h = 2R/√3 and r = 2/3R.25. Let r and h be the dimensions shown in the ﬁgure,then the surface area is S = 2πrh + 2πr2.But r2+h22= R2thus h = 2√R2 − r2 soS = 4πr√R2 − r2 + 2πr2for 0 ≤ r ≤ R,dSdr=4π(R2− 2r2)√R2 − r2+ 4πr;dSdr= 0 whenR2− 2r2√R2 − r2= −r (1)R2− 2r2= −r√R2 − r2R4− 4R2r2+ 4r4= r2(R2− r2)5r4− 5R2r2+ R4= 0h2hrRand using the quadratic formula r2=5R2±√25R4 − 20R410=5 ±√510R2, r =5 ±√510R, ofwhich only r =5 +√510R satisﬁes (1). If r = 0,5 +√510R, 0 then S = 0, (5 +√5)πR2, 2πR2sothe surface area is greatest when r =5 +√510R and, from h = 2 R2 − r2, h = 25 −√510R.
• 196 Chapter 526. Let R and H be the radius and height of the cone, andr and h the radius and height of the cylinder (see ﬁg-ure), then the volume of the cylinder is V = πr2h.By similar triangles (see ﬁgure)H − hH=rRthush =HR(R − r) so V = πHR(R − r)r2= πHR(Rr2− r3)for 0 ≤ r ≤ R.dVdr= πHR(2Rr−3r2) = πHRr(2R−3r),dVdr= 0 for 0 < r < R when r = 2R/3. Ifr = 0, 2R/3, R then V = 0, 4πR2H/27, 0 so the maxi-mum volume is4πR2H27=4913πR2H =49· (volume ofcone).RrHh27. From (13), S = 2πr2+ 2πrh. But V = πr2h thus h = V/(πr2) and so S = 2πr2+ 2V/r for r > 0.dS/dr = 4πr − 2V/r2, dS/dr = 0 if r = 3V/(2π). Since d2S/dr2= 4π + 4V/r3> 0, the minimumsurface area is achieved when r = 3V/2π and so h = V/(πr2) = [V/(πr3)]r = 2r.28. V = πr2h where S = 2πr2+ 2πrh so h =S − 2πr22πr, V =12(Sr − 2πr3) for r > 0.dVdr=12(S − 6πr2) = 0 if r = S/(6π),d2Vdr2= −6πr < 0 so V is maximum whenr = S/(6π) and h =S − 2πr22πr=S − 2πr22πr2r =S − S/3S/3r = 2r, thus the height is equal to thediameter of the base.29. The surface area is S = πr2+ 2πrhwhere V = πr2h = 500 so h = 500/(πr2)and S = πr2+ 1000/r for r > 0;dS/dr = 2πr − 1000/r2= (2πr3− 1000)/r2,dS/dr = 0 when r = 3500/π, d2S/dr2> 0for r > 0 so S is minimum when r = 3500/π cm andh =500πr2=500ππ5002/3= 3500/π cmrh30. The total area of material used isA = Atop + Abottom + Aside = (2r)2+ (2r)2+ 2πrh = 8r2+ 2πrh.The volume is V = πr2h thus h = V/(πr2) so A = 8r2+ 2V/r for r > 0,dA/dr = 16r −2V/r2= 2(8r3−V )/r2, dA/dr = 0 when r = 3√V /2. This is the only critical point,d2A/dr2> 0 there so the least material is used when r = 3√V /2,rh=rV/(πr2)=πVr3and, forr =3√V /2,rh=πVV8=π8.31. Let x be the length of each side of the squares and y the height of the frame, then the volume isV = x2y. The total length of the wire is L thus 8x + 4y = L, y = (L − 8x)/4 soV = x2(L − 8x)/4 = (Lx2− 8x3)/4 for 0 ≤ x ≤ L/8. dV/dx = (2Lx − 24x2)/4, dV/dx = 0 for0 < x < L/8 when x = L/12. If x = 0, L/12, L/8 then V = 0, L3/1728, 0 so the volume is greatestwhen x = L/12 and y = L/12.
• Exercise Set 5.5 19732. (a) Let x = diameter of the sphere, y = length of an edge of the cube. The combined volume isV =16πx3+ y3and the surface area is S = πx2+ 6y2= constant. Thus y =(S − πx2)1/261/2and V =π6x3+(S − πx2)3/263/2for 0 ≤ x ≤Sπ;dVdx=π2x2−3π63/2x(S − πx2)1/2=π2√6x(√6x − S − πx2).dVdx= 0 when x = 0, or when√6x =√S − πx2, 6x2= S − πx2, x2=S6 + π, x =S6 + π. If x = 0,S6 + π,Sπ,then V =S3/263/2,S3/26√6 + π,S3/26√πso that V is smallest when x =S6 + π, and hence wheny =S6 + π, thus x = y.(b) From Part (a), the sum of the volumes is greatest when there is no cube.33. Let h and r be the dimensions shown in the ﬁgure, then thevolume is V =13πr2h. But r2+ h2= L2thus r2= L2− h2so V =13π(L2−h2)h =13π(L2h−h3) for 0 ≤ h ≤ L.dVdh=13π(L2− 3h2).dVdh= 0 when h = L/√3. If h = 0, L/√3, 0then V = 0,2π9√3L3, 0 so the volume is as large as possiblewhen h = L/√3 and r = 2/3L.hLr34. Let r and h be the radius and height of the cone (see ﬁgure).The slant height of any such cone will be R, the radius ofthe circular sheet. Refer to the solution of Exercise 33 toﬁnd that the largest volume is2π9√3R3. h Rr35. The area of the paper is A = πrL = πr√r2 + h2, butV =13πr2h = 10 thus h = 30/(πr2)so A = πr r2 + 900/(π2r4).To simplify the computations let S = A2,S = π2r2r2+900π2r4= π2r4+900r2for r > 0,dSdr= 4π2r3−1800r3=4(π2r6− 450)r3, dS/dr = 0 whenr = 6450/π2, d2S/dr2> 0, so S and hence A is leastwhen r = 6450/π2 cm, h =30π3π2/450 cm.h Lr
• 198 Chapter 536. The area of the triangle is A =12hb. By similar trian-gles (see ﬁgure)b/2h=R√h2 − 2Rh, b =2Rh√h2 − 2Rhso A =Rh2√h2 − 2Rhfor h > 2R,dAdh=Rh2(h − 3R)(h2 − 2Rh)3/2,dAdh= 0 for h > 2R when h = 3R, by the ﬁrst deriva-tive test A is minimum when h = 3R. If h = 3R thenb = 2√3R (the triangle is equilateral).Rh − RRhbb/2h2 − 2Rh37. The volume of the cone is V =13πr2h. By similar tri-angles (see ﬁgure)rh=R√h2 − 2Rh, r =Rh√h2 − 2Rhso V =13πR2 h3h2 − 2Rh=13πR2 h2h − 2Rfor h > 2R,dVdh=13πR2 h(h − 4R)(h − 2R)2,dVdh= 0 for h > 2R whenh = 4R, by the ﬁrst derivative test V is minimumwhen h = 4R. If h = 4R then r =√2R.rRh − RRhh2 − 2Rh38. The area is (see ﬁgure)A =12(2 sin θ)(4 + 4 cos θ)= 4(sin θ + sin θ cos θ)for 0 ≤ θ ≤ π/2;dA/dθ = 4(cos θ − sin2θ + cos2θ)= 4(cos θ − [1 − cos2θ] + cos2θ)= 4(2 cos2θ + cos θ − 1)= 4(2 cos θ − 1)(cos θ + 1)dA/dθ = 0 when θ = π/3 for 0 < θ < π/2. If θ = 0, π/3, π/2 thenA = 0, 3√3, 4 so the maximum area is 3√3.42 cos θθ4 cos θ2 sin θ239. Let b and h be the dimensions shown in the ﬁgure,then the cross-sectional area is A =12h(5 + b). Buth = 5 sin θ and b = 5 + 2(5 cos θ) = 5 + 10 cos θ soA =52sin θ(10 + 10 cos θ) = 25 sin θ(1 + cos θ) for0 ≤ θ ≤ π/2.dA/dθ = −25 sin2θ + 25 cos θ(1 + cos θ)= 25(− sin2θ + cos θ + cos2θ)= 25(−1 + cos2θ + cos θ + cos2θ)= 25(2 cos2θ + cos θ − 1) = 25(2 cos θ − 1)(cos θ + 1).dA/dθ = 0 for 0 < θ < π/2 when cos θ = 1/2, θ = π/3.If θ = 0, π/3, π/2 then A = 0, 75√3/4, 25 so the cross-sectional area is greatest when θ = π/3.b5 cos θθ555h = 5 sin θ
• Exercise Set 5.5 19940. I = kcos φ2, k the constant of proportionality. If h is the height of the lamp above the table thencos φ = h/ and =√h2 + r2 so I = kh3= kh(h2 + r2)3/2for h > 0,dIdh= kr2− 2h2(h2 + r2)5/2,dIdh= 0when h = r/√2, by the ﬁrst derivative test I is maximum when h = r/√2.41. Let L, L1, and L2 be as shown in the ﬁgure, thenL = L1 + L2 = 8 csc θ + sec θ,dLdθ= −8 csc θ cot θ + sec θ tan θ, 0 < θ < π/2= −8 cos θsin2θ+sin θcos2 θ=−8 cos3θ + sin3θsin2θ cos2 θ;dLdθ= 0 if sin3θ = 8 cos3θ, tan3θ = 8, tan θ = 2 which givesthe absolute minimum for L because limθ→0+L = limθ→π/2−L = +∞.If tan θ = 2, then csc θ =√5/2 and sec θ =√5 so L = 8(√5/2) +√5 = 5√5 ft.18θL2L1L42. Let x = number of steers per acrew = average market weight per steerT = total market weight per acrethen T = xw where w = 2000 − 50(x − 20) = 3000 − 50xso T = x(3000 − 50x) = 3000x − 50x2for 0 ≤ x ≤ 60,dT/dx = 3000 − 100x and dT/dx = 0 when x = 30. If x = 0, 30, 60 then T = 0, 45000, 0 so thetotal market weight per acre is largest when 30 steers per acre are allowed.43. (a) The daily proﬁt isP = (revenue) − (production cost) = 100x − (100, 000 + 50x + 0.0025x2)= −100, 000 + 50x − 0.0025x2for 0 ≤ x ≤ 7000, so dP/dx = 50 − 0.005x and dP/dx = 0 when x = 10, 000. Because 10,000is not in the interval [0, 7000], the maximum proﬁt must occur at an endpoint. When x = 0,P = −100, 000; when x = 7000, P = 127, 500 so 7000 units should be manufactured and solddaily.(b) Yes, because dP/dx > 0 when x = 7000 so proﬁt is increasing at this production level.(c) dP/dx = 15 when x = 7000, so P(7001) − P(7000) ≈ 15, and the marginal proﬁt is \$15.44. (a) R(x) = px but p = 1000 − x so R(x) = (1000 − x)x(b) P(x) = R(x) − C(x) = (1000 − x)x − (3000 + 20x) = −3000 + 980x − x2(c) P (x) = 980 − 2x, P (x) = 0 for 0 < x < 500 when x = 490; test the points 0, 490, 500 to ﬁndthat the proﬁt is a maximum when x = 490.(d) P(490) = 237,100(e) p = 1000 − x = 1000 − 490 = 510.45. The proﬁt isP = (proﬁt on nondefective) − (loss on defective) = 100(x − y) − 20y = 100x − 120ybut y = 0.01x + 0.00003x2so P = 100x − 120(0.01x + 0.00003x2) = 98.8x − 0.0036x2for x > 0,dP/dx = 98.8 − 0.0072x, dP/dx = 0 when x = 98.8/0.0072 ≈ 13, 722, d2P/dx2< 0 so the proﬁt ismaximum at a production level of about 13,722 pounds.
• 200 Chapter 546. To cover 1 mile requires 1/v hours, and 1/(10 − 0.07v) gallons of diesel fuel, so the total cost tothe client is C =15v+1.5010 − 0.07v,dCdv=0.0315v2+ 21v − 1500v2(0.07v − 10)2. By the second derivative test, Chas a minimum of 50.6 cents/mile at v = 65.08 miles per hour.47. The distance between the particles is D = (1 − t − t)2 + (t − 2t)2 =√5t2 − 4t + 1 for t ≥ 0. Forconvenience, we minimize D2instead, so D2= 5t2− 4t + 1, dD2/dt = 10t − 4, which is 0 whent = 2/5. d2D2/dt2> 0 so D2and hence D is minimum when t = 2/5. The minimum distance isD = 1/√5.48. The distance between the particles is D = (2t − t)2 + (2 − t2)2 =√t4 − 3t2 + 4 for t ≥ 0. Forconvenience we minimize D2instead so D2= t4− 3t2+ 4, dD2/dt = 4t3− 6t = 4t(t2− 3/2), whichis 0 for t > 0 when t = 3/2. d2D2/dt2= 12t2− 6 > 0 when t = 3/2 so D2and hence D isminimum there. The minimum distance is D =√7/2.49. Let P(x, y) be a point on the curve x2+ y2= 1. The distance between P(x, y) and P0(2, 0) isD = (x − 2)2 + y2, but y2= 1 − x2so D = (x − 2)2 + 1 − x2 =√5 − 4x for −1 ≤ x ≤ 1,dDdx= −2√5 − 4xwhich has no critical points for −1 < x < 1. If x = −1, 1 then D = 3, 1 so theclosest point occurs when x = 1 and y = 0.50. Let P(x, y) be a point on y =√x, then the distance D between P and (2, 0) isD = (x − 2)2 + y2 = (x − 2)2 + x =√x2 − 3x + 4, for 0 ≤ x ≤ 3. For convenience we ﬁnd theextrema for D2instead, so D2= x2− 3x + 4, dD2/dx = 2x − 3 = 0 when x = 3/2. If x = 0, 3/2, 3then D2= 4, 7/4, 4 so D = 2,√7/2, 2. The points (0, 0) and (3,√3) are at the greatest distance,and (3/2, 3/2) the shortest distance from (2, 0).51. (a) Draw the line perpendicular to the given line that passes through Q.(b) Let Q : (x0, y0). If (x, mx + b) is the point on the graph closest to Q, then the distancesquared d2= D = (x − x0)2+ (mx + b − y0)2is minimized. ThendD/dx = 2(x − x0) + 2m(mx + b − y0) = 0at the point (x, mx + b) which minimizes the distance.On the other hand, the line connecting the point Q with the line y = mx+b is perpendicular tothis line provided the connecting line has slope −1/m. But this condition is (y−y0)/(x−x0) =−1/m, or m(y −y0) = −(x−x0). Since y = mx+b the condition becomes m(mx+b−y0) =−(x − x0), which is the equivalent to the expression above which minimizes the distance.52. (a) Draw the line from the point Q through the center P of the circle. The nearer and fartherpoints are the points on C that are, respectively closest to and farthest from Q.(b) Let Q : (xQ, yQ) and P : (xP , yP ), and let the equation of the circle be (x − xP )2+ (y −yP )2= c2where c is constant. Let T be the square of the distance from (x, y) to Q:T = (x − xQ)2+ (y − yQ)2. It is desired to ﬁnd the minimum (and maximum) value of Twhen (x, y) lies on the circle.This occurs when the derivative of T is zero, i.e. 2(x − xQ) + 2(y − yQ)(dy/dx) = 0. Notethat dy/dx here expresses the slope of the line tangent to the circle at the point (x, y). Anequivalent expression is dy/dx = −(x − xQ)/(y − yQ), which is the negative reciprocal of theslope of the line connecting (x, y) with Q. Thus the minimum and maximum distances areachieved on this line.
• Exercise Set 5.5 20153. (a) The line through (x, y) and the origin has slope y/x, and the negative reciprocal is −x/y. If(x, y) is a point on the ellipse, then x2− xy + y2= 4 and, diﬀerentiating,2x − x(dy/dx) − y + 2y(dy/dx) = 0.For the desired points we have dy/dx = −x/y, and inserting that into the previous equationresults in 2x + x2/y − y − 2x = 0, 2xy + x2− y2− 2xy = 0, x2− y2= 0, y = ±x. Insertingthis into the equation of the ellipse we have x2∓ x2+ x2= 4 with solutions y = x = ±2or y = −x = ±2/√3. Thus there are four solutions, (2, 2), (−2, −2), (2/√3, −2/√3) and(−2/√3, 2/√3).(b) In general, the shortest/longest distance from a point to a curve is taken on the line connectingthe point to the curve which is perpendicular to the tangent line at the point in question.54. The tangent line to the ellipse at (x, y) has slope dy/dx,where x2− xy + y2= 4. This yields2x−y −xdy/dx+2ydy/dx = 0, dy/dx = (2x−y)/(x−2y). The line through (x, y) that also passes throughthe origin has slope y/x. Check to see if the twoslopes are negative reciprocals: (2x − y)/(x − 2y) =−x/y; y(2x − y) = −x(x − 2y); x = ±y, so thepoints lie on the line y = x or on y = −x.yx–2 –1 1 2–2–11255. If P(x0, y0) is on the curve y = 1/x2, then y0 = 1/x20. At P the slope of the tangent line is −2/x30so its equation is y −1x20= −2x30(x − x0), or y = −2x30x +3x20. The tangent line crosses the y-axisat3x20, and the x-axis at32x0. The length of the segment then is L =9x40+94x20 for x0 > 0. Forconvenience, we minimize L2instead, so L2=9x40+94x20,dL2dx0= −36x50+92x0 =9(x60 − 8)2x50, whichis 0 when x60 = 8, x0 =√2.d2L2dx20> 0 so L2and hence L is minimum when x0 =√2, y0 = 1/2.56. If P(x0, y0) is on the curve y = 1 − x2, then y0 = 1 − x20. At P the slope of the tangent line is−2x0 so its equation is y − (1 − x20) = −2x0(x − x0), or y = −2x0x + x20 + 1. The y-intercept isx20 + 1 and the x-intercept is12(x0 + 1/x0) so the area A of the triangle isA =14(x20 + 1)(x0 + 1/x0) =14(x30 + 2x0 + 1/x0) for 0 ≤ x0 ≤ 1.dA/dx0 =14(3x20 + 2 − 1/x20) =14(3x40 + 2x20 − 1)/x20 which is 0 when x20 = −1 (reject), or whenx20 = 1/3 so x0 = 1/√3. d2A/dx20 =14(6x0 + 2/x30) > 0 at x0 = 1/√3 so a relative minimum andhence the absolute minimum occurs there.57. At each point (x, y) on the curve the slope of the tangent line is m =dydx= −2x(1 + x2)2for anyx,dmdx=2(3x2− 1)(1 + x2)3,dmdx= 0 when x = ±1/√3, by the ﬁrst derivative test the only relativemaximum occurs at x = −1/√3, which is the absolute maximum because limx→±∞m = 0. Thetangent line has greatest slope at the point (−1/√3, 3/4).
• 202 Chapter 558. Let x be how far P is upstream from where the manstarts (see ﬁgure), then the total time to reach T ist = (time from M to P) + (time from P to T)=√x2 + 1rR+1 − xrWfor 0 ≤ x ≤ 1,where rR and rW are the rates at which he can rowand walk, respectively.MTP11x(a) t =√x2 + 13+1 − x5,dtdx=x3√x2 + 1−15sodtdx= 0 when 5x = 3 x2 + 1,25x2= 9(x2+ 1), x2= 9/16, x = 3/4. If x = 0, 3/4, 1 then t = 8/15, 7/15,√2/3 so the timeis a minimum when x = 3/4 mile.(b) t =√x2 + 14+1 − x5,dtdx=x4√x2 + 1−15sodtdx= 0 when x = 4/3 which is not in theinterval [0, 1]. Check the endpoints to ﬁnd that the time is a minimum when x = 1 (he shouldrow directly to the town).59. With x and y as shown in the ﬁgure, the maximumlength of pipe will be the smallest value of L = x + y.By similar trianglesy8=x√x2 − 16, y =8x√x2 − 16soL = x +8x√x2 − 16for x > 4,dLdx= 1 −128(x2 − 16)3/2,dLdx= 0 when(x2− 16)3/2= 128x2− 16 = 1282/3= 16(22/3)x2= 16(1 + 22/3)x = 4(1 + 22/3)1/2,d2L/dx2= 384x/(x2− 16)5/2> 0 if x > 4 so L is smallest when x = 4(1 + 22/3)1/2.For this value of x, L = 4(1 + 22/3)3/2ft.84xyx2 – 1660. s = (x1 − ¯x)2+ (x2 − ¯x)2+ · · · + (xn − ¯x)2,ds/d¯x = −2(x1 − ¯x) − 2(x2 − ¯x) − · · · − 2(xn − ¯x),ds/d¯x = 0 when(x1 − ¯x) + (x2 − ¯x) + · · · + (xn − ¯x) = 0(x1 + x2 + · · · xn) − (¯x + ¯x + · · · + ¯x) = 0(x1 + x2 + · · · + xn) − n¯x = 0¯x =1n(x1 + x2 + · · · + xn),d2s/d¯x2= 2 + 2 + · · · + 2 = 2n > 0, so s is minimum when ¯x =1n(x1 + x2 + · · · + xn).
• Exercise Set 5.5 20361. Let x = distance from the weaker light source, I = the intensity at that point, and k the constantof proportionality. ThenI =kSx2+8kS(90 − x)2if 0 < x < 90;dIdx= −2kSx3+16kS(90 − x)3=2kS[8x3− (90 − x)3]x3(90 − x)3= 18kS(x − 30)(x2+ 2700)x3(x − 90)3,which is 0 when x = 30;dIdx< 0 if x < 30, anddIdx> 0 if x > 30, so the intensity is minimum at adistance of 30 cm from the weaker source.62. If f(x0) is a maximum then f(x) ≤ f(x0) for all x in some open interval containing x0 thusf(x) ≤ f(x0) because√x is an increasing function, so f(x0) is a maximum of f(x) at x0.The proof is similar for a minimum value, simply replace ≤ by ≥.xyA(2, 1)B(5, 4)P(x, 0)52x – 2 5 – x␣ ␤␪63. θ = π − (α + β)= π − cot−1(x − 2) − cot−1 5 − x4,dθdx=11 + (x − 2)2+−1/41 + (5 − x)2/16= −3(x2− 2x − 7)[1 + (x − 2)2][16 + (5 − x)2]dθ/dx = 0 when x =2 ±√4 + 282= 1 ± 2√2,only 1 + 2√2 is in [2, 5]; dθ/dx > 0 for x in [2, 1 + 2√2),dθ/dx < 0 for x in (1 + 2√2, 5], θ is maximum when x = 1 + 2√2.␪␣␤x10264. θ = α − β= cot−1(x/12) − cot−1(x/2)dθdx= −12144 + x2+24 + x2=10(24 − x2)(144 + x2)(4 + x2)dθ/dx = 0 when x =√24 = 2√6, bythe ﬁrst derivative test θ ismaximum there.65. Let v = speed of light in the medium. The total time required for the light to travel from A to Pto B ist = (total distance from A to P to B)/v =1v( (c − x)2 + a2 + x2 + b2),dtdx=1v−c − x(c − x)2 + a2+x√x2 + b2anddtdx= 0 whenx√x2 + b2=c − x(c − x)2 + a2. But x/√x2 + b2 = sin θ2 and(c − x)/ (c − x)2 + a2 = sin θ1 thus dt/dx = 0 when sin θ2 = sin θ1 so θ2 = θ1.
• 204 Chapter 566. The total time required for the light to travel from A to P to B ist = (time from A to P) + (time from P to B) =√x2 + a2v1+(c − x)2 + b2v2,dtdx=xv1√x2 + a2−c − xv2 (c − x)2 + b2but x/√x2 + a2 = sin θ1 and(c − x)/ (c − x)2 + b2 = sin θ2 thusdtdx=sin θ1v1−sin θ2v2sodtdx= 0 whensin θ1v1=sin θ2v2.67. (a) The rate at which the farmer walks is analogous to the speed of light in Fermat’s principle.(b) the best path occurs when θ1 = θ2(see ﬁgure).x 1 − x3414θ2θ1HouseBarn(c) by similar triangles,x/(1/4) = (1 − x)/(3/4)3x = 1 − x4x = 1x = 1/4 mi.EXERCISE SET 5.61. f(x) = x2− 2, f (x) = 2x, xn+1 = xn −x2n − 22xnx1 = 1, x2 = 1.5, x3 = 1.416666667, . . . , x5 = x6 = 1.4142135622. f(x) = x2− 5, f (x) = 2x, xn+1 = xn −x2n − 52xnx1 = 2, x2 = 2.25, x3 = 2.236111111, x4 = 2.2360679779, x4 = x5 = 2.23606797753. f(x) = x3− 6, f (x) = 3x2, xn+1 = xn −x3n − 63x2nx1 = 2, x2 = 1.833333333, x3 = 1.817263545, . . . , x5 = x6 = 1.8171205934. xn− a = 05. f(x) = x3− 2x − 2, f (x) = 3x2− 2, xn+1 = xn −x3n − 2xn − 23x2n − 2x1 = 2, x2 = 1.8, x3 = 1.7699481865, x4 = 1.7692926629, x5 = x6 = 1.76929235426. f(x) = x3+ x − 1, f (x) = 3x2+ 1, xn+1 = xn −x3n + xn − 13x2n + 1x1 = 1, x2 = 0.75, x3 = 0.686046512, . . . , x5 = x6 = 0.6823278047. f(x) = x5+ x4− 5, f (x) = 5x4+ 4x3, xn+1 = xn −x5n + x4n − 55x4n + 4x3nx1 = 1, x2 = 1.333333333, x3 = 1.239420573, . . . , x6 = x7 = 1.224439550
• Exercise Set 5.6 2058. f(x) = x5− 3x + 3, f (x) = 5x4− 3, xn+1 = xn −x5n − 3xn + 35x4n − 3x1 = −1.5, x2 = −1.49579832, x3 = −1.49577135 = x4 = x59. f(x) = x4+ x2− 4, f (x) = 4x3+ 2x,xn+1 = xn −x4n + x2n − 44x3n + 2xnx1 = 1, x2 = 1.3333, x3 = 1.2561, x4 = 1.24966, . . . ,x7 = x8 = 1.249621068;by symmetry, x = −1.249621068 is the other solution.16–5–2.2 2.210. f(x) = x5− 5x3− 2, f (x) = 5x4− 15x2,xn+1 = xn −x5n − 5x3n − 25x4n − 15x2nx1 = 2, x2 = 2.5,x3 = 2.327384615, . . . , x7 = x8 = 2.27379173210–20–2.5 2.511. f(x) = 2 cos x − x, f (x) = −2 sin x − 1xn+1 = xn −2 cos x − x−2 sin x − 1x1 = 1, x2 = 1.03004337, x3 = 1.02986654,x4 = x5 = 1.029866538–6O o12. f(x) = sin x − x2,f (x) = cos x − 2x,xn+1 = xn −sin xn − x2ncos xn − 2xnx1 = 1, x2 = 0.891395995,x3 = 0.876984845, . . . , x5 = x6 = 0.8767262150.3–1.30 1.513. f(x) = x − tan x,f (x) = 1 − sec2x = − tan2x, xn+1 = xn +xn − tan xntan2xnx1 = 4.5, x2 = 4.493613903, x3 = 4.493409655,x4 = x5 = 4.493409458100–1006 i
• 206 Chapter 514. f(x) = 1 + exsin x, f (x) = ex(cos x + sin x)xn+1 = xn −1 + exnsin xnexn (cos xn + sin xn)x1 = 3, x2 = 3.2249, x3 = 3.1847,. . . ,x10 = x11 = 3.1830630123–220 c15. The graphs of y = x3and y = 1 − x intersect nearthe point x = 0.7. Let f(x) = x3+ x − 1, so thatf (x) = 3x2+ 1, andxn+1 = xn −x3+ x − 13x2 + 1.If x1 = 0.7 then x2 = 0.68259109,x3 = 0.68232786, x4 = x5 = 0.68232780.4–1–1 216. The graphs of y = sin x and y = x3− 2x2+ 1 intersectat points near x = −0.8 and x = 0.6 and x = 2. Letf(x) = sin x−x3+2x2−1, then f (x) = cos x−3x2+4x,soxn+1 = xn −cos x − 3x2+ 4xsin x − x3 + 2x2 + 1.If x1 = −0.8, then x2 = −0.783124811,x3 = −0.782808234,x4 = x5 = −0.782808123; if x1 = 0.6, thenx2 = 0.568003853, x3 = x4 = 0.568025739 ; if x1 = 2, thenx2 = 1.979461151, x3 = 1.979019264, x4 = x5 = 1.979019061–1.5 2.5–1217. The graphs of y = x2and y =√2x + 1 intersect atpoints near x = −0.5 and x = 1; x2=√2x + 1,x4− 2x − 1 = 0. Let f(x) = x4− 2x − 1, thenf (x) = 4x3− 2 soxn+1 = xn −x4n − 2xn − 14x3n − 2.If x1 = −0.5, then x2 = −0.475,x3 = −0.474626695,x4 = x5 = −0.474626618; ifx1 = 1, then x2 = 2,x3 = 1.633333333, . . . , x8 = x9 = 1.395336994.40–0.5 218. The graphs of y = 18 x3− 1 and y = cos x − 2 intersectwhen x = 0 and near the point x = −2. Let f(x) =18 x3+ 1 − cos x so that f (x) = 38 x2+ sin x. Thenxn+1 = xn −x3/8 + 1 − cos x3x2/8 + sin x.If x1 = −2 then x2 = −2.70449471,x3 = −2.46018026, . . . , x6 = x7 = −2.406293820–3–3 2
• Exercise Set 5.6 20719. The graphs of y = 1 and y = exsin x intersect near thepoints x = 1 and x = 3. Let f(x) = 1−exsin x, f (x) =−ex(cos x + sin x), andxn+1 = xn +1 − exsin xex(cos x + sin x). If x1 = 1 thenx2 = 0.65725814, x3 = 0.59118311, . . . , x5 = x6 =0.58853274, and if x1 = 3 then x2 = 3.10759324, x3 =3.09649396, . . . , x5 = x6 = 3.09636393.800 3.220. The graphs of y = e−xand y = ln x intersect near x = 1.3; letf(x) = e−x− ln x, f (x) = −e−x− 1/x, x1 = 1.3,xn+1 = xn +e−xn− ln xne−xn + 1/xn, x2 = 1.309759929,x4 = x5 = 1.3097995861–40 221. (a) f(x) = x2− a, f (x) = 2x, xn+1 = xn −x2n − a2xn=12xn +axn(b) a = 10; x1 = 3, x2 = 3.166666667, x3 = 3.162280702, x4 = x5 = 3.16227766022. (a) f(x) =1x− a, f (x) = −1x2, xn+1 = xn(2 − axn)(b) a = 17; x1 = 0.05, x2 = 0.0575, x3 = 0.058793750, x5 = x6 = 0.05882352923. f (x) = x3+ 2x − 5; solve f (x) = 0 to ﬁnd the critical points. Graph y = x3and y = −2x + 5 tosee that they intersect at a point near x = 1.25; f (x) = 3x2+ 2 so xn+1 = xn −x3n + 2xn − 53x2n + 2.x1 = 1.25, x2 = 1.3317757009, x3 = 1.3282755613, x4 = 1.3282688557, x5 = 1.3282688557 so theminimum value of f(x) occurs at x ≈ 1.3282688557 because f (x) > 0; its value is approximately−4.098859123.24. From a rough sketch of y = x sin x we see that the maximum occurs at a point near x = 2, whichwill be a point where f (x) = x cos x + sin x = 0. f (x) = 2 cos x − x sin x soxn+1 = xn −xn cos xn + sin xn2 cos xn − xn sin xn= xn −xn + tan xn2 − xn tan xn.x1 = 2, x2 = 2.029048281, x3 = 2.028757866, x4 = x5 = 2.028757838; the maximum value isapproximately 1.819705741.25. A graphing utility shows that there are two inﬂection points at x ≈ 0.25, −1.25. These pointsare the zeros of f (x) = (x4+ 4x3+ 8x2+ 4x − 1)e−x(x2 + 1)3. It is equivalent to ﬁnd the zeros ofg(x) = x4+4x3+8x2+4x−1. One root is x = −1 by inspection. Since g (x) = 4x3+12x2+16x+4,Newton’s Method becomesxn+1 = xn −x4n + 4x3n + 8x2n + 4xn − 14x3n + 12x2n + 16xn + 4With x0 = 0.25, x1 = 0.18572695, x2 = 0.179563312, x3 = 0.179509029,x4 = x5 = 0.179509025. So the points of inﬂection are at x ≈ 0.18, x = −1.
• 208 Chapter 526. f (x) = −2 tan−1x +1 − 2xx2 + 1= 0 for x = x1 ≈ 0.2451467013, f(x1) ≈ 0.122536352127. Let f(x) be the square of the distance between (1, 0) and any point (x, x2) on the parabola, thenf(x) = (x − 1)2+ (x2− 0)2= x4+ x2− 2x + 1 and f (x) = 4x3+ 2x − 2. Solve f (x) = 0 to ﬁndthe critical points; f (x) = 12x2+ 2 so xn+1 = xn −4x3n + 2xn − 212x2n + 2= xn −2x3n + xn − 16x2n + 1.x1 = 1, x2 = 0.714285714, x3 = 0.605168701, . . . , x6 = x7 = 0.589754512; the coordinates areapproximately (0.589754512, 0.347810385).28. The area is A = xy = x cos x so dA/dx = cos x − x sin x. Find x so that dA/dx = 0;d2A/dx2= −2 sin x − x cos x so xn+1 = xn +cos xn − xn sin xn2 sin xn + xn cos xn= xn +1 − xn tan xn2 tan xn + xn.x1 = 1, x2 = 0.864536397, x3 = 0.860339078, x4 = x5 = 0.860333589; y ≈ 0.652184624.29. (a) Let s be the arc length, and L the length of the chord, then s = 1.5L. But s = rθ andL = 2r sin(θ/2) so rθ = 3r sin(θ/2), θ − 3 sin(θ/2) = 0.(b) Let f(θ) = θ − 3 sin(θ/2), then f (θ) = 1 − 1.5 cos(θ/2) so θn+1 = θn −θn − 3 sin(θn/2)1 − 1.5 cos(θn/2).θ1 = 3, θ2 = 2.991592920, θ3 = 2.991563137, θ4 = θ5 = 2.991563136 rad so θ ≈ 171◦.30. r2(θ − sin θ)/2 = πr2/4 so θ − sin θ − π/2 = 0. Let f(θ) = θ − sin θ − π/2, then f (θ) = 1 − cos θso θn+1 =θn − sin θn − π/21 − cos θn.θ1 = 2, θ2 = 2.339014106, θ3 = 2.310063197, . . . , θ5 = θ6 = 2.309881460 rad; θ ≈ 132◦.31. If x = 1, then y4+ y = 1, y4+ y − 1 = 0. Graph z = y4and z = 1 − y to see that they intersectnear y = −1 and y = 1. Let f(y) = y4+ y − 1, then f (y) = 4y3+ 1 so yn+1 = yn −y4n + yn − 14y3n + 1.If y1 = −1, then y2 = −1.333333333, y3 = −1.235807860, . . . , y6 = y7 = −1.220744085;if y1 = 1, then y2 = 0.8, y3 = 0.731233596, . . . , y6 = y7 = 0.724491959.32. If x = 1, then 2y − cos y = 0. Graph z = 2y and z = cos y to see that they intersect near y = 0.5.Let f(y) = 2y − cos y, then f (y) = 2 + sin y so yn+1 = yn −2yn − cos yn2 + sin yn.y1 = 0.5, y2 = 0.450626693, y3 = 0.450183648, y4 = y5 = 0.450183611.33. S(25) = 250,000 =5000i(1 + i)25− 1 ; set f(i) = 50i−(1+i)25+1, f (i) = 50−25(1+i)24; solvef(i) = 0. Set i0 = .06 and ik+1 = ik − 50i − (1 + i)25+ 1 / 50 − 25(1 + i)24. Then i1 = 0.05430,i2 = 0.05338, i3 = 0.05336, . . . , i = 0.053362.34. (a) x1 = 2, x2 = 5.3333,x3 = 11.055, x4 = 22.293,x5 = 44.6760.500 15(b) x1 = 0.5, x2 = −0.3333, x3 = 0.0833, x4 = −0.0012, x5 = 0.0000 (and xn = 0 for n ≥ 6)
• Exercise Set 5.7 20935. (a) x1 x2 x3 x4 x5 x6 x7 x8 x9 x100.5000 −0.7500 0.2917 −1.5685 −0.4654 0.8415 −0.1734 2.7970 1.2197 0.1999(b) The sequence xn must diverge, since if it did converge then f(x) = x2+ 1 = 0 would have asolution. It seems the xn are oscillating back and forth in a quasi-cyclical fashion.36. (a) xn+1 = xn, i.e. the constant sequence xn is generated.(b) This is equivalent to f(xn) = 0 as in part (a).(c) xn = xn+2 = xn+1 −f(xn+1)f (xn+1)= xn −f(xn)f (xn)−f(xn+1)f (xn+1), so f(xn+1) = −f (xn+1)f (xn)f(xn)37. (a) |xn+1 − xn| ≤ |xn+1 − c| + |c − xn| < 1/n + 1/n = 2/n(b) The closed interval [c − 1, c + 1] contains all of the xn, since |xn − c| < 1/n. Let M be anupper bound for |f (x)| on [c − 1, c + 1]. Since xn+1 = xn − f(xn)/f (xn) it follows that|f(xn)| ≤ |f (xn)||xn+1 − xn| < M|xn+1 − xn| < 2M/n.(c) Assume that f(c) = 0. The sequence xn converges to c, since |xn − c| < 1/n. By thecontinuity of f, f(c) = f( limn→+∞xn) = limn→+∞f(xn).Let = |f(c)|/2. Choose N such that |f(xn) − f(c)| < /2 for n > N. Then|f(xn) − f(c)| < |f(c)|/2 for n > N, so −|f(c)|/2 < f(xn) − f(c) < |f(c)/2|.If f(c) > 0 then f(xn) > f(c) − |f(c)|/2 = f(c)/2.If f(c) < 0, then f(xn) < f(c) + |f(c)|/2 = −|f(c)|/2, or |f(xn)| > |f(c)|/2.(d) From (b) it follows that limn→+∞f(xn) = 0. From (c) it follows that if f(c) = 0 thenlimn→+∞f(xn) = 0, a contradiction. The conclusion, then, is that f(c) = 0.EXERCISE SET 5.71. f(3) = f(5) = 0; f (x) = 2x − 8, 2c − 8 = 0, c = 4, f (4) = 02. f(0) = f(2) = 0, f (x) = 3x2− 6x + 2, 3c2− 6c + 2 = 0; c =6 ±√36 − 246= 1 ±√3/33. f(π/2) = f(3π/2) = 0, f (x) = − sin x, − sin c = 0, c = π4. f(−1) = f(3) = 0; f (1) = 0; f (x) = 2(1 − x)/(4 + 2x − x2); 2(1 − c) = 0, c = 15. f(0) = f(4) = 0, f (x) =12−12√x,12−12√c= 0, c = 16. f(1) = f(3) = 0, f (x) = −2x3+43x2, −2c3+43c2= 0, −6 + 4c = 0, c = 3/27. (f(5) − f(−3))/(5 − (−3)) = 1; f (x) = 2x − 1; 2c − 1 = 1, c = 18. f(−1) = −6, f(2) = 6, f (x) = 3x2+ 1, 3c2+ 1 =6 − (−6)2 − (−1)= 4, c2= 1, c = ±1 of which onlyc = 1 is in (−1, 2)9. f(0) = 1, f(3) = 2, f (x) =12√x + 1,12√c + 1=2 − 13 − 0=13,√c + 1 = 3/2, c + 1 = 9/4, c = 5/410. f(4) = 15/4, f(3) = 8/3, solve f (c) = (15/4−8/3)/1 = 13/12; f (x) = 1+1/x2, f (c) = 1+1/c2=13/12, c2= 12, c = ±2√3, but −2√3 is not in the interval, so c = 2√3.
• 210 Chapter 511. f(−5) = 0, f(3) = 4, f (x) = −x√25 − x2, −c√25 − c2=4 − 03 − (−5)=12, −2c =√25 − c2,4c2= 25 − c2, c2= 5, c = −√5(we reject c =√5 because it does not satisfy the equation −2c =√25 − c2)12. f(2) = 1, f(5) = 1/4, f (x) = −1/(x − 1)2, −1(c − 1)2=1/4 − 15 − 2= −14, (c − 1)2= 4, c − 1 = ±2,c = −1 (reject), or c = 313. (a) f(−2) = f(1) = 0The interval is [−2, 1](b) c = −1.296–2–2 1(c) x0 = −1, x1 = −1.5, x2 = −1.32, x3 = −1.290, x4 = −1.288584314. (a) m =f(−2) − f(1)−2 − 1=0 + 3−3= −1 so y + 3 = −(x − 1), y = −x − 2(b) f (x) = 3x2− 4 = −1 has solutions x = ±1; discard x = 1, so c = −1(c) y − (3) = −(x − (−1)) or y = −x + 2(d) 4–4–3 215. (a) f (x) = sec2x, sec2c = 0 has no solution (b) tan x is not continuous on [0, π]16. (a) f(−1) = 1, f(8) = 4, f (x) =23x−1/323c−1/3=4 − 18 − (−1)=13, c1/3= 2, c = 8 which is not in (−1, 8).(b) x2/3is not diﬀerentiable at x = 0, which is in (−1, 8).17. (a) Two x-intercepts of f determine two solutions a and b of f(x) = 0; by Rolle’s Theorem thereexists a point c between a and b such that f (c) = 0, i.e. c is an x-intercept for f .(b) f(x) = sin x = 0 at x = nπ, and f (x) = cos x = 0 at x = nπ + π/2, which lies between nπand (n + 1)π, (n = 0, ±1, ±2, . . .)18.f(x1) − f(x0)x1 − x0is the average rate of change of y with respect to x on the interval [x0, x1]. Bythe Mean-Value Theorem there is a value c in (x0, x1) such that the instantaneous rate of changef (c) =f(x1) − f(x0)x1 − x0.
• Exercise Set 5.7 21119. Let s(t) be the position function of the automobile for 0 ≤ t ≤ 5, then by the Mean-Value Theoremthere is at least one point c in (0, 5) wheres (c) = v(c) = [s(5) − s(0)]/(5 − 0) = 4/5 = 0.8 mi/min = 48 mi/h.20. Let T(t) denote the temperature at time with t = 0 denoting 11 AM, then T(0) = 76 andT(12) = 52.(a) By the Mean-Value Theorem there is a value c between 0 and 12 such thatT (c) = [T(12) − T(0)]/(12 − 0) = (52 − 76)/(12) = −2◦F/h.(b) Assume that T(t1) = 88◦F where 0 < t1 < 12, then there is at least one point c in (t1, 12)where T (c) = [T(12)−T(t1)]/(12−t1) = (52−88)/(12−t1) = −36/(12−t1). But 12−t1 < 12so T (c) < −3◦F.21. Let f(t) and g(t) denote the distances from the ﬁrst and second runners to the starting point,and let h(t) = f(t) − g(t). Since they start (at t = 0) and ﬁnish (at t = t1) at the same time,h(0) = h(t1) = 0, so by Rolle’s Theorem there is a time t2 for which h (t2) = 0, i.e. f (t2) = g (t2);so they have the same velocity at time t2.22. f(x) = x − (2 − x) ln(2 − x); solve f(x) = 0 in (0, 1). f(0) = −2 ln 2 < 0; f(1) = 1 > 0. By theIntermediate-Value Theorem (Theorem 2.5.7) there exists x in (0, 1) such that f(x) = 0.23. (a) By the Constant Diﬀerence Theorem f(x) − g(x) = k for some k; since f(x0) = g(x0), k = 0,so f(x) = g(x) for all x.(b) Set f(x) = sin2x + cos2x, g(x) = 1; then f (x) = 2 sin x cos x − 2 cos x sin x = 0 = g (x).Since f(0) = 1 = g(0), f(x) = g(x) for all x.24. (a) By the Constant Diﬀerence Theorem f(x) − g(x) = k for some k; since f(x0) − g(x0) = c,k = c, so f(x) − g(x) = c for all x.(b) Set f(x) = (x − 1)3, g(x) = (x2+ 3)(x − 3). Thenf (x) = 3(x − 1)2, g (x) = (x2+ 3) + 2x(x − 3) = 3x2− 6x + 3 = 3(x2− 2x + 1) = 3(x − 1)2,so f (x) = g (x) and hence f(x) − g(x) = k. Expand f(x) and g(x) to geth(x) = f(x) − g(x) = (x3− 3x2+ 3x − 1) − (x3− 3x2+ 3x − 9) = 8.(c) h(x) = x3− 3x2+ 3x − 1 − (x3− 3x2+ 3x − 9) = 825. By the Constant Diﬀerence Theorem it follows that f(x) = g(x) + c; since g(1) = 0 and f(1) = 2we get c = 2; f(x) = xex− ex+ 2.26. By the Constant Diﬀerence Theorem f(x) = tan−1x + C and2 = f(1) = tan−1(1) + C = π/4 + C, C = 2 − π/4, f(x) = tan−1x + 2 − π/4.27. (a) If x, y belong to I and x < y then for some c in I,f(y) − f(x)y − x= f (c),so |f(x)−f(y)| = |f (c)||x−y| ≤ M|x−y|; if x > y exchange x and y; if x = y the inequalityalso holds.(b) f(x) = sin x, f (x) = cos x, |f (x)| ≤ 1 = M, so |f(x) − f(y)| ≤ |x − y| or| sin x − sin y| ≤ |x − y|.28. (a) If x, y belong to I and x < y then for some c in I,f(y) − f(x)y − x= f (c),so |f(x) − f(y)| = |f (c)||x − y| ≥ M|x − y|; if x > y exchange x and y; ifx = y the inequality also holds.(b) If x and y belong to (−π/2, π/2) and f(x) = tan x, then |f (x)| = sec2x ≥ 1 and| tan x − tan y| ≥ |x − y|(c) y lies in (−π/2, π/2) if and only if −y does; use Part (b) and replace y with −y
• 212 Chapter 529. (a) Let f(x) =√x. By the Mean-Value Theorem there is a number c between x and y such that√y −√xy − x=12√c<12√xfor c in (x, y), thus√y −√x <y − x2√x(b) multiply through and rearrange to get√xy <12(x + y).30. Suppose that f(x) has at least two distinct real solutions r1 and r2 in I. Thenf(r1) = f(r2) = 0 so by Rolle’s Theorem there is at least one number between r1 and r2 wheref (x) = 0, but this contradicts the assumption that f (x) = 0, so f(x) = 0 must have fewer thantwo distinct solutions in I.31. (a) If f(x) = x3+ 4x − 1 then f (x) = 3x2+ 4 is never zero, so by Exercise 30 f has at most onereal root; since f is a cubic polynomial it has at least one real root, so it has exactly one realroot.(b) Let f(x) = ax3+ bx2+ cx + d. If f(x) = 0 has at least two distinct real solutions r1 and r2,then f(r1) = f(r2) = 0 and by Rolle’s Theorem there is at least one number between r1 andr2 where f (x) = 0. But f (x) = 3ax2+ 2bx + c = 0 forx = (−2b ±√4b2 − 12ac)/(6a) = (−b ±√b2 − 3ac)/(3a), which are not real if b2− 3ac < 0so f(x) = 0 must have fewer than two distinct real solutions.32. f (x) =12√x,12√c=√4 −√34 − 3= 2 −√3. But14<12√c<12√3for c in (3, 4) so14< 2 −√3 <12√3, 0.25 < 2 −√3 < 0.29, −1.75 < −√3 < −1.71, 1.71 <√3 < 1.75.33. By the Mean-Value Theorem on the interval [0, x],tan−1x − tan−10x − 0=tan−1xx=11 + c2for c in (0, x), but11 + x2<11 + c2< 1 for c in (0, x) so11 + x2<tan−1xx< 1,x1 + x2< tan−1x < x.34. (a)ddx[f2(x) − g2(x)] = 2f(x)f (x) − 2g(x)g (x) = 2f(x)g(x) − 2g(x)f(x) = 0, so f2− g2isconstant.(b) f (x) = 12 (ex− e−x) = g(x), g (x) = 12 (ex+ e−x) = f(x)35. (a)ddx[f2(x) + g2(x)] = 2f(x)f (x) + 2g(x)g (x) = 2f(x)g(x) + 2g(x)[−f(x)] = 0,so f2(x) + g2(x) is constant.(b) f(x) = sin x and g(x) = cos x36. Let h = f − g, then h is continuous on [a, b], diﬀerentiable on (a, b), and h(a) = f(a) − g(a) = 0,h(b) = f(b) − g(b) = 0. By Rolle’s Theorem there is some c in (a, b) where h (c) = 0. Buth (c) = f (c) − g (c) so f (c) − g (c) = 0, f (c) = g (c).37. yxc]
• Exercise Set 5.8 21338. (a) Suppose f (x) = 0 more than once in (a, b), say at c1 and c2. Then f (c1) = f (c2) = 0 andby using Rolle’s Theorem on f , there is some c between c1 and c2 where f (c) = 0, whichcontradicts the fact that f (x) > 0 so f (x) = 0 at most once in (a, b).(b) If f (x) > 0 for all x in (a, b), then f is concave up on (a, b) and has at most one relativeextremum, which would be a relative minimum, on (a, b).39. (a) similar to the proof of Part (a) with f (c) < 0(b) similar to the proof of Part (a) with f (c) = 040. Let x = x0 be suﬃciently near x0 so that there exists (by the Mean-Value Theorem) a number c(which depends on x) between x and x0, such thatf(x) − f(x0x − x0= f (c).Since c is between x and x0 it follows thatf (x0) = limx→x0f(x) − f(x0)x − x0(by deﬁnition of derivative)= limx→x0f (c) (by the Mean-Value Theorem)= limx→x0f (x) (since lim f (x) exists and c is between x and x0).41. If f is diﬀerentiable at x = 1, then f is continuous there;limx→1+f(x) = limx→1−f(x) = f(1) = 3, a + b = 3; limx→1+f (x) = a andlimx→1−f (x) = 6 so a = 6 and b = 3 − 6 = −3.42. (a) limx→0−f (x) = limx→0−2x = 0 and limx→0+f (x) = limx→0+2x = 0; f (0) does not exist because f isnot continuous at x = 0.(b) limx→0−f (x) = limx→0+f (x) = 0 and f is continuous at x = 0, so f (0) = 0;limx→0−f (x) = limx→0−(2) = 2 and limx→0+f (x) = limx→0+6x = 0, so f (0) does not exist.43. From Section 3.2 a function has a vertical tangent line at a point of its graph if the slopes ofsecant lines through the point approach +∞ or −∞. Suppose f is continuous at x = x0 andlimx→x+0f(x) = +∞. Then a secant line through (x1, f(x1)) and (x0, f(x0)), assuming x1 > x0, willhave slopef(x1) − f(x0)x1 − x0. By the Mean Value Theorem, this quotient is equal to f (c) for somec between x0 and x1. But as x1 approaches x0, c must also approach x0, and it is given thatlimc→x+0f (c) = +∞, so the slope of the secant line approaches +∞. The argument can be alteredappropriately for x1 < x0, and/or for f (c) approaching −∞.EXERCISE SET 5.81. (a) positive, negative, slowing down (b) positive, positive, speeding up(c) negative, positive, slowing down2. (a) positive, slowing down (b) negative, slowing down(c) positive, speeding up
• 214 Chapter 53. (a) left because v = ds/dt < 0 at t0(b) negative because a = d2s/dt2and the curve is concave down at t0(d2s/dt2< 0)(c) speeding up because v and a have the same sign(d) v < 0 and a > 0 at t1 so the particle is slowing down because v and a have opposite signs.4. (a) III(b) I(c) II5.t (s)s (m)6. (a) when s ≥ 0, so 0 < t < 2 and 4 < t ≤ 8 (b) when the slope is zero, at t = 3(c) when s is decreasing, so 0 ≤ t < 37.1 2 3 4 5 6–10–5051015t|v|6ta–15158. (a) v ≈ (30 − 10)/(15 − 10) = 20/5 = 4 m/s(b)tvta2525(1) (2)9. (a) At 60 mi/h the tangent line seems to pass through the points (0, 20) and (16, 100). Thus theacceleration would bev1 − v0t1 − t08860=100 − 2016 − 08860≈ 7.3 ft/s2.(b) The maximum acceleration occurs at maximum slope, so when t = 0.10. (a) At 60 mi/h the tangent line seems to pass through the points (0, 20) and (17.75, 100). Thusthe acceleration would be (recalling that 60 mi/h is the same as 88 ft/s)v1 − v0t1 − t08860=100 − 2017.75 − 08860≈ 6.6 ft/s2.(b) The maximum acceleration occurs at maximum slope, so when t = 0.
• Exercise Set 5.8 21511. (a) t 1 2 3 4 5s 0.71 1.00 0.71 0.00 −0.71v 0.56 0.00 −0.56 −0.79 −0.56a −0.44 −0.62 −0.44 0.00 0.44(b) to the right at t = 1, stopped at t = 2, otherwise to the left(c) speeding up at t = 3; slowing down at t = 1, 5; neither at t = 2, 412. (a) t 1 2 3 4 5s 0.37 2.16 4.03 4.68 4.21v 1.10 2.16 1.34 0 −0.84a 1.84 0 −1.34 −1.17 −0.51(b) to the right at t = 1, 2, 3, stopped at t = 4, to the left at t = 5(c) speeding up at t = 1, 5; slowing down at t = 3; stopped at t = 4, neither at t = 213. (a) v(t) = 3t2− 6t, a(t) = 6t − 6(b) s(1) = −2 ft, v(1) = −3 ft/s, speed = 3 ft/s, a(1) = 0 ft/s2(c) v = 0 at t = 0, 2(d) for t ≥ 0, v(t) changes sign at t = 2, and a(t) changes sign at t = 1; so the particle is speedingup for 0 < t < 1 and 2 < t and is slowing down for 1 < t < 2(e) total distance = |s(2) − s(0)| + |s(5) − s(2)| = | − 4 − 0| + |50 − (−4)| = 58 ft14. (a) v(t) = 4t3− 8t, a(t) = 12t2− 8(b) s(1) = 1 ft, v(1) = −4 ft/s, speed = 4 ft/s, a(1) = 0 ft/s2(c) v = 0 at t = 0,√2(d) for t ≥ 0, v(t) changes sign at t =√2, and a(t) changes sign at t =√6/3. The particle isspeeding up for 0 < t <√6/3 and√2 < t and slowing down for√6/3 < t <√2.(e) total distance = |s(√2) − s(0)| + |s(5) − s(√2)| = |0 − 4| + |529 − 0| = 533 ft15. (a) s(t) = 9 − 9 cos(πt/3), v(t) = 3π sin(πt/3), a(t) = π2cos(πt/3)(b) s(1) = 9/2 ft, v(1) = 3π√3/2 ft/s, speed = 3π√3/2 ft/s, a(1) = π2/2 ft/s2(c) v = 0 at t = 0, 3(d) for 0 < t < 5, v(t) changes sign at t = 3 and a(t) changes sign at t = 3/2, 9/2; so the particleis speeding up for 0 < t < 3/2 and 3 < t < 9/2 and slowing down for 3/2 < t < 3 and9/2 < t < 5(e) total distance = |s(3) − s(0)| + |s(5) − s(3)| = |18 − 0| + |9/2 − 18| = 18 + 27/2 = 63/2 ft16. (a) v(t) =4 − t2(t2 + 4)2, a(t) =2t(t2− 12)(t2 + 4)3(b) s(1) = 1/5 ft, v(1) = 3/25 ft/s, speed = 3/25 ft/s, a(1) = −22/125 ft/s2(c) v = 0 at t = 2(d) a changes sign at t = 2√3, so the particle is speeding up for 2 < t < 2√3 and it is slowingdown for 0 < t < 2 and for 2√3 < t(e) total distance = |s(2) − s(0)| + |s(5) − s(2)| =14− 0 +529−14=1958ft
• 216 Chapter 517. (a) s(t) = (t2+ 8)e−t/3ft, v(t) = −13 t2+ 2t − 83 e−t/3ft/s, a(t) = 19 t2− 43 t + 269 e−t/3ft/s2(b) s(1) = 9e−1/3, v(1) = −e−1/3, a(1) = 53 e−1/3(c) v = 0 for t = 2, 4(d) v changes sign at t = 2, 4 and a changes sign at t = 6 ±√10, so the particle is speeding upfor 2 < t < 6 −√10 and 4 < t < 6 +√10, and slowing down for 0 < t < 2, 6 −√10 < t < 4and t > 6 +√10(e) total distance = |s(2) − s(0)| + |s(4) − s(2)| + |s(4)|= |12e−2/3− 8e−1/3| + |24e−4/3− 12e−2/3| + |33e−5/3− 24e−4/3| ≈ 8.33 ft18. (a) s(t) = 14 t2− ln(t + 1), v(t) =t2+ t − 22(t + 1), a(t) =t2+ 2t + 32(t + 1)2(b) s(1) = 14 − ln 2 ft, v(1) = 0 ft/s, speed = 0 ft/s, a(1) = 34 ft/s2(c) v = 0 for t = 1(d) v changes sign at t = 1 and a does not change sign, so the particle is slowing down for0 < t < 1 and speeding up for t > 1(e) total distance = |s(5) − s(1)| + |s(1) − s(0)| = |25/4 − ln 6 − (1/4 − ln 2)| + |1/4 − ln 2| =23/4 + ln(2/3) ≈ 5.345 ft19. v(t) =5 − t2(t2 + 5)2, a(t) =2t(t2− 15)(t2 + 5)30.2500 20s(t)0.2–0.050 20v(t)0.01–0.150 10a(t)(a) v = 0 at t =√5 (b) s =√5/10 at t =√5(c) a changes sign at t =√15, so the particle is speeding up for√5 < t <√15 and slowing downfor 0 < t <√5 and√15 < t20. v(t) = (1 − t)e−t, a(t) = (t − 2)e−t200 2s(t)1–0.20 2v(t)0.5–20 3a(t)(a) v = 0 at t = 1 (b) s = 1/e at t = 1(c) a changes sign at t = 2, so the particle is speeding up for 1 < t < 2 and slowing down for0 < t < 1 and 2 < t21. s = −4t + 3v = −4a = 030Not speeding up,not slowing downt = 3/4–3t = 3/2 t = 0 s
• Exercise Set 5.8 21722. s = 5t2− 20v = 10ta = 10200Speeding up–20t = 0 s23. s = t3− 9t2+ 24tv = 3(t − 2)(t − 4)a = 6(t − 3)2018160t = 2 (Stopped)t = 0(Stopped) t = 4t = 3sSpeeding upSlowing downSlowing down24. s = t3− 6t2+ 9t + 1v = 3(t − 1)(t − 3)a = 6(t − 2)531t = 1 (Stopped)t = 0(Stopped) t = 3 t = 2sSpeeding upSlowing down25. s = 16te−t2/8v = (−4t2+ 16)e−t2/8a = t(−12 + t2)e−t2/8t = 2√30 10 20Speeding upSlowing downt = +∞t = 0t = 2 (Stopped)s26. s = t + 25/(t + 2);v = (t − 3)(t + 7)/(t + 2)2a = 50/(t + 2)312.5108sSlowing downt = 3t = 0Speeding up27. s =cos t, 0 ≤ t ≤ 2π1, t > 2πv =− sin t, 0 ≤ t ≤ 2π0, t > 2πa =− cos t, 0 ≤ t < 2π0, t > 2π10-1sSpeeding upt = ct = c/2t = 3c/2t = 2ct = 0(Stoppedpermanently)Slowing down28. s =2t(t − 2)2, 0 ≤ t ≤ 313 − 7(t − 4)2, t > 3v =6t2− 16t + 8, 0 ≤ t ≤ 3−14t + 56, t > 3a =12t − 16, 0 ≤ t < 3−14, t > 3 1364/27 60t = 4 (Stopped)t = 2/3 (Stopped)t = 0t = 2(Stopped)t = 3t = 4/3sSpeeding upSlowing down29. (a) v = 10t − 22, speed = |v| = |10t − 22|. d|v|/dt does not exist at t = 2.2 which is the onlycritical point. If t = 1, 2.2, 3 then |v| = 12, 0, 8. The maximum speed is 12 ft/s.(b) the distance from the origin is |s| = |5t2− 22t| = |t(5t − 22)|, but t(5t − 22) < 0 for1 ≤ t ≤ 3 so |s| = −(5t2− 22t) = 22t − 5t2, d|s|/dt = 22 − 10t, thus the only critical point ist = 2.2. d2|s|/dt2< 0 so the particle is farthest from the origin when t = 2.2. Its position iss = 5(2.2)2− 22(2.2) = −24.2.
• 218 Chapter 530. v = −200t(t2 + 12)2, speed = |v| =200t(t2 + 12)2for t ≥ 0.d|v|dt=600(4 − t2)(t2 + 12)3= 0 when t = 2, whichis the only critical point in (0, +∞). By the ﬁrst derivative test there is a relative maximum, andhence an absolute maximum, at t = 2. The maximum speed is 25/16 ft/s to the left.31. s = ln(3t2− 12t + 13)v =6t − 123t2 − 12t + 13a = −6(3t2− 12t + 11)(3t2 − 12t + 13)2(a) a = 0 when t = 2 ±√3/3;s(2 −√3/3) = ln 2; s(2 +√3/3) = ln 2;v(2 −√3/3 = −√3; v(2 +√3) =√3(b) v = 0 when t = 2s(2) = 0; a(2) = 632. s = t3− 6t2+ 1, v = 3t2− 12t, a = 6t − 12.(a) a = 0 when t = 2; s = −15, v = −12.(b) v = 0 when 3t2− 12t = 3t(t − 4) = 0, t = 0 or t = 4. If t = 0, then s = 1 and a = −12; ift = 4, then s = −31 and a = 12.33. (a) 1.500 5(b) v =2t√2t2 + 1, limt→+∞v =2√2=√234. (a) a =dvdt=dvdsdsdt= vdvdsbecause v =dsdt(b) v =32√3t + 7=32s;dvds= −32s2; a = −94s3= −9/50035. (a) s1 = s2 if they collide, so 12 t2− t + 3 = −14 t2+ t + 1, 34 t2− 2t + 2 = 0 which has no realsolution.(b) Find the minimum value of D = |s1 − s2| = 34 t2− 2t + 2 . From Part (a), 34 t2− 2t + 2is never zero, and for t = 0 it is positive, hence it is always positive, so D = 34 t2− 2t + 2.dDdt= 32 t − 2 = 0 when t = 43 .d2Ddt2> 0 so D is minimum when t = 43 , D = 23 .(c) v1 = t − 1, v2 = −12t + 1. v1 < 0 if 0 ≤ t < 1, v1 > 0 if t > 1; v2 < 0 if t > 2, v2 > 0 if0 ≤ t < 2. They are moving in opposite directions during the intervals 0 ≤ t < 1 and t > 2.36. (a) sA − sB = 20 − 0 = 20 ft(b) sA = sB, 15t2+ 10t + 20 = 5t2+ 40t, 10t2− 30t + 20 = 0, (t − 2)(t − 1) = 0, t = 1 or t = 2 s.(c) vA = vB, 30t+10 = 10t+40, 20t = 30, t = 3/2 s. When t = 3/2, sA = 275/4 and sB = 285/4so car B is ahead of car A.37. r(t) = v2(t), r (t) = 2v(t)v (t)/[2 v(t)] = v(t)a(t)/|v(t)| so r (t) > 0 (speed is increasing) if vand a have the same sign, and r (t) < 0 (speed is decreasing) if v and a have opposite signs.If v(t) > 0 then r(t) = v(t) and r (t) = a(t), so if a(t) > 0 then the particle is speeding up and aand v have the same sign; if a(t) < 0, then the particle is slowing down, and a and v have oppositesigns.
• Review Exercises 219If v(t) < 0 then r(t) = −v(t), r (t) = −a(t), and if a(t) > 0 then the particle is speeding up and aand v have opposite signs; if a(t) < 0 then the particle is slowing down and a and v have the samesign.REVIEW EXERCISES3. f (x) = 2x − 5f (x) = 2(a) [5/2, +∞) (b) (−∞, 5/2](c) (−∞, +∞) (d) none(e) none4. f (x) = 4x(x2− 4)f (x) = 12(x2− 4/3)(a) [−2, 0], [2, +∞) (b) (−∞, −2], [0, 2](c) (−∞, −2/√3), (2/√3, +∞) (d) (−2/√3, 2/√3)(e) −2/√3, 2/√35. f (x) =4x(x2 + 2)2f (x) = −43x2− 2(x2 + 2)3(a) [0, +∞) (b) (−∞, 0](c) (− 2/3, 2/3) (d) (−∞, − 2/3), ( 2/3, +∞)(e) − 2/3, 2/36. f (x) = 13 (x + 2)−2/3f (x) = −29 (x + 2)−5/3(a) (−∞, +∞) (b) none(c) (−∞, −2) (d) (−2, +∞)(e) −27. f (x) =4(x + 1)3x2/3f (x) =4(x − 2)9x5/3(a) [−1, +∞) (b) (−∞, −1](c) (−∞, 0), (2, +∞) (d) (0, 2)(e) 0, 28. f (x) =4(x − 1/4)3x2/3f (x) =4(x + 1/2)9x5/3(a) [1/4, +∞) (b) (−∞, 1/4](c) (−∞, −1/2), (0, +∞) (d) (−1/2, 0)(e) −1/2, 09. f (x) = −2xex2f (x) =2(2x2− 1)ex2(a) (−∞, 0] (b) [0, +∞)(c) (−∞, −√2/2), (√2/2, +∞) (d) (−√2/2,√2/2)(e) −√2/2,√2/210. f (x) =2x1 + x4f (x) = −2(−1 + 3x4)(1 + x4)2(a) [0, +∞) (b) (−∞, 0](c)(−1/31/4, 1/31/4) (d)(−∞, −1/31/4), (1/31/4, +∞)(e)−1/31/4, 1/31/4
• 220 Chapter 511. f (x) = − sin xf (x) = − cos x(a) [π, 2π] (b) [0, π](c) (π/2, 3π/2) (d) (0, π/2), (3π/2, 2π)(e) π/2, 3π/21–10 o12. f (x) = sec2xf (x) = 2 sec2x tan x(a) (−π/2, π/2) (b) none(c) (0, π/2) (d) (−π/2, 0)(e) 010–10^ 613. f (x) = cos 2xf (x) = −2 sin 2x(a) [0, π/4], [3π/4, π] (b) [π/4, 3π/4](c) (π/2, π) (d) (0, π/2)(e) π/20.5–0.50 p14. f (x) = −2 cos x sin x − 2 cos x = −2 cos x(1 + sin x)f (x) = 2 sin x (sin x + 1) − 2 cos2x = 2 sin x(sin x + 1) − 2 + 2 sin2x = 4(1 + sin x)(sin x − 1/2)Note: 1 + sin x ≥ 0(a) [π/2, 3π/2] (b) [0, π/2], [3π/2, 2π](c) (π/6, 5π/6) (d) (0, π/6), (5π/6, 2π)(e) π/6, 5π/62–20 o15. (a)24xy (b)24xy (c)24xy16. (a) p(x) = x3− x (b) p(x) = x4− x2(c) p(x) = x5− x4− x3+ x2(d) p(x) = x5− x317. f (x) = 2ax + b; f (x) > 0 or f (x) < 0 on [0, +∞) if f (x) = 0 has no positive solution, so thepolynomial is always increasing or always decreasing on [0, +∞) provided −b/2a ≤ 0.
• Review Exercises 22118. f (x) = 3ax2+ 2bx + c; f (x) > 0 or f (x) < 0 on (−∞, +∞) if f (x) = 0 has no real solutions sofrom the quadratic formula (2b)2− 4(3a)c < 0, 4b2− 12ac < 0, b2− 3ac < 0. If b2− 3ac = 0, thenf (x) = 0 has only one real solution at, say, x = c so f is always increasing or always decreasingon both (−∞, c] and [c, +∞), and hence on (−∞, +∞) because f is continuous everywhere. Thusf is always increasing or decreasing if b2− 3ac ≤ 0.19. The maximum increase in yseems to occur near x = −1, y = 1/4.–2 –1 1 20.250.5xy20. y =ax1 + ax+ky =axln a(1 + ax+k)2y = −ax(ln a)2(ax+k− 1)(1 + ax+k)3y = 0 when x = −k and y changes sign there22. (a) False; an example is y =x33−x22on [−2, 2]; x = 0 is a relative maximum and x = 1 is arelative minimum, but y = 0 is not the largest value of y on the interval, nor is y = −16thesmallest.(b) true(c) False; for example y = x3on (−1, 1) which has a critical number but no relative extrema24. (a) f (x) = 3x2+ 6x − 9 = 3(x + 3)(x − 1), f (x) = 0 when x = −3, 1 (stationary points).(b) f (x) = 4x(x2− 3), f (x) = 0 when x = 0, ±√3 (stationary points).25. (a) f (x) = (2 − x2)/(x2+ 2)2, f (x) = 0 when x = ±√2 (stationary points).(b) f (x) = 8x/(x2+ 1)2, f (x) = 0 when x = 0 (stationary point).26. (a) f (x) =4(x + 1)3x2/3, f (x) = 0 when x = −1 (stationary point), f (x) does not exist whenx = 0.(b) f (x) =4(x − 3/2)3x2/3, f (x) = 0 when x = 3/2 (stationary point), f (x) does not exist whenx = 0.27. (a) f (x) =7(x − 7)(x − 1)3x2/3; critical numbers at x = 0, 1, 7;neither at x = 0, relative maximum at x = 1, relative minimum at x = 7 (First DerivativeTest)(b) f (x) = 2 cos x(1 + 2 sin x); critical numbers at x = π/2, 3π/2, 7π/6, 11π/6;relative maximum at x = π/2, 3π/2, relative minimum at x = 7π/6, 11π/6(c) f (x) = 3 −3√x − 12; critical numbers at x = 5; relative maximum at x = 528. (a) f (x) =x − 918x3/2, f (x) =27 − x36x5/2; critical number at x = 9;f (9) > 0, relative minimum at x = 9
• 222 Chapter 5(b) f (x) = 2x3− 4x2, f (x) = 2x3+ 8x3;critical number at x = 41/3, f (41/3) > 0, relative minimum at x = 41/3(c) f (x) = sin x(2 cos x + 1), f (x) = 2 cos2x − 2 sin2x + cos x; critical numbers atx = 2π/3, π, 4π/3; f (2π/3) < 0, relative maximum at x = 2π/3; f (π) > 0, relativeminimum at x = π; f (4π/3) < 0, relative maximum at x = 4π/329. limx→−∞f(x) = +∞, limx→+∞f(x) = +∞f (x) = x(4x2− 9x + 6), f (x) = 6(2x − 1)(x − 1)relative minimum at x = 0,points of inﬂection when x = 1/2, 1,no asymptotesyx12341 2(0,1))((1,2)12 ,231630. limx→−∞f(x) = −∞, limx→+∞f(x) = +∞f(x) = x3(x − 2)2, f (x) = x2(5x − 6)(x − 2),f (x) = 4x(5x2− 12x + 6)critical numbers at x = 0,8 ± 2√315relative maximum at x =8 − 2√315≈ −0.63relative minimum at x =8 + 2√315≈ 3.83points of inﬂection at x = 0,6 ±√665≈ 0, −0.42, 2.82no asymptotesyx–200–100–1 1 2 3 4(0,0)(–0.42, 0.16)(2.82, –165.00)(3.83, –261.31)(–0.63, 0.27)31. limx→±∞f(x) doesn’t existf (x) = 2x sec2(x2+ 1),f (x) = 2 sec2(x2+ 1) 1 + 4x2tan(x2+ 1)critical number at x = 0; relative minimum at x = 0point of inﬂection when 1 + 4x2tan(x2+ 1) = 0vertical asymptotes at x = ± π(n + 12 ) − 1, n = 0, 1, 2, . . .yx–4–224–2 –1 1 232. limx→−∞f(x) = −∞, limx→+∞f(x) = +∞f (x) = 1 + sin x, f (x) = cos xcritical numbers at x = 2nπ + π/2, n = 0, ±1, ±2, . . .,no extrema because f ≥ 0 and by Exercise 59 of Section 5.1,f is increasing on (−∞, +∞)inﬂections points at x = nπ + π/2, n = 0, ±1, ±2, . . .no asymptotesyx–6–4–224-c c
• Review Exercises 22333. f (x) = 2x(x + 5)(x2 + 2x + 5)2, f (x) = −22x3+ 15x2− 25(x2 + 2x + 5)3critical numbers at x = −5, 0;relative maximum at x = −5,relative minimum at x = 0points of inﬂection at x = −7.26, −1.44, 1.20horizontal asymptote y = 1 as x → ±∞yx0.20.40.60.81–20 –10 10 2034. f (x) = 33x2− 25x4, f (x) = −63x2− 50x5critical numbers at x = ±5√3/3;relative maximum at x = −5√3/3,relative minimum at x = +5√3/3inﬂection points at x = ±5 2/3horizontal asymptote of y = 0 as x → ±∞,vertical asymptote x = 0yx–55–435. limx→−∞f(x) = +∞, limx→+∞f(x) = −∞f (x) =x−2xifx ≤ 0x > 0critical number at x = 0, no extremainﬂection point at x = 0 (f changes concavity)no asymptotesxy–212–2 136. f (x) =5 − 3x3(1 + x)1/3(3 − x)2/3,f (x) =−329(1 + x)4/3(3 − x)5/3critical number at x = 5/3;relative maximum at x = 5/3cusp at x = −1;point of inﬂection at x = 3oblique asymptote y = −x as x → ±∞yx–3–124–4 –2 237. f (x) = 3x2+ 5; no relative extrema because there are no critical numbers.38. f (x) = 4x(x2− 1); critical numbers x = 0, 1, −1f (x) = 12x2− 4; f (0) < 0, f (1) > 0, f (−1) > 0relative minimum of 6 at x = 1, −1, relative maximum of 7 at x = 039. f (x) = 45 x−1/5; critical number x = 0; relative minimum of 0 at x = 0 (ﬁrst derivative test)40. f (x) = 2 + 23 x−1/3; critical numbers x = 0, −1/27relative minimum of 0 at x = 0, relative maximum of 1/27 at x = −1/27
• 224 Chapter 541. f (x) = 2x/(x2+ 1)2; critical number x = 0; relative minimum of 0 at x = 042. f (x) = 2/(x + 2)2; no critical numbers (x = −2 is not in the domain of f) no relative extrema43. f (x) = 2x/(1 + x2); critical point at x = 0; relative minimum of 0 at x = 0 (ﬁrst derivative test)44. f (x) = x(2 + x)ex; critical points at x = 0, −2; relative minimum of 0 at x = 0 and relativemaximum of 4/e2at x = −2 (ﬁrst derivative test)45. (a) 40–40–5 5(b) f (x) = x2−1400, f (x) = 2xcritical points at x = ±120;relative maximum at x = −120,relative minimum at x =120(c) The ﬁner details can be seen whengraphing over a much smaller x-window.0.0001–0.0001–0.1 0.146. (a) 200–200–5 5(b) critical points at x = ±√2, 32 , 2;relative maximum at x = −√2,relative minimum at x =√2,relative maximum at x = 32 ,relative minimum at x = 2(c) 10–4–2.2 3.5–2.909–2.9121.3 1.647. (a) 6–6–5 5(b) Divide y = x2+ 1 into y = x3− 8to get the asymptote ax + b = x
• Review Exercises 22548. cos x − (sin y)dydx= 2dydx;dydx= 0 when cos x = 0. Use the ﬁrst derivative test:dydx=cos x2 + sin yand 2 + sin y > 0, so critical points when cos x = 0, relative maxima when x = 2nπ + π/2, relativeminima when x = 2nπ − π/2, n = 0, ±1, ±2, . . .49. f(x) =(2x − 1)(x2+ x − 7)(2x − 1)(3x2 + x − 1)=x2+ x − 73x2 + x − 1, x = 1/2horizontal asymptote: y = 1/3,vertical asymptotes: x = (−1 ±√13)/6yx–55–4 2 450. (a) f(x) =(x − 2)(x2+ x + 1)(x2− 2)(x − 2)(x2 − 2)2(x2 + 1)=x2+ x + 1(x2 − 2)(x2 + 1)(b)–1 15xyx = ¤x = –¤53. (a) If f has an absolute extremum at a point of (a, b) then it must, by Theorem 5.4.3, be at acritical point of f; since f is diﬀerentiable on (a, b) the critical point is a stationary point.(b) It could occur at a critical point which is not a stationary point: for example, f(x) = |x| on[−1, 1] has an absolute minimum at x = 0 but is not diﬀerentiable there.54. (a) f (x) = −1/x2= 0, no critical points; by inspection M = −1/2 at x = −2; m = −1 atx = −1(b) f (x) = 3x2− 4x3= 0 at x = 0, 3/4; f(−1) = −2, f(0) = 0, f(3/4) = 27/256,f(3/2) = −27/16, so m = −2 at x = −1, M = 27/256 at x = 3/4(c) f (x) = 1 − sec2x, f (x) = 0 for x in (−π/4, π/4) when x = 0; f(−π/4) = 1 − π/4, f(0) = 0,f(π/4) = π/4 − 1 so the maximum value is 1 − π/4 at x = −π/4 and the minimum value isπ/4 − 1 at x = π/4.(d) critical point at x = 2; m = −3 at x = 3, M = 0 at x = 255. (a) f (x) = 2x − 3; critical point x = 3/2. Minimum value f(3/2) = −13/4, no maximum.(b) No maximum or minimum because limx→+∞f(x) = +∞ and limx→−∞f(x) = −∞.(c) limx→0+f(x) = limx→+∞f(x) = +∞ and f (x) =ex(x − 2)x3, stationary point at x = 2; byTheorem 5.4.4 f(x) has an absolute minimum at x = 2, and m = e2/4.(d) f (x) = (1 + ln x)xx, critical point at x = 1/e; limx→0+f(x) = limx→0+ex ln x= 1,limx→+∞f(x) = +∞; no absolute maximum, absolute minimum m = e−1/eat x = 1/e56. (a) f (x) = 10x3(x − 2), critical points at x = 0, 2; limx→3−f(x) = 88, so f(x) has no maximum;m = −9 at x = 2(b) limx→2−f(x) = +∞ so no maximum; f (x) = 1/(x−2)2, so f (x) is never zero, thus no minimum
• 226 Chapter 5(c) f (x) = 23 − x2(x2 + 3)2, critical point at x =√3. Since limx→0+f(x) = 0, f(x) has no minimum,and M =√3/3 at x =√3.(d) f (x) =x(7x − 12)3(x − 2)2/3, critical points at x = 12/7, 2; m = f(12/7) =14449−271/3≈ −1.9356at x = 12/7, M = 9 at x = 357. (a) (x2− 1)2can never be less than zero because it is thesquare of x2− 1; the minimum value is 0 for x = ±1,no maximum because limx→+∞f(x) = +∞.100–2 2(b) f (x) = (1 − x2)/(x2+ 1)2; critical point x = 1. Maxi-mum value f(1) = 1/2, minimum value 0 because f(x)is never less than zero on [0, +∞) and f(0) = 0.0.500 20(c) f (x) = 2 sec x tan x − sec2x = (2 sin x − 1)/ cos2x,f (x) = 0 for x in (0, π/4) when x = π/6; f(0) = 2,f(π/6) =√3, f(π/4) = 2√2−1 so the maximum valueis 2 at x = 0 and the minimum value is√3 at x = π/6.21.50 3(d) f (x) = 1/2 + 2x/(x2+ 1),f (x) = 0 on [−4, 0] for x = −2 ±√3if x = −2 −√3, −2 +√3 thenf(x) = −1 −√3/2 + ln 4 + ln(2 +√3) ≈ 0.84,−1 +√3/2 + ln 4 + ln(2 −√3) ≈ −0.06,absolute maximum at x = −2 −√3,absolute minimum at x = −2 +√31–0.5–4 058. Let f(x) = sin−1x − x for 0 ≤ x ≤ 1. f(0) = 0 and f (x) =1√1 − x2− 1 =1 −√1 − x2√1 − x2. Notethat√1 − x2 ≤ 1, so f (x) ≥ 0. Thus we know that f is increasing. Since f(0) = 0, it follows thatf(x) ≥ 0 for 0 ≤ x ≤ 1.
• Review Exercises 22759. (a) 2.1–0.5–10 10(b) minimum: (−2.111985, −0.355116)maximum: (0.372591, 2.012931).60. Let k be the amount of light admitted per unit area of clear glass. The total amount of lightadmitted by the entire window isT = k · (area of clear glass) +12k · (area of blue glass) = 2krh +14πkr2.But P = 2h + 2r + πr which gives 2h = P − 2r − πr soT = kr(P − 2r − πr) +14πkr2= k Pr − 2 + π −π4r2= k Pr −8 + 3π4r2for 0 < r <P2 + π,dTdr= k P −8 + 3π2r ,dTdr= 0 when r =2P8 + 3π.This is the only critical point and d2T/dr2< 0 there so the most light is admitted whenr = 2P/(8 + 3π) ft.61. If one corner of the rectangle is at (x, y) with x > 0, y > 0, then A = 4xy, y = 3 1 − (x/4)2,A = 12x 1 − (x/4)2 = 3x√16 − x2,dAdx= 68 − x2√16 − x2, critical point at x = 2√2. Since A = 0when x = 0, 4 and A > 0 otherwise, there is an absolute maximum A = 24 at x = 2√2.62. (a) y x–2–1.5–1–0.50.2 0.6 1(b) The distance between the boat and the origin is x2 + y2, where y = (x10/3− 1)/(2x2/3).The minimum distance is 0.8247 mi when x = 0.6598 mi. The boat gets swept downstream.63. V = x(12 − 2x)2for 0 ≤ x ≤ 6;dV/dx = 12(x − 2)(x − 6), dV/dx = 0when x = 2 for 0 < x < 6. If x = 0, 2, 6then V = 0, 128, 0 so the volume is largestwhen x = 2 in.xxxxxxxx121212 – 2x12 – 2x
• 228 Chapter 565. x = −2.11491, 0.25410, 1.86081 66. x = 2.356194567. At the point of intersection, x3= 0.5x − 1,x3− 0.5x + 1 = 0. Let f(x) = x3− 0.5x + 1. Bygraphing y = x3and y = 0.5x − 1 it is evident thatthere is only one point of intersection and it occursin the interval [−2, −1]; note that f(−2) < 0 andf(−1) > 0. f (x) = 3x2− 0.5 soxn+1 = xn −x3n − 0.5x + 13x2n − 0.5;x1 = −1, x2 = −1.2,x3 = −1.166492147, . . . ,x5 = x6 = −1.1653730432–2–2 268. Solve φ − 0.0167 sin φ = 2π(90)/365 to get φ = 1.565978 sor = 150 × 106(1 − 0.0167 cos φ) = 149.988 × 106km.69. Solve φ − 0.0934 sin φ = 2π(1)/1.88 to get φ = 3.325078 sor = 228 × 106(1 − 0.0934 cos φ) = 248.938 × 106km.70. Yes; by the Mean-Value Theorem there is a point c in (a, b) such that f (c) =f(b) − f(a)b − a= 0.71. (a) yes; f (0) = 0(b) no, f is not diﬀerentiable on (−1, 1)(c) yes, f ( π/2) = 072. (a) no, f is not diﬀerentiable on (−2, 2)(b) yes,f(3) − f(2)3 − 2= −1 = f (1 +√2)(c) limx→1−f(x) = 2, limx→1+f(x) = 2 so f is continuous on [0, 2]; limx→1−f (x) = limx→1−−2x = −2 andlimx→1+f (x) = limx→1+(−2/x2) = −2, so f is diﬀerentiable on (0, 2);andf(2) − f(0)2 − 0= −1 = f (√2).73. f(x) = x6− 2x2+ x satisﬁes f(0) = f(1) = 0, so by Rolle’s Theorem f (c) = 0 for some c in (0, 1).74. If f (x) = g (x), then f(x) = g(x) + k. Let x = 1,f(1) = g(1) + k = (1)3− 4(1) + 6 + k = 3 + k = 2, so k = −1. f(x) = x3− 4x + 5.75. (a) If a = k, a constant, then v = kt + b where b is constant;so the velocity changes sign at t = −b/k.vtb– b/ k
• Review Exercises 229(b) Consider the equation s = 5 − t3/6, v = −t2/2, a = −t.Then for t > 0, a is decreasing and av > 0, so theparticle is speeding up.vt76. s(t) = t/(2t2+ 8), v(t) = (4 − t2)/2(t2+ 4)2, a(t) = t(t2− 12)/(t2+ 4)3(a) 0.20–0.0500.15–0.1000.2500 202020s(t) v(t) a(t)(b) v changes sign at t = 2(c) s = 1/8, v = 0, a = −1/32(d) a changes sign at t = 2√3, so the particle is speeding up for 2 < t < 2√3, and it is slowingdown for 0 < t < 2 and 2√3 < t(e) v(0) = 1/5, limt→+∞v(t) = 0, v(t) has one t-intercept at t =√5 and v(t) has one criticalpoint at t =√15. Consequently the maximum velocity occurs when t = 0 and the minimumvelocity occurs when t =√15.77. (a) v = −2t(t4+ 2t2− 1)(t4 + 1)2, a = 23t8+ 10t6− 12t4− 6t2+ 1(t4 + 1)3(b) st12vt–0.20.22at12(c) It is farthest from the origin at approximately t = 0.64 (when v = 0) and s = 1.2(d) Find t so that the velocity v = ds/dt > 0. The particle is moving in the positive directionfor 0 ≤ t ≤ 0.64 s.(e) It is speeding up when a, v > 0 or a, v < 0, so for 0 ≤ t < 0.36 and 0.64 < t < 1.1, otherwiseit is slowing down.(f) Find the maximum value of |v| to obtain: maximum speed = 1.05 m/s when t = 1.10 s.78. No; speeding up means the velocity and acceleration have the same sign, i.e. av > 0; the velocity isincreasing when the acceleration is positive, i.e. a > 0. These are not the same thing. An exampleis s = t − t2at t = 1, where v = −1 and a = −2, so av > 0 but a < 0.
• 230CHAPTER 6IntegrationEXERCISE SET 6.11. Endpoints 0,1n,2n, . . . ,n − 1n, 1; using right endpoints,An =1n+2n+ · · · +n − 1n+ 11nn 2 5 10 50 100An 0.853553 0.749739 0.710509 0.676095 0.6714632. Endpoints 0,1n,2n, . . . ,n − 1n, 1; using right endpoints,An =nn + 1+nn + 2+nn + 3+ · · · +n2n − 1+121nn 2 5 10 50 100An 0.583333 0.645635 0.668771 0.688172 0.6906533. Endpoints 0,πn,2πn, . . . ,(n − 1)πn, π; using right endpoints,An = [sin(π/n) + sin(2π/n) + · · · + sin(π(n − 1)/n) + sin π]πnn 2 5 10 50 100An 1.57080 1.93376 1.98352 1.99935 1.999844. Endpoints 0,π2n,2π2n, . . . ,(n − 1)π2n,π2; using right endpoints,An = [cos(π/2n) + cos(2π/2n) + · · · + cos((n − 1)π/2n) + cos(π/2)]π2nn 2 5 10 50 100An 0.555359 0.834683 0.919400 0.984204 0.9921205. Endpoints 1,n + 1n,n + 2n, . . . ,2n − 1n, 2; using right endpoints,An =nn + 1+nn + 2+ · · · +n2n − 1+121nn 2 5 10 50 100An 0.583333 0.645635 0.668771 0.688172 0.6906536. Endpoints −π2, −π2+πn, −π2+2πn, . . . , −π2+(n − 1)πn,π2; using right endpoints,An = cos −π2+πn+ cos −π2+2πn+ · · · + cos −π2+(n − 1)πn+ cosπ2πnn 2 5 10 50 100An 1.57080 1.93376 1.98352 1.99936 1.99985
• Exercise Set 6.1 2317. Endpoints 0,1n,2n, . . . ,n − 1n, 1; using right endpoints,An = 1 −1n2+ 1 −2n2+ · · · + 1 −n − 1n2+ 0 1nn 2 5 10 50 100An 0.433013 0.659262 0.726130 0.774567 0.7801068. Endpoints −1, −1 +2n, −1 +4n, . . . , −1 +2(n − 1)n, 1; using right endpoints,An = 1 −n − 2n2+ 1 −n − 4n2+ · · · + 1 −n − 2n2+ 0 2nn 2 5 10 50 100An 1 1.423837 1.518524 1.566097 1.5691369. Endpoints −1, −1 +2n, −1 +4n, . . . , 1 −2n, 1; using right endpoints,An = e−1+ 2n + e−1+ 4n + e−1+ 6n + . . . + e1− 2n + e1 2nn 2 5 10 50 100An 3.718281 2.851738 2.59327 2.39772 2.3504010. Endpoints 1, 1 +1n, 1 +2n, . . . , 2 −1n, 2; using right endpoints,An = ln 1 +1n+ ln 1 +2n+ . . . + ln 2 −1n+ ln 21nn 2 5 10 50 100An 0.549 0.454 0.421 0.393 0.39011. Endpoints 0,1n,2n, . . . ,n − 1n, 1; using right endpoints,An = sin−1 1n+ sin−1 2n+ . . . + sin−1 n − 1n+ sin−1(1)1nn 2 5 10 50 100An 1.04729 0.75089 0.65781 0.58730 0.5789412. Endpoints 0,1n,2n, . . . ,n − 1n, 1; using right endpoints,An = tan−1 1n+ tan−1 2n+ . . . + tan−1 n − 1n+ tan−1(1)1nn 2 5 10 50 100An 0.62452 0.51569 0.47768 0.44666 0.4427413. 3(x − 1) 14. 5(x − 2) 15. x(x + 2) 16.32(x − 1)2
• 232 Chapter 617. (x + 3)(x − 1) 18.32x(x − 2)19. The area in Exercise 17 is always 3 less than the area in Exercise 15. The regions are identicalexcept that the area in Exercise 15 has the extra trapezoid with vertices at (0, 0), (1, 0), (0, 2), (1, 4)(with area 3).20. (a) The region in question is a trapezoid, and the area of a trapezoid is12(h1 + h2)w.(b) From Part (a), A (x) =12[f(a) + f(x)] + (x − a)12f (x)=12[f(a) + f(x)] + (x − a)12f(x) − f(a)x − a= f(x)21. A(6) represents the area between x = 0 and x = 6; A(3) represents the area between x = 0and x = 3; their diﬀerence A(6) − A(3) represents the area between x = 3 and x = 6, andA(6) − A(3) =13(63− 33) = 63.22. A(9) = 93/3, A(−3) = (−3)3/3, and the area between x = −3 and x = 9 is given by A(9)−A(−3) =(93− (−3)3)/3 = 252.23. f(x) = A (x) = 2x; 0 = A(a) = a2− 4, so take a = 2 (or a = −2), A(x) = x2− 4,A (x) = 2x = f(x), so a = ±2, f(x) = 2x24. f(x) = A (x) = 2x − 1, 0 = A(a) = a2− a, so take a = 0 (or a = 1).25. B is also the area between the graph of f(x) =√x and the interval [0, 1] on the y−axis, so A + Bis the area of the square.26. If the plane is rotated about the line y = x then A becomes B and vice versa.EXERCISE SET 6.21. (a)x√1 + x2dx = 1 + x2 + C (b) (x + 1)exdx = xex+ C2. (a)ddx(sin x − x cos x + C) = cos x − cos x + x sin x = x sin x(b)ddxx√1 − x2+ C =√1 − x2 + x2/√1 − x21 − x2=1(1 − x2)3/25.ddxx3 + 5 =3x22√x3 + 5so3x22√x3 + 5dx = x3 + 5 + C6.ddxxx2 + 3=3 − x2(x2 + 3)2so3 − x2(x2 + 3)2dx =xx2 + 3+ C7.ddxsin 2√x =cos (2√x)√xsocos (2√x)√xdx = sin 2√x + C8.ddx[sin x − x cos x] = x sin x so x sin x dx = sin x − x cos x + C
• Exercise Set 6.2 2339. (a) x9/9 + C (b)712x12/7+ C (c)29x9/2+ C10. (a)35x5/3+ C (b) −15x−5+ C = −15x5+ C (c) 8x1/8+ C11. 5x +23x5dx = 5x dx +231x5dx =52x2+23−141x4C =52x2−16x4+ C12. x−1/2− 3x7/5+19dx = x−1/2dx − 3 x7/5dx +19dx = 2x1/2− 3512x12/5+19x + C13. x−3− 3x1/4+ 8x2dx = x−3dx − 3 x1/4dx + 8 x2dx = −12x−2−125x5/4+83x3+ C14.10y3/4− 3√y +4√ydy = 101y3/4dy − 3√y dy + 41√ydy= 10(4)y1/4−34y4/3+ 4(2)y1/2+ C = 40y1/4−34y4/3+ 8√y + C15. (x + x4)dx = x2/2 + x5/5 + C 16. (4 + 4y2+ y4)dy = 4y +43y3+15y5+ C17. x1/3(4 − 4x + x2)dx = (4x1/3− 4x4/3+ x7/3)dx = 3x4/3−127x7/3+310x10/3+ C18. (2 − x + 2x2− x3)dx = 2x −12x2+23x3−14x4+ C19. (x + 2x−2− x−4)dx = x2/2 − 2/x + 1/(3x3) + C20. (t−3− 2)dt = −12t−2− 2t + C 21.2x+ 3exdx = 2 ln |x| + 3ex+ C22.12t−1−√2etdt =12ln |t| −√2et+ C23. [3 sin x − 2 sec2x] dx = −3 cos x − 2 tan x + C24. [csc2t − sec t tan t] dt = − cot t − sec t + C25. (sec2x + sec x tan x)dx = tan x + sec x + C26. csc x(sin x + cot x) dx = (1 + csc x cot x) dx = x − csc x + C27.sec θcos θdθ = sec2θ dθ = tan θ + C 28. sin y dy = − cos y + C29. sec x tan x dx = sec x + C 30. (φ + 2 csc2φ)dφ = φ2/2 − 2 cot φ + C
• 234 Chapter 631. (1 + sin θ)dθ = θ − cos θ + C 32.12sec2x +12dx =12tan x +12x + C33.12√1 − x2−31 + x2dx =12sin−1x − 3 tan−1x + C34.4x√x2 − 1+1 + x + x31 + x2dx = 4 sec−1x+ x +1x2 + 1dx = 4 sec−1x+12x2+tan−1x+C35.1 − sin x1 − sin2xdx =1 − sin xcos2 xdx = sec2x − sec x tan x dx = tan x − sec x + C36.11 + cos 2xdx =12 cos2 xdx =12sec2x dx =12tan x + C37. yx–55c/4 c/238.–42yx1 239. f (x) = m = − sin x so f(x) = (− sin x)dx = cos x + C; f(0) = 2 = 1 + Cso C = 1, f(x) = cos x + 140. f (x) = m = (x + 1)2, so f(x) = (x + 1)2dx =13(x + 1)3+ C;f(−2) = 8 =13(−2 + 1)3+ C = −13+ C, = 8 +13=253, f(x) =13(x + 1)3+25341. (a) y(x) = x1/3dx =34x4/3+ C, y(1) =34+ C = 2, C =54; y(x) =34x4/3+54(b) y(t) = (sin t + 1) dt = − cos t + t + C, yπ3= −12+π3+ C = 1/2, C = 1 −π3;y(t) = − cos t + t + 1 −π3(c) y(x) = (x1/2+ x−1/2)dx =23x3/2+ 2x1/2+ C, y(1) = 0 =83+ C, C = −83,y(x) =23x3/2+ 2x1/2−8342. (a) y(x) =18x−3dx = −116x−2+ C, y(1) = 0 = −116+ C, C =116; y(x) = −116x−2+116(b) y(t) = (sec2t − sin t) dt = tan t + cos t + C, y(π4) = 1 = 1 +√22+ C, C = −√22;y(t) = tan t + cos t −√22(c) y(x) = x7/2dx =29x9/2+ C, y(0) = 0 = C, C = 0; y(x) =29x9/2
• Exercise Set 6.2 23543. (a) y = 4exdx = 4ex+ C, 1 = y(0) = 4 + C, C = −3, y = 4ex− 3(b) y(t) = t−1dt = ln |t| + C, y(−1) = C = 5, C = 5; y(t) = ln |t| + 544. (a) y =3√1 − t2dt = 3 sin−1t + C, y√32= 0 = π + C, C = −π, y = 3 sin−1t − π(b)dydx= 1 −2x2 + 1, y = 1 −2x2 + 1dx = x − 2 tan−1x + C,y(1) =π2= 1 − 2π4+ C, C = π − 1, y = x − 2 tan−1x + π − 145. f (x) =23x3/2+ C1; f(x) =415x5/2+ C1x + C246. f (x) = x2/2 + sin x + C1, use f (0) = 2 to get C1 = 2 so f (x) = x2/2 + sin x + 2,f(x) = x3/6 − cos x + 2x + C2, use f(0) = 1 to get C2 = 2 so f(x) = x3/6 − cos x + 2x + 247. dy/dx = 2x + 1, y = (2x + 1)dx = x2+ x + C; y = 0 when x = −3so (−3)2+ (−3) + C = 0, C = −6 thus y = x2+ x − 648. dy/dx = x2, y = x2dx = x3/3 + C; y = 2 when x = −1 so (−1)3/3 + C = 2, C = 7/3thus y = x3/3 + 7/349. dy/dx = 6xdx = 3x2+ C1. The slope of the tangent line is −3 so dy/dx = −3 when x = 1.Thus 3(1)2+ C1 = −3, C1 = −6 so dy/dx = 3x2− 6, y = (3x2− 6)dx = x3− 6x + C2. If x = 1,then y = 5 − 3(1) = 2 so (1)2− 6(1) + C2 = 2, C2 = 7 thus y = x3− 6x + 7.50. (a) f(x) =13x2sin 3x −227sin 3x +29x cos 3x − 0.251607(b) f(x) = 4 + x2 +4√4 + x2− 651. (a)2y–2 2x(b) yx4–1 1(c) f(x) = x2/2 − 152. (a)56xy (b)1yx1(c) y = (ex+ 1)/2
• 236 Chapter 653. This slope ﬁeld is zero along the y-axis,and so corresponds to (b).–3 –1 1 3–10–5510xy54. This slope ﬁeld is independent of y, is near zerofor large negative values of x, and is very largefor large positive x. It must correspond to (d).–4 –2 2 4–10–5510xy55. This slope ﬁeld has a negative valuealong the y-axis, and thus correspondsto (c).–2 1 3–9–339xy56. This slope ﬁeld appears to be constant(approximately 2), and thus correspondsto diﬀerential equation (a).–3 –1 2 4–10510xy57. (a) F (x) = G (x) = 3x + 4(b) F(0) = 16/6 = 8/3, G(0) = 0, so F(0) − G(0) = 8/3(c) F(x) = (9x2+ 24x + 16)/6 = 3x2/2 + 4x + 8/3 = G(x) + 8/358. (a) F (x) = G (x) = 10x/(x2+ 5)2(b) F(0) = 0, G(0) = −1, so F(0) − G(0) = 1(c) F(x) =x2x2 + 5=(x2+ 5) − 5x2 + 5= 1 −5x2 + 5= G(x) + 159. (a) For x = 0, F (x) = G (x) = 1. But if I is an interval containing 0 then neither F nor G hasa derivative at 0, so neither F nor G is an antiderivative on I.(b) Suppose G(x) = F(x) + C for some C. Then F(1) = 4 and G(1) = 4 + C, so C = 0, butF(−1) = −2 and G(−1) = −1, a contradiction.(c) No, because neither F nor G is an antiderivative on (−∞, +∞).60. (a) Neither F nor G is diﬀerentiable at x = 0. For x > 0, F (x) = 1/x, G (x) = 1/x, and forx < 0, F (x) = 1/x, G (x) = 1/x.(b) F(1) = 0, G(1) = 2; F(−1) = 0, G(−1) = 1, so F(x) = G(x) + C is impossible.(c) The hypotheses of the Theorem are violated by any interval containing 0.61. (sec2x − 1)dx = tan x − x + C 62. (csc2x − 1)dx = − cot x − x + C63. (a)12(1 − cos x)dx =12(x − sin x) + C (b)12(1 + cos x) dx =12(x + sin x) + C
• Exercise Set 6.3 23764. For x > 0,ddx[sec−1x] =1|x|√x2 − 1, and for x < 0,ddx[sec−1|x|] =ddx[sec−1(−x)] = (−1)1|x|√x2 − 1=1x√x2 − 1which yields formula (14) in both cases.65. v =10872√273T−1/2dT =1087√273T1/2+C, v(273) = 1087 = 1087+C so C = 0, v =1087√273T1/2ft/s66. dT/dx = C1, T = C1x + C2; T = 25 when x = 0 so C2 = 25, T = C1x + 25. T = 85 when x = 50so 50C1 + 25 = 85, C1 = 1.2, T = 1.2x + 25EXERCISE SET 6.31. (a) u23du = u24/24 + C = (x2+ 1)24/24 + C(b) − u3du = −u4/4 + C = −(cos4x)/4 + C(c) 2 sin u du = −2 cos u + C = −2 cos√x + C(d)38u−1/2du =34u1/2+ C =344x2 + 5 + C2. (a)14sec2u du =14tan u + C =14tan(4x + 1) + C(b)14u1/2du =16u3/2+ C =16(1 + 2y2)3/2+ C(c)1πu1/2du =23πu3/2+ C =23πsin3/2(πθ) + C(d) u4/5du =59u9/5+ C =59(x2+ 7x + 3)9/5+ C3. (a) − u du = −12u2+ C = −12cot2x + C(b) u9du =110u10+ C =110(1 + sin t)10+ C(c)12cos u du =12sin u + C =12sin 2x + C(d)12sec2u du =12tan u + C =12tan x2+ C4. (a) (u − 1)2u1/2du = (u5/2− 2u3/2+ u1/2)du =27u7/2−45u5/2+23u3/2+ C=27(1 + x)7/2−45(1 + x)5/2+23(1 + x)3/2+ C(b) csc2u du = − cot u + C = − cot(sin x) + C
• 238 Chapter 6(c) sin u du = − cos u + C = − cos(x − π) + C(d)duu2= −1u+ C = −1x5 + 1+ C5. (a)1udu = ln |u| + C = ln | ln x| + C(b) −15eudu = −15eu+ C = −15e−5x+ C(c) −131udu = −13ln |u| + C = −13ln |1 + cos 3θ| + C(d)duu= ln u + C = ln(1 + ex) + C6. (a) u = x3,13du1 + u2=13tan−1(x3) + C(b) u = ln x,1√1 − u2du = sin−1(ln x) + C(c) u = 3x,1u√u2 − 1du = sec−1(3x) + C(d) u =√x, 2du1 + u2= 2 tan−1u + C = 2 tan−1(√x) + C9. u = 4x − 3,14u9du =140u10+ C =140(4x − 3)10+ C10. u = 5 + x4,14√u du =16u3/2+ C =16(5 + x4)3/2+ C11. u = 7x,17sin u du = −17cos u + C = −17cos 7x + C12. u = x/3, 3 cos u du = 3 sin u + C = 3 sin(x/3) + C13. u = 4x, du = 4dx;14sec u tan u du =14sec u + C =14sec 4x + C14. u = 5x, du = 5dx;15sec2u du =15tan u + C =15tan 5x + C15. u = 2x, du = 2dx;12eudu =12eu+ C =12e2x+ C16. u = 2x, du = 2dx;121udu =12ln |u| + C =12ln |2x| + C17. u = 2x,121√1 − u2du =12sin−1(2x) + C18. u = 4x,1411 + u2du =14tan−1(4x) + C
• Exercise Set 6.3 23919. u = 7t2+ 12, du = 14t dt;114u1/2du =121u3/2+ C =121(7t2+ 12)3/2+ C20. u = 4 − 5x2, du = −10x dx; −110u−1/2du = −15u1/2+ C = −154 − 5x2 + C21. u = 1 − 2x, du = −2dx, −31u3du = (−3) −121u2+ C =321(1 − 2x)2+ C22. u = x3+ 3x, du = (3x2+ 3) dx,131√udu =23x3 + 3x + C23. u = 5x4+ 2, du = 20x3dx,120duu3du = −1401u2+ C = −140(5x4 + 2)2+ C24. u =1x, du = −1x2dx, −13sin u du =13cos u + C =13cos1x+ C25. u = sin x, du = cos x dx; eudu = eu+ C = esin x+ C26. u = x4, du = 4x3dx;14eudu =14eu+ C =14ex4+ C27. u = −2x3, du = −6x2, −16eudu = −16eu+ C = −16e−2x3+ C28. u = ex− e−x, du = (ex+ e−x)dx,1udu = ln |u| + C = ln ex− e−x+ C29. u = ex,11 + u2du = tan−1(ex) + C 30. u = t2,121u2 + 1du =12tan−1(t2) + C31. u = 5/x, du = −(5/x2)dx; −15sin u du =15cos u + C =15cos(5/x) + C32. u =√x, du =12√xdx; 2 sec2u du = 2 tan u + C = 2 tan√x + C33. u = cos 3t, du = −3 sin 3t dt, −13u4du = −115u5+ C = −115cos53t + C34. u = sin 2t, du = 2 cos 2t dt;12u5du =112u6+ C =112sin62t + C35. u = x2, du = 2x dx;12sec2u du =12tan u + C =12tan x2+ C36. u = 1 + 2 sin 4θ, du = 8 cos 4θ dθ;181u4du = −1241u3+ C = −1241(1 + 2 sin 4θ)3+ C37. u = 2 − sin 4θ, du = −4 cos 4θ dθ; −14u1/2du = −16u3/2+ C = −16(2 − sin 4θ)3/2+ C38. u = tan 5x, du = 5 sec25x dx;15u3du =120u4+ C =120tan45x + C
• 240 Chapter 639. u = tan x,1√1 − u2du = sin−1(tan x) + C40. u = cos θ, −1u2 + 1du = − tan−1(cos θ) + C41. u = sec 2x, du = 2 sec 2x tan 2x dx;12u2du =16u3+ C =16sec32x + C42. u = sin θ, du = cos θ dθ; sin u du = − cos u + C = − cos(sin θ) + C43. e−xdx; u = −x, du = −dx; − eudu = −eu+ C = −e−x+ C44. ex/2dx; u = x/2, du = dx/2; 2 eudu = 2eu+ C = 2ex/2+ C = 2√ex + C45. u = 2√x, du =1√xdx; ,1eudu = −e−u+ C = −e−2√x+ C46. u = 2y + 1, du =1√2y + 1dy; eudu = eu+ C = e√2y+1+ C47. u = 2y + 1, du = 2dy;14(u − 1)1√udu =16u3/2−12√u + C =16(2y + 1)3/2−122y + 1 + C48. u = 4 − x, du = −dx;− (4 − u)√u du = −83u3/2+25u5/2+ C =25(4 − x)5/2−83(4 − x)3/2+ C49. sin22θ sin 2θ dθ = (1 − cos22θ) sin 2θ dθ; u = cos 2θ, du = −2 sin 2θ dθ,−12(1 − u2)du = −12u +16u3+ C = −12cos 2θ +16cos32θ + C50. sec23θ = tan23θ + 1, u = 3θ, du = 3dθsec43θ dθ =13(tan2u + 1) sec2u du =19tan3u +13tan u + C =19tan33θ +13tan 3θ + C51. 1 +1tdt = t + ln |t| + C52. e2 ln x= eln x2= x2, x > 0, so e2 ln xdx = x2dx =13x3+ C53. ln(ex) + ln(e−x) = ln(exe−x) = ln 1 = 0 so [ln(ex) + ln(e−x)]dx = C54.cos xsin xdx; u = sin x, du = cos xdx;1udu = ln |u| + C = ln | sin x| + C55. (a) sin−1(x/3) + C (b) (1/√5) tan−1(x/√5) + C (c) (1/√π) sec−1(x/√π) + C
• Exercise Set 6.3 24156. (a) u = ex,14 + u2du =12tan−1(ex/2) + C(b) u = 2x,121√9 − u2du =12sin−1(2x/3) + C,(c) u =√5y,1u√u2 − 3du =1√3sec−1(√5y/√3) + C57. u = a + bx, du = b dx,(a + bx)ndx =1bundu =(a + bx)n+1b(n + 1)+ C58. u = a + bx, du = b dx, dx =1bdu1bu1/ndu =nb(n + 1)u(n+1)/n+ C =nb(n + 1)(a + bx)(n+1)/n+ C59. u = sin(a + bx), du = b cos(a + bx)dx1bundu =1b(n + 1)un+1+ C =1b(n + 1)sinn+1(a + bx) + C61. (a) with u = sin x, du = cos x dx; u du =12u2+ C1 =12sin2x + C1;with u = cos x, du = − sin x dx; − u du = −12u2+ C2 = −12cos2x + C2(b) because they diﬀer by a constant:12sin2x + C1 − −12cos2x + C2 =12(sin2x + cos2x) + C1 − C2 = 1/2 + C1 − C262. (a) First method: (25x2− 10x + 1)dx =253x3− 5x2+ x + C1;second method:15u2du =115u3+ C2 =115(5x − 1)3+ C2(b)115(5x − 1)3+ C2 =115(125x3− 75x2+ 15x − 1) + C2 =253x3− 5x2+ x −115+ C2;the answers diﬀer by a constant.63. y =√5x + 1 dx =215(5x + 1)3/2+ C; −2 = y(3) =21564 + C,so C = −2 −21564 = −15815, and y =215(5x + 1)3/2−1581564. y = (2 + sin 3x) dx = 2x −13cos 3x + C and0 = yπ3=2π3+13+ C, C = −2π + 13, y = 2x −13cos 3x −2π + 1365. y = − e2tdt = −12e2t+ C, 6 = y(0) = −12+ C, y = −12e2t+13266. y =125 + 9t2dt =115tan−1 35t + C,π30= y −53= −115π4+ C,C =π60, y =115tan−1 35t +π60
• 242 Chapter 667. (a) u = x2+ 1, du = 2x dx;121√udu =√u + C = x2 + 1 + C(b) 50–5 568. (a) u = x2+ 1, du = 2x dx;121udu =12ln u + C =12ln(x2+ 1) + C(b) yx4–4 469. f (x) = m =√3x + 1, f(x) = (3x + 1)1/2dx =29(3x + 1)3/2+ Cf(0) = 1 =29+ C, C =79, so f(x) =29(3x + 1)3/2+7970. p(t) = (3 + 0.12t)3/2dt =103(3 + 0.12t)5/2+ C;100 = p(0) =10335/2+ C, C = 100 − 10 · 33/2≈ 48.038 so thatp(5) =103(3 + 5 · (0.12))5/2+ 100 − 10 ∗ 33/2≈ 130.005 so that the population at the beginning ofthe year 2010 is approximately 130,005.71. u = a sin θ, du = a cos θ dθ;du√a2 − u2= aθ + C = sin−1 ua+ C72. If u > 0 then u = a sec θ, du = a sec θ tan θ dθ,duu√u2 − a2=1aθ =1asec−1 ua+ CEXERCISE SET 6.41. (a) 1 + 8 + 27 = 36 (b) 5 + 8 + 11 + 14 + 17 = 55(c) 20 + 12 + 6 + 2 + 0 + 0 = 40 (d) 1 + 1 + 1 + 1 + 1 + 1 = 6(e) 1 − 2 + 4 − 8 + 16 = 11 (f) 0 + 0 + 0 + 0 + 0 + 0 = 02. (a) 1 + 0 − 3 + 0 = −2 (b) 1 − 1 + 1 − 1 + 1 − 1 = 0(c) π2+ π2+ · · · + π2= 14π2(14 terms)(d) 24+ 25+ 26= 112(e)√1 +√2 +√3 +√4 +√5 +√6(f) 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 = 1
• Exercise Set 6.4 2433.10k=1k 4.20k=13k 5.10k=12k6.8k=1(2k − 1) 7.6k=1(−1)k+1(2k − 1) 8.5k=1(−1)k+1 1k9. (a)50k=12k (b)50k=1(2k − 1)10. (a)5k=1(−1)k+1ak (b)5k=0(−1)k+1bk (c)nk=0akxk(d)5k=0a5−kbk11.12(100)(100 + 1) = 5050 12. 7100k=1k +100k=11 =72(100)(101) + 100 = 35,45013.16(20)(21)(41) = 2870 14.20k=1k2−3k=1k2= 2870 − 14 = 285615.30k=1k(k2− 4) =30k=1(k3− 4k) =30k=1k3− 430k=1k =14(30)2(31)2− 4 ·12(30)(31) = 214,36516.6k=1k −6k=1k3=12(6)(7) −14(6)2(7)2= −42017.nk=13kn=3nnk=1k =3n·12n(n + 1) =32(n + 1)18.n−1k=1k2n=1nn−1k=1k2=1n·16(n − 1)(n)(2n − 1) =16(n − 1)(2n − 1)19.n−1k=1k3n2=1n2n−1k=1k3=1n2·14(n − 1)2n2=14(n − 1)220.nk=15n−2kn=5nnk=11 −2nnk=1k =5n(n) −2n·12n(n + 1) = 4 − n22.n(n + 1)2= 465, n2+ n − 930 = 0, (n + 31)(n − 30) = 0, n = 30.23.1 + 2 + 3 + · · · + nn2=nk=1kn2=1n2nk=1k =1n2·12n(n + 1) =n + 12n; limn→+∞n + 12n=1224.12+ 22+ 32+ · · · + n2n3=nk=1k2n3=1n3nk=1k2=1n3·16n(n + 1)(2n + 1) =(n + 1)(2n + 1)6n2;limn→+∞(n + 1)(2n + 1)6n2= limn→+∞16(1 + 1/n)(2 + 1/n) =1325.nk=15kn2=5n2nk=1k =5n2·12n(n + 1) =5(n + 1)2n; limn→+∞5(n + 1)2n=52
• 244 Chapter 626.n−1k=12k2n3=2n3n−1k=1k2=2n3·16(n − 1)(n)(2n − 1) =(n − 1)(2n − 1)3n2;limn→+∞(n − 1)(2n − 1)3n2= limn→+∞13(1 − 1/n)(2 − 1/n) =2327. (a)5j=02j(b)6j=12j−1(c)7j=22j−228. (a)5k=1(k + 4)2k+8(b)13k=9(k − 4)2k29. (a) 2 +3n43n, 2 +6n43n, 2 +9n43n, . . . , 2 +3(n − 1)n43n, (2 + 3)4 3nWhen [2, 5] is subdivided into n equal intervals, the endpoints are 2, 2 +3n, 2 + 2 ·3n, 2 + 3 ·3n, . . . , 2 + (n − 1)3n, 2 + 3 = 5, and the right endpoint approximation to the area under thecurve y = x4is given by the summands above.(b)n−1k=02 + k ·3n43ngives the left endpoint approximation.30. n is the number of elements of the partition, x∗k is an arbitrary point in the k-th interval,k = 0, 1, 2, . . . , n − 1, n, and ∆x is the width of an interval in the partition.In the usual deﬁnition of area, the parts above the curve are given a + sign, and the parts belowthe curve are given a − sign. These numbers are then replaced with their absolute values andsummed.In the deﬁnition of net signed area, the parts given above are summed without considering absolutevalues. In this case there could be lots of cancellation of ’positive’ areas with ’negative’ areas.31. Endpoints 2, 3, 4, 5, 6; ∆x = 1;(a) Left endpoints:4k=1f(x∗k)∆x = 7 + 10 + 13 + 16 = 46(b) Midpoints:4k=1f(x∗k)∆x = 8.5 + 11.5 + 14.5 + 17.5 = 52(c) Right endpoints:4k=1f(x∗k)∆x = 10 + 13 + 16 + 19 = 5832. Endpoints 1, 3, 5, 7, 9, ∆x = 2;(a) Left endpoints:4k=1f(x∗k)∆x = 1 +13+15+172 =352105(b) Midpoints:4k=1f(x∗k)∆x =12+14+16+182 =2512(c) Right endpoints:4k=1f(x∗k)∆x =13+15+17+192 =496315
• Exercise Set 6.4 24533. Endpoints: 0, π/4, π/2, 3π/4, π; ∆x = π/4(a) Left endpoints:4k=1f(x∗k)∆x = 1 +√2/2 + 0 −√2/2 (π/4) = π/4(b) Midpoints:4k=1f(x∗k)∆x = [cos(π/8) + cos(3π/8) + cos(5π/8) + cos(7π/8)] (π/4)= [cos(π/8) + cos(3π/8) − cos(3π/8) − cos(π/8)] (π/4) = 0(c) Right endpoints:4k=1f(x∗k)∆x =√2/2 + 0 −√2/2 − 1 (π/4) = −π/434. Endpoints −1, 0, 1, 2, 3; ∆x = 1(a)4k=1f(x∗k)∆x = −3 + 0 + 1 + 0 = −2 (b)4k=1f(x∗k)∆x = −54+34+34+154= 4(c)4k=1f(x∗k)∆x = 0 + 1 + 0 − 3 = −235. (a) 0.718771403, 0.705803382, 0.698172179(b) 0.668771403, 0.680803382, 0.688172179(c) 0.692835360, 0.693069098, 0.69313468236. (a) 0.761923639, 0.712712753, 0.684701150(b) 0.584145862, 0.623823864, 0.649145594(c) 0.663501867, 0.665867079, 0.66653834637. (a) 4.884074734, 5.115572731, 5.248762738(b) 5.684074734, 5.515572731, 5.408762738(c) 5.34707029, 5.338362719, 5.33464441638. (a) 0.919403170, 0.960215997, 0.984209789(b) 1.076482803, 1.038755813, 1.015625715(c) 1.001028824, 1.000257067, 1.00004112539. ∆x =3n, x∗k = 1 +3nk; f(x∗k)∆x =12x∗k∆x =121 +3nk3n=321n+3n2knk=1f(x∗k)∆x =32nk=11n+nk=13n2k =321 +3n2·12n(n + 1) =321 +32n + 1nA = limn→+∞321 +321 +1n=321 +32=15440. ∆x =5n, x∗k = 0 + k5n; f(x∗k)∆x = (5 − x∗k)∆x = 5 −5nk5n=25n−25n2knk=1f(x∗k)∆x =nk=125n−25n2nk=1k = 25 −25n2·12n(n + 1) = 25 −252n + 1nA = limn→+∞25 −2521 +1n= 25 −252=252
• 246 Chapter 641. ∆x =3n, x∗k = 0 + k3n; f(x∗k)∆x = 9 − 9k2n23nnk=1f(x∗k)∆x =nk=19 − 9k2n23n=27nnk=11 −k2n2= 27 −27n3nk=1k2A = limn→+∞27 −27n3nk=1k2= 27 − 2713= 1842. ∆x =3n, x∗k = k3nf(x∗k)∆x = 4 −14(x∗k)2∆x = 4 −149k2n23n=12n−27k24n3nk=1f(x∗k)∆x =nk=112n−274n3nk=1k2= 12 −274n3·16n(n + 1)(2n + 1) = 12 −98(n + 1)(2n + 1)n2A = limn→+∞12 −981 +1n2 +1n= 12 −98(1)(2) = 39/443. ∆x =4n, x∗k = 2 + k4nf(x∗k)∆x = (x∗k)3∆x = 2 +4nk34n=32n1 +2nk3=32n1 +6nk +12n2k2+8n3k3nk=1f(x∗k)∆x =32nnk=11 +6nnk=1k +12n2nk=1k2+8n3nk=1k3=32nn +6n·12n(n + 1) +12n2·16n(n + 1)(2n + 1) +8n3·14n2(n + 1)2= 32 1 + 3n + 1n+ 2(n + 1)(2n + 1)n2+ 2(n + 1)2n2A = limn→+∞32 1 + 3 1 +1n+ 2 1 +1n2 +1n+ 2 1 +1n2= 32[1 + 3(1) + 2(1)(2) + 2(1)2] = 32044. ∆x =2n, x∗k = −3 + k2n; f(x∗k)∆x = [1 − (x∗k)3]∆x = 1 − −3 +2nk32n=2n28 −54nk +36n2k2−8n3k3nk=1f(x∗k)∆x =2n28n − 27(n + 1) + 6(n + 1)(2n + 1)n− 2(n + 1)2nA = limn→+∞2 28 − 27 1 +1n+ 6 1 +1n2 +1n− 2 1 +1n2= 2(28 − 27 + 12 − 2) = 22
• Exercise Set 6.4 24745. ∆x =3n, x∗k = 1 + (k − 1)3nf(x∗k)∆x =12x∗k∆x =121 + (k − 1)3n3n=123n+ (k − 1)9n2nk=1f(x∗k)∆x =12nk=13n+9n2nk=1(k − 1) =123 +9n2·12(n − 1)n =32+94n − 1nA = limn→+∞32+941 −1n=32+94=15446. ∆x =5n, x∗k =5n(k − 1)f(x∗k)∆x = (5 − x∗k)∆x = 5 −5n(k − 1)5n=25n−25n2(k − 1)nk=1f(x∗k)∆x =25nnk=11 −25n2nk=1(k − 1) = 25 −252n − 1nA = limn→+∞25 −2521 −1n= 25 −252=25247. ∆x =3n, x∗k = 0 + (k − 1)3n; f(x∗k)∆x = (9 − 9(k − 1)2n2)3nnk=1f(x∗k)∆x =nk=19 − 9(k − 1)2n23n=27nnk=11 −(k − 1)2n2= 27 −27n3nk=1k2+54n3nk=1k −27n2A = limn→+∞= 27 − 2713+ 0 + 0 = 1848. ∆x =3n, x∗k = (k − 1)3nf(x∗k)∆x = 4 −14(x∗k)2∆x = 4 −149(k − 1)2n23n=12n−27k24n3+27k2n3−274n3nk=1f(x∗k)∆x =nk=112n−274n3nk=1k2+272n3nk=1k −274n3nk=11= 12 −274n3·16n(n + 1)(2n + 1) +272n3n(n + 1)2−274n2= 12 −98(n + 1)(2n + 1)n2+274n+274n2−274n2A = limn→+∞12 −981 +1n2 +1n+ 0 + 0 − 0 = 12 −98(1)(2) = 39/449. Endpoints 0,4n,8n, . . . ,4(n − 1)n,4nn= 4, and midpoints2n,6n,10n, . . . ,4n − 6n,4n − 2n. Approxi-mate the area with the sumnk=124k − 2n4n=16n22n(n + 1)2− n → 16 as n → +∞.
• 248 Chapter 650. Endpoints 1, 1 +4n, 1 +8n, . . . , 1 +4(n − 1)n, 1 + 4 = 5, and midpoints1 +2n, 1 +6n, 1 +10n, . . . , 1 +4(n − 1) − 2n,4n − 2n. Approximate the area with the sumnk=16 − 1 +4k − 2n4n=nk=154n−16n2k +8n2= 20 −16n2n(n + 1)2+8n= 20 − 8 = 12,which happens to be exact.51. ∆x =1n, x∗k =2k − 12nf(x∗k)∆x =(2k − 1)2(2n)21n=k2n3−kn3+14n3nk=1f(x∗k)∆x =1n3nk=1k2−1n3nk=1k +14n3nk=11Using Theorem 6.4.4,A = limn→+∞nk=1f(x∗k)∆x =13+ 0 + 0 =1352. ∆x =2n, x∗k = −1 +2k − 1nf(x∗k)∆x = −1 +2k − 1n22n=8k2n3−8kn3+2n3−2nnk=1f(x∗k)∆x =8n3nk=1k2−8n3nk=1k +2n2− 2A = limn→+∞nk=1f(x∗k)∆x =83+ 0 + 0 − 2 =2353. ∆x =2n, x∗k = −1 +2knf(x∗k)∆x = −1 +2kn2n= −2n+ 4kn2nk=1f(x∗k)∆x = −2 +4n2nk=1k = −2 +4n2n(n + 1)2= −2 + 2 +2nA = limn→+∞nk=1f(x∗k)∆x = 0The area below the x-axis cancels the area above the x-axis.54. ∆x =3n, x∗k = −1 +3knf(x∗k)∆x = −1 +3kn3n= −3n+9n2knk=1f(x∗k)∆x = −3 +9n2n(n + 1)2
• Exercise Set 6.4 249A = limn→+∞nk=1f(x∗k)∆x = −3 +92+ 0 =32The area below the x-axis cancels the area above the x-axis that lies to the right of the line x = 1;the remaining area is a trapezoid of width 1 and heights 1, 2, hence its area is1 + 22=32.55. ∆x =2n, x∗k =2knf(x∗k) =2kn2− 12n=8k2n3−2nnk=1f(x∗k)∆x =8n3nk=1k2−2nnk=11 =8n3n(n + 1)(2n + 1)6− 2A = limn→+∞nk=1f(x∗k)∆x =166− 2 =2356. ∆x =2n, x∗k = −1 +2knf(x∗k)∆x = −1 +2kn32n= −2n+ 12kn2− 24k2n3+ 16k3n4nk=1f(x∗k)∆x = −2 +12n2n(n + 1)2−24n3n(n + 1)(2n + 1)6+16n4n(n + 1)22A = limn→+∞nk=1f(x∗k) = −2 +122−486+1622= 057. ∆x =b − an, x∗k = a +b − an(k − 1)f(x∗k)∆x = mx∗k∆x = m a +b − an(k − 1)b − an= m(b − a)an+b − an2(k − 1)nk=1f(x∗k)∆x = m(b − a) a +b − a2·n − 1nA = limn→+∞m(b − a) a +b − a21 −1n= m(b − a)b + a2=12m(b2− a2)58. ∆x =b − an, x∗k = a +kn(b − a)f(x∗k)∆x =man(b − a) +mkn2(b − a)2nk=1f(x∗k)∆x = ma(b − a) +mn2(b − a)2 n(n + 1)2A = limn→+∞nk=1f(x∗k)∆x = ma(b − a) +m2(b − a)2= m(b − a)a + b2
• 250 Chapter 659. (a) With x∗k as the right endpoint, ∆x =bn, x∗k =bnkf(x∗k)∆x = (x∗k)3∆x =b4n4k3,nk=1f(x∗k)∆x =b4n4nk=1k3=b44(n + 1)2n2A = limn→+∞b441 +1n2= b4/4(b) ∆x =b − an, x∗k = a +b − ankf(x∗k)∆x = (x∗k)3∆x = a +b − ank3b − an=b − ana3+3a2(b − a)nk +3a(b − a)2n2k2+(b − a)3n3k3nk=1f(x∗k)∆x = (b − a) a3+32a2(b − a)n + 1n+12a(b − a)2 (n + 1)(2n + 1)n2+14(b − a)3 (n + 1)2n2A = limn→+∞nk=1f(x∗k)∆x= (b − a) a3+32a2(b − a) + a(b − a)2+14(b − a)3=14(b4− a4)60. Let A be the area of the region under the curve and above the interval 0 ≤ x ≤ 1 on the x-axis,and let B be the area of the region between the curve and the interval 0 ≤ y ≤ 1 on the y-axis.Together A and B form the square of side 1, so A + B = 1.But B can also be considered as the area between the curve x = y2and the interval 0 ≤ y ≤ 1 onthe y-axis. By Exercise 51 above, B =13, so A = 1 −13=23.61. If n = 2m then 2m + 2(m − 1) + · · · + 2 · 2 + 2 = 2mk=1k = 2 ·m(m + 1)2= m(m + 1) =n2+ 2n4;if n = 2m + 1 then (2m + 1) + (2m − 1) + · · · + 5 + 3 + 1 =m+1k=1(2k − 1)= 2m+1k=1k −m+1k=11 = 2 ·(m + 1)(m + 2)2− (m + 1) = (m + 1)2=n2+ 2n + 1462. 50·30+49·29+· · ·+22·2+21·1 =30k=1k(k+20) =30k=1k2+2030k=1k =30 · 31 · 616+2030 · 312= 18,75563. (35− 34) + (36− 35) + · · · + (317− 316) = 317− 3464. 1 −12+12−13+ · · · +150−151=505165.122−112+132−122+ · · · +1202−1192=1202− 1 = −39940066. (22− 2) + (23− 22) + · · · + (2101− 2100) = 2101− 2
• Exercise Set 6.4 25167. (a)nk=11(2k − 1)(2k + 1)=12nk=112k − 1−12k + 1=121 −13+13−15+15−17+ · · · +12n − 1−12n + 1=121 −12n + 1=n2n + 1(b) limn→+∞n2n + 1=1268. (a)nk=11k(k + 1)=nk=11k−1k + 1= 1 −12+12−13+13−14+ · · · +1n−1n + 1= 1 −1n + 1=nn + 1(b) limn→+∞nn + 1= 169.ni=1(xi − ¯x) =ni=1xi −ni=1¯x =ni=1xi − n¯x but ¯x =1nni=1xi thusni=1xi = n¯x soni=1(xi − ¯x) = n¯x − n¯x = 070. S − rS =nk=0ark−nk=0ark+1= (a + ar + ar2+ · · · + arn) − (ar + ar2+ ar3+ · · · + arn+1)= a − arn+1= a(1 − rn+1)so (1 − r)S = a(1 − rn+1), hence S = a(1 − rn+1)/(1 − r)71. (a)19k=03k+1=19k=03(3k) =3(1 − 320)1 − 3=32(320− 1)(b)25k=02k+5=25k=0252k=25(1 − 226)1 − 2= 231− 25(c)100k=0(−1)−12k=(−1)(1 − (−1/2)101)1 − (−1/2)= −23(1 + 1/2101)72. (a) 1.999023438, 1.999999046, 2.000000000; 2 (b) 2.831059456, 2.990486364, 2.999998301; 373. both are valid 74. none is valid75.nk=1(ak − bk) = (a1 − b1) + (a2 − b2) + · · · + (an − bn)= (a1 + a2 + · · · + an) − (b1 + b2 + · · · + bn) =nk=1ak −nk=1bk
• 252 Chapter 676. (a)nk=11 means add 1 to itself n times, which gives the result.(b)1n2nk=1k =1n2n(n + 1)2=12+12n, so limn→+∞1n2nk=1k =12(c)1n3nk=1k2=1n3n(n + 1)(2n + 1)6=26+36n+16n2, so limn→+∞1n3nk=1k2=13(d)1n4nk=1k3=1n4n(n + 1)22=14+12n+14n2, so limn→+∞1n4nk=1k3=14EXERCISE SET 6.51. (a) (4/3)(1) + (5/2)(1) + (4)(2) = 71/6 (b) 22. (a) (√2/2)(π/2) + (−1)(3π/4) + (0)(π/2) + (√2/2)(π/4) = 3(√2 − 2)π/8(b) 3π/43. (a) (−9/4)(1) + (3)(2) + (63/16)(1) + (−5)(3) = −117/16(b) 34. (a) (−8)(2) + (0)(1) + (0)(1) + (8)(2) = 0 (b) 25.2−1x2dx 6.21x3dx7.3−34x(1 − 3x)dx 8.π/20sin2x dx9. (a) limmax ∆xk→0nk=12x∗k∆xk; a = 1, b = 2 (b) limmax ∆xk→0nk=1x∗kx∗k + 1∆xk; a = 0, b = 110. (a) limmax ∆xk→0nk=1x∗k ∆xk, a = 1, b = 2(b) limmax ∆xk→0nk=1(1 + cos x∗k) ∆xk, a = −π/2, b = π/211. (a) A =12(3)(3) = 9/23xyA(b) −A = −12(1)(1 + 2) = −3/2–2 –1 xyA
• Exercise Set 6.5 253(c) −A1 + A2 = −12+ 8 = 15/2–14xyA1A2(d) −A1 + A2 = 0–55xyA1A212. (a) A =12(1)(2) = 121xyA(b) A =12(2)(3/2 + 1/2) = 2–1 11xyA(c) −A = −12(1/2)(1) = −1/421xyA(d) A1 − A2 = 1 − 1/4 = 3/421xyA2A113. (a) A = 2(5) = 10yx125A(b) 0; A1 = A2 by symmetry6c xyA1A2(c) A1 + A2 =12(5)(5/2) +12(1)(1/2)= 13/2–152xy32A1A2(d)12[π(1)2] = π/2yx1–1 1A14. (a) A = (6)(5) = 30–10 –56xyA(b) −A1 + A2 = 0 becauseA1 = A2 by symmetry\$4xyA1A2
• 254 Chapter 6(c) A1 + A2 =12(2)(2) +12(1)(1) = 5/222xyA1A2(d)14π(2)2= πyx22A15. (a)0−2f(x) dx =0−2(x + 2) dxTriangle of height 2 and width 2, above x-axis, so answer is 2.(b)2−2f(x) dx =0−2(x + 2) dx +02(2 − x) dxTwo triangles of height 2 and base 2; answer is 4.(c)60|x − 2| dx =20(2 − x) dx +62(x − 2) dxTriangle of height 2 and base 2 together with a triangle with height 4 and base 4, so 2+8 = 10.(d)6−4f(x) dx =−2−4(x + 2) dx +0−2(x + 2) dx +20(2 − x) dx +62(x − 2) dxTriangle of height 2 and base 2, below axis, plus a triangle of height 2, base 2 above axis,another of height 2 and base 2 above axis, and a triangle of height 4 and base 4, above axis.Thus f(x) = −2 + 2 + 2 + 8 = 10.16. (a)102x dx =area of a triangle with height 2 and base 1, so 1.(b)1−12x dx =0−12x dx +102x dxTwo triangles of height 2 and base 1 on opposite sides of the x-axis, so they cancel to yield 0.(c)1012 dxRectangle of height 2 and base 9, area = 18(d)11/22x dx +512 dxTrapezoid of width 1/2 and heights 1 and 2, together with a rectangle of height 2 and base4, so 1/21 + 22+ 2 · 4 = 3/4 + 8 = 35/417. (a) 0.8 (b) −2.6 (c) −1.8 (d) −0.318. (a) 10 (b) −94 (c) −84 (d) −7519.2−1f(x)dx + 22−1g(x)dx = 5 + 2(−3) = −120. 341f(x)dx −41g(x)dx = 3(2) − 10 = −4
• Exercise Set 6.5 25521.51f(x)dx =50f(x)dx −10f(x)dx = 1 − (−2) = 322.−23f(x)dx = −3−2f(x)dx = −1−2f(x)dx +31f(x)dx = −(2 − 6) = 423. 43−1dx − 53−1xdx = 4 · 4 − 5(−1/2 + (3 · 3)/2) = −424.2−2dx − 32−2|x|dx = 4 · 1 − 3(2)(2 · 2)/2 = −825.10xdx + 2101 − x2dx = 1/2 + 2(π/4) = (1 + π)/226.0−32dx +0−39 − x2dx = 2 · 3 + (π(3)2)/4 = 6 + 9π/427. (a)√x > 0, 1 − x < 0 on [2, 3] so the integral is negative(b) x2> 0, 3 − cos x > 0 for all x so the integral is positive28. (a) x4> 0,√3 − x > 0 on [−3, −1] so the integral is positive(b) x3− 9 < 0, |x| + 1 > 0 on [−2, 2] so the integral is negative29.10025 − (x − 5)2dx = π(5)2/2 = 25π/2 30.309 − (x − 3)2dx = π(3)2/4 = 9π/431.10(3x + 1)dx = 5/2 32.2−24 − x2dx = π(2)2/2 = 2π33. (a) The graph of the integrand is the horizontal line y = C. At ﬁrst, assume that C > 0. Thenthe region is a rectangle of height C whose base extends from x = a to x = b. ThusbaC dx = (area of rectangle) = C(b − a).If C ≤ 0 then the rectangle lies below the axis and its integral is the negative area, i.e.−|C|(b − a) = C(b − a).(b) Since f(x) = C, the Riemann sum becomeslimmax ∆xk→0nk=1f(x∗k)∆xk = limmax ∆xk→0nk=1C = limmax ∆xk→0C(b − a) = C(b − a).By Deﬁnition 6.5.1,baf(x) dx = C(b − a).34. (a) f is continuous on [−1, 1] so f is integrable there by Part (a) of Theorem 6.5.8(b) |f(x)| ≤ 1 so f is bounded on [−1, 1], and f has one point of discontinuity, so by Part (b) ofTheorem 6.5.8 f is integrable on [−1, 1](c) f is not bounded on [-1,1] because limx→0f(x) = +∞, so f is not integrable on [0,1](d) f(x) is discontinuous at the point x = 0 because limx→0sin1xdoes not exist. f is continuouselsewhere. −1 ≤ f(x) ≤ 1 for x in [−1, 1] so f is bounded there. By Part (b), Theorem 6.5.8,f is integrable on [−1, 1].
• 256 Chapter 635. For any partition of [0, 1] we havenk=1f(x∗k)∆xk =nk=2∆xk = 1 − ∆x1 or we havenk=1f(x∗k)∆xk =nk=1∆xk = 1.This is because f(x) = 1 for all x except possibly x∗1, which lies in the interval [0, x1] and could be0. In any event, since in the limit the maximum size of the ∆xk goes to zero, the two possibilitiesare 1 in the limit, and thus10f(x) dx = 1.36. Each subinterval of a partition of [a, b] contains both rational and irrational numbers. If all x∗k arechosen to be rational thennk=1f(x∗k)∆xk =nk=1(1)∆xk =nk=1∆xk = b − a so limmax ∆xk→0nk=1f(x∗k)∆xk = b − a.If all x∗k are irrational then limmax ∆xk→0nk=1f(x∗k)∆xk = 0. Thus f is not integrable on [a, b] becausethe preceding limits are not equal.37. On [0,π4] the minimum value of the integrand is 0 and the maximum is sinπ4=√22. Onπ4,5π6the minimum value is sin5π6=12and the maximum is sinπ2= 1. On5π6, π theminimum value is 0 and the maximum value is12. Thus the minimum value of the Riemann sumsis 0 ·π4+12·7π12+ 0 ·π6=7π24, and the maximum is√22·π4+ 1 ·7π12+12·π6=√28+23π.38. For the smallest, ﬁnd x∗k so that f(x∗k) is minimum on each subinterval: x∗1 = 1, x∗2 = 3/2, x∗3 = 3so (2)(1) + (7/4)(2) + (4)(1) = 9.5. For the largest, ﬁnd x∗k so that f(x∗k) is maximum on eachsubinterval: x∗1 = 0, x∗2 = 3, x∗3 = 4 so (4)(1) + (4)(2) + (8)(1) = 20.39. ∆xk =4k2n2−4(k − 1)2n2=4n2(2k − 1), x∗k =4k2n2,f(x∗k) =2kn, f(x∗k)∆xk =8kn3(2k − 1) =8n3(2k2− k),nk=1f(x∗k)∆xk =8n3nk=1(2k2− k) =8n313n(n + 1)(2n + 1) −12n(n + 1) =43(n + 1)(4n − 1)n2,limn→+∞nk=1f(x∗k)∆xk = limn→+∞431 +1n4 −1n=163.40. For any partition of [a, b] use the right endpoints to form the sumnk=1f(x∗k)∆xk. Since f(x∗k) = 0for each k, the sum is zero and so isbaf(x) dx = limn→+∞nk=1f(x∗k)∆xk.
• Exercise Set 6.6 25741. With f(x) = g(x) then f(x) − g(x) = 0 for a < x ≤ b. By Theorem 6.5.4(b)baf(x) dx =ba[(f(x) − g(x) + g(x)]dx =ba[f(x) − g(x)]dx +bag(x)dx.But the ﬁrst term on the right hand side is zero (from Exercise 40), sobaf(x) dx =bag(x) dx42. Choose any large positive integer N and any partition of [0, a]. Then choose x∗1 in the ﬁrst intervalso small that f(x∗1)∆x1 > N. For example choose x∗1 < ∆x1/N. Then with this partition andchoice of x∗1,nk=1f(x∗k)∆xk > f(x∗1)∆x1 > N. This shows that the sum is dependent on partitionand/or points, so Deﬁnition 6.5.1 is not satisﬁed.EXERCISE SET 6.61. (a)20(2 − x)dx = (2x − x2/2)20= 4 − 4/2 = 2(b)1−12dx = 2x1−1= 2(1) − 2(−1) = 4(c)31(x + 1)dx = (x2/2 + x)31= 9/2 + 3 − (1/2 + 1) = 62. (a)50xdx = x2/250= 25/2 (b)935dx = 5x93= 5(9) − 5(3) = 30(c)2−1(x + 3)dx = (x2/2 + 3x)2−1= 4/2 + 6 − (1/2 − 3) = 21/23.32x3dx = x4/432= 81/4 − 16/4 = 65/4 4.1−1x4dx = x5/51−1= 1/5 − (−1)/5 = 2/55.413√x dx = 2x3/241= 16 − 2 = 14 6.271x−2/3dx = 3x1/3271= 3(3 − 1) = 67.ln 20e2xdx =12e2xln 20=12(4 − 1) =328.511xdx = ln x51= ln 5 − ln 1 = ln 59.1−2(x2− 6x + 12) dx =13x3− 3x2+ 12x1−2=13− 3 + 12 − −83− 12 − 24 = 4810.2−14x(1 − x2) dx = (2x2− x4)2−1= 8 − 16 − (2 − 1) = −911.411x2dx = −x−141= −14+ 1 =3412.21x−6dx = −15x521= 31/16013.45x5/294= 844/5 14.411x√xdx = −2√x41= −22+21= 1
• 258 Chapter 615. − cos θπ/2−π/2= 0 16. tan θπ/40= 117. sin xπ/4−π/4=√2 18. (x2− sec xπ/30=π29− 119. 5ex3ln 2= 5e3− 5(2) = 5e3− 10 20. (ln x)/211/2= (ln 2)/221. sin−1x1/√20= sin−1(1/√2) − sin−10 = π/422. tan−1x1−1= tan−11 − tan−1(−1) = π/4 − (−π/4) = π/223. sec−1x2√2= sec−12 − sec−1√2 = π/3 − π/4 = π/1224. − sec−1x−2/√3−√2= − sec−1(−2/√3) + sec−1(−√2) = −5π/6 + 3π/4 = −π/1225. 2√t − 2t3/241= −12 26. 8√y +43y3/2−23y3/294= 10819/32427.12x2− 2 cot xπ/2π/6= π2/9 + 2√3 28. a1/2x −23x3/24aa= −53a3/229. (a)1−1|2x − 1| dx =1/2−1(1 − 2x) dx +11/2(2x − 1) dx = (x − x2)1/2−1+ (x2− x)11/2=52(b)π/20cos x dx +3π/4π/2(− cos x)dx = sin xπ/20− sin x3π/4π/2= 2 −√2/230. (a)0−1√2 − x dx +20√2 + x dx = −23(2 − x)3/20−1+23(2 + x)3/220= −23(2√2 − 3√3) +23(8 − 2√2) =23(8 − 4√2 + 3√3)(b)π/30(cos x − 1/2) dx +π/2π/3(1/2 − cos x) dx= (sin x − x/2)π/30+ (x/2 − sin x)π/2π/3= (√3/2 − π/6) + π/4 − 1 − (π/6 −√3/2) =√3 − π/12 − 131. (a)0−1(1−ex)dx+10(ex−1)dx = (x−ex)0−1+(ex−x)10= −1−(−1−e−1)+e−1−1 = e+1/e−2(b)212 − xxdx +42x − 2xdx = 2 ln x21− 1 + 2 − 2 ln x42= 2 ln 2 + 1 − 2 ln 4 + 2 ln 2 = 1
• Exercise Set 6.6 25932. (a) The function f(x) = x2− 1 −15x2 + 1is an even function and changes sign at x = 2, thus3−3|f(x)| dx = 230|f(x)| dx = −220f(x) dx + 232f(x) dx=283− 30 tan−1(3) + 60 tan−1(2)(b)√3/201√1 − x2−√2 dx = −√2/201√1 − x2−√2 dx +√3/2√2/21√1 − x2−√2 dx= −2 sin−1√22+ sin−1√32−√2√32−√22+ 1 = −2π4+π3−√3√2+ 2= 2 −√3√2−π633. (a) 17/6 (b) F(x) =12x2, x ≤ 113x3+16, x > 134. (a)10√x dx +411x2dx =23x3/210−1x41= 17/12(b) F(x) =23x3/2, x < 1−1x+53, x ≥ 135. 0.665867079;311x2dx = −1x31= 2/336. 1.000257067;π/20sin x dx = − cos xπ/20= 137. 3.106017890;1−1sec2x dx = tan x1−1= 2 tan 1 ≈ 3.11481545038. 1.098242635;311xdx = ln x31= ln 3 ≈ 1.09861228939. A =30(x2+ 1)dx =13x3+ x30= 1240. A =10(x − x2) dx =12x2−13x310=12−13=1641. A =2π/303 sin x dx = −3 cos x2π/30= 9/242. A = −−1−2x3dx = −14x4−1−2= 15/4
• 260 Chapter 643. Area = −10(x2− x) dx +21(x2− x) dx = 5/6 + 1/6 = 1212xyA1 A244. Area =π0sin x dx −3π/2πsin x dx = 2 + 1 = 3–116i xyA1A245. Area = −5/40(2√x + 1 − 3) dx +35/4(2√x + 1 − 3) dx= 7/12 + 11/12 = 3/22 3–11xyA1A246. Area =11/2(x2− 1)/x2dx +21(x2− 1)/x2dx = 1/2 + 1/2 = 12–3–11xyA1A247. Area = −0−1(ex− 1) dx +10(ex− 1) dx = 1/e + e − 2A1A2–11–112xy
• Exercise Set 6.6 26148. Area = −21x − 2xdx +10x − 2xdx = 2 ln 2 − 1 + 2 ln 2 − 2 ln 3 + 1 = 4 ln 2 − 2 ln 31 3–3–2–1xyA2A149. (a) A =0.801√1 − x2dx = sin−1x0.80= sin−1(0.8)(b) The calculator was in degree mode instead of radian mode; the correct answer is 0.93.50. (a) the area is positive(b)5−21100x3−120x2−125x +15dx =1400x4−160x3−150x2+15x5−2=343120051. (a) the area between the curve and the x-axis breaks into equal parts, one above and one belowthe x-axis, so the integral is zero(b)1−1x3dx =14x41−1=14(14− (−1)4) = 0;π/2−π/2sin xdx = − cos xπ/2−π/2= − cos(π/2) + cos(−π/2) = 0 + 0 = 0(c) The area on the left side of the y-axis is equal to the area on the right side, soa−af(x)dx = 2a0f(x)dx(d)1−1x2dx =13x31−1=13(13− (−1)3) =23= 210x2dx;π/2−π/2cos xdx = sin xπ/2−π/2= sin(π/2) − sin(−π/2) = 1 + 1 = 2 = 2π/20cos xdx52. The numerator is an odd function and the denominator is an even function, so the integrand is anodd function and the integral is zero.53. (a) F (x) = 3x2− 3(b)x1(3t2− 3) dt = (t3− 3t)x1= x3− 3x + 2, andddx(x3− 3x + 2) = 3x2− 354. (a) cos 2x (b) F(x) =12sin 2txπ/4=12sin 2x −12, F (x) = cos 2x55. (a) sin x2(b) e√x56. (a)11 +√x(b) ln x
• 262 Chapter 657. −xcos x58. |u|59. F (x) =√x2 + 9, F (x) =x√x2 + 9(a) 0 (b) 5 (c)4560. F (x) = tan−1x, F (x) =11 + x2(a) 0 (b) π/3 (c) 1/461. (a) F (x) =x − 3x2 + 7= 0 when x = 3, which is a relative minimum, and hence the absoluteminimum, by the ﬁrst derivative test.(b) increasing on [3, +∞), decreasing on (−∞, 3](c) F (x) =7 + 6x − x2(x2 + 7)2=(7 − x)(1 + x)(x2 + 7)2; concave up on (−1, 7), concave down on (−∞, −1)and on (7, +∞)62. Ft23–20 –10 2063. (a) (0, +∞) because f is continuous there and 1 is in (0, +∞)(b) at x = 1 because F(1) = 064. (a) (−3, 3) because f is continuous there and 1 is in (−3, 3)(b) at x = 1 because F(1) = 065. (a)30√x dx =23x3/230= 2√3 = f(x∗)(3 − 0), so f(x∗) =2√3, x∗=43(b)0−12(x2+ x) dx =13x3+12x20−12= 504, so f(x∗)(0 − (−12)) = 504, x2+ x = 42, x∗= 666. (a) fave =12ππ−πsin x dx = 0; sin x∗= 0, x∗= −π, 0, π(b) fave =12311x2dx =13;1(x∗)2=13, x∗=√367.√2 ≤√x3 + 2 ≤√29, so 3√2 ≤30x3 + 2dx ≤ 3√2968. Let f(x) = x sin x, f(0) = f(1) = 0, f (x) = sin x + x cos x = 0 when x = − tan x, x ≈ 2.0288,so f has an absolute maximum at x ≈ 2.0288; f(2.0288) ≈ 1.8197, so 0 ≤ x sin x ≤ 1.82 and0 ≤π0x sin xdx ≤ 1.82π = 5.72
• Exercise Set 6.6 26369. (a) cF(x)ba= cF(b) − cF(a) = c[F(b) − F(a)] = c F(x)ba(b) F(x) + G(x)ba= [F(b) + G(b)] − [F(a) + G(a)]= [F(b) − F(a)] + [G(b) − G(a)] = F(x)ba+ G(x)ba(c) F(x) − G(x)ba= [F(b) − G(b)] − [F(a) − G(a)]= [F(b) − F(a)] − [G(b) − G(a)] = F(x)ba− G(x)ba71. (a) the increase in height in inches, during the ﬁrst ten years(b) the change in the radius in centimeters, during the time interval t = 1 to t = 2 seconds(c) the change in the speed of sound in ft/s, during an increase in temperature from t = 32◦Fto t = 100◦F(d) the displacement of the particle in cm, during the time interval t = t1 to t = t2 seconds72. (a)10V (t)dt gal(b) the change f(x1) − f(x2) in the values of f over the interval73. (a) amount of water = (rate of ﬂow)(time) = 4t gal, total amount = 4(30) = 120 gal(b) amount of water =600(4 + t/10)dt = 420 gal(c) amount of water =1200(10 +√t)dt = 1200 + 160√30 ≈ 2076.36 gal74. (a) The maximum value of R occurs at 4:30 P.M. when t = 0.(b)600100(1 − 0.0001t2)dt = 5280 cars75.nk=1π4nsec2 πk4n=nk=1f(x∗k)∆x where f(x) = sec2x, x∗k =πk4nand ∆x =π4nfor 0 ≤ x ≤π4.Thus limn→+∞nk=1π4nsec2 πk4n= limn→+∞nk=1f(x∗k)∆x =π/40sec2x dx = tan xπ/40= 176.nn2 + k2=11 + k2/n21nsonk=1nn2 + k2=nk=1f(x∗k)∆x where f(x) =11 + x2, x∗k =kn, and ∆x =1nfor 0 ≤ x ≤ 1. Thus limn→+∞nk=1nn2 + k2= limn→+∞nk=1f(x∗k)∆x =1011 + x2dx =π4.77. Let f be continuous on a closed interval [a, b] and let F be an antiderivative of f on [a, b]. ByTheorem 5.7.2,F(b) − F(a)b − a= F (x∗) for some x∗in (a, b). By Theorem 6.6.1,baf(x) dx = F(b) − F(a), i.e.baf(x) dx = F (x∗)(b − a) = f(x∗)(b − a).
• 264 Chapter 6EXERCISE SET 6.71. (a) displ = s(3) − s(0)=30v(t)dt =20(1 − t)dt +32(t − 3)dt = (t − t2/2)20+ (t2/2 − 3t)32= −1/2;dist =30|v(t)|dt = (t − t2/2)10+ (t2/2 − t)21− (t2/2 − 3t)32= 3/2(b) displ = s(3) − s(0)=30v(t)dt =10tdt +21dt +32(5 − 2t)dt = t2/210+ t21+ (5t − t2)32= 3/2;dist =10tdt +21dt +5/22(5 − 2t)dt +35/2(2t − 5)dt= t2/210+ t21+ (5t − t2)5/22+ (t2− 5t)35/2= 22. vt–112 4 8 103. (a) v(t) = 20 +t0a(u)du; add areas of the small blocks to getv(4) ≈ 20 + 1.4 + 3.0 + 4.7 + 6.2 = 35.3 m/s(b) v(6) = v(4) +64a(u)du ≈ 35.3 + 7.5 + 8.6 = 51.4 m/s4. (a) negative, because v is decreasing(b) speeding up when av > 0, so 2 < t < 5; slowing down when 1 < t < 2(c) negative, because the area between the graph of v(t) and the t-axis appears to be greaterwhere v < 0 compared to where v > 05. (a) s(t) = t3− t2+ C; 1 = s(0) = C, so s(t) = t3− t2+ 1(b) v(t) = − cos 3t + C1; 3 = v(0) = −1 + C1, C1 = 4, so v(t) = − cos 3t + 4. Thens(t) = −13sin 3t + 4t + C2; 3 = s(0) = C2, so s(t) = −13sin 3t + 4t + 36. (a) s(t) = t − cos t + C1; −3 = s(0) = −1 + C1, C1 = −2, so s(t) = t − cos t − 2(b) v(t) =13t3−32t2+ t + C1; 0 = v(0) = C1, so C1 = 0, v(t) =13t3−32t2+ t.Then s(t) =112t4−12t3+12t2+ C2; 0 = s(0) = +C2,,so C2 = 0, s(t) = 112 t4− 12 t3− 12 t2− 112
• Exercise Set 6.7 2657. (a) s(t) =32t2+ t + C; 4 = s(2) = 6 + 2 + C, C = −4 and s(t) =32t2+ t − 4(b) v(t) = −t−1+ C1, 0 = v(1) = −1 + C1, C1 = 1 andv(t) = −t−1+ 1 so s(t) = − ln t + t + C2, 2 = s(1) = 1 + C2,C2 = 1 and s(t) = − ln t + t + 18. (a) s(t) = t2/3dt =35t5/3+ C, s(8) = 0 =3532 + C, C = −965, s(t) =35t5/3−965(b) v(t) =√tdt =23t3/2+ C1, v(4) = 1 =238 + C1, C1 = −133, v(t) =23t3/2−133,s(t) =23t3/2−133dt =415t5/2−133t + C2,s(4) = −5 =41532 −1334 + C2 = −445+ C2,C2 =195, s(t) =415t5/2−133t +1959. (a) displacement = s(π/2) − s(0) =π/20sin tdt = − cos tπ/20= 1 mdistance =π/20| sin t|dt = 1 m(b) displacement = s(2π) − s(π/2) =2ππ/2cos tdt = sin t2ππ/2= −1 mdistance =2ππ/2| cos t|dt = −3π/2π/2cos tdt +2π3π/2cos tdt = 3 m10. (a) displacement =20(3t − 2) dt = 2distance =20|3t − 2| dt = −2/30(3t − 2) dt +22/3(3t − 2) dt =23+83=103m(b) displacement =20|1 − 2t| dt =52mdistance =20|1 − 2t| dt =52m11. (a) v(t) = t3− 3t2+ 2t = t(t − 1)(t − 2)displacement =30(t3− 3t2+ 2t)dt = 9/4 mdistance =30|v(t)|dt =10v(t)dt +21−v(t)dt +32v(t)dt = 11/4 m(b) displacement =30(√t − 2)dt = 2√3 − 6 mdistance =30|v(t)|dt = −30v(t)dt = 6 − 2√3 m
• 266 Chapter 612. (a) displacement =40(t −√t) dt =83mdistance =40|t −√t| dt = 3 m(b) displacement =301√t + 1dt = 2 mdistance =301√t + 1dt = 2 m13. v = 3t − 1displacement =20(3t − 1) dt = 4 mdistance =20|3t − 1| dt =133m14. v(t) =12t2− 2tdisplacement =5112t2− 2t dt = −10/3 mdistance =5112t2− 2t dt =41−12t2− 2t dt +5412t2− 2t dt = 17/3 m15. v = (1/√3t + 1 dt =23√3t + 1 + C; v(0) = 4/3 so C = 2/3, v = 23√3t + 1 + 2/3displacement =5123√3t + 1 dt =29627mdistance =5123√3t + 1 dt =29627m16. v(t) = − cos t + 2displacement =π/2π/4(− cos t + 2)dt = (π +√2 − 2)/2 mdistance =π/2π/4| − cos t + 2|dt =π/2π/4(− cos t + 2)dt = (π +√2 − 2)/2 m17. (a) s = sin12πt dt = −2πcos12πt + Cs = 0 when t = 0 which gives C =2πso s = −2πcos12πt +2π.a =dvdt=π2cos12πt. When t = 1 : s = 2/π, v = 1, |v| = 1, a = 0.(b) v = −3 t dt = −32t2+ C1, v = 0 when t = 0 which gives C1 = 0 so v = −32t2s = −32t2dt = −12t3+ C2, s = 1 when t = 0 which gives C2 = 1 so s = −12t3+ 1.When t = 1 : s = 1/2, v = −3/2, |v| = 3/2, a = −3.
• Exercise Set 6.7 26718. (a) s = cosπt3dt =3πsinπt3+ Cs = 0 when t =32which gives C = −3πso s =3πsinπt3−3πa =dvdt= −π3sinπt3; when t = 1 : s =3π√32−3π, v =12, a = −π3√32(b) v = 4e2t−2dt = 2e2t−2+ C, v =2e2− 3 when t = 0, hence C = −3 so v = 2e2t−2− 3. Thens(t) = v(t) dt = e2t−2− 3t + C , s = e−2when t = 0, so C = 0 and s(t) = e2t−2− 3t.When t = 1, s(1) = −2, v(1) = −1, a(1) = 4.19. By inspection the velocity is positive for t > 0, and during the ﬁrst second the particle is at most5/2 cm from the starting position. For T > 1 the displacement of the particle during the timeinterval [0, T] is given byT0v(t) dt = 5/2 +T1(6√t − 1/t) dt = 5/2 + (4t3/2− ln t)T1= −3/2 + 4T3/2− ln T,and the displacement equals 4 cm if 4T3/2− ln T = 11/2, T ≈ 1.272 s20. The displacement of the particle during the time interval [0, T] is given byT0v(t)dt = 3 tan−1T − 0.25T2. The particle is 2 cm from its starting position when3 tan−1T − 0.25T2= 2 or when 3 tan−1T − 0.25T2= −2; solve for T to getT = 0.90, 2.51, and 4.95 s.21. s(t) = (20t2− 110t + 120) dt =203t3− 55t2+ 120t + C. But s = 0 when t = 0, so C = 0 ands =203t3− 55t2+ 120t. Moreover, a(t) =ddtv(t) = 40t − 110.180–400130–130018000 666s(t) v(t) a(t)22. a(t) = 4t − 30, v(t) = 2t2− 30t + 3, s(t) =23t3− 15t2+ 3t − 5;500–1101100–1100070–3000 252525s(t) v(t) a(t)
• 268 Chapter 65.5–1.50 523. (a) positive on (0, 0.74) and (2.97, 5), negative on (0.75, 2.97)(b) For 0 < T < 5 the displacement isdisp = T/2 − sin(T) + T cos(T)1.500 124. (a) the displacement is positive on (0, 1)(b) For 0 < T < 1 the displacement isdisp =1π2+12T −1π2cos πT −1πT sin πT0.500 525. (a) the displacement is positive on (0, 5)(b) For 0 < T < 5 the displacement isdisp =12T + (T + 1)e−T− 10.1–0.30 126. (a) the displacement is negative on (0, 1)(b) For 0 < T < 1 the displacement isdisp = −1200ln25+12T2−1200ln T +110−14T2+120T27. (a) a(t) =0, t < 4−10, t > 4a t–10–52 4 12(b) v(t) =25, t < 465 − 10t, t > 4vt–40–20202 4 6 8 10 12(c) x(t) =25t, t < 465t − 5t2− 80, t > 4, so x(8) = 120, x(12) = −20(d) x(6.5) = 131.25
• Exercise Set 6.7 26928. (a) From (11) t =v − v0a; from that and (10)s − s0 = v0v − v0a+12a(v − v0)2a2; multiply through by a to geta(s − s0) = v0(v − v0) +12(v − v0)2= (v − v0) v0 +12(v − v0) =12(v2− v20). Thusa =v2− v202(s − s0).(b) Put the last result of Part (a) into the ﬁrst equation of Part (a) to obtaint =v − v0a= (v − v0)2(s − s0)v2 − v20=2(s − s0)v + v0.(c) From (11) v0 = v − at; use this in (10) to gets − s0 = (v − at)t +12at2= vt −12at2This expression contains no v0 terms and so diﬀers from (10).29. (a) a = −1.5 mi/h/s = −33/15 ft/s2(b) a = 30 km/h/min = 1/7200 km/s230. Take t = 0 when deceleration begins, then a = −11 so v = −11t + C1, but v = 88 when t = 0which gives C1 = 88 thus v = −11t + 88, t ≥ 0(a) v = 45 mi/h = 66 ft/s, 66 = −11t + 88, t = 2 s(b) v = 0 (the car is stopped) when t = 8 ss = v dt = (−11t + 88)dt = −112t2+ 88t + C2, and taking s = 0 when t = 0, C2 = 0 sos = −112t2+ 88t. At t = 8, s = 352. The car travels 352 ft before coming to a stop.31. a = a0 ft/s2, v = a0t + v0 = a0t + 132 ft/s, s = a0t2/2 + 132t + s0 = a0t2/2 + 132t ft; s = 200 ftwhen v = 88 ft/s. Solve 88 = a0t + 132 and 200 = a0t2/2 + 132t to get a0 = −1215when t =2011,so s = −12.1t2+ 132t, v = −1215t + 132.(a) a0 = −1215ft/s2(b) v = 55 mi/h =2423ft/s when t =7033s(c) v = 0 w hen t =6011s32. dv/dt = 5, v = 5t + C1, but v = v0 when t = 0 so C1 = v0, v = 5t + v0. From ds/dt = v = 5t + v0we get s = 5t2/2 + v0t + C2 and, with s = 0 when t = 0, C2 = 0 so s = 5t2/2 + v0t. s = 60 whent = 4 thus 60 = 5(4)2/2 + v0(4), v0 = 5 m/s33. Suppose s = s0 = 0, v = v0 = 0 at t = t0 = 0; s = s1 = 120, v = v1 at t = t1; and s = s2,v = v2 = 12 at t = t2. From Exercise 28(a),2.6 = a =v21 − v202(s1 − s0), v21 = 2as1 = 5.2(120) = 624. Applying the formula again,−1.5 = a =v22 − v212(s2 − s1), v22 = v21 − 3(s2 − s1), sos2 = s1 − (v22 − v21)/3 = 120 − (144 − 624)/3 = 280 m.
• 270 Chapter 634. a(t) =4, t < 20, t > 2, so, with v0 = 0, v(t) =4t, t < 28, t > 2and,since s0 = 0, s(t) =2t2, t < 28t − 8, t > 2s = 100 when 8t − 8 = 100, t = 108/8 = 13.5 s35. The truck’s velocity is vT = 50 and its position is sT = 50t + 2500. The car’s acceleration isaC = 4 ft/s2, so vC = 4t, sC = 2t2(initial position and initial velocity of the car are bothzero). sT = sC when 50t + 2500 = 2t2, 2t2− 50t − 2500 = 2(t + 25)(t − 50) = 0, t = 50 s andsC = sT = 2t2= 5000 ft.36. Let t = 0 correspond to the time when the leader is 100 m from the ﬁnish line; let s = 0 cor-respond to the ﬁnish line. Then vC = 12, sC = 12t − 115; aL = 0.5 for t > 0, vL = 0.5t + 8,sL = 0.25t2+ 8t − 100. sC = 0 at t = 115/12 ≈ 9.58 s, and sL = 0 at t = −16 + 4√41 ≈ 9.61,so the challenger wins.37. s = 0 and v = 112 when t = 0 so v(t) = −32t + 112, s(t) = −16t2+ 112t(a) v(3) = 16 ft/s, v(5) = −48 ft/s(b) v = 0 when the projectile is at its maximum height so −32t + 112 = 0, t = 7/2 s,s(7/2) = −16(7/2)2+ 112(7/2) = 196 ft.(c) s = 0 when it reaches the ground so −16t2+ 112t = 0, −16t(t − 7) = 0, t = 0, 7 of whicht = 7 is when it is at ground level on its way down. v(7) = −112, |v| = 112 ft/s.38. s = 112 when t = 0 so s(t) = −16t2+ v0t + 112. But s = 0 when t = 2 thus−16(2)2+ v0(2) + 112 = 0, v0 = −24 ft/s.39. (a) s(t) = 0 when it hits the ground, s(t) = −16t2+ 16t = −16t(t − 1) = 0 when t = 1 s.(b) The projectile moves upward until it gets to its highest point where v(t) = 0,v(t) = −32t + 16 = 0 when t = 1/2 s.40. (a) s(t) = s0 − 12 gt2= 800 − 16t2ft, s(t) = 0 when t =80016= 5√2(b) v(t) = −32t and v(5√2) = −160√2 ≈ 226.27 ft/s = 154.28 mi/h41. s(t) = s0 + v0t − 12 gt2= 60t − 4.9t2m and v(t) = v0 − gt = 60 − 9.8t m/s(a) v(t) = 0 when t = 60/9.8 ≈ 6.12 s(b) s(60/9.8) ≈ 183.67 m(c) another 6.12 s; solve for t in s(t) = 0 to get this result, or use the symmetry of the parabolas = 60t − 4.9t2about the line t = 6.12 in the t-s plane(d) also 60 m/s, as seen from the symmetry of the parabola (or compute v(6.12))42. (a) they are the same(b) s(t) = v0t − 12 gt2and v(t) = v0 − gt; s(t) = 0 when t = 0, 2v0/g;v(0) = v0 and v(2v0/g) = v0 − g(2v0/g) = −v0 so the speed is the sameat launch (t = 0) and at return (t = 2v0/g).43. s(t) = −4.9t2+ 49t + 150 and v(t) = −9.8t + 49(a) the projectile reaches its maximum height when v(t) = 0, −9.8t + 49 = 0, t = 5 s(b) s(5) = −4.9(5)2+ 49(5) + 150 = 272.5 m(c) the projectile reaches its starting point when s(t) = 150, −4.9t2+ 49t + 150 = 150,−4.9t(t − 10) = 0, t = 10 s
• Exercise Set 6.8 271(d) v(10) = −9.8(10) + 49 = −49 m/s(e) s(t) = 0 when the projectile hits the ground, −4.9t2+ 49t + 150 = 0 when (use the quadraticformula) t ≈ 12.46 s(f) v(12.46) = −9.8(12.46) + 49 ≈ −73.1, the speed at impact is about 73.1 m/s44. take s = 0 at the water level and let h be the height of the bridge, then s = h and v = 0 whent = 0 so s(t) = −16t2+ h(a) s = 0 when t = 4 thus −16(4)2+ h = 0, h = 256 ft(b) First, ﬁnd how long it takes for the stone to hit the water (ﬁnd t for s = 0) : −16t2+ h = 0,t =√h/4. Next, ﬁnd how long it takes the sound to travel to the bridge: this time is h/1080because the speed is constant at 1080 ft/s. Finally, use the fact that the total of these twotimes must be 4 s:h1080+√h4= 4, h + 270√h = 4320, h + 270√h − 4320 = 0, and bythe quadratic formula√h =−270 ± (270)2 + 4(4320)2, reject the negative value to get√h ≈ 15.15, h ≈ 229.5 ft.45. If g = 32 ft/s2, s0 = 7 and v0 is unknown, then s(t) = 7 + v0t − 16t2and v(t) = v0 − 32t; s = smaxwhen v = 0, or t = v0/32; and smax = 208 yields208 = s(v0/32) = 7 + v0(v0/32) − 16(v0/32)2= 7 + v20/64, so v0 = 8√201 ≈ 113.42 ft/s.46. s = 1000+v0t− 12 (32)t2= 1000+v0t−16t2; s = 0 when t = 5, so v0 = −(1000+16·52)/5 = −280ft/s.EXERCISE SET 6.81. (a)1251u3du (b)32259√u du(c)1ππ/2−π/2cos u du (d)21(u + 1)u5du2. (a)127−3u8du (b)5/23/21√udu(c)10u2du (d)1243(u − 3)u1/2du3. (a)121−1eudu (b)21u du4. (a)π/3π/4√u du (b)1/20du√1 − u25. u = 2x + 1,1231u3du =18u431= 10 or18(2x + 1)410= 106. u = 4x − 2,1462u3du =116u462= 80, or116(4x − 2)421= 80
• 272 Chapter 67. u = 2x − 1,121−1u3du = 0, because u3is odd on [−1, 1].8. u = 4 − 3x, −13−21u8du = −127u9−21= 19, or −127(4 − 3x)921= 199. u = 1 + x,91(u − 1)u1/2du =91(u3/2− u1/2)du =25u5/2−23u3/291= 1192/15,or25(1 + x)5/2−23(1 + x)3/280= 1192/1510. u = 1 − x,41(1 − u)√u du =23u3/2−25u5/241= −116/15 or23(1 − x)3/2−25(1 − x)5/20−3= −116/1511. u = x/2, 8π/40sin u du = −8 cos uπ/40= 8 − 4√2, or − 8 cos(x/2)π/20= 8 − 4√212. u = 3x,23π/20cos u du =23sin uπ/20= 2/3, or23sin 3xπ/60= 2/313. u = x2+ 2,1236u−3du = −14u236= −1/48, or −141(x2 + 2)2−1−2= −1/4814. u =14x −14, 4π/4−π/4sec2u du = 4 tan uπ/4−π/4= 8, or 4 tan14x −141+π1−π= 815. u = ex+ 4, du = exdx, u = e− ln 3+ 4 =13+ 4 =133when x = − ln 3u = eln 3+ 4 = 3 + 4 = 7 when x = ln 3,713/31udu = ln u713/3= ln(7) − ln(13/3) = ln(21/13), orln(ex+ 4)ln 3− ln 3= ln 7 − ln(13/3) = ln 21/1316. u = 3 − 4ex, du = −4exdx, u = −1 when x = 0, u = −17 when x = ln 5−14−17−1u du = −18u2−17−1= −36, or −18(3 − 4ex)2ln 50= −3617. u =√x, 2√311u2 + 1du = 2 tan−1u√31= 2(tan−1√3 − tan−11) = 2(π/3 − π/4) = π/6 or2 tan−1√x31= π/6
• Exercise Set 6.8 27318. u = e−x, −√3/21/21√1 − u2du = − sin−1u√3/21/2= − sin−1√32+ sin−1 12= −π3+π6= −π6or− sin−1e−xln(2/√3)ln 2= −π3+π6= −π/619.135−525 − u2 du =1312π(5)2=256π 20.124016 − u2 du =1214π(4)2= 2π21. −12011 − u2 du =12101 − u2 du =12·14[π(1)2] = π/822.3−39 − u2du = π(3)2/2 =92π23.10sin πxdx = −1πcos πx10= −1π(−1 − 1) = 2/π24. A =π/803 cos 2x dx =32sin 2xπ/80= 3√2/425.1−19(x + 2)2dx = −9(x + 2)−11−1= −913− 1 = 626. A =10dx(3x + 1)2= −13(3x + 1)10=1427. A =1/601√1 − 9x2dx =131/201√1 − u2du =13sin−1u1/20= π/1828. x = sin y, A =π/20sin y dy = − cos yπ/20= 129. u = 2x − 1,12911√udu =√u91= 2 30.215(5x − 1)3/221= 38/1531.23(x3+ 9)1/21−1=23(√10 − 2√2) 32. u = cos x + 1, b10u5du =b633. u = x2+ 4x + 7,122812u−1/2du = u1/22812=√28 −√12 = 2(√7 −√3)34.211(x − 3)2dx = −1x − 321= 1/235. 2 sin2xπ/40= 1 36.23(tan x)3/2π/40= 2/3 37.52sin(x2)√π0= 0
• 274 Chapter 638. u =√x, 22ππsin u du = −2 cos u2ππ= −439. u = 3θ,13π/3π/4sec2u du =13tan uπ/3π/4= (√3 − 1)/340. u = cos 2θ, −121/211udu =12ln u11/2= ln√241. u = 4 − 3y, y =13(4 − u), dy = −13du−1271416 − 8u + u2u1/2du =12741(16u−1/2− 8u1/2+ u3/2)du=12732u1/2−163u3/2+25u5/241= 106/40542. u = 5 + x,94u − 5√udu =94(u1/2− 5u−1/2)du =23u3/2− 10u1/294= 8/343.12ln(2x + e)e0=12(ln(3e) − ln e) =ln 3244. −12e−x2√21= (e−1− e−2)/245. u =√3x2,12√3√301√4 − u2du =12√3sin−1 u2√30=12√3π3=π6√346. u =√x, 2√211√4 − u2du = 2 sin−1 u2√21= 2(π/4 − π/6) = π/647. u = 3x,13√3011 + u2du =13tan−1u√30=13π3=π948. u = x2,123113 + u2du =12√3tan−1 u√331=12√3(π/3 − π/6) =π12√349. (b)π/60sin4x(1 − sin2x) cos x dx =15sin5x −17sin7xπ/60=1160−1896=23448050. (b)π/4−π/4tan2x(sec2x − 1) dx =13tan3xπ/4−π/4−π/4−π/4(sec2x − 1) dx=23+ (− tan x + x)π/4−π/4=23− 2 +π2= −43+π251. (a) u = 3x + 1,1341f(u)du = 5/3 (b) u = 3x,1390f(u)du = 5/3(c) u = x2, 1/204f(u)du = −1/240f(u)du = −1/2
• Exercise Set 6.8 27552. u = 1 − x,10xm(1 − x)ndx = −01(1 − u)mundu =10un(1 − u)mdu =10xn(1 − x)mdx53. sin x = cos(π/2 − x),π/20sinnx dx =π/20cosn(π/2 − x)dx = −0π/2cosnu du (u = π/2 − x)=π/20cosnu du =π/20cosnx dx (by replacing u by x)54. u = 1 − x, −01(1 − u)undu =10(1 − u)undu =10(un− un+1)du =1n + 1−1n + 2=1(n + 1)(n + 2)55. y(t) = (802.137) e1.528tdt = 524.959e1.528t+ C; y(0) = 750 = 524.959 + C, C = 225.041,y(t) = 524.959e1.528t+ 225.041, y(12) = 48,233,500,00056. s(t) = (25 + 10e−0.05t)dt = 25t − 200e−0.05t+ C(a) s(10) − s(0) = 250 − 200(e−0.5− 1) = 450 − 200/√e ≈ 328.69 ft(b) yes; without it the distance would have been 250 ft57.k0e2xdx = 3,12e2xk0= 3,12(e2k− 1) = 3, e2k= 7, k =12ln 758. The area is given by201/(1 + kx2)dx = (1/√k) tan−1(2√k) = 0.6; solve for k to getk = 5.081435.59. (a)10sin πxdx = 2/π60. Let u = t − x, then du = −dx andt0f(t − x)g(x)dx = −0tf(u)g(t − u)du =t0f(u)g(t − u)du;the result follows by replacing u by x in the last integral.61. (a) I = −0af(a − u)f(a − u) + f(u)du =a0f(a − u) + f(u) − f(u)f(a − u) + f(u)du=a0du −a0f(u)f(a − u) + f(u)du, I = a − I so 2I = a, I = a/2(b) 3/2 (c) π/462. x =1u, dx = −1u2du, I =1−111 + 1/u2(−1/u2)du = −1−11u2 + 1du = −I so I = 0 which isimpossible because11 + x2is positive on [−1, 1]. The substitution u = 1/x is not valid because uis not continuous for all x in [−1, 1].
• 276 Chapter 663. (a) Let u = −x thena−af(x)dx = −−aaf(−u)du =a−af(−u)du = −a−af(u)duso, replacing u by x in the latter integral,a−af(x)dx = −a−af(x)dx, 2a−af(x)dx = 0,a−af(x)dx = 0The graph of f is symmetric about the origin so0−af(x)dx is the negative ofa0f(x)dxthusa−af(x)dx =0−af(x) dx +a0f(x)dx = 0(b)a−af(x)dx =0−af(x)dx +a0f(x)dx, let u = −x in0−af(x)dx to get0−af(x)dx = −0af(−u)du =a0f(−u)du =a0f(u)du =a0f(x)dxsoa−af(x)dx =a0f(x)dx +a0f(x)dx = 2a0f(x)dxThe graph of f(x) is symmetric about the y-axis so there is as much signed area to the leftof the y-axis as there is to the right.64. (a) By Exercise 63(a),1−1x cos(x2) dx = 0(b) u = x − π/2, du = dx, sin(u + π/2) = sin u, cos(u + π/2) = − sin uπ0sin8x cos5x dx =π/2−π/2sin8u(− sin5u) du = −π/2−π/2sin13u du = 0 by Exercise 63(a).EXERCISE SET 6.91. (a) yt1231 2 3(b) yt1230.5 1(c) yt1231 e22. yt12312332
• Exercise Set 6.9 2773. (a) ln tac1= ln(ac) = ln a + ln c = 7 (b) ln t1/c1= ln(1/c) = −5(c) ln ta/c1= ln(a/c) = 2 − 5 = −3 (d) ln ta31= ln a3= 3 ln a = 64. (a) ln t√a1= ln a1/2=12ln a = 9/2 (b) ln t2a1= ln 2 + 9(c) ln t2/a1= ln 2 − 9 (d) ln ta2= 9 − ln 25. ln 5 ≈ 1.603210678; ln 5 = 1.609437912; magnitude of error is < 0.00636. ln 3 ≈ 1.098242635; ln 3 = 1.098612289; magnitude of error is < 0.00047. (a) x−1, x > 0 (b) x2, x = 0(c) −x2, −∞ < x < +∞ (d) −x, −∞ < x < +∞(e) x3, x > 0 (f) ln x + x, x > 0(g) x − 3√x, −∞ < x < +∞ (h)exx, x > 08. (a) f(ln 3) = e−2 ln 3= eln(1/9)= 1/9(b) f(ln 2) = eln 2+ 3e− ln 2= 2 + 3eln(1/2)= 2 + 3/2 = 7/29. (a) 3π= eπ ln 3(b) 2√2= e√2 ln 210. (a) π−x= e−x ln π(b) x2x= e2x ln x11. (a) y = 2x, limx→+∞1 +12xx= limx→+∞1 +12x2x 1/2= limy→+∞1 +1yy 1/2= e1/2(b) y = 2x, limy→0(1 + y)2/y= limy→0(1 + y)1/y2= e212. (a) y = x/3, limy→+∞1 +1yy 3= limy→+∞1 +1yy 3= e3(b) limx→0(1 + x)1/3x= limx→0(1 + x)1/x1/3= e1/313. g (x) = x2− x 14. g (x) = 1 − cos x15. (a)1x3(3x2) =3x(b) eln x 1x= 116. (a) 2x√x2 + 1 (b) −1x2sin1x17. F (x) =sin xx2 + 1, F (x) =(x2+ 1) cos x − 2x sin x(x2 + 1)2(a) 0 (b) 0 (c) 1
• 278 Chapter 618. F (x) =√3x2 + 1, F (x) =3x√3x2 + 1(a) 0 (b)√13 (c) 6/√1319. (a)ddxx21t√1 + tdt = x21 + x2(2x) = 2x31 + x2(b)x21t√1 + tdt = −23(x2+ 1)3/2+25(x2+ 1)5/2−4√21520. (a)ddxaxf(t)dt = −ddxxaf(t)dt = −f(x)(b)ddxag(x)f(t)dt = −ddxg(x)af(t)dt = −f(g(x))g (x)21. (a) − cos x3(b) −tan2x1 + tan2xsec2x = − tan2x22. (a) −1(x2 + 1)2(b) − cos3 1x−1x2=cos3(1/x)x223. −33x − 19x2 + 1+ 2xx2− 1x4 + 124. If f is continuous on an open interval I and g(x), h(x), and a are in I theng(x)h(x)f(t)dt =ah(x)f(t)dt +g(x)af(t)dt = −h(x)af(t)dt +g(x)af(t)dtsoddxg(x)h(x)f(t)dt = −f(h(x))h (x) + f(g(x))g (x)25. (a) sin2(x3)(3x2) − sin2(x2)(2x) = 3x2sin2(x3) − 2x sin2(x2)(b)11 + x(1) −11 − x(−1) =21 − x226. F (x) =15x(5)−1x(1) = 0 so F(x) is constant on (0, +∞). F(1) = ln 5 so F(x) = ln 5 for all x > 0.27. from geometry,30f(t)dt = 0,53f(t)dt = 6,75f(t)dt = 0; and107f(t)dt=107(4t − 37)/3dt = −3(a) F(0) = 0, F(3) = 0, F(5) = 6, F(7) = 6, F(10) = 3(b) F is increasing where F = f is positive, so on [3/2, 6] and [37/4, 10], decreasing on [0, 3/2]and [6, 37/4](c) critical points when F (x) = f(x) = 0, so x = 3/2, 6, 37/4; maximum 15/2 at x = 6, minimum−9/4 at x = 3/2
• Exercise Set 6.9 279(d) F(x)x–22462 4 6 8 1028. fave =110 − 0100f(t)dt =110F(10) = 0.329. x < 0 : F(x) =x−1(−t)dt = −12t2x−1=12(1 − x2),x ≥ 0 : F(x) =0−1(−t)dt +x0t dt =12+12x2; F(x) =(1 − x2)/2, x < 0(1 + x2)/2, x ≥ 030. 0 ≤ x ≤ 2 : F(x) =x0t dt =12x2,x > 2 : F(x) =20t dt +x22 dt = 2 + 2(x − 2) = 2x − 2; F(x) =x2/2, 0 ≤ x ≤ 22x − 2, x > 231. y(x) = 2 +x12t2+ 1tdt = 2 + (t2+ ln t)x1= x2+ ln x + 132. y(x) =x1(t1/2+ t−1/2)dt =23x3/2−23+ 2x1/2− 2 =23x3/2+ 2x1/2−8333. y(x) = 1 +xπ/4(sec2t − sin t)dt = tan x + cos x −√2/234. y(x) = 1 +xe1x ln xdx = 1 + ln ln txe= 1 + ln ln x35. P(x) = P0 +x0r(t)dt individuals 36. s(T) = s1 +T1v(t)dt37. II has a minimum at x = 12, and I has a zero there, so I could be the derivative of II; on the otherhand I has a minimum near x = 1/3, but II is not zero there, so II could not be the derivative ofI, so I is the graph of f(x) and II is the graph ofx0f(t) dt.38. (b) limk→01k(xk− 1) =ddtxtt=0= ln x39. (a) where f(t) = 0; by the First Derivative Test, at t = 3(b) where f(t) = 0; by the First Derivative Test, at t = 1, 5(c) at t = 0, 1 or 5; from the graph it is evident that it is at t = 5(d) at t = 0, 3 or 5; from the graph it is evident that it is at t = 3
• 280 Chapter 6(e) F is concave up when F = f is positive, i.e. where f is increasing, so on (0, 1/2) and (2, 4);it is concave down on (1/2, 2) and (4, 5)(f) F(x)x–1–0.50.511 2 3 540. (a)x–11–4 –2 2 4erf(x)(c) erf (x) > 0 for all x, so there are no relative extrema(e) erf (x) = −4xe−x2/√π changes sign only at x = 0 so that is the only point of inﬂection(g) limx→+∞erf(x) = +1, limx→−∞erf(x) = −141. C (x) = cos(πx2/2), C (x) = −πx sin(πx2/2)(a) cos t goes from negative to positive at 2kπ − π/2, and from positive to negative att = 2kπ + π/2, so C(x) has relative minima when πx2/2 = 2kπ − π/2, x = ±√4k − 1,k = 1, 2, . . ., and C(x) has relative maxima when πx2/2 = (4k + 1)π/2, x = ±√4k + 1,k = 0, 1, . . ..(b) sin t changes sign at t = kπ, so C(x) has inﬂection points at πx2/2 = kπ, x = ±√2k,k = 1, 2, . . .; the case k = 0 is distinct due to the factor of x in C (x), but x changes sign atx = 0 and sin(πx2/2) does not, so there is also a point of inﬂection at x = 042. Let F(x) =x1ln tdt, F (x) = limh→0F(x + h) − F(x)h= limh→01hx+hxln tdt; but F (x) = ln x solimh→01hx+hxln tdt = ln x43. Diﬀerentiate: f(x) = 2e2x, so 4 +xaf(t)dt = 4 +xa2e2tdt = 4 + e2txa= 4 + e2x− e2a= e2xprovided e2a= 4, a = (ln 4)/2.44. (a) The area under 1/t for x ≤ t ≤ x + 1 is less than the area of the rectangle with altitude 1/xand base 1, but greater than the area of the rectangle with altitude 1/(x + 1) and base 1.(b)x+1x1tdt = ln tx+1x= ln(x + 1) − ln x = ln(1 + 1/x), so1/(x + 1) < ln(1 + 1/x) < 1/x for x > 0.(c) from Part (b), e1/(x+1)< eln(1+1/x)< e1/x, e1/(x+1)< 1 + 1/x < e1/x,ex/(x+1)< (1 + 1/x)x< e; by the Squeezing Theorem, limx→+∞(1 + 1/x)x= e.
• Review Exercise Set 281(d) Use the inequality ex/(x+1)< (1 + 1/x)xto get e < (1 + 1/x)x+1so(1 + 1/x)x< e < (1 + 1/x)x+1.45. From Exercise 44(d) e − 1 +15050< y(50), and from the graph y(50) < 0.060.200 10046. F (x) = f(x), thus F (x) has a value at each x in I because f is continuous on I so F is continuouson I because a function that is diﬀerentiable at a point is also continuous at that pointREVIEW EXERCISE SET3. −14x2+83x3/2+ C 4. u4/4 − u2+ 7u + C5. −4 cos x + 2 sin x + C 6. (sec x tan x + 1)dx = sec x + x + C7. 3x1/3− 5ex+ C 8.34ln x − tan x + C9. tan−1x + 2 sin−1x + C 10. 12 sec−1x + x − 13 x3+ C11. (a) y(x) = 2√x −23x3/2+ C; y(1) = 0, so C = −43, y(x) = 2√x −23x3/2−43(b) y(x) = sin x − 5ex+ C, y(0) = 0 = −5 + C, C = 5, y(x) = sin x − 5ex+ 512. The direction ﬁeld is clearly an even function, which means that the solution is even, its derivativeis odd. Since sin x is periodic and the direction ﬁeld is not, that eliminates all but x, the solutionof which is the family y = x2/2 + C.13. (a) If u = sec x, du = sec x tan xdx, sec2x tan xdx = udu = u2/2 + C1 = (sec2x)/2 + C1;if u = tan x, du = sec2xdx, sec2x tan xdx = udu = u2/2 + C2 = (tan2x)/2 + C2.(b) They are equal only if sec2x and tan2x diﬀer by a constant, which is true.14.12sec2xπ/40=12(2 − 1) = 1/2 and 12 tan2xπ/40=12(1 − 0) = 1/215. u = x2− 1, du = 2x dx,12duu√u2 − 1=12sec−1u + C =12sec−1(x2− 1) + C
• 282 Chapter 616. 1 + x−2/3 dx = x−1/3x2/3 + 1 dx; u = x2/3+ 1, du =23x−1/3dx32u1/2du = u3/2+ C = (x2/3+ 1)3/2+ C17. u = 5 + 2 sin 3x, du = 6 cos 3xdx;16√udu =13u1/2+ C =13√5 + 2 sin 3x + C18. u = 3 +√x, du =12√xdx; 2√udu =43u3/2+ C =43(3 +√x)3/2+ C19. u = ax3+ b, du = 3ax2dx;13au2du = −13au+ C = −13a2x3 + 3ab+ C20. u = ax2, du = 2axdx;12asec2udu =12atan u + C =12atan(ax2) + C21. (a) 1 · 2 + 2 · 3 + · · · + n(n + 1) =nk=1k(k + 1) =nk=1k2+nk=1k=16n(n + 1)(2n + 1) +12n(n + 1) =13n(n + 1)(n + 2)(b)n−1k=19n−kn2=9nn−1k=11 −1n2n−1k=1k =9n(n − 1) −1n2·12(n − 1)(n) =172n − 1n;limn→+∞172n − 1n=172(c)3i=12j=1i +2j=1j =3i=12i +12(2)(3) = 23i=1i +3i=13 = 2 ·12(3)(4) + (3)(3) = 2122. (a)14k=0(k + 4)(k + 1) (b)19k=5(k − 1)(k − 4)23. For 1 ≤ k ≤ n the k-th L-shaped strip consists of the corner square, a strip above and a strip tothe left for a combined area of 1 + (k − 1) + (k − 1) = 2k − 1, so the total area isnk=1(2k − 1) = n2.24. 1 + 3 + 5 + · · · + (2n − 1) =nk=1(2k − 1) = 2nk=1k −nk=11 = 2 ·12n(n + 1) − n = n225. left endpoints: x∗k = 1, 2, 3, 4;4k=1f(x∗k)∆x = (2 + 3 + 2 + 1)(1) = 8right endpoints: x∗k = 2, 3, 4, 5;4k=1f(x∗k)∆x = (3 + 2 + 1 + 2)(1) = 8
• Review Exercise Set 28326. (a) x∗k = 0, 1, 2, 3, 44k=1f(x∗k)∆x = e0+ e1+ e2+ e3+ e4(1) = (1 − e5)/(1 − e) = 85.791(b) x∗k = 1, 2, 3, 4, 54k=1f(x∗k)∆x = e1+ e2+ e3+ e4+ e5(1) = e(1 − e5)/(1 − e) = 233.204(c) x∗k = 1/2, 3/2, 5/2, 7/2, 9/24k=1f(x∗k)∆x = e1/2+ e3/2+ e5/2+ e7/2+ e9/2(1) = e1/2(1 − e5)/(1 − e) = 141.44627. limn→+∞nk=144kn−4kn24n= limn→+∞64n3nk=1(kn − k2)= limn→+∞64n3n2(n + 1)2−n(n + 1)(2n + 1)6= limn→+∞646n3[n3− n] =32328. limn→+∞nk=125(k − 1)n−25(k − 1)2n25n=125629. 0.351220577, 0.420535296, 0.386502483 30. 1.63379940, 1.805627583, 1.71756608732. Since y = exand y = ln x are inverse functions, their graphsare symmetric with respect to the line y = x; consequently theareas A1 and A3 are equal (see ﬁgure). But A1 + A2 = e, soe1ln xdx +10exdx = A2 + A3 = A2 + A1 = eyx1e1 eA3A1A233. (a)12+14=34(b) −1 −12= −32(c) 5 −1 −34= −354(d) −2(e) not enough information (f) not enough information34. (a)12+ 2 =52(b) not enough information(c) not enough information (d) 4(2) − 312=13235. (a)1−1dx +1−11 − x2 dx = 2(1) + π(1)2/2 = 2 + π/2(b)13(x2+ 1)3/230− π(3)2/4 =13(103/2− 1) − 9π/4(c) u = x2, du = 2xdx;12101 − u2du =12π(1)2/4 = π/8
• 284 Chapter 636.12yx0.20.40.60.810.2 0.6 137. The rectangle with vertices (0, 0), (π, 0), (π, 1) and (0, 1) has area π and is much too large; sois the triangle with vertices (0, 0), (π, 0) and (π, 1) which has area π/2; 1 − π is negative; so theanswer is 35π/128.38. (a)1nnk=1kn=nk=1f(x∗k)∆x where f(x) =√x, x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thuslimn→+∞1nnk=1kn=10x1/2dx =23(b)1nnk=1kn4=nk=1f(x∗k)∆x where f(x) = x4, x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thuslimn→+∞1nnk=1kn4=10x4dx =15(c)nk=1ek/nn=nk=1f(x∗k)∆x where f(x) = ex, x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thuslimn→+∞nk=1ek/nn= limn→+∞nk=1f(x∗k)∆x =10exdx = e − 1.39. (a)bank=1fk(x)dx =nk=1bafk(x)dx(b) yes; substitute ckfk(x) for fk(x) in part (a), and then usebackfk(x)dx = ckbafk(x)dxfrom Theorem 6.5.440. f(x) = ex, [a, b] = [0, 1], ∆x =1n; limn→+∞nk=1f(x∗k)1n=10exdx = e − 141. The left endpoint of the top boundary is ((b − a)/2, h) and the right endpoint of the top boundaryis ((b + a)/2, h) sof(x) =2hx/(b − a), x < (b − a)/2h, (b − a)/2 < x < (b + a)/22h(x − b)/(a − b), x > (a + b)/2The area of the trapezoid is given by(b−a)/202hxb − adx+(b+a)/2(b−a)/2hdx+b(b+a)/22h(x − b)a − bdx = (b−a)h/4+ah+(b−a)h/4 = h(a+b)/2.
• Review Exercise Set 28542. Since f(x) =1xis positive and increasing on the interval [1, 2], the left endpoint approximationoverestimates the integral of1xand the right endpoint approximation underestimates it.(a) For n = 5 this becomes0.211.2+11.4+11.6+11.8+12.0<211xdx < 0.211.0+11.2+11.4+11.6+11.8(b) For general n the left endpoint approximation to211xdx = ln 2 is1nnk=111 + (k − 1)/n=nk=11n + k − 1=n−1k=01n + kand the right endpoint approximation isnk=11n + k. This yieldsnk=11n + k<211xdx <n−1k=01n + kwhich is the desired inequality.(c) By telescoping, the diﬀerence is1n−12n=12nso12n≤ 0.1, n ≥ 5(d) n ≥ 1, 00043.91√xdx =23x3/291=23(27 − 1) = 52/3 44.41x−3/5dx =52x2/541=52(42/5− 1)45.31exdx = ex31= e3− e 46.e311xdx = ln xe31= 3 − ln 1 = 347.13x3− 2x2+ 7x0−3= 48 48.12x2+15x52−1= 81/1049.31x−2dx = −1x31= 2/3 50. 3x5/3+4x81= 179/251.12x2− sec x10= 3/2 − sec(1) 52. 6√t −103t3/2+2√t41= −55/353.3/20(3 − 2x)dx +23/2(2x − 3)dx = (3x − x2)3/20+ (x2− 3x)23/2= 9/4 + 1/4 = 5/254.π/60(1/2 − sin x) dx +π/2π/6(sin x − 1/2) dx= (x/2 + cos x)π/60− (cos x + x/2)π/2π/6= (π/12 +√3/2) − 1 − π/4 + (√3/2 + π/12) =√3 − π/12 − 155. A =21(−x2+ 3x − 2)dx = −13x3+32x2− 2x21= 1/656. With b = 1.618034, area =b0(x + x2+ x3) dx ≈ 1.007514.57. (a) x3+ 1 (b) F(x) =14t4+ tx1=14x4+ x −54; F (x) = x3+ 1
• 286 Chapter 658. (a) F (x) =1√x(b) F(x) = 2√tx4= 2√x − 2; F (x) =1√x59. ex260.xcos x261. |x − 1|62. cos√x 63.cos x1 + sin3x64.(ln√x)22√x66. (a) F (x) =x2− 3x2 + 7; increasing on (−∞, −√3], [√3, +∞), decreasing on [−√3,√3](b) F (x) =20x(x2 + 7)2; concave down on (−∞, 0), concave up on (0, +∞)(c) limx→±∞F(x) = ∓∞, so F has no absolute extrema.(d)–4 4–0.50.5xyF(x)67. F (x) =11 + x2+11 + (1/x)2(−1/x2) = 0 so F is constant on (0, +∞).68. (−3, 3) because f is continuous there and 1 is in (−3, 3)69. (a) The domain is (−∞, +∞); F(x) is 0 if x = 1, positive if x > 1, and negative if x < 1, becausethe integrand is positive, so the sign of the integral depends on the orientation (forwards orbackwards).(b) The domain is [−2, 2]; F(x) is 0 if x = −1, positive if −1 < x ≤ 2, and negative if−2 ≤ x < −1; same reasons as in Part (a).70. F(x) =x−1t√2 + t3dt, F (x) =x√2 + x3, so F is increasing on [1, 3]; Fmax = F(3) ≈ 1.152082854and Fmin = F(1) ≈ −0.0764949314171. (a) fave =1330x1/2dx = 2√3/3;√x∗ = 2√3/3, x∗=43(b) fave =1e − 1e11xdx =1e − 1ln xe1=1e − 1;1x∗=1e − 1, x∗= e − 172. Mar 1 to Jun 7 is 14 weeks, so w(t) =t0s7ds =t214, so the weight on June 7 will be 14 gm.73. If the acceleration a = const, then v(t) = at + v0, s(t) = 12 at2+ v0t + s0.74. (a) no, since the velocity curve is not a straight line(b) 25 < t < 40 (c) 141.5 ft (d) 3.54 ft/s(e) no since the velocity is positive and the acceleration is never negative(f) need the position at any one given time (e.g. s0)
• Review Exercise Set 28775. s(t) = (t3− 2t2+ 1)dt =14t4−23t3+ t + C,s(0) =14(0)4−23(0)3+ 0 + C = 1, C = 1, s(t) =14t4−23t3+ t + 176. v(t) = 4 cos 2t dt = 2 sin 2t + C1, v(0) = 2 sin 0 + C1 = −1, C1 = −1,v(t) = 2 sin 2t − 1, s(t) = (2 sin 2t − 1)dt = − cos 2t − t + C2,s(0) = − cos 0 − 0 + C2 = −3, C2 = −2, s(t) = − cos 2t − t − 277. s(t) = (2t − 3)dt = t2− 3t + C, s(1) = (1)2− 3(1) + C = 5, C = 7, s(t) = t2− 3t + 778. v(t) = (cos t − 2t) dt = sin t − t2+ v0; but v0 = 0 so v(t) = sin t − t2s(t) = v(t)dt = − cos t − t3/3 + C : s(0) = 0 = −1 + C, C = 1, s(t) = − cos t − t3/3 + 179. displacement = s(6) − s(0) =60(2t − 4)dt = (t2− 4t)60= 12 mdistance =60|2t − 4|dt =20(4 − 2t)dt +62(2t − 4)dt = (4t − t2)20+ (t2− 4t)62= 20 m80. displacement =50|t − 3|dt =30−(t − 3)dt +53(t − 3)dt = 13/2 mdistance =50|t − 3|dt = 13/2 m81. displacement =3112−1t2dt = 1/3 mdistance =31|v(t)|dt = −√21v(t)dt +3√2v(t)dt = 10/3 − 2√2 m82. displacement =943t−1/2dt = 6 mdistance =94|v(t)|dt =94v(t)dt = 6 m83. v(t) = −2t + 3displacement =41(−2t + 3)dt = −6 mdistance =41| − 2t + 3|dt =3/21(−2t + 3)dt +43/2(2t − 3)dt = 13/2 m84. v(t) =25√5t + 1 +85displacement =3025√5t + 1 +85dt =475(5t + 1)3/2+85t30= 204/25 mdistance =30|v(t)|dt =30v(t)dt = 204/25 m
• 288 Chapter 685. A = A1 + A2 =10(1 − x2)dx +31(x2− 1)dx = 2/3 + 20/3 = 22/386. A = A1 + A2 =0−11 −√x + 1 dx +10[√x + 1 − 1] dx= x −23(x + 1)3/20−1+23(x + 1)3/2− x10= −23+ 1 +4√23− 1 −23= 4√2 − 1387. A = A1 + A2 =11/21 − xxdx +21x − 1xdx = −12− ln 2 + (1 − ln 2) = 1/288. s(t) =203t3− 50t2+ 50t + s0, s(0) = 0 gives s0 = 0, so s(t) =203t3− 50t2+ 50t, a(t) = 40t − 100150–1000150–10005000 66 6s(t) v(t) a(t)89. Take t = 0 when deceleration begins, then a = −10 so v = −10t + C1, but v = 88 when t = 0which gives C1 = 88 thus v = −10t + 88, t ≥ 0(a) v = 45 mi/h = 66 ft/s, 66 = −10t + 88, t = 2.2 s(b) v = 0 (the car is stopped) when t = 8.8 ss = v dt = (−10t + 88)dt = −5t2+ 88t + C2, and taking s = 0 when t = 0, C2 = 0 sos = −5t2+ 88t. At t = 8.8, s = 387.2. The car travels 387.2 ft before coming to a stop.90. dv/dt = 3, v = 3t + C1, but v = v0 when t = 0 so C1 = v0, v = 3t + v0. From ds/dt = v = 3t + v0we get s = 3t2/2 + v0t + C2 and, with s = 0 when t = 0, C2 = 0 so s = 3t2/2 + v0t. s = 40 whent = 4 thus 40 = 3(4)2/2 + v0(4), v0 = 4 m/s91. (a) Use the second and then the ﬁrst of the given formulae to getv2= v20 − 2v0gt + g2t2= v20 − 2g(v0t − 12 gt2) = v20 − 2g(s − s0).(b) Add v0 to both sides of the second formula: 2v0 − gt = v0 + v, v0 − 12 gt = 12 (v0 + v);from the ﬁrst formula s = s0 + t(v0 − 12 gt) = s0 + 12 (v0 + v)t(c) Add v to both sides of the second formula: 2v + gt = v0 + v, v + 12 gt = 12 (v0 + v); fromPart (b), s = s0 + 12 (v0 + v)t = s0 + vt + 12 gt292. v0 = 0 and g = 9.8, so v2= −19.6(s − s0) (see Exercise 91); since v = 24 when s = 0 it followsthat 19.6s0 = 242or s0 = 29.39 m.93. u = 2x + 1,1231u4du =110u531= 121/5, or110(2x + 1)510= 121/5
• Review Exercise Set 28994. u = 4 − x,49(u − 4)u1/2du =49(u3/2− 4u1/2)du =25u5/2−83u3/249= −506/15or25(4 − x)5/2−83(4 − x)3/20−5= −506/1595.23(3x + 1)1/210= 2/396. u = x2,π012sin u du = −12cos uπ0= 197.13πsin3πx10= 098. u = ln x, du = (1/x)dx;211udu = ln u21= ln 299.10e−x/2dx = 2(1 − 1/√e)100. u = 3x/2, du = 3/2dx,16√3011 + u2du =16tan−1u√30=118π101. (a) limx→+∞1 +1xx 2= limx→+∞1 +1xx 2= e2(b) y = 3x, limy→01 +1yy/3= limy→01 +1yy 1/3= e1/3102. (a) y(x) = 2 +x1t1/3dt = 2 +34t4/3x1=54+34x4/3(b) y(x) =x0tet2dt =12ex2−12103. Diﬀerentiate: f(x) = 3e3x, so 2 +xaf(t)dt = 2 +xa3e3tdt = 2 + e3txa= 2 + e3x− e3a= e3xprovided e3a= 2, a = (ln 2)/3.
• 290CHAPTER 7Applications of the Deﬁnite Integral inGeometry, Science, and EngineeringEXERCISE SET 7.11. A =2−1(x2+ 1 − x)dx = (x3/3 + x − x2/2)2−1= 9/22. A =40(√x + x/4)dx = (2x3/2/3 + x2/8)40= 22/33. A =21(y − 1/y2)dy = (y2/2 + 1/y)21= 14. A =20(2 − y2+ y)dy = (2y − y3/3 + y2/2)20= 10/35. (a) A =40(4x − x2)dx = 32/3 (b) A =160(√y − y/4)dy = 32/351(4, 16)y = 4xy = x2xy(4, 4)(1, –2)xyy2 = 4xy = 2x – 46. Eliminate x to get y2= 4(y + 4)/2, y2− 2y − 8 = 0,(y − 4)(y + 2) = 0; y = −2, 4 with correspondingvalues of x = 1, 4.(a) A =10[2√x − (−2√x)]dx +41[2√x − (2x − 4)]dx=104√xdx +41(2√x − 2x + 4)dx = 8/3 + 19/3 = 9(b) A =4−2[(y/2 + 2) − y2/4]dy = 9
• Exercise Set 7.1 2917. A =11/4(√x − x2)dx = 49/19214(1, 1)xyy = x2y = √x8. A =20[0 − (x3− 4x)]dx=20(4x − x3)dx = 42xyy = 2x3 – 4x9. A =π/2π/4(0 − cos 2x)dx= −π/2π/4cos 2x dx = 1/23 6–11xyy = cos 2x10. Equate sec2x and 2 to get sec2x = 2,12xyy = sec2 x(#, 2) (3, 2)sec x = ±√2, x = ±π/4A =π/4−π/4(2 − sec2x)dx = π − 211. A =3π/4π/4sin y dy =√239xyx = sin y12. A =2−1[(x + 2) − x2]dx = 9/2(2, 4)(–1, 1)xyy = x2x = y – 213. A =ln 20e2x− exdx=12e2x− exln 20= 1/224xyln 2y = e2xy = ex14. A =e1dyy= ln ye1= 11/e 11exy
• 292 Chapter 715. A =1−121 + x2− |x| dx= 21021 + x2− x dx= 4 tan−1x − x210= π − 1–1 112xy16.1√1 − x2= 2, x = ±√32, soA =√3/2−√3/22 −1√1 − x2dx= 2 − sin−1x√3/2−√3/2= 2√3 − 23 π0.511.52xy23– 231 – x21y =y = 2(–5, 8)(5, 6)y = 3 – xy = 1 + xy = – x + 715xy17. y = 2 + |x − 1| =3 − x, x ≤ 11 + x, x ≥ 1,A =1−5−15x + 7 − (3 − x) dx+51−15x + 7 − (1 + x) dx=1−545x + 4 dx +516 −65x dx= 72/5 + 48/5 = 2418. A =2/50(4x − x)dx+12/5(−x + 2 − x)dx=2/503x dx +12/5(2 − 2x)dx = 3/5(1, 1)2585( , )xyy = 4xy = xy = –x + 219. A =10(x3− 4x2+ 3x)dx+31[−(x3− 4x2+ 3x)]dx= 5/12 + 32/12 = 37/124–8–1 4
• Exercise Set 7.1 2939–2–1 320. Equate y = x3− 2x2and y = 2x2− 3xto get x3− 4x2+ 3x = 0,x(x − 1)(x − 3) = 0; x = 0, 1, 3with corresponding values of y = 0, −1.9.A =10[(x3− 2x2) − (2x2− 3x)]dx+31[(2x3− 3x) − (x3− 2x2)]dx=10(x3− 4x2+ 3x)dx +31(−x3+ 4x2− 3x)dx=512+83=371221. From the symmetry of the regionA = 25π/4π/4(sin x − cos x)dx = 4√21–10 o22. The region is symmetric about the origin soA = 220|x3− 4x|dx = 83.1–3.1–3 323. A =0−1(y3− y)dy +10−(y3− y)dy= 1/21–1–1 124. A =10y3− 4y2+ 3y − (y2− y) dy+41y2− y − (y3− 4y2+ 3y) dy= 7/12 + 45/4 = 71/64.10–2.2 12.1
• 294 Chapter 725. The curves meet when x =√ln 2, soA =√ln 20(2x − xex2) dx = x2−12ex2√ln 20= ln 2 −120.5 10.511.522.5xy26. The curves meet for x = e−2√2/3, e2√2/3thusA =e2√2/3e−2√2/33x−1x 1 − (ln x)2dx= 3 ln x − sin−1(ln x)e2√2/3e−2√2/3= 4√2 − 2 sin−1 2√231 2 35101520xy27. The area is given byk0(1/ 1 − x2 − x)dx = sin−1k − k2/2 = 1; solve for k to getk = 0.997301.28. The curves intersect at x = a = 0 and x = b = 0.838422 so the area isba(sin 2x − sin−1x)dx ≈ 0.174192.29. Solve 3−2x = x6+2x5−3x4+x2to ﬁnd the real roots x = −3, 1; from a plot it is seen that the lineis above the polynomial when −3 < x < 1, so A =1−3(3−2x−(x6+2x5−3x4+x2)) dx = 9152/10530. Solve x5− 2x3− 3x = x3to ﬁnd the roots x = 0, ±126 + 2√21. Thus, by symmetry,A = 2√(6+2√21)/20(x3− (x5− 2x3− 3x)) dx =274+74√2131.k02√ydy =9k2√ydyk0y1/2dy =9ky1/2dy23k3/2=23(27 − k3/2)k3/2= 27/2k = (27/2)2/3= 9/ 3√4y = 9y = kxy
• Exercise Set 7.1 29532.k0x2dx =2kx2dx13k3=13(8 − k3)k3= 4k = 3√42xyx = √yx = k33. (a) A =20(2x − x2)dx = 4/3(b) y = mx intersects y = 2x − x2where mx = 2x − x2, x2+ (m − 2)x = 0, x(x + m − 2) = 0 sox = 0 or x = 2 − m. The area below the curve and above the line is2−m0(2x − x2− mx)dx =2−m0[(2 − m)x − x2]dx =12(2 − m)x2−13x32−m0=16(2 − m)3so (2 − m)3/6 = (1/2)(4/3) = 2/3, (2 − m)3= 4, m = 2 − 3√4.34. The line through (0, 0) and (5π/6, 1/2) is y =35πx;A =5π/60sin x −35πx dx =√32−524π + 1c1 12c56( ),xyy = sin x35. The curves intersect at x = 0 and, by Newton’s Method, at x ≈ 2.595739080 = b, soA ≈b0(sin x − 0.2x)dx = − cos x + 0.1x2b0≈ 1.18089833436. By Newton’s Method, the points of intersection are at x ≈ ±0.824132312, so withb = 0.824132312 we have A ≈ 2b0(cos x − x2)dx = 2(sin x − x3/3)b0≈ 1.09475360937. By Newton’s Method the points of intersection are x = x1 ≈ 0.4814008713 andx = x2 ≈ 2.363938870, and A ≈x2x1ln xx− (x − 2) dx ≈ 1.189708441.38. By Newton’s Method the points of intersection are x = ±x1 where x1 ≈ 0.6492556537, thusA ≈ 2x1021 + x2− 3 + 2 cos x dx ≈ 0.82624788839. The x-coordinates of the points of intersection are a ≈ −0.423028 and b ≈ 1.725171; the area isba(2 sin x − x2+ 1)dx ≈ 2.542696.40. Let (a, k), where π/2 < a < π, be the coordinates of the point of intersection of y = k withy = sin x. Thus k = sin a and if the shaded areas are equal,a0(k − sin x)dx =a0(sin a − sin x) dx = a sin a + cos a − 1 = 0Solve for a to get a ≈ 2.331122, so k = sin a ≈ 0.724611.
• 296 Chapter 741.600(v2(t) − v1(t)) dt = s2(60) − s2(0) − (s1(60) − s1(0)), but they are even at time t = 60, sos2(60) = s1(60). Consequently the integral gives the diﬀerence s1(0)−s2(0) of their starting pointsin meters.42. Since a1(0) = a2(0) = 0, A =T0(a2(t)−a1(t)) dt = v2(T)−v1(T) is the diﬀerence in the velocitiesof the two cars at time T.43. (a) It gives the area of the region that is between f and g when f(x) > g(x) minus the area ofthe region between f and g when f(x) < g(x), for a ≤ x ≤ b.(b) It gives the area of the region that is between f and g for a ≤ x ≤ b.44. (b) limn→+∞10(x1/n− x) dx = limn→+∞nn + 1x(n+1)/n−x2210= limn→+∞nn + 1−12= 1/2aaxy45. Solve x1/2+ y1/2= a1/2for y to gety = (a1/2− x1/2)2= a − 2a1/2x1/2+ xA =a0(a − 2a1/2x1/2+ x)dx = a2/646. Solve for y to get y = (b/a)√a2 − x2 for the upper half of the ellipse; make use of symmetry toget A = 4a0baa2 − x2dx =4baa0a2 − x2dx =4ba·14πa2= πab.47. Let A be the area between the curve and the x-axis and AR the area of the rectangle, thenA =b0kxmdx =km + 1xm+1b0=kbm+1m + 1, AR = b(kbm) = kbm+1, so A/AR = 1/(m + 1).EXERCISE SET 7.21. V = π3−1(3 − x)dx = 8π 2. V = π10[(2 − x2)2− x2]dx= π10(4 − 5x2+ x4)dx= 38π/153. V = π2014(3 − y)2dy = 13π/6 4. V = π21/2(4 − 1/y2)dy = 9π/25. V =20x4dx = 32/52xyy = x2
• Exercise Set 7.2 2976. V =π/3π/4sec2x dx =√3 − 13 4-2-112xyy = sec x7. V = ππ/2π/4cos x dx = (1 −√2/2)π3 6–11xyy = √cos x8. V = π10[(x2)2− (x3)2]dx= π10(x4− x6)dx = 2π/3511 (1, 1)y = x2y = x3xy9. V = π4−4[(25 − x2) − 9]dx= 2π40(16 − x2)dx = 256π/35xyy = √25 – x2y = 310. V = π3−3(9 − x2)2dx= π3−3(81 − 18x2+ x4)dx = 1296π/5–3 39xyy = 9 – x211. V = π40[(4x)2− (x2)2]dx= π40(16x2− x4)dx = 2048π/15416 (4, 16)xyy = x2y = 4x12. V = ππ/40(cos2x − sin2x)dx= ππ/40cos 2x dx = π/23–11xyy = cos xy = sin x
• 298 Chapter 713. V = πln 30e2xdx =π2e2xln 30= 4π 14. V = π10e−4xdx =π4(1 − e−4)1–11xy15. V =2−2π14 + x2dx =π2tan−1(x/2)2−2= π2/416. V =10πe6x1 + e6xdx =π6ln(1 + e6x)10=π6(ln(1 + e6) − ln 2)17. V =10y1/32dy =350.250.750.250.75xy18. V =10(1 − y2)2dy = 1 −23+15=8150.250.750.250.75xy19. V = π3−1(1 + y)dy = 8π32xyx = √1 + y20. V = π30[22− (y + 1)]dy= π30(3 − y)dy = 9π/23 (2, 3)xy y = x2 – 1x = √y + 1
• Exercise Set 7.2 29921. V = π3π/4π/4csc2y dy = 2π–2 –1 1 2369xyx = csc y22. V = π10(y − y4)dy = 3π/10–1 1–11(1, 1)xyx = √yx = y223. V = π2−1[(y + 2)2− y4]dy = 72π/5(4, 2)xyx = y2x = y + 2(1, –1)24. V = π1−1(2 + y2)2− (1 − y2)2dy= π1−1(3 + 6y2)dy = 10π1 2–11xy x = 2 + y2x = 1 – y225. V =10πe2ydy =π2e2− 1 26. V =20π1 + y2dy = π tan−1227. V = πa−ab2a2(a2− x2)dx = 4πab2/3–a abxybay = √a2– x228. V = π2b1x2dx = π(1/b − 1/2);π(1/b − 1/2) = 3, b = 2π/(π + 6)29. V = π0−1(x + 1)dx+ π10[(x + 1) − 2x]dx= π/2 + π/2 = π–1 11xyy = √2x(1, √2)y = √x + 1
• 300 Chapter 730. V = π40x dx + π64(6 − x)2dx= 8π + 8π/3 = 32π/34 6xyy = √x y = 6 – x31. Partition the interval [a, b] with a = x0 < x1 < x2 < . . . < xn−1 < xn = b. Let x∗k be an arbitrarypoint of [xk−1, xk]. The disk in question is obtained by revolving about the line y = k the rectan-gle for which xk−1 < x < xk, and y lies between y = k and y = f(x); the volume of this disk is∆Vk = π(f(x∗k) − k)2∆xk, and the total volume is given by V = πba(f(x) − k)2dx.32. Assume for c < y < d that k ≤ v(y) ≤ w(y) (A similar proof holds for k ≥ v(y) ≥ w(y)). Partitionthe interval [c, d] withc = y0 < y1 < y2 < . . . < yn−1 < yn = d. Let y∗k be an arbitrary point of [yk−1, yk]. The washerin question is the region obtained by revolving the strip v(y∗k) < x < w(y∗k), yk−1 < y < yk aboutthe line x = k. The volume of this washer is ∆V = π[(v(y∗k) − k)2− (w(y∗k) − k)2]∆yk, and thevolume of the solid obtained by rotating R isV = πdc[(v(y) − k)2− (w(y) − k)2] dy33. (a) Intuitively, it seems that a line segment which is revolved about a line which is perpendicularto the line segment will generate a larger area, the farther it is from the line. This is becausethe average point on the line segment will be revolved through a circle with a greater radius,and thus sweeps out a larger circle.Consider the line segment which connects a point (x, y) on the curve y =√3 − x to the point(x, 0) beneath it. If this line segment is revolved around the x-axis we generate an area πy2.If on the other hand the segment is revolved around the line y = 2 then the area of theresulting (inﬁnitely thin) washer is π[22−(2−y)2]. So the question can be reduced to askingwhether y2≥ [22−(2−y)2], y2≥ 4y −y2, or y ≥ 2. In the present case the curve y =√3 − xalways satisﬁes y ≤ 2, so V2 has the larger volume.(b) The volume of the solid generated by revolving the area around the x-axis isV1 = π3−1(3 − x) dx = 8π, and the volume generated by revolving the area around the liney = 2 is V2 = π3−1[22− (2 −√3 − x)2] dx =403π34. (a) In general, points in the region R are farther from the y-axis than they are from the linex = 2.5, so by the reasoning in Exercise 33(a) the former should generate a larger volumethan the latter, i.e. the volume mentioned in Exercise 4 will be greater than that gotten byrevolving about the line x = 2.5.(b) The original volume V1 of Exercise 4 is given byV1 = π21/2(4 − 1/y2) dy = 9π/2, and the other volumeV2 = π21/21y− 2.52− (2 − 2.5)2dy =212− 10 ln 2 π ≈ 3.568528194π,and thus V1 is the larger volume.
• Exercise Set 7.2 30135. V = π30(9 − y2)2dy= π30(81 − 18y2+ y4)dy= 648π/593xyx = y236. V = π90[32− (3 −√x)2]dx= π90(6√x − x)dx= 135π/29xyy = √xy = 311xyx = y2x = yy = –137. V = π10[(√x + 1)2− (x + 1)2]dx= π10(2√x − x − x2)dx = π/238. V = π10[(y + 1)2− (y2+ 1)2]dy= π10(2y − y2− y4)dy = 7π/1511xyx = y2x = yx = –139. A(x) = π(x2/4)2= πx4/16,V =200(πx4/16)dx = 40, 000π ft340. V = π10(x − x4)dx = 3π/1041. V =10(x − x2)2dx=10(x2− 2x3+ x4)dx = 1/30Square(1, 1)1y = xy = x2xy42. A(x) =12π12√x2=18πx,V =4018πx dx = π4xyy = √x
• 302 Chapter 743. On the upper half of the circle, y =√1 − x2, so:(a) A(x) is the area of a semicircle of radius y, soA(x) = πy2/2 = π(1 − x2)/2; V =π21−1(1 − x2) dx = π10(1 − x2) dx = 2π/31–1y = √1 – x2 xyy(b) A(x) is the area of a square of side 2y, soA(x) = 4y2= 4(1 − x2); V = 41−1(1 − x2) dx = 810(1 − x2) dx = 16/31–1y = √1 – x2xy2y(c) A(x) is the area of an equilateral triangle with sides 2y, soA(x) =√34(2y)2=√3y2=√3(1 − x2);V =1−1√3(1 − x2) dx = 2√310(1 − x2) dx = 4√3/3xy1–1y = √1 – x22y2y2y44. The base of the dome is a hexagon of side r. An equation of the circle of radius r that lies in avertical x-y plane and passes through two opposite vertices of the base hexagon is x2+ y2= r2.A horizontal, hexagonal cross section at height y above the base has areaA(y) =3√32x2=3√32(r2− y2), hence the volume is V =r03√32(r2− y2) dy =√3r3.45. The two curves cross at x = b ≈ 1.403288534, soV = πb0((2x/π)2− sin16x) dx + ππ/2b(sin16x − (2x/π)2) dx ≈ 0.710172176.46. Note that π2sin x cos3x = 4x2for x = π/4. From the graph it is apparent that this is the ﬁrstpositive solution, thus the curves don’t cross on (0, π/4) andV = ππ/40[(π2sin x cos3x)2− (4x2)2] dx =148π5+172560π6
• Exercise Set 7.2 30347. V = πe1(1 − (ln y)2) dy = π48. V =tan 10π[x2− x2tan−1x] dx =π6[tan21 − ln(1 + tan21)]49. (a) V = πrr−h(r2− y2) dy = π(rh2− h3/3) =13πh2(3r − h)rhr xyx2 + y2 = r2(b) By the Pythagorean Theorem,r2= (r − h)2+ ρ2, 2hr = h2+ ρ2; from Part (a),V =πh3(3hr − h2) =πh332(h2+ ρ2) − h2)=16πh(h2+ 3ρ2)50. Find the volume generated by revolvingthe shaded region about the y-axis.V = π−10+h−10(100 − y2)dy =π3h2(30 − h)Find dh/dt when h = 5 given that dV/dt = 1/2.V =π3(30h2− h3),dVdt=π3(60h − 3h2)dhdt,12=π3(300 − 75)dhdt,dhdt= 1/(150π) ft/min10h – 10–10hxyx = √100 – y251. (b) ∆x =510= 0.5; {y0, y1, · · · , y10} = {0, 2.00, 2.45, 2.45, 2.00, 1.46, 1.26, 1.25, 1.25, 1.25, 1.25};left = π9i=0yi22∆x ≈ 11.157;right = π10i=1yi22∆x ≈ 11.771; V ≈ average = 11.464 cm3r2xyx = √r2 – y2√3r2√3r2–52. If x = r/2 then from y2= r2− x2we get y = ±√3r/2as limits of integration; for −√3 ≤ y ≤√3,A(y) = π[(r2− y2) − r2/4] = π(3r2/4 − y2), thusV = π√3r/2−√3r/2(3r2/4 − y2)dy= 2π√3r/20(3r2/4 − y2)dy =√3πr3/2
• 304 Chapter 753. (a)h–4xy0 ≤ h < 2h – 4(b)–4–2h2 ≤ h ≤ 4h – 4xyIf the cherry is partially submerged then 0 ≤ h < 2 as shown in Figure (a); if it is totally sub-merged then 2 ≤ h ≤ 4 as shown in Figure (b). The radius of the glass is 4 cm and that of thecherry is 1 cm so points on the sections shown in the ﬁgures satisfy the equations x2+ y2= 16and x2+ (y + 3)2= 1. We will ﬁnd the volumes of the solids that are generated when the shadedregions are revolved about the y-axis.For 0 ≤ h < 2,V = πh−4−4[(16 − y2) − (1 − (y + 3)2)]dy = 6πh−4−4(y + 4)dy = 3πh2;for 2 ≤ h ≤ 4,V = π−2−4[(16 − y2) − (1 − (y + 3)2)]dy + πh−4−2(16 − y2)dy= 6π−2−4(y + 4)dy + πh−4−2(16 − y2)dy = 12π +13π(12h2− h3− 40)=13π(12h2− h3− 4)soV =3πh2if 0 ≤ h < 213π(12h2− h3− 4) if 2 ≤ h ≤ 454. x = h ± r2 − y2,V = πr−r(h + r2 − y2)2− (h − r2 − y2)2dy= 4πhr−rr2 − y2dy= 4πh12πr2= 2π2r2hxy(x – h2) + y2= r2xhu55. tan θ = h/x so h = x tan θ,A(y) =12hx =12x2tan θ =12(r2− y2) tan θbecause x2= r2− y2,V =12tan θr−r(r2− y2)dy= tan θr0(r2− y2)dy =23r3tan θ
• Exercise Set 7.3 30556. A(x) = (x tan θ)(2 r2 − x2)= 2(tan θ)x r2 − x2,V = 2 tan θr0x r2 − x2dx=23r3tan θyx√r2 – x2x tan u57. Each cross section perpendicular to they-axis is a square soA(y) = x2= r2− y2,18V =r0(r2− y2)dyV = 8(2r3/3) = 16r3/3rx = √r2 – y2xy58. The regular cylinder of radius r and height h has the same circular cross sections as do those ofthe oblique clinder, so by Cavalieri’s Principle, they have the same volume: πr2h.EXERCISE SET 7.31. V =212πx(x2)dx = 2π21x3dx = 15π/22. V =√202πx( 4 − x2 − x)dx = 2π√20(x 4 − x2 − x2)dx =8π3(2 −√2)3. V =102πy(2y − 2y2)dy = 4π10(y2− y3)dy = π/34. V =202πy[y − (y2− 2)]dy = 2π20(y2− y3+ 2y)dy = 16π/35. V =102π(x)(x3)dx= 2π10x4dx = 2π/5–1 1–11xyy = x36. V =942πx(√x)dx= 2π94x3/2dx = 844π/5–9 –4 4 9123xyy = √x7. V =312πx(1/x)dx = 2π31dx = 4π–3 –1 1 3y = x1xy
• 306 Chapter 78. V =√π/202πx cos(x2)dx = π/√2√p2xyy = cos (x2)9. V =212πx[(2x − 1) − (−2x + 3)]dx= 8π21(x2− x)dx = 20π/3(2, 3)(2, –1)(1, 1)xy10. V =202πx(2x − x2)dx= 2π20(2x2− x3)dx =83π2xyy = 2x – x211. V = 2π10xx2 + 1dx= π ln(x2+ 1)10= π ln 2–1 11xyy = 1x2 + 112. V =√312πxex2dx = πex2√31= π(e3− e)-√3 -1 1 √31020xyy = ex213. V =102πy3dy = π/21xyx = y214. V =322πy(2y)dy = 4π32y2dy = 76π/323xyx = 2y15. V =102πy(1 −√y)dy= 2π10(y − y3/2)dy = π/5 1xyy = √x
• Exercise Set 7.3 30716. V =412πy(5 − y − 4/y)dy= 2π41(5y − y2− 4)dy = 9π(1, 4)(4, 1)xyx = 5 – yx = 4/y17. V = 2π21xexdx = 2π(x − 1)ex21= 2πe218. V = 2ππ/20x cos xdx = π2− 2π19. The volume is given by 2πk0x sin x dx = 2π(sin k − k cos k) = 8; solve for k to getk = 1.736796.20. (a)ba2πx[f(x) − g(x)]dx (b)dc2πy[f(y) − g(y)]dy21. (a) V =102πx(x3− 3x2+ 2x)dx = 7π/30(b) much easier; the method of slicing would require that x be expressed in terms of y.–1 1xyy = x3– 3x2+ 2x22. Let a = x0 < x1 < x2 < . . . < xn−1 < xn = b be a partition of [a, b]. Let x∗k be the midpoint of[xk−1, xk].Revolve the strip xk−1 < x < xk, 0 < y < f(x∗k) about the line x = k. The result is acylindrical shell, a large coin with a very large hole through the center. The volume of the shell is∆Vk = 2π(x−k)f(x∗k)∆xk, just as the volume of a ring of average radius r, height y and thicknessh is 2πryh. Summing these volumes of cylindrical shells and taking the limit as max∆xk goes tozero, we obtain V = 2πba(x − k)f(x) dx23. V =212π(x + 1)(1/x3)dx= 2π21(x−2+ x−3)dx = 7π/4–1 x 21xyy = 1/x3x + 124. V =102π(1 − y)y1/3dy= 2π10(y1/3− y4/3)dy = 9π/141xy1 – y x = y1/3
• 308 Chapter 725. x =hr(r − y) is an equation of the linethrough (0, r) and (h, 0) soV =r02πyhr(r − y) dy=2πhrr0(ry − y2)dy = πr2h/3xy(0, r)(h, 0)26. V =k/402π(k/2 − x)2√kxdx= 2π√kk/40(kx1/2− 2x3/2)dx = 7πk3/60xyx = k/2k/2 – xx = k/4y = √kxy = –√kx27. Let the sphere have radius R, the hole radius r. By the Pythagorean Theorem, r2+ (L/2)2= R2.Use cylindrical shells to calculate the volume of the solid obtained by rotating about the y-axisthe region r < x < R, −√R2 − x2 < y <√R2 − x2:V =Rr(2πx)2 R2 − x2 dx = −43π(R2− x2)3/2Rr=43π(L/2)3,so the volume is independent of R.28. V =a−a2π(b − x)(2 a2 − x2)dx= 4πba−aa2 − x2dx − 4πa−ax a2 − x2dx= 4πb · (area of a semicircle of radius a) − 4π(0)= 2π2a2ba–axy√a2 – x2–√a2 – x2b – xx = b29. Vx = πb1/21x2dx = π(2 − 1/b), Vy = 2πb1/2dx = π(2b − 1);Vx = Vy if 2 − 1/b = 2b − 1, 2b2− 3b + 1 = 0, solve to get b = 1/2 (reject) or b = 1.30. (a) V = 2πb1x1 + x4dx = π tan−1(x2)b1= π tan−1(b2) −π4(b) limb→+∞V = ππ2−π4=14π2
• Exercise Set 7.4 309EXERCISE SET 7.41. (a)dydx= 2, L =21√1 + 4dx =√5(b)dxdy=12, L =421 + 1/4 dy = 2√5/2 =√52.dxdt= 1,dydt= 5, L =1012 + 52 dt =√263. f (x) =92x1/2, 1 + [f (x)]2= 1 +814x,L =101 + 81x/4 dx =82431 +814x3/2 10= (85√85 − 8)/2434. g (y) = y(y2+ 2)1/2, 1 + [g (y)]2= 1 + y2(y2+ 2) = y4+ 2y2+ 1 = (y2+ 1)2,L =10(y2 + 1)2dy =10(y2+ 1)dy = 4/35.dydx=23x−1/3, 1 +dydx2= 1 +49x−2/3=9x2/3+ 49x2/3,L =81√9x2/3 + 43x1/3dx =1184013u1/2du, u = 9x2/3+ 4=127u3/24013=127(40√40 − 13√13) =127(80√10 − 13√13)or (alternate solution)x = y3/2,dxdy=32y1/2, 1 +dxdy2= 1 +94y =4 + 9y4,L =12414 + 9y dy =1184013u1/2du =127(80√10 − 13√13)6. f (x) =14x3− x−3, 1 + [f (x)]2= 1 +116x6−12+ x−6=116x6+12+ x−6=14x3+ x−32,L =3214x3 + x−32dx =3214x3+ x−3dx = 595/1447. x = g(y) =124y3+ 2y−1, g (y) =18y2− 2y−2,1 + [g (y)]2= 1 +164y4−12+ 4y−4=164y4+12+ 4y−4=18y2+ 2y−22,L =4218y2+ 2y−2dy = 17/6
• 310 Chapter 78. g (y) =12y3−12y−3, 1 + [g (y)]2= 1 +14y6−12+14y−6=12y3+12y−32,L =4112y3+12y−3dy = 2055/649. (dx/dt)2+ (dy/dt)2= (t2)2+ (t)2= t2(t2+ 1), L =10t(t2+ 1)1/2dt = (2√2 − 1)/310. (dx/dt)2+ (dy/dt)2= [2(1 + t)]2+ [3(1 + t)2]2= (1 + t)2[4 + 9(1 + t)2],L =10(1 + t)[4 + 9(1 + t)2]1/2dt = (80√10 − 13√13)/2711. (dx/dt)2+ (dy/dt)2= (−2 sin 2t)2+ (2 cos 2t)2= 4, L =π/202 dt = π12. (dx/dt)2+ (dy/dt)2= (− sin t + sin t + t cos t)2+ (cos t − cos t + t sin t)2= t2,L =π0t dt = π2/213. (dx/dt)2+ (dy/dt)2= [et(cos t − sin t)]2+ [et(cos t + sin t)]2= 2e2t,L =π/20√2etdt =√2(eπ/2− 1)14. (dx/dt)2+ (dy/dt)2= (2etcos t)2+ (−2etsin t)2= 4e2t, L =412etdt = 2(e4− e)15. dy/dx =sec x tan xsec x= tan x, 1 + (y )2 =√1 + tan2x = sec x when 0 < x < π/4, soL =π/40sec x dx = ln(1 +√2)16. dy/dx =cos xsin x= cot x, 1 + (y )2 =√1 + cot2x = csc x when π/4 < x < π/2, soL =π/2π/4csc x dx = − ln(√2 − 1) = − ln√2 − 1√2 + 1(√2 + 1) = ln(1 +√2)17. (a)(–1, 1)(8, 4)xy (b) dy/dx does not exist at x = 0.(c) x = g(y) = y3/2, g (y) =32y1/2,L =101 + 9y/4 dy (portion for − 1 ≤ x ≤ 0)+401 + 9y/4 dy (portion for 0 ≤ x ≤ 8)=827138√13 − 1 +827(10√10 − 1) = (13√13 + 80√10 − 16)/27
• Exercise Set 7.4 31118. For (4), express the curve y = f(x) in the parametric form x = t, y = f(t) so dx/dt = 1 anddy/dt = f (t) = f (x) = dy/dx. For (5), express x = g(y) as x = g(t), y = t sodx/dt = g (t) = g (y) = dx/dy and dy/dt = 1.19. (a) The function y = f(x) = x2is inverse to the functionx = g(y) =√y : f(g(y)) = y for 1/4 ≤ y ≤ 4, andg(f(x)) = x for 1/2 ≤ x ≤ 2. Geometrically thismeans that the graphs of y = f(x) and x = g(y) aresymmetric to each other with respect to the line y = xand hence have the same arc length.1 2 3 41234xy(b) L1 =21/21 + (2x)2 dx and L2 =41/41 +12√x2dxMake the change of variables x =√y in the ﬁrst integral to obtainL1 =41/41 + (2√y)212√ydy =41/412√y2+ 1 dy = L2(c) L1 =21/21 + (2y)2 dy, L2 =41/41 +12√y2dy(d) For L1, ∆x =320, xk =12+ k320=3k + 1020, and thusL1 ≈10k=1(∆x)2 + [f(xk) − (f(xk−1)]2=10k=13202+(3k + 10)2 − (3k + 7)24002≈ 4.072396336For L2, ∆x =1540=38, xk =14+3k8=3k + 28, and thusL2 ≈10k=1382+3k + 28−3k − 182≈ 4.071626502(e) The expression for L1 is better, perhaps because L1 has in general a smaller slope and soapproximations of the true slope are better.(f) For L1, ∆x =320, the midpoint is x∗k =12+ k −12320=6k + 1740, and thusL1 ≈10k=13201 + 26k + 17402≈ 4.072396336.For L2, ∆x =1540, and the midpoint is x∗k =14+ k −121540=6k + 116, and thusL2 ≈10k=115401 + 46k + 116−1≈ 4.066160149(g) L1 =21/21 + (2x)2 dx ≈ 4.0729, L2 =41/41 +12√x2dx ≈ 4.0729
• 312 Chapter 720. (a) The function y = f(x) = x8/3is inverse to the functionx = g(y) = y3/8: f(g(y)) = y for 10−8≤ y ≤ 1 andg(f(x)) = x for 10−3≤ x ≤ 1. Geometrically thismeans that the graphs of y = f(x) and x = g(y) aresymmetric to each other with respect to the line y = x.0.2 0.6 10.20.61xy(b) L1 =110−31 +83x5/32dx,L2 =110−81 +38x−5/82dx;In the expression for L1 make the change of variable y = x8/3. ThenL1 =110−81 +83y5/8238y−5/8dy =110−838y−5/82+ 1 dy = L2(c) L1 =110−31 +83y5/32dy, L2 =110−81 +38y−5/82dy;(d) For L1, ∆x =99910000, xk =11000+ k99910000, and thusL1 ≈10k=1(∆x)2 + [f(xk) − f(xk−1)]2=10k=1999100002+11000+999k100008/3−11000+999(k − 1)100008/3 2≈ 1.524983407For L2, ∆y =999999991000000000, yk = 10−8+ k999999991000000000, and thusL2 ≈10k=1(∆y)2 + [g(yk) − g(yk−1)]2≈ 1.518667833(e) The expression for L1 is better, perhaps because the second curve has a very steep slope forsmall values of x, and approximations of such large numbers are less accurate.(f) For L1, ∆x =99910000, the midpoint is x∗k = 10−3+ k −1299910000, and thusL1 ≈10k=1999100001 +83(x∗k)5/32≈ 1.524166463.For L2, ∆y =999999991000000000, the midpoint is y∗k = 10−8+ k −12999999991000000000and thusL2 ≈10k=11 + (g (y∗k)2 ∆y ≈ 1.347221106(g) L1 =110−31 +83x5/32≈ 1.525898203, L2 =110−81 +38y−5/82dy ≈ 1.526898203
• Exercise Set 7.4 31321. (a) The function y = f(x) = tan x is inverse to the func-tion x = g(y) = tan−1x : f(g(y)) = y for 0 ≤ y ≤√3,and g(f(x)) = x for 0 ≤ x ≤ π/3. Geometrically thismeans that the graphs of y = f(x) and x = g(y) aresymmetric to each other with respect to the line y = x.0.5 1 1.5 20.511.52xy(b) L1 =π/301 + sec4 x dx, L2 =√301 +1(1 + x2)2dx;In the expression for L1 make the change of variabley = tan x to obtainL1 =√301 + ( 1 + y2)411 + y2dy =√301(1 + y2)2+ 1 dy = L2(c) L1 =π/301 + sec4 y dy, L2 =√301 +1(1 + y2)2dy;(d) For L1, ∆xk =π30, xk = kπ30, and thusL1 ≈10k=1(∆xk)2 + [f(xk) − (f(xk−1)]2=10k=1π302+ [tan(kπ/30) − tan((k − 1)π/30)]2 ≈ 2.056603923For L2, ∆xk =√310, xk = k√310, and thusL2 ≈10k=1√3102+ tan−1k√310− tan−1(k − 1)√3102≈ 2.056724591(e) The expression for L2 is slightly more accurate. The slope of tan x is on average greater thanthe slope of tan−1x as indicated in the graph in Part (a).(f) For L1, ∆xk =π30, the midpoint is x∗k = k −12π30, and thusL1 ≈10k=1π301 + sec4 k −12π30≈ 2.050944217.For L2, ∆xk =√310, and the midpoint is x∗k = k −12√310, and thusL2 ≈10k=1√3101 +1((x∗k)2 + 1)2≈ 2.057065139(g) L1 =π/301 + sec4 x dx ≈ 2.0570L2 =√301 +1(12 + y2)2dx ≈ 2.0570
• 314 Chapter 722. 0 ≤ m ≤ f (x) ≤ M, so m2≤ [f (x)]2≤ M2, and 1 + m2≤ 1 + [f (x)]2≤ 1 + M2; thus√1 + m2 ≤ 1 + [f (x)]2 ≤√1 + M2,ba1 + m2dx ≤ba1 + [f (x)]2dx ≤ba1 + M2 dx, and(b − a) 1 + m2 ≤ L ≤ (b − a) 1 + M223. f (x) = sec x tan x, 0 ≤ sec x tan x ≤ 2√3 for 0 ≤ x ≤ π/3 soπ3≤ L ≤π3√13.24. The distance is4.601 + (2.09 − 0.82x)2 dx ≈ 6.65 m25. L =π01 + (k cos x)2 dx 1 2 1.84 1.83 1.8323.8202 5.2704 5.0135 4.9977 5.0008kLExperimentation yields the values in the table, which by the Intermediate-Value Theorem showthat the true solution k to L = 5 lies between k = 1.83 and k = 1.832, so k = 1.83 to two decimalplaces.26. (a)100 200–1.6–1.2–0.8–0.4xy (b) The maximum deﬂection occurs atx = 96 inches (the midpointof the beam) and is about 1.42 in.(c) The length of the centerline is19201 + (dy/dx)2 dx = 192.026 in.27. y = 0 at x = b = 30.585; distance =b01 + (12.54 − 0.82x)2 dx = 196.306 yd28. (a) (dx/dθ)2+ (dy/dθ)2= (a(1 − cos θ))2+ (a sin θ)2= a2(2 − 2 cos θ), soL =2π0(dx/dθ)2 + (dy/dθ)2 dθ = a2π02(1 − cos θ) dθ29. (a) Use the interval 0 ≤ φ < 2π.(b) (dx/dφ)2+ (dy/dφ)2= (−3a cos2φ sin φ)2+ (3a sin2φ cos φ)2= 9a2cos2φ sin2φ(cos2φ + sin2φ) = (9a2/4) sin22φ, soL = (3a/2)2π0| sin 2φ| dφ = 6aπ/20sin 2φ dφ = −3a cos 2φπ/20= 6a30. (dx/dt)2+ (dy/dt)2= (−a sin t)2+ (b cos t)2= a2sin2t + b2cos2t= a2(1 − cos2t) + b2cos2t = a2− (a2− b2) cos2t= a21 −a2− b2a2cos2t = a2[1 − k2cos2t],L =2π0a 1 − k2 cos2 t dt = 4aπ/201 − k2 cos2 t dt
• Exercise Set 7.5 31531. (a) (dx/dt)2+ (dy/dt)2= 4 sin2t + cos2t = 4 sin2t + (1 − sin2t) = 1 + 3 sin2t,L =2π01 + 3 sin2t dt = 4π/201 + 3 sin2t dt(b) 9.69(c) distance traveled =4.81.51 + 3 sin2t dt ≈ 5.16 cmEXERCISE SET 7.51. S =102π(7x)√1 + 49dx = 70π√210x dx = 35π√22. f (x) =12√x, 1 + [f (x)]2= 1 +14xS =412π√x 1 +14xdx = 2π41x + 1/4dx = π(17√17 − 5√5)/63. f (x) = −x/√4 − x2, 1 + [f (x)]2= 1 +x24 − x2=44 − x2,S =1−12π 4 − x2(2/ 4 − x2)dx = 4π1−1dx = 8π4. y = f(x) = x3for 1 ≤ x ≤ 2, f (x) = 3x2,S =212πx31 + 9x4dx =π27(1 + 9x4)3/221= 5π(29√145 − 2√10)/275. S =202π(9y + 1)√82dy = 2π√8220(9y + 1)dy = 40π√826. g (y) = 3y2, S =102πy31 + 9y4dy = π(10√10 − 1)/277. g (y) = −y/ 9 − y2, 1 + [g (y)]2=99 − y2, S =2−22π 9 − y2 ·39 − y2dy = 6π2−2dy = 24π8. g (y) = −(1 − y)−1/2, 1 + [g (y)]2=2 − y1 − y,S =0−12π(2 1 − y)√2 − y√1 − ydy = 4π0−12 − y dy = 8π(3√3 − 2√2)/39. f (x) =12x−1/2−12x1/2, 1 + [f (x)]2= 1 +14x−1−12+14x =12x−1/2+12x1/22,S =312π x1/2−13x3/2 12x−1/2+12x1/2dx =π331(3 + 2x − x2)dx = 16π/9
• 316 Chapter 710. f (x) = x2−14x−2, 1 + [f (x)]2= 1 + x4−12+116x−4= x2+14x−22,S =212π13x3+14x−1x2+14x−2dx = 2π2113x5+13x +116x−3dx = 515π/6411. x = g(y) =14y4+18y−2, g (y) = y3−14y−3,1 + [g (y)]2= 1 + y6−12+116y−6= y3+14y−32,S =212π14y4+18y−2y3+14y−3dy =π1621(8y7+ 6y + y−5)dy = 16,911π/102412. x = g(y) =√16 − y; g (y) = −12√16 − y, 1 + [g (y)]2=65 − 4y4(16 − y),S =1502π 16 − y65 − 4y4(16 − y)dy = π15065 − 4y dy = (65√65 − 5√5)π613. f (x) = cos x, 1 + [f (x)]2= 1 + cos2x,S =π02π sin x 1 + cos2 x dx = 2π(√2 + ln(√2 + 1)) ≈ 14.4214. x = g(y) = tan y, g (y) = sec2y, 1 + [g (y)]2= 1 + sec4y;S =π/402π tan y 1 + sec4 y dy ≈ 3.8415. f (x) = ex, 1 + [f (x)]2= 1 + e2x, S =102πex1 + e2x dx ≈ 22.9416. x = g(y) = ln y, g (y) = 1/y, 1 + [g (y)]2= 1 + 1/y2; S =e12π 1 + 1/y2 ln y dy ≈ 7.05496560817. n = 20, a = 0, b = π, ∆x = (b − a)/20 = π/20, xk = kπ/20,S ≈ π20k=1[sin(k − 1)π/20 + sin kπ/20] (π/20)2 + [sin(k − 1)π/20 − sin kπ/20]2 ≈ 14.3939449618. n = 20, a = 0, b = e, ∆x = (b − a)/20 = (1 − e)/20, xk = k(1 − e)/20,S = πe1[ln yk−1 + ln yk] (∆yk)2 + [ln yk−1 − ln yk]2 ≈ 7.05284689119. Revolve the line segment joining the points (0, 0) and (h, r) about the x-axis. An equation of theline segment is y = (r/h)x for 0 ≤ x ≤ h soS =h02π(r/h)x 1 + r2/h2dx =2πrh2r2 + h2h0x dx = πr r2 + h220. f(x) =√r2 − x2, f (x) = −x/√r2 − x2, 1 + [f (x)]2= r2/(r2− x2),S =r−r2π r2 − x2(r/ r2 − x2)dx = 2πrr−rdx = 4πr2
• Exercise Set 7.5 31721. g(y) = r2 − y2, g (y) = −y/ r2 − y2, 1 + [g (y)]2= r2/(r2− y2),(a) S =rr−h2π r2 − y2 r2/(r2 − y2) dy = 2πrrr−hdy = 2πrh(b) From Part (a), the surface area common to two polar caps of height h1 > h2 is2πrh1 − 2πrh2 = 2πr(h1 − h2).22. (a) length of arc of sector = circumference of base of cone,θ = 2πr, θ = 2πr/ ; S = area of sector =122(2πr/ ) = πr(b) S = πr2 2 − πr1 1 = πr2( 1 + ) − πr1 1 = π[(r2 − r1) 1 + r2 ];Using similar triangles 2/r2 = 1/r1, r1 2 = r2 1,r1( 1 + ) = r2 1, (r2 − r1) 1 = r1so S = π (r1 + r2 ) = π (r1 + r2) .r1r2l2l1l23. S =ba2π[f(x) + k] 1 + [f (x)]2dx24. 2πk 1 + [f (x)]2 ≤ 2πf(x) 1 + [f (x)]2 ≤ 2πK 1 + [f (x)]2, soba2πk 1 + [f (x)]2dx ≤ba2πf(x) 1 + [f (x)]2dx ≤ba2πK 1 + [f (x)]2dx,2πkba1 + [f (x)]2dx ≤ S ≤ 2πKba1 + [f (x)]2dx, 2πkL ≤ S ≤ 2πKL25. Note that 1 ≤ sec x ≤ 2 for 0 ≤ x ≤ π/3. Let L be the arc length of the curve y = tan xfor 0 < x < π/3. Then L =π/301 + sec2 x dx, and by Exercise 24, and the inequaities above,2πL ≤ S ≤ 4πL. But from the inequalities for sec x above, we can show that√2π/3 ≤ L ≤√5π/3.Hence, combining the two sets of inequalities, 2π(√2π/3) ≤ 2πL ≤ S ≤ 4πL ≤ 4π√5π/3. Toobtain the inequalities in the text, observe that2π23< 2π√2π3≤ 2πL ≤ S ≤ 4πL ≤ 4π√5π3<4π23√13.26. (a) 1 ≤ 1 + [f (x)]2 so 2πf(x) ≤ 2πf(x) 1 + [f (x)]2,ba2πf(x)dx ≤ba2πf(x) 1 + [f (x)]2dx, 2πbaf(x)dx ≤ S, 2πA ≤ S(b) 2πA = S if f (x) = 0 for all x in [a, b] so f(x) is constant on [a, b].27. Let a = t0 < t1 < . . . < tn−1 < tn = b be a partition of [a, b]. Then the lateral area of the frustumof slant height = ∆x2k + ∆y2k and radii y(t1) and y(t2) is π(y(tk) + y(tk−1)) . Thus the area ofthe frustum Sk is given by Sk = π(y(tk−1) + y(tk)) [(x(tk) − x(tk−1)]2 + [y(tk) − y(tk−1)]2 withthe limit as max ∆tk → 0 of S =ba2πy(t) [x (t)2 + y (t)2 dt
• 318 Chapter 728. Let a = t0 < t1 < . . . < tn−1 < tn = b be a partition of [a, b]. Then the lateral area of the frustumof slant height = ∆x2k + ∆y2k and radii x(t1) and x(t2) is π(x(tk) + x(tk−1)) . Thus the area ofthe frustum Sk is given by Sk = π(x(tk−1) + x(tk)) [(x(tk) − x(tk−1)]2 + [y(tk) − y(tk−1)]2 withthe limit as max ∆tk → 0 of S =ba2πx(t) [x (t)2 + y (t)2 dt29. x = 2t, y = 2, (x )2+ (y )2= 4t2+ 4S = 2π40(2t) 4t2 + 4dt = 8π40t t2 + 1dt =8π3(17√17 − 1)30. x = −2 cos t sin t, y = 5 cos t, (x )2+ (y )2= 4 cos2t sin2t + 25 cos2t,S = 2ππ/205 sin t 4 cos2 t sin2t + 25 cos2 t dt =π6(145√29 − 625)31. x = 1, y = 4t, (x )2+ (y )2= 1 + 16t2, S = 2π10t 1 + 16t2 dt =π24(17√17 − 1)32. x = −2 sin t cos t, y = 2 sin t cos t, (x )2+ (y )2= 8 sin2t cos2tS = 2ππ/20cos2t 8 sin2t cos2 t dt = 4√2ππ/20cos3t sin t dt =√2π33. x = −r sin t, y = r cos t, (x )2+ (y )2= r2,S = 2ππ0r sin t√r2 dt = 2πr2π0sin t dt = 4πr234.dxdφ= a(1 − cos φ),dydφ= a sin φ,dxdφ2+dydφ2= 2a2(1 − cos φ)S = 2π2π0a(1 − cos φ) 2a2(1 − cos φ) dφ = 2√2πa22π0(1 − cos φ)3/2dφ,but 1 − cos φ = 2 sin2 φ2so (1 − cos φ)3/2= 2√2 sin3 φ2for 0 ≤ φ ≤ π and, taking advantage of thesymmetry of the cycloid, S = 16πa2π0sin3 φ2dφ = 64πa2/3.35. x = et(cos t − sin t), y = et(cos t + sin t), (x )2+ (y )2= 2e2tS = 2ππ/20(etsin t)√2e2tdt = 2√2ππ/20e2tsin t dt= 2√2π15e2t(2 sin t − cos t)π/20=2√25π(2eπ+ 1)36. For (4), express the curve y = f(x) in the parametric form x = t, y = f(t) so dx/dt = 1 anddy/dt = f (t) = f (x) = dy/dx. For (5), express x = g(y) as x = g(t), y = t sodx/dt = g (t) = g (y) = dx/dy and dy/dt = 1.
• Exercise Set 7.6 319EXERCISE SET 7.61. (a) fave =14 − 0402x dx = 4(b) 2x∗= 4, x∗= 2(c)2 448xy2. (a) fave =12 − 020x2dx = 4/3(b) (x∗)2= 4/3, x∗= ±2/√3,but only 2/√3 is in [0, 2](c)24xy233. fave =13 − 1313x dx =34x231= 64. fave =18 − (−1)8−1x1/3dx =1934x4/38−1=545. fave =1ππ0sin x dx = −1πcos xπ0=2π6. fave =3ππ/30sec x tan x dx =3πsec xπ/30=3π7. fave =1e − 1e11xdx =1e − 1(ln e − ln 1) =1e − 18. fave =11 + ln 5ln 5−1exdx =11 + ln 5(5 − e−1)9. fave =1√3 − 1√31dx1 + x2=1√3 − 1tan−1x√31=1√3 − 1π3−π4=1√3 − 1π1210. fave = 20−1/2dx√1 − x2= 2 sin−1x0−1/2=π311.12 − 020x(5x2 + 1)2dx = −1211015x2 + 120=12112. fave =11/4 − (−1/4)1/4−1/4sec2πxdx =2πtan πx1/4−1/4=4π13. fave =1440e−2xdx = −18e−2x40=18(1 − e−8)
• 320 Chapter 714. fave =2ln 311/√3du1 + u2=2ln 3tan−1u11/√3=2ln 3π4−π6=π6 ln 315. (a) 15 [f(0.4) + f(0.8) + f(1.2) + f(1.6) + f(2.0)] = 15 [0.48 + 1.92 + 4.32 + 7.68 + 12.00] = 5.28(b)1203[(0.1)2+ (0.2)2+ . . . + (1.9)2+ (2.0)2] =861200= 4.305(c) fave =12203x2dx =12x320= 4(d) Parts (a) and (b) can be interpreted as being two Riemann sums (n = 5, n = 20) for theaverage, using right endpoints. Since f is increasing, these sums overestimate the integral.16. (a)41472520≈ 1.645634921 (b)388477567232792560≈ 1.668771403(c) fave =211 +1xdx = (x + ln x)21= 1 + ln 2 ≈ 1.693147181(d) Parts (a) and (b) can be interpreted as being two Riemann sums (n = 5, n = 10) for theaverage, using right endpoints. Since f is decreasing, these sums underestimate the integral.17. (a)30v(t) dt =20(1 − t) dt +32(t − 3) dt = −12, so vave = −16(b)30v(t) dt =10t dt +21dt +32(−2t + 5) dt =12+ 1 + 0 =32, so vave =1218. Find v = f(t) such that50f(t) dt = 10, f(t) ≥ 0, f (5) = f (0) = 0. Let f(t) = ct(5 − t); then50ct(5−t) dt =52ct2−13ct350= c1252−1253=125c6= 10, c =1225, so v = f(t) =1225t(5−t)satisﬁes all the conditions.19. Linear means f(αx1 + βx2) = αf(x1) + βf(x2), so fa + b2=12f(a) +12f(b) =f(a) + f(b)2.20. Suppose a(t) represents acceleration, and that a(t) = a0 for a ≤ t ≤ b. Then the velocity is givenby v(t) = a0t + v0, and the average velocity =1b − aba(a0t + v0) dt =a02(b + a) + v0, and thevelocity at the midpoint is va + b2= a0a + b2+ v0 which proves the result.21. (a) vave =14 − 141(3t3+ 2)dt =137894=2634(b) vave =s(4) − s(1)4 − 1=100 − 73= 3122. (a) aave =15 − 050(t + 1)dt = 7/2(b) aave =v(π/4) − v(0)π/4 − 0=√2/2 − 1π/4= (2√2 − 4)/π
• Exercise Set 7.6 32123. time to ﬁll tank = (volume of tank)/(rate of ﬁlling) = [π(3)25]/(1) = 45π, weight of water in tankat time t = (62.4) (rate of ﬁlling)(time) = 62.4t,weightave =145π45π062.4t dt = 1404π lb24. (a) If x is the distance from the cooler end, then the temperature is T(x) = (15 + 1.5x)◦C, andTave =110 − 0100(15 + 1.5x)dx = 22.5◦C(b) By the Mean-Value Theorem for Integrals there exists x∗in [0, 10] such thatf(x∗) =110 − 0100(15 + 1.5x)dx = 22.5, 15 + 1.5x∗= 22.5, x∗= 525.300100(1 − 0.0001t2)dt = 2910 cars, so an average of291030= 97 cars/min.26. Vave =27500010 − 0100e−0.17tdt = −161764.7059e−0.17t100= \$132, 212.9627. (a)17[0.74 + 0.65 + 0.56 + 0.45 + 0.35 + 0.25 + 0.16] = 0.4514285714(b)1770[0.5 + 0.5 sin(0.213x + 2.481) dx = 0.461428. (a) J0(1) =1ππ0cos(sin t) dt (b) 0.7651976866(c)1 2 4 6 7 8–0.50.51y = J0(x)xy (d) J0(x) = 0 if x = 2.40482629. Solve for k :k0√3x dx = 6k, so√323x3/2k0=23√3k3/2= 6k, k = (3√3)2= 2730. Solve for k :12kk−kdxk2 + x2= π, k =1231. (a) V 2rms =11/f − 01/f0V 2p sin2(2πft)dt =12fV 2p1/f0[1 − cos(4πft)]dt=12fV 2p t −14πfsin(4πft)1/f0=12V 2p , so Vrms = Vp/√2(b) Vp/√2 = 120, Vp = 120√2 ≈ 169.7 V32. w (t) = kt, w(t) = kt2/2 + w0; wave =1265226(kt2/2 + w0) dt =126k6t35226+ w0 =23663k + w0Solve 2366k/3 + w0 = kt2/2 + w0 for t, t = 2 · 2366/3, so t ≈ 39.716, so during the 40th week.
• 322 Chapter 7EXERCISE SET 7.71. (a) W = F · d = 30(7) = 210 ft·lb(b) W =61F(x) dx =61x−2dx = −1x61= 5/6 ft·lb2. W =50F(x) dx =2040 dx −52403(x − 5) dx = 80 + 60 = 140 J3. Since W =baF(x) dx = the area under the curve, it follows that d < 2.5 since the area increasesfaster under the left part of the curve. In fact, Wd =d0F(x) dx = 40d, andW =50F(x) dx = 140, so d = 7/4.4. W d =d0F(x) dx, so total work =baF(x) dx whereas Fave =1b − abaF(x) dx, so that theaverage is the work divided by the length of the interval.5. The calculus book has displacement zero, so no work is done holding it.6. One Newton is the same as 0.445 lb, so 40N is 17.8 lb.distance traveled =502t dt +155(15 − t) dt = 25 + 50 ft. The force is a constant 17.8 lb, so thework done is 17.8 · 75 = 1335 ft·lb.7. distance traveled =50v(t) dt =504t5dt =25t250= 10 ft. The force is a constant 10 lb, so thework done is 10 · 10 = 100 ft·lb.8. (a) F(x) = kx, F(0.05) = 0.05k = 45, k = 900 N/m(b) W =0.030900x dx = 0.405 J (c) W =0.100.05900x dx = 3.375 J9. F(x) = kx, F(0.2) = 0.2k = 100, k = 500 N/m, W =0.80500xdx = 160 J10. F(x) = kx, F(1/2) = k/2 = 6, k = 12 N/m, W =2012x dx = 24 J11. W =10kx dx = k/2 = 10, k = 20 lb/ft12. W =60(9 − x)62.4(25π)dx= 1560π60(9 − x)dx = 56,160π ft·lb 9 - xx0695
• Exercise Set 7.7 32313. W =60(9 − x)ρ(25π)dx = 900πρ ft·lb14. r/10 = x/15, r = 2x/3,W =100(15 − x)62.4(4πx2/9)dx=83.23π100(15x2− x3)dx= 208, 000π/3 ft·lb15 – xx0101510r3 – xx0234w(x)15. w/4 = x/3, w = 4x/3,W =20(3 − x)(9810)(4x/3)(6)dx= 7848020(3x − x2)dx= 261, 600 J320–23 – xxw(x)216. w = 2 4 − x2W =2−2(3 − x)(50)(2 4 − x2)(10)dx= 30002−24 − x2dx − 10002−2x 4 − x2dx= 3000[π(2)2/2] − 0 = 6000π ft·lb0109 10 – xx20 1517. (a) W =90(10 − x)62.4(300)dx= 18,72090(10 − x)dx= 926,640 ft·lb(b) to empty the pool in one hour would require926,640/3600 = 257.4 ft·lb of work per secondso hp of motor = 257.4/550 = 0.46818. W =90x(62.4)(300) dx = 18,72090x dx = (81/2)18,720 = 758,160 ft·lb19. W =100015(100 − x)dx= 75, 000 ft·lb1000100 – xxPulleyChain
• 324 Chapter 720. The total time of winding the rope is (20 ft)/(2 ft/s) = 10 s. During the time interval from time tto time t + ∆t the work done is ∆W = F(t) · ∆x.The distance ∆x = 2∆t, and the force F(t) is given by the weight w(t) of the bucket, rope andwater at time t. The bucket and its remaining water together weigh (3+20)−t/2 lb, and the ropeis 20 − 2t ft long and weighs 4(20 − 2t) oz or 5 − t/2 lb. Thus at time t the bucket, water and ropetogether weigh w(t) = 23 − t/2 + 5 − t/2 = 28 − t lb.The amount of work done in the time interval from time t to time t + ∆t is thus∆W = (28 − t)2∆t, and the total work done isW = limn→+∞(28 − t)2∆t =100(28 − t)2 dt = 2(28t − t2/2)100= 460 ft·lb.30000xRocket21. When the rocket is x ft above the groundtotal weight = weight of rocket + weight of fuel= 3 + [40 − 2(x/1000)]= 43 − x/500 tons,W =30000(43 − x/500)dx = 120, 000 ft·tons0–a axBA22. Let F(x) be the force needed to holdcharge A at position x, thenF(x) =c(a − x)2, F(−a) =c4a2= k,so c = 4a2k.W =0−a4a2k(a − x)−2dx = 2ak J23. (a) 150 = k/(4000)2, k = 2.4 × 109, w(x) = k/x2= 2,400,000,000/x2lb(b) 6000 = k/(4000)2, k = 9.6 × 1010, w(x) = 9.6 × 1010/(x + 4000)2lb(c) W =500040009.6(1010)x−2dx = 4,800,000 mi·lb = 2.5344 × 1010ft·lb24. (a) 20 = k/(1080)2, k = 2.3328 × 107, weight = w(x + 1080) = 2.3328 · 107/(x + 1080)2lb(b) W =10.80[2.3328 · 107/(x + 1080)2] dx = 213.86 mi·lb = 1,129,188 ft·lb25. W =12mv2f −12mv2i =124.00 × 105(v2f − 202). But W = F · d = (6.40 × 105) · (3.00 × 103), so19.2 × 108= 2.00 × 105v2f − 8.00 × 107, 19200 = 2v2f − 800, vf = 100 m/s.26. W = F · d = (2.00 × 105)(2.00 × 105) = 4 × 1010J; from the Work-Energy Relationship (5),v2f = 2W/m + v2i = 8 · 1010/(2 · 103) + 108≈ 11.832 m/s.27. (a) The kinetic energy would have decreased by12mv2=124 · 106(15000)2= 4.5 × 1014J(b) (4.5 × 1014)/(4.2 × 1015) ≈ 0.107 (c)100013(0.107) ≈ 8.24 bombs
• Exercise Set 7.8 325EXERCISE SET 7.81. (a) F = ρhA = 62.4(5)(100) = 31,200 lbP = ρh = 62.4(5) = 312 lb/ft2(b) F = ρhA = 9810(10)(25) = 2,452,500 NP = ρh = 9810(10) = 98.1 kPa2. (a) F = PA = 6 · 105(160) = 9.6 × 107N (b) F = PA = 100(60) = 6000 lb3. F =2062.4x(4)dx= 249.620x dx = 499.2 lb20 4x4. F =319810x(4)dx= 39,24031x dx= 156,960 N3104x5. F =509810x(2 25 − x2)dx= 19,62050x(25 − x2)1/2dx= 8.175 × 105N505x y = √25 – x2y2√25 – x26. By similar trianglesw(x)4=2√3 − x2√3, w(x) =2√3(2√3 − x),F =2√3062.4x2√3(2√3 − x) dx=124.8√32√30(2√3x − x2)dx = 499.2 lb2√30 44 4xw(x)7. By similar trianglesw(x)6=10 − x8w(x) =34(10 − x),F =1029810x34(10 − x) dx= 7357.5102(10x − x2)dx = 1,098,720 N1020xw(x)86
• 326 Chapter 78. w(x) = 16 + 2u(x), butu(x)4=12 − x8so u(x) =12(12 − x),w(x) = 16 + (12 − x) = 28 − x,F =12462.4x(28 − x)dx= 62.4124(28x − x2)dx = 77,209.6 lb.12441640xw(x)u(x)9. Yes: if ρ2 = 2ρ1 then F2 =baρ2h(x)w(x) dx =ba2ρ1h(x)w(x) dx = 2baρ1h(x)w(x) dx = 2F1.10. F =2050x(2 4 − x2)dx= 10020x(4 − x2)1/2dx= 800/3 lb20xy2√4 – x2y = √4 – x20xx√2a√2a√2a/2w1(x)w2(x)aaaa11. Find the forces on the upper and lower halves and add them:w1(x)√2a=x√2a/2, w1(x) = 2xF1 =√2a/20ρx(2x)dx = 2ρ√2a/20x2dx =√2ρa3/6,w2(x)√2a=√2a − x√2a/2, w2(x) = 2(√2a − x)F2 =√2a√2a/2ρx[2(√2a − x)]dx = 2ρ√2a√2a/2(√2ax − x2)dx =√2ρa3/3,F = F1 + F2 =√2ρa3/6 +√2ρa3/3 = ρa3/√2 lb12. If a constant vertical force is applied to a ﬂat plate which is horizontal and the magnitude of theforce is F, then, if the plate is tilted so as to form an angle θ with the vertical, the magnitude ofthe force on the plate decreases to F cos θ.Suppose that a ﬂat surface is immersed, at an angle θ with the vertical, in a ﬂuid of weight densityρ, and that the submerged portion of the surface extends from x = a to x = b along an x-axiswhose positive diretion is not necessarily down, but is slanted.Following the derivation of equation (8), we divide the interval [a, b] into n subintervalsa = x0 < x1 < . . . < xn−1 < xn = b. Then the magnitude Fk of the force on the plate satisﬁes theinequalities ρh(xk−1)Ak cos θ ≤ Fk ≤ ρh(xk)Ak cos θ, or equivalently thath(xk−1) ≤Fk sec θρAk≤ h(xk). Following the argument in the text we arrive at the desired equationF =baρh(x)w(x) sec θ dx.
• Exercise Set 7.8 327161044044√17xh(x)13.√162 + 42 =√272 = 4√17 is theother dimension of the bottom.(h(x) − 4)/4 = x/(4√17)h(x) = x/√17 + 4,sec θ = 4√17/16 =√17/4F =4√17062.4(x/√17 + 4)10(√17/4) dx= 156√174√170(x/√17 + 4)dx= 63,648 lb14. If we lower the water level by y ft then the force F1 is computed as in Exercise 13, but with h(x)replaced by h1(x) = x/√17 + 4 − y, and we obtainF1 = F − y4√17062.4(10)√17/4 dx = F − 624(17)y = 63,648 − 10,608y.If F1 = F/2 then 63,648/2 = 63,648 − 10,608y, y = 63,648/(2 · 10,608) = 3,so the water level should be reduced by 3 ft.15. h(x) = x sin 60◦=√3x/2,θ = 30◦, sec θ = 2/√3,F =10009810(√3x/2)(200)(2/√3) dx= 200 · 98101000x dx= 9810 · 1003= 9.81 × 109Nxh(x)200100060°16. F =h+2hρ0x(2)dx= 2ρ0h+2hx dx= 4ρ0(h + 1)h + 2h022xh17. (a) From Exercise 16, F = 4ρ0(h + 1) so (assuming that ρ0 is constant) dF/dt = 4ρ0(dh/dt)which is a positive constant if dh/dt is a positive constant.(b) If dh/dt = 20 then dF/dt = 80ρ0 lb/min from Part (a).18. (a) Let h1 and h2 be the maximum and minimum depths of the disk Dr. The pressure P(r) onone side of the disk satisﬁes inequality (5):ρh1 ≤ P(r) ≤ ρh2. Butlimr→0+h1 = limr→0+h2 = h, and henceρh = limr→0+ρh1 ≤ limr→0+P(r) ≤ limr→0+ρh2 = ρh, so limr→0+P(r) = ρh.(b) The disks Dr in Part (a) have no particular direction (the axes of the disks have arbitrarydirection). Thus P, the limiting value of P(r), is independent of direction.
• 328 Chapter 7EXERCISE SET 7.91. (a) sinh 3 ≈ 10.0179(b) cosh(−2) ≈ 3.7622(c) tanh(ln 4) = 15/17 ≈ 0.8824(d) sinh−1(−2) ≈ −1.4436(e) cosh−13 ≈ 1.7627(f) tanh−1 34≈ 0.97302. (a) csch(−1) ≈ −0.8509(b) sech(ln 2) = 0.8(c) coth 1 ≈ 1.3130(d) sech−1 12≈ 1.3170(e) coth−13 ≈ 0.3466(f) csch−1(−√3) ≈ −0.54933. (a) sinh(ln 3) =12(eln 3− e− ln 3) =123 −13=43(b) cosh(− ln 2) =12(e− ln 2+ eln 2) =1212+ 2 =54(c) tanh(2 ln 5) =e2 ln 5− e−2 ln 5e2 ln 5 + e−2 ln 5=25 − 1/2525 + 1/25=312313(d) sinh(−3 ln 2) =12(e−3 ln 2− e3 ln 2) =1218− 8 = −63164. (a)12(eln x+ e− ln x) =12x +1x=x2+ 12x, x > 0(b)12(eln x− e− ln x) =12x −1x=x2− 12x, x > 0(c)e2 ln x− e−2 ln xe2 ln x + e−2 ln x=x2− 1/x2x2 + 1/x2=x4− 1x4 + 1, x > 0(d)12(e− ln x+ eln x) =121x+ x =1 + x22x, x > 05. sinh x0 cosh x0 tanh x0 coth x0 sech x0 csch x02 √5 2/√5 √5/2 1/√5 1/23/4 5/4 3/5 5/3 4/5 4/34/3(a)(b)(c) 5/3 4/5 5/4 3/5 3/4(a) cosh2x0 = 1 + sinh2x0 = 1 + (2)2= 5, cosh x0 =√5(b) sinh2x0 = cosh2x0 − 1 =2516− 1 =916, sinh x0 =34(because x0 > 0)(c) sech2x0 = 1 − tanh2x0 = 1 −452= 1 −1625=925, sech x0 =35,cosh x0 =1sech x0=53, fromsinh x0cosh x0= tanh x0 we get sinh x0 =5345=436.ddxcschx =ddx1sinh x= −cosh xsinh2x= − coth x csch x for x = 0ddxsech x =ddx1cosh x= −sinh xcosh2x= − tanh x sech x for all xddxcoth x =ddxcosh xsinh x=sinh2x − cosh2xsinh2x= − csch2x for x = 0
• Exercise Set 7.9 3297. (a) y = sinh−1x if and only if x = sinh y; 1 =dydxdxdy=dydxcosh y; soddx[sinh−1x] =dydx=1cosh y=11 + sinh2y=1√1 + x2for all x.(b) Let x ≥ 1. Then y = cosh−1x if and only if x = cosh y; 1 =dydxdxdy=dydxsinh y, soddx[cosh−1x] =dydx=1sinh y=1cosh2y − 1=1x2 − 1for x ≥ 1.(c) Let −1 < x < 1. Then y = tanh−1x if and only if x = tanh y; thus1 =dydxdxdy=dydxsech2y =dydx(1 − tanh2y) = 1 − x2, soddx[tanh−1x] =dydx=11 − x2.9. 4 cosh(4x − 8) 10. 4x3sinh(x4) 11. −1xcsch2(ln x)12. 2sech22xtanh 2x13.1x2csch(1/x) coth(1/x) 14. −2e2xsech(e2x) tanh(e2x)15.2 + 5 cosh(5x) sinh(5x)4x + cosh2(5x)16. 6 sinh2(2x) cosh(2x)17. x5/2tanh(√x) sech2(√x) + 3x2tanh2(√x)18. −3 cosh(cos 3x) sin 3x 19.11 + x2/913= 1/ 9 + x220.11 + 1/x2(−1/x2) = −1|x|√x2 + 121. 1/ (cosh−1x)√x2 − 122. 1/ (sinh−1x)2 − 1√1 + x2 23. −(tanh−1x)−2/(1 − x2)24. 2(coth−1x)/(1 − x2) 25.sinh xcosh2x − 1=sinh x| sinh x|=1, x > 0−1, x < 026. (sech2x)/ 1 + tanh2x 27. −ex2x√1 − x+ exsech−1x28. 10(1 + x csch−1x)9−x|x|√1 + x2+ csch−1x31.17sinh7x + C 32.12sinh(2x − 3) + C 33.23(tanh x)3/2+ C34. −13coth(3x) + C 35. ln(cosh x) + C 36. −13coth3x + C37. −13sech3xln 3ln 2= 37/375 38. ln(cosh x)ln 30= ln 5 − ln 339. u = 3x,131√1 + u2du =13sinh−13x + C
• 330 Chapter 740. x =√2u,√2√2u2 − 2du =1√u2 − 1du = cosh−1(x/√2) + C41. u = ex,1u√1 − u2du = − sech−1(ex) + C42. u = cos θ, −1√1 + u2du = − sinh−1(cos θ) + C43. u = 2x,duu√1 + u2= −csch−1|u| + C = −csch−1|2x| + C44. x = 5u/3,5/3√25u2 − 25du =131√u2 − 1du =13cosh−1(3x/5) + C45. tanh−1x1/20= tanh−1(1/2) − tanh−1(0) =12ln1 + 1/21 − 1/2=12ln 346. sinh−1t√30= sinh−1√3 − sinh−10 = ln(√3 + 2)49. A =ln 30sinh 2x dx =12cosh 2xln 30=12[cosh(2 ln 3) − 1],but cosh(2 ln 3) = cosh(ln 9) =12(eln 9+ e− ln 9) =12(9 + 1/9) = 41/9 so A =12[41/9 − 1] = 16/9.50. V = πln 20sech2x dx = π tanh xln 20= π tanh(ln 2) = 3π/551. V = π50(cosh22x − sinh22x)dx = π50dx = 5π52.10cosh ax dx = 2,1asinh ax10= 2,1asinh a = 2, sinh a = 2a;let f(a) = sinh a − 2a, then an+1 = an −sinh an − 2ancosh an − 2, a1 = 2.2, . . . , a4 = a5 = 2.177318985.53. y = sinh x, 1 + (y )2= 1 + sinh2x = cosh2xL =ln 20cosh x dx = sinh xln 20= sinh(ln 2) =12(eln 2− e− ln 2) =122 −12=3454. y = sinh(x/a), 1 + (y )2= 1 + sinh2(x/a) = cosh2(x/a)L =x10cosh(x/a)dx = a sinh(x/a)x10= a sinh(x1/a)55. (a) limx→+∞sinh x = limx→+∞12(ex− e−x) = +∞ − 0 = +∞(b) limx→−∞sinh x = limx→−∞12(ex− e−x) = 0 − ∞ = −∞(c) limx→+∞tanh x = limx→+∞ex− e−xex + e−x= 1
• Exercise Set 7.9 331(d) limx→−∞tanh x = limx→−∞ex− e−xex + e−x= −1(e) limx→+∞sinh−1x = limx→+∞ln(x + x2 + 1) = +∞(f) limx→1−tanh−1x = limx→1−12[ln(1 + x) − ln(1 − x)] = +∞56. limx→±∞tanh x = limx→±∞sinh xcosh x= limx→±∞ex− e−xex + e−x= ±157. sinh(−x) =12(e−x− ex) = −12(ex− e−x) = − sinh xcosh(−x) =12(e−x+ ex) =12(ex+ e−x) = cosh x58. (a) cosh x + sinh x =12(ex+ e−x) +12(ex− e−x) = ex(b) cosh x − sinh x =12(ex+ e−x) −12(ex− e−x) = e−x(c) sinh x cosh y + cosh x sinh y =14(ex− e−x)(ey+ e−y) +14(ex+ e−x)(ey− e−y)=14[(ex+y− e−x+y+ ex−y− e−x−y) + (ex+y+ e−x+y− ex−y− e−x−y)]=12[e(x+y)− e−(x+y)] = sinh(x + y)(d) Let y = x in Part (c).(e) The proof is similar to Part (c), or: treat x as variable and y as constant, and diﬀerentiatethe result in Part (c) with respect to x.(f) Let y = x in Part (e).(g) Use cosh2x = 1 + sinh2x together with Part (f).(h) Use sinh2x = cosh2x − 1 together with Part (f).59. (a) Divide cosh2x − sinh2x = 1 by cosh2x.(b) tanh(x + y) =sinh x cosh y + cosh x sinh ycosh x cosh y + sinh x sinh y=sinh xcosh x+sinh ycosh y1 +sinh x sinh ycosh x cosh y=tanh x + tanh y1 + tanh x tanh y(c) Let y = x in Part (b).60. (a) Let y = cosh−1x; then x = cosh y =12(ey+ e−y), ey− 2x + e−y= 0, e2y− 2xey+ 1 = 0,ey=2x ±√4x2 − 42= x ± x2 − 1. To determine which sign to take, note that y ≥ 0so e−y≤ ey, x = (ey+ e−y)/2 ≤ (ey+ ey)/2 = ey, hence ey≥ x thus ey= x +√x2 − 1,y = cosh−1x = ln(x +√x2 − 1).(b) Let y = tanh−1x; then x = tanh y =ey− e−yey + e−y=e2y− 1e2y + 1, xe2y+ x = e2y− 1,1 + x = e2y(1 − x), e2y= (1 + x)/(1 − x), 2y = ln1 + x1 − x, y =12ln1 + x1 − x.
• 332 Chapter 761. (a)ddx(cosh−1x) =1 + x/√x2 − 1x +√x2 − 1= 1/ x2 − 1(b)ddx(tanh−1x) =ddx12(ln(1 + x) − ln(1 − x)) =1211 + x+11 − x= 1/(1 − x2)62. Let y = sech−1x then x = sech y = 1/ cosh y, cosh y = 1/x, y = cosh−1(1/x); the proofs for theremaining two are similar.63. If |u| < 1 then, by Theorem 7.8.6,du1 − u2= tanh−1u + C.For |u| > 1,du1 − u2= coth−1u + C = tanh−1(1/u) + C.64. (a)ddx(sech−1|x|) =ddx(sech−1√x2) = −1√x2√1 − x2x√x2= −1x√1 − x2(b) Similar to solution of Part (a)65. (a) limx→+∞(cosh−1x − ln x) = limx→+∞[ln(x + x2 − 1) − ln x]= limx→+∞lnx +√x2 − 1x= limx→+∞ln(1 + 1 − 1/x2) = ln 2(b) limx→+∞cosh xex= limx→+∞ex+ e−x2ex= limx→+∞12(1 + e−2x) = 1/266. For |x| < 1, y = tanh−1x is deﬁned and dy/dx = 1/(1 − x2) > 0; y = 2x/(1 − x2)2changes signat x = 0, so there is a point of inﬂection there.67. Let x = −u/a,1√u2 − a2du = −aa√x2 − 1dx = − cosh−1x + C = − cosh−1(−u/a) + C.− cosh−1(−u/a) = − ln(−u/a + u2/a2 − 1) = lna−u +√u2 − a2u +√u2 − a2u +√u2 − a2= ln u + u2 − a2 − ln a = ln |u + u2 − a2| + C1so1√u2 − a2du = ln u + u2 − a2 + C2.68. Using sinh x + cosh x = ex(Exercise 58a), (sinh x + cosh x)n= (ex)n= enx= sinh nx + cosh nx.69.a−aetxdx =1tetxa−a=1t(eat− e−at) =2 sinh attfor t = 0.70. (a) y = sinh(x/a), 1 + (y )2= 1 + sinh2(x/a) = cosh2(x/a)L = 2b0cosh(x/a) dx = 2a sinh(x/a)b0= 2a sinh(b/a)(b) The highest point is at x = b, the lowest at x = 0,so S = a cosh(b/a) − a cosh(0) = a cosh(b/a) − a.
• Exercise Set 7.9 33371. From Part (b) of Exercise 70, S = a cosh(b/a) − a so 30 = a cosh(200/a) − a. Let u = 200/a,then a = 200/u so 30 = (200/u)[cosh u − 1], cosh u − 1 = 0.15u. If f(u) = cosh u − 0.15u − 1,then un+1 = un −cosh un − 0.15un − 1sinh un − 0.15; u1 = 0.3, . . . , u4 = u5 = 0.297792782 ≈ 200/a soa ≈ 671.6079505. From Part (a), L = 2a sinh(b/a) ≈ 2(671.6079505) sinh(0.297792782) ≈ 405.9 ft.72. From Part (a) of Exercise 70, L = 2a sinh(b/a) so 120 = 2a sinh(50/a), a sinh(50/a) = 60. Letu = 50/a, then a = 50/u so (50/u) sinh u = 60, sinh u = 1.2u. If f(u) = sinh u − 1.2u, thenun+1 = un −sinh un − 1.2uncosh un − 1.2; u1 = 1, . . . , u5 = u6 = 1.064868548 ≈ 50/a so a ≈ 46.95415231.From Part (b), S = a cosh(b/a) − a ≈ 46.95415231[cosh(1.064868548) − 1] ≈ 29.2 ft.73. Set a = 68.7672, b = 0.0100333, c = 693.8597, d = 299.2239.(a) 6500–300 300(b) L = 2d01 + a2b2 sinh2bx dx= 1480.2798 ft(c) x = 283.6249 ft (d) 82◦74. (a)112tr (b) r = 1 when t ≈ 0.673080 s.(c) dr/dt = 4.48 m/s.75. (a) When the bow of the boat is at the point (x, y) and the person has walked a distance D,then the person is located at the point (0, D), the line segment connecting (0, D) and (x, y)has length a; thus a2= x2+ (D − y)2, D = y +√a2 − x2 = a sech−1(x/a).(b) Find D when a = 15, x = 10: D = 15 sech−1(10/15) = 15 ln1 + 5/92/3≈ 14.44 m.(c) dy/dx = −a2x√a2 − x2+x√a2 − x2=1√a2 − x2−a2x+ x = −1xa2 − x2,1 + [y ]2= 1 +a2− x2x2=a2x2; with a = 15 and x = 5, L =155225x2dx = −225x155= 30 m.
• 334 Chapter 7REVIEW EXERCISE SET6. (a) A =20(2 + x − x2) dx (b) A =20√y dy +42[(√y − (y − 2)] dy(c) V = π20[(2 + x)2− x4] dx(d) V = 2π20y√y dy + 2π42y[√y − (y − 2)] dy(e) V = 2π20x(2 + x − x2) dx (f) V = π20y dy +42π(y − (y − 2)2) dy(g) V = π20[(2 + x + 3)2− (x2+ 3)2] dx (h) V = 2π20[2 + x − x2](5 − x) dx7. (a) A =ba(f(x) − g(x)) dx +cb(g(x) − f(x)) dx +dc(f(x) − g(x)) dx(b) A =0−1(x3− x) dx +10(x − x3) dx +21(x3− x) dx =14+14+94=1148. distance = |v| dt, so(a) distance =600(3t − t2/20) dt = 1800 ft.(b) If T ≤ 60 then distance =T0(3t − t2/20) dt =32T2−160T3ft.9. Find where the curves cross: set x3= 4x2, by observation x = 2 is a solution. ThenV = π20[(x2+ 4)2− (x3)2] dx =4352105π.10. V = 2L/20π16R2L4(x2− L2/4)2=4π15LR211. V =41√x −1√x2dx = 2 ln 2 +3212. (a) π10(sin−1x)2dx. (b) 2ππ/20y(1 − sin y)dy.13. By implicit diﬀerentiationdydx= −yx1/3, so 1 +dydx2= 1 +yx2/3=x2/3+ y2/3x2/3=4x2/3,L =−1−82(−x)1/3dx = 9.14. (a) L =ln 1001 + (ex)2 dx (b) L =1011 +1y2dy15. A = 2π169√25 − x 4 +125 − xdx = 653/2− 373/2 π6
• Review Exercise Set 33516. (a) S =8/2702πx 1 + x−4/3 dx (b) S =202πy3271 + y4/81 dy(c) S =20π(y + 2) 1 + y4/81 dy17. For 0 < x < 3 the area between the curve and the x-axis consists of two triangles of equal areabut of opposite signs, hence 0. For 3 < x < 5 the area is a rectangle of width 2 and height 3.For 5 < x < 7 the area consists of two triangles of equal area but opposite sign, hence 0; and for7 < x < 10 the curve is given by y = (4t − 37)/3 and107(4t − 37)/3 dt = −3. Thus the desiredaverage is110(0 + 6 + 0 − 3) = 0.3.18. fave =1ln 2 − ln(1/2)ln 2ln(1/2)(ex+ e−x) dx =12 ln 2ln 2− ln 2(ex+ e−x) dx =32 ln 219. A cross section of the solid, perpendicular to the x-axis, has area equal to π(sec x)2, and the averageof these cross sectional areas is given by A =1π/3π/30π(sec x)2dx =3ππ tan xπ/30= 3√320. The average rate of change of f over [a, b] is1b − abaf (x) dx; but this is precisely the same asthe average value of f over [a, b].21. (a) F = kx,12= k14, k = 2, W =1/40kx dx = 1/16 J(b) 25 =L0kx dx = kL2/2, L = 5 m22. F = 30x + 2000, W =1500(30x + 2000) dx = 15 · 1502+ 2000 · 150 = 637,500 lb·ft23. (a) F =10ρx3 dx N042h(x) = 1 + xxw(x)(b) By similar trianglesw(x)4=x2, w(x) = 2x, soF =41ρ(1 + x)2x dx lb/ft2.(c) A formula for the parabola is y =8125x2− 10,so F =0−109810|y|21258(y + 10) dy N.24. y = a cosh ax, y = a2sinh ax = a2y
• 336 Chapter 725. (a) cosh 3x = cosh(2x + x) = cosh 2x cosh x + sinh 2x sinh x= (2 cosh2x − 1) cosh x + (2 sinh x cosh x) sinh x= 2 cosh3x − cosh x + 2 sinh2x cosh x= 2 cosh3x − cosh x + 2(cosh2x − 1) cosh x = 4 cosh3x − 3 cosh x(b) from Theorem 7.8.2 with x replaced byx2: cosh x = 2 cosh2 x2− 1,2 cosh2 x2= cosh x + 1, cosh2 x2=12(cosh x + 1),coshx2=12(cosh x + 1) (because coshx2> 0)(c) from Theorem 7.8.2 with x replaced byx2: cosh x = 2 sinh2 x2+ 1,2 sinh2 x2= cosh x − 1, sinh2 x2=12(cosh x − 1), sinhx2= ±12(cosh x − 1)
• 337CHAPTER 8Principles of Integral ValuationEXERCISE SET 8.11. u = 4 − 2x, du = −2dx, −12u3du = −18u4+ C = −18(4 − 2x)4+ C2. u = 4 + 2x, du = 2dx,32√u du = u3/2+ C = (4 + 2x)3/2+ C3. u = x2, du = 2xdx,12sec2u du =12tan u + C =12tan(x2) + C4. u = x2, du = 2xdx, 2 tan u du = −2 ln | cos u | + C = −2 ln | cos(x2)| + C5. u = 2 + cos 3x, du = −3 sin 3xdx, −13duu= −13ln |u| + C = −13ln(2 + cos 3x) + C6. u =23x, du =23dx,16du1 + u2=16tan−1u + C =16tan−1 23x + C7. u = ex, du = exdx, sinh u du = cosh u + C = cosh ex+ C8. u = ln x, du =1xdx, sec u tan u du = sec u + C = sec(ln x) + C9. u = tan x, du = sec2xdx, eudu = eu+ C = etan x+ C10. u = x2, du = 2xdx,12du√1 − u2=12sin−1u + C =12sin−1(x2) + C11. u = cos 5x, du = −5 sin 5xdx, −15u5du = −130u6+ C = −130cos65x + C12. u = sin x, du = cos x dx,duu√u2 + 1= − ln1 +√1 + u2u+ C = − ln1 + 1 + sin2xsin x+ C13. u = ex, du = exdx,du√4 + u2= ln u + u2 + 4 + C = ln ex+ e2x + 4 + C14. u = tan−1x, du =11 + x2dx, eudu = eu+ C = etan−1x+ C15. u =√x − 1, du =12√x − 1dx, 2 eudu = 2eu+ C = 2e√x−1+ C16. u = x2+ 2x, du = (2x + 2)dx,12cot u du =12ln | sin u| + C =12ln sin |x2+ 2x| + C17. u =√x, du =12√xdx, 2 cosh u du = 2 sinh u + C = 2 sinh√x + C
• 338 Chapter 818. u = ln x, du =dxx,duu2= −1u+ C = −1ln x+ C19. u =√x, du =12√xdx,2 du3u= 2 e−u ln 3du = −2ln 3e−u ln 3+ C = −2ln 33−√x+ C20. u = sin θ, du = cos θdθ, sec u tan u du = sec u + C = sec(sin θ) + C21. u =2x, du = −2x2dx, −12csch2u du =12coth u + C =12coth2x+ C22.dx√x2 − 4= ln x + x2 − 4 + C23. u = e−x, du = −e−xdx, −du4 − u2= −14ln2 + u2 − u+ C = −14ln2 + e−x2 − e−x+ C24. u = ln x, du =1xdx, cos u du = sin u + C = sin(ln x) + C25. u = ex, du = exdx,exdx√1 − e2x=du√1 − u2= sin−1u + C = sin−1ex+ C26. u = x−1/2, du = −12x3/2dx, − 2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2) + C27. u = x2, du = 2xdx,12ducsc u=12sin u du = −12cos u + C = −12cos(x2) + C28. 2u = ex, 2du = exdx,2du√4 − 4u2= sin−1u + C = sin−1(ex/2) + C29. 4−x2= e−x2ln 4, u = −x2ln 4, du = −2x ln 4 dx = −x ln 16 dx,−1ln 16eudu = −1ln 16eu+ C = −1ln 16e−x2ln 4+ C = −1ln 164−x2+ C30. 2πx= eπx ln 2, 2πxdx =1π ln 2eπx ln 2+ C =1π ln 22πx+ C31. (a) u = sin x, du = cos x dx, u du =12u2+ C =12sin2x + C(b) sin x cos x dx =12sin 2x dx = −14cos 2x + C = −14(cos2x − sin2x) + C(c) −14(cos2x − sin2x) + C = −14(1 − sin2x − sin2x) + C = −14+12sin2x + C,and this is the same as the answer in part (a) except for the constants.32. (a) sech 2x =1cosh 2x=1cosh2x + sinh2x(now multiply top and bottom by sech2x)=sech2x1 + tanh2x
• Exercise Set 8.2 339(b) sech2x dx =sech2x1 + tanh2xdx = tan−1(tanh x) + C, or, replacing 2x with x,sechx dx = tan−1(tanh(x/2)) + C(c) sech x =1cosh x=2ex + e−x=2exe2x + 1(d) sech x dx = 2exe2x + 1dx = 2 tan−1(ex) + C33. (a)sec2xtan x=1cos2 x tan x=1cos x sin x(b) csc 2x =1sin 2x=12 sin x cos x=12sec2xtan x, so csc 2x dx =12ln tan x + C(c) sec x =1cos x=1sin(π/2 − x)= csc(π/2 − x), sosec x dx = csc(π/2 − x) dx = −12ln tan(π/2 − x) + CEXERCISE SET 8.21. u = x, dv = e−2xdx, du = dx, v = −12 e−2x;xe−2xdx = −12xe−2x+12e−2xdx = −12xe−2x−14e−2x+ C2. u = x, dv = e3xdx, du = dx, v =13e3x; xe3xdx =13xe3x−13e3xdx =13xe3x−19e3x+ C3. u = x2, dv = exdx, du = 2x dx, v = ex; x2exdx = x2ex− 2 xexdx.For xexdx use u = x, dv = exdx, du = dx, v = exto getxexdx = xex− ex+ C1 so x2exdx = x2ex− 2xex+ 2ex+ C4. u = x2, dv = e−2xdx, du = 2x dx, v = −12e−2x; x2e−2xdx = −12x2e−2x+ xe−2xdxFor xe−2xdx use u = x, dv = e−2xdx to getxe−2xdx = −12xe−2x+12e−2xdx = −12xe−2x−14e−2x+ Cso x2e−2xdx = −12x2e−2x−12xe−2x−14e−2x+ C5. u = x, dv = sin 3x dx, du = dx, v = −13cos 3x;x sin 3x dx = −13x cos 3x +13cos 3x dx = −13x cos 3x +19sin 3x + C
• 340 Chapter 86. u = x, dv = cos 2x dx, du = dx, v =12sin 2x;x cos 2x dx =12x sin 2x −12sin 2x dx =12x sin 2x +14cos 2x + C7. u = x2, dv = cos x dx, du = 2x dx, v = sin x; x2cos x dx = x2sin x − 2 x sin x dxFor x sin x dx use u = x, dv = sin x dx to getx sin x dx = −x cos x + sin x + C1 so x2cos x dx = x2sin x + 2x cos x − 2 sin x + C8. u = x2, dv = sin x dx, du = 2x dx, v = − cos x;x2sin x dx = −x2cos x + 2 x cos x dx; for x cos x dx use u = x, dv = cos x dx to getx cos x dx = x sin x + cos x + C1 so x2sin x dx = −x2cos x + 2x sin x + 2 cos x + C9. u = ln x, dv = x dx, du =1xdx, v =12x2; x ln x dx =12x2ln x −12x dx =12x2ln x −14x2+ C10. u = ln x, dv =√x dx, du =1xdx, v =23x3/2;√x ln x dx =23x3/2ln x −23x1/2dx =23x3/2ln x −49x3/2+ C11. u = (ln x)2, dv = dx, du = 2ln xxdx, v = x; (ln x)2dx = x(ln x)2− 2 ln x dx.Use u = ln x, dv = dx to get ln x dx = x ln x − dx = x ln x − x + C1 so(ln x)2dx = x(ln x)2− 2x ln x + 2x + C12. u = ln x, dv =1√xdx, du =1xdx, v = 2√x;ln x√xdx = 2√x ln x−21√xdx = 2√x ln x−4√x+C13. u = ln(3x − 2), dv = dx, du =33x − 2dx, v = x; ln(3x − 2)dx = x ln(3x − 2) −3x3x − 2dxbut3x3x − 2dx = 1 +23x − 2dx = x +23ln(3x − 2) + C1 soln(3x − 2)dx = x ln(3x − 2) − x −23ln(3x − 2) + C14. u = ln(x2+ 4), dv = dx, du =2xx2 + 4dx, v = x; ln(x2+ 4)dx = x ln(x2+ 4) − 2x2x2 + 4dxbutx2x2 + 4dx = 1 −4x2 + 4dx = x − 2 tan−1 x2+ C1 soln(x2+ 4)dx = x ln(x2+ 4) − 2x + 4 tan−1 x2+ C
• Exercise Set 8.2 34115. u = sin−1x, dv = dx, du = 1/√1 − x2dx, v = x;sin−1x dx = x sin−1x − x/ 1 − x2dx = x sin−1x + 1 − x2 + C16. u = cos−1(2x), dv = dx, du = −2√1 − 4x2dx, v = x;cos−1(2x)dx = x cos−1(2x) +2x√1 − 4x2dx = x cos−1(2x) −121 − 4x2 + C17. u = tan−1(3x), dv = dx, du =31 + 9x2dx, v = x;tan−1(3x)dx = x tan−1(3x) −3x1 + 9x2dx = x tan−1(3x) −16ln(1 + 9x2) + C18. u = tan−1x, dv = x dx, du =11 + x2dx, v =12x2; x tan−1x dx =12x2tan−1x −12x21 + x2dxbutx21 + x2dx = 1 −11 + x2dx = x − tan−1x + C1 sox tan−1x dx =12x2tan−1x −12x +12tan−1x + C19. u = ex, dv = sin x dx, du = exdx, v = − cos x; exsin x dx = −excos x + excos x dx.For excos x dx use u = ex, dv = cos x dx to get excos x = exsin x − exsin x dx soexsin x dx = −excos x + exsin x − exsin x dx,2 exsin x dx = ex(sin x − cos x) + C1, exsin x dx =12ex(sin x − cos x) + C20. u = e3x, dv = cos 2x dx, du = 3e3xdx, v =12sin 2x;e3xcos 2x dx =12e3xsin 2x −32e3xsin 2x dx. Use u = e3x, dv = sin 2x dx to gete3xsin 2x dx = −12e3xcos 2x +32e3xcos 2x dx, soe3xcos 2x dx =12e3xsin 2x +34e3xcos 2x −94e3xcos 2x dx,134e3xcos 2x dx =14e3x(2 sin 2x+3 cos 2x)+C1, e3xcos 2x dx =113e3x(2 sin 2x+3 cos 2x)+C21. u = eax, dv = sin bx dx, du = aeaxdx, v = −1bcos bx (b = 0);eaxsin bx dx = −1beaxcos bx +abeaxcos bx dx. Use u = eax, dv = cos bx dx to geteaxcos bx dx =1beaxsin bx −abeaxsin bx dx soeaxsin bx dx = −1beaxcos bx +ab2eaxsin bx −a2b2eaxsin bx dx,eaxsin bx dx =eaxa2 + b2(a sin bx − b cos bx) + C
• 342 Chapter 822. From Exercise 21 with a = −3, b = 5, x = θ, answer =e−3θ√34(−3 sin 5θ − 5 cos 5θ) + C23. u = sin(ln x), dv = dx, du =cos(ln x)xdx, v = x;sin(ln x)dx = x sin(ln x) − cos(ln x)dx. Use u = cos(ln x), dv = dx to getcos(ln x)dx = x cos(ln x) + sin(ln x)dx sosin(ln x)dx = x sin(ln x) − x cos(ln x) − sin(ln x)dx,sin(ln x)dx =12x[sin(ln x) − cos(ln x)] + C24. u = cos(ln x), dv = dx, du = −1xsin(ln x)dx, v = x;cos(ln x)dx = x cos(ln x) + sin(ln x)dx. Use u = sin(ln x), dv = dx to getsin(ln x)dx = x sin(ln x) − cos(ln x)dx socos(ln x)dx = x cos(ln x) + x sin(ln x) − cos(ln x)dx,cos(ln x)dx =12x[cos(ln x) + sin(ln x)] + C25. u = x, dv = sec2x dx, du = dx, v = tan x;x sec2x dx = x tan x − tan x dx = x tan x −sin xcos xdx = x tan x + ln | cos x| + C26. u = x, dv = tan2x dx = (sec2x − 1)dx, du = dx, v = tan x − x;x tan2x dx = x tan x − x2− (tan x − x)dx= x tan x − x2+ ln | cos x| +12x2+ C = x tan x −12x2+ ln | cos x| + C27. u = x2, dv = xex2dx, du = 2x dx, v =12ex2;x3ex2dx =12x2ex2− xex2dx =12x2ex2−12ex2+ C28. u = xex, dv =1(x + 1)2dx, du = (x + 1)exdx, v = −1x + 1;xex(x + 1)2dx = −xexx + 1+ exdx = −xexx + 1+ ex+ C =exx + 1+ C29. u = x, dv = e2xdx, du = dx, v =12e2x;20xe2xdx =12xe2x20−1220e2xdx = e4−14e2x20= e4−14(e4− 1) = (3e4+ 1)/4
• Exercise Set 8.2 34330. u = x, dv = e−5xdx, du = dx, v = −15e−5x;10xe−5xdx = −15xe−5x10+1510e−5xdx= −15e−5−125e−5x10= −15e−5−125(e−5− 1) = (1 − 6e−5)/2531. u = ln x, dv = x2dx, du =1xdx, v =13x3;e1x2ln x dx =13x3ln xe1−13e1x2dx =13e3−19x3e1=13e3−19(e3− 1) = (2e3+ 1)/932. u = ln x, dv =1x2dx, du =1xdx, v = −1x;e√eln xx2dx = −1xln xe√e+e√e1x2dx= −1e+1√eln√e −1xe√e= −1e+12√e−1e+1√e=3√e − 42e33. u = ln(x + 2), dv = dx, du =1x + 2dx, v = x;1−1ln(x + 2)dx = x ln(x + 2)1−1−1−1xx + 2dx = ln 3 + ln 1 −1−11 −2x + 2dx= ln 3 − [x − 2 ln(x + 2)]1−1= ln 3 − (1 − 2 ln 3) + (−1 − 2 ln 1) = 3 ln 3 − 234. u = sin−1x, dv = dx, du =1√1 − x2dx, v = x;√3/20sin−1x dx = x sin−1x√3/20−√3/20x√1 − x2dx =√32sin−1√32+ 1 − x2√3/20=√32π3+12− 1 =π√36−1235. u = sec−1√θ, dv = dθ, du =12θ√θ − 1dθ, v = θ;42sec−1√θdθ = θ sec−1√θ42−12421√θ − 1dθ = 4 sec−12 − 2 sec−1√2 −√θ − 142= 4π3− 2π4−√3 + 1 =5π6−√3 + 136. u = sec−1x, dv = x dx, du =1x√x2 − 1dx, v =12x2;21x sec−1x dx =12x2sec−1x21−1221x√x2 − 1dx=12[(4)(π/3) − (1)(0)] −12x2 − 121= 2π/3 −√3/2
• 344 Chapter 837. u = x, dv = sin 2x dx, du = dx, v = −12cos 2x;π0x sin 2x dx = −12x cos 2xπ0+12π0cos 2x dx = −π/2 +14sin 2xπ0= −π/238.π0(x + x cos x)dx =12x2π0+π0x cos x dx =π22+π0x cos x dx;u = x, dv = cos x dx, du = dx, v = sin xπ0x cos x dx = x sin xπ0−π0sin x dx = cos xπ0= −2 soπ0(x + x cos x)dx = π2/2 − 239. u = tan−1√x, dv =√xdx, du =12√x(1 + x)dx, v =23x3/2;31√x tan−1√xdx =23x3/2tan−1√x31−1331x1 + xdx=23x3/2tan−1√x31−13311 −11 + xdx=23x3/2tan−1√x −13x +13ln |1 + x|31= (2√3π − π/2 − 2 + ln 2)/340. u = ln(x2+ 1), dv = dx, du =2xx2 + 1dx, v = x;20ln(x2+ 1)dx = x ln(x2+ 1)20−202x2x2 + 1dx = 2 ln 5 − 2201 −1x2 + 1dx= 2 ln 5 − 2(x − tan−1x)20= 2 ln 5 − 4 + 2 tan−1241. t =√x, t2= x, dx = 2t dt(a) e√xdx = 2 tetdt; u = t, dv = etdt, du = dt, v = et,e√xdx = 2tet− 2 etdt = 2(t − 1)et+ C = 2(√x − 1)e√x+ C(b) cos√x dx = 2 t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t,cos√x dx = 2t sin t − 2 sin tdt = 2t sin t + 2 cos t + C = 2√x sin√x + 2 cos√x + C
• Exercise Set 8.2 34542. Let q1(x), q2(x), q3(x) denote successive antiderivatives of q(x),so that q3(x) = q2(x), q2(x) = q1(x), q1(x) = q(x). Let p(x) = ax2+ bx + c.Repeated RepeatedDiﬀerentiation Antidiﬀerentiationax2+ bx + c q(x)+2ax + b q1(x)−2a q2(x)+0 q3(x)Then p(x)q(x) dx = (ax2+ bx + c)q1(x) − (2ax + b)q2(x) + 2aq3(x) + C. Check:ddx[(ax2+bx + c)q1(x) − (2ax + b)q2(x) + 2aq3(x)]= (2ax + b)q1(x) + (ax2+ bx + c)q(x) − 2aq2(x) − (2ax + b)q1(x) + 2aq2(x) = p(x)q(x)43. Repeated RepeatedDiﬀerentiation Antidiﬀerentiation3x2− x + 2 e−x+6x − 1 −e−x−6 e−x+0 −e−x(3x2− x + 2)e−x= −(3x2− x + 2)e−x− (6x − 1)e−x− 6e−x+ C = −e−x[3x2+ 5x + 7] + C44. Repeated RepeatedDiﬀerentiation Antidiﬀerentiationx2+ x + 1 sin x+2x + 1 − cos x−2 − sin x+0 cos x(x2+ x + 1) sin x dx = −(x2+ x + 1) cos x + (2x + 1) sin x + 2 cos x + C= −(x2+ x − 1) cos x + (2x + 1) sin x + C
• 346 Chapter 845. Repeated RepeatedDiﬀerentiation Antidiﬀerentiation4x4sin 2x+16x3−12cos 2x−48x2−14sin 2x+96x18cos 2x−96116sin 2x+0 −132cos 2x4x4sin 2x dx = (−2x4+ 6x2− 3) cos 2x + −(4x3+ 6x) sin 2x + C46. Repeated RepeatedDiﬀerentiation Antidiﬀerentiationx3√2x + 1+3x2 13(2x + 1)3/2−6x115(2x + 1)5/2+61105(2x + 1)7/2−01945(2x + 1)9/2x3√2x + 1 dx =13x3(2x + 1)3/2−15x2(2x + 1)5/2+235x(2x + 1)7/2−2315(2x + 1)9/2+ C47. (a) We perform a single integration by parts:u = cos x, dv = sin x dx, du = − sin x dx, v = − cos x,sin x cos x dx = − cos2x − sin x cos x dx. Thus2 sin x cos x dx = − cos2x + C, sin x cos x dx = −12cos2x + C(b) u = sin x, du = cos x dx, sin x cos x dx = u du =12u2+ C =12sin2x + C48. (a) u = x2, dv =x√x2 + 1, du = 2x dx, v = x2 + 1,10x3√x2 + 1dx = x2x2 + 110−102x x2 + 1 dx =√2 −23(x2+ 1)3/210= −13√2 +23
• Exercise Set 8.2 347(b) u = x2 + 1, du =x√x2 + 1dx,√21(u2− 1) du =13u3− u√21=23√2 −√2 −13+ 1 = −13√2 +23.49. (a) A =e1ln x dx = (x ln x − x)e1= 1(b) V = πe1(ln x)2dx = π (x(ln x)2− 2x ln x + 2x)e1= π(e − 2)50. A =π/20(x − x sin x)dx =12x2π/20−π/20x sin x dx =π28− (−x cos x + sin x)π/20= π2/8 − 151. V = 2ππ0x sin x dx = 2π(−x cos x + sin x)π0= 2π252. V = 2ππ/20x cos x dx = 2π(cos x + x sin x)π/20= π(π − 2)53. distance =π0t3sin tdt;Repeated RepeatedDiﬀerentiation Antidiﬀerentiationt3sin t+3t2− cos t−6t − sin t+6 cos t−0 sin tπ0t3sin t dx = [(−t3cos t + 3t2sin t + 6t cos t − 6 sin t)]π0= π3− 6π54. u = 2t, dv = sin(kωt)dt, du = 2dt, v = −1kωcos(kωt); the integrand is an even function of t soπ/ω−π/ωt sin(kωt) dt = 2π/ω0t sin(kωt) dt = −2kωt cos(kωt)π/ω0+ 2π/ω01kωcos(kωt) dt=2π(−1)k+1kω2+2k2ω2sin(kωt)π/ω0=2π(−1)k+1kω255. (a) sin4x dx = −14sin3x cos x +34sin2x dx= −14sin3x cos x +34−12sin x cos x +12x + C= −14sin3x cos x −38sin x cos x +38x + C
• 348 Chapter 8(b)π/20sin5x dx = −15sin4x cos xπ/20+45π/20sin3x dx=45−13sin2x cos xπ/20+23π/20sin x dx= −815cos xπ/20=81556. (a) cos5x dx =15cos4x sin x +45cos3x dx =15cos4x sin x +4513cos2x sin x +23sin x + C=15cos4x sin x +415cos2x sin x +815sin x + C(b) cos6x dx =16cos5x sin x +56cos4x dx=16cos5x sin x +5614cos3x sin x +34cos2x dx=16cos5x sin x +524cos3x sin x +5812cos x sin x +12x + C,16cos5x sin x +524cos3x sin x +516cos x sin x +516xπ/20= 5π/3257. u = sinn−1x, dv = sin x dx, du = (n − 1) sinn−2x cos x dx, v = − cos x;sinnx dx = − sinn−1x cos x + (n − 1) sinn−2x cos2x dx= − sinn−1x cos x + (n − 1) sinn−2x (1 − sin2x)dx= − sinn−1x cos x + (n − 1) sinn−2x dx − (n − 1) sinnx dx,n sinnx dx = − sinn−1x cos x + (n − 1) sinn−2x dx,sinnx dx = −1nsinn−1x cos x +n − 1nsinn−2x dx58. (a) u = secn−2x, dv = sec2x dx, du = (n − 2) secn−2x tan x dx, v = tan x;secnx dx = secn−2x tan x − (n − 2) secn−2x tan2x dx= secn−2x tan x − (n − 2) secn−2x (sec2x − 1)dx= secn−2x tan x − (n − 2) secnx dx + (n − 2) secn−2x dx,(n − 1) secnx dx = secn−2x tan x + (n − 2) secn−2x dx,secnx dx =1n − 1secn−2x tan x +n − 2n − 1secn−2x dx
• Exercise Set 8.2 349(b) tannx dx = tann−2x (sec2x − 1) dx = tann−1x sec2x dx − tann−2x dx=1n − 1tann−1x − tann−2x dx(c) u = xn, dv = exdx, du = nxn−1dx, v = ex; xnexdx = xnex− n xn−1exdx59. (a) tan4x dx =13tan3x − tan2x dx =13tan3x − tan x + dx =13tan3x − tan x + x + C(b) sec4x dx =13sec2x tan x +23sec2x dx =13sec2x tan x +23tan x + C(c) x3exdx = x3ex− 3 x2exdx = x3ex− 3 x2ex− 2 xexdx= x3ex− 3x2ex+ 6 xex− exdx = x3ex− 3x2ex+ 6xex− 6ex+ C60. (a) u = 3x,x2e3xdx =127u2eudu =127u2eu− 2 ueudu =127u2eu−227ueu− eudu=127u2eu−227ueu+227eu+ C =13x2e3x−29xe3x+227e3x+ C(b) u = −√x,10xe−√xdx = 2−10u3eudu,u3eudu = u3eu− 3 u2eudu = u3eu− 3 u2eu− 2 ueudu= u3eu− 3u2eu+ 6 ueu− eudu = u3eu− 3u2eu+ 6ueu− 6eu+ C,2−10u3eudu = 2(u3− 3u2+ 6u − 6)eu−10= 12 − 32e−161. u = x, dv = f (x)dx, du = dx, v = f (x);1−1x f (x)dx = xf (x)1−1−1−1f (x)dx= f (1) + f (−1) − f(x)1−1= f (1) + f (−1) − f(1) + f(−1)62. (a) u dv = uv − v du = x(sin x + C1) + cos x − C1x + C2 = x sin x + cos x + C2;the constant C1 cancels out and hence plays no role in the answer.(b) u(v + C1) − (v + C1)du = uv + C1u − v du − C1u = uv − v du
• 350 Chapter 863. u = ln(x + 1), dv = dx, du =dxx + 1, v = x + 1;ln(x + 1) dx = u dv = uv − v du = (x + 1) ln(x + 1) − dx = (x + 1) ln(x + 1) − x + C64. u = ln(3x − 2), dv = dx, du =3dx3x − 2, v = x −23;ln(3x − 2) dx = u dv = uv − v du = x −23ln(3x − 2) − x −231x − 2/3dx= x −23ln(3x − 2) − x −23+ C65. u = tan−1x, dv = x dx, du =11 + x2dx, v =12(x2+ 1)x tan−1x dx = u dv = uv − v du =12(x2+ 1) tan−1x −12dx=12(x2+ 1) tan−1x −12x + C66. u =1ln x, dv = 1x dx, du = − 1x(ln x)2 dx, v = ln x1x ln xdx = 1 +1x ln xdx.This seems to imply that 1 = 0, but recall that both sides represent a function plus an arbitraryconstant; these two arbitrary constants will take care of the 1.67. (a) u = f(x), dv = dx, du = f (x), v = x;baf(x) dx = xf(x)ba−baxf (x) dx = bf(b) − af(a) −baxf (x) dx(b) Substitute y = f(x), dy = f (x) dx, x = a when y = f(a), x = b when y = f(b),baxf (x) dx =f(b)f(a)x dy =f(b)f(a)f−1(y) dy(c) From a = f−1(α) and b = f−1(β) we getbf(b) − af(a) = βf−1(β) − αf−1(α); thenβαf−1(x) dx =βαf−1(y) dy =f(b)f(a)f−1(y) dy,which, by Part (b), yieldsaba = f –1(a) b = f –1(b)xyA1A2βαf−1(x) dx = bf(b) − af(a) −baf(x) dx= βf−1(β) − αf−1(α) −f−1(β)f−1(α)f(x) dxNote from the ﬁgure that A1 =βαf−1(x) dx, A2 =f−1(β)f−1(α)f(x) dx, andA1 + A2 = βf−1(β) − αf−1(α), a “picture proof”.
• Exercise Set 8.3 35168. (a) Use Exercise 67(c);1/20sin−1x dx =12sin−1 12−0·sin−10−sin−1(1/2)sin−1(0)sin x dx =12sin−1 12−π/60sin x dx(b) Use Exercise 67(b);e2eln x dx = e2ln e2− e ln e −ln e2ln ef−1(y) dy = 2e2− e −21eydy = 2e2− e −21exdxEXERCISE SET 8.31. u = cos x, − u3du = −14cos4x + C2. u = sin 3x,13u5du =118sin63x + C3. sin25θ =12(1 − cos 10θ) dθ =12θ −120sin 10θ + C4. cos23x dx =12(1 + cos 6x)dx =12x +112sin 6x + C5. sin3aθ dθ = sin aθ(1 − cos2aθ) dθ = −1acos aθ −13acos3aθ + C (a = 0)6. cos3at dt = (1 − sin2at) cos at dt= cos at dt − sin2at cos at dt =1asin at −13asin3at + C (a = 0)7. u = sin ax,1au du =12asin2ax + C, a = 08. sin3x cos3x dx = sin3x(1 − sin2x) cos x dx= (sin3x − sin5x) cos x dx =14sin4x −16sin6x + C9. sin2t cos3t dt = sin2t(1 − sin2t) cos t dt = (sin2t − sin4t) cos t dt=13sin3t −15sin5t + C10. sin3x cos2x dx = (1 − cos2x) cos2x sin x dx= (cos2x − cos4x) sin x dx = −13cos3x +15cos5x + C11. sin2x cos2x dx =14sin22x dx =18(1 − cos 4x)dx =18x −132sin 4x + C
• 352 Chapter 812. sin2x cos4x dx =18(1 − cos 2x)(1 + cos 2x)2dx =18(1 − cos22x)(1 + cos 2x)dx=18sin22x dx +18sin22x cos 2x dx =116(1 − cos 4x)dx +148sin32x=116x −164sin 4x +148sin32x + C13. sin 2x cos 3x dx =12(sin 5x − sin x)dx = −110cos 5x +12cos x + C14. sin 3θ cos 2θdθ =12(sin 5θ + sin θ)dθ = −110cos 5θ −12cos θ + C15. sin x cos(x/2)dx =12[sin(3x/2) + sin(x/2)]dx = −13cos(3x/2) − cos(x/2) + C16. u = cos x, − u1/3du = −34cos4/3x + C17.π/20cos3x dx =π/20(1 − sin2x) cos x dx= sin x −13sin3xπ/20=2318.π/20sin2(x/2) cos2(x/2)dx =14π/20sin2x dx =18π/20(1 − cos 2x)dx=18x −12sin 2xπ/20= π/1619.π/30sin43x cos33x dx =π/30sin43x(1 − sin23x) cos 3x dx =115sin53x −121sin73xπ/30= 020.π−πcos25θ dθ =12π−π(1 + cos 10θ)dθ =12θ +110sin 10θπ−π= π21.π/60sin 4x cos 2x dx =12π/60(sin 2x + sin 6x)dx = −14cos 2x −112cos 6xπ/60= [(−1/4)(1/2) − (1/12)(−1)] − [−1/4 − 1/12] = 7/2422.2π0sin2kx dx =122π0(1 − cos 2kx)dx =12x −12ksin 2kx2π0= π −14ksin 4πk (k = 0)23.12tan(2x − 1) + C 24. −15ln | cos 5x| + C25. u = e−x, du = −e−xdx; − tan u du = ln | cos u| + C = ln | cos(e−x)| + C26.13ln | sin 3x| + C 27.14ln | sec 4x + tan 4x| + C
• Exercise Set 8.3 35328. u =√x, du =12√xdx; 2 sec u du = 2 ln | sec u + tan u| + C = 2 ln sec√x + tan√x + C29. u = tan x, u2du =13tan3x + C30. tan5x(1 + tan2x) sec2x dx = (tan5x + tan7x) sec2x dx =16tan6x +18tan8x + C31. tan 4x(1 + tan24x) sec24x dx = (tan 4x + tan34x) sec24x dx =18tan24x +116tan44x + C32. tan4θ(1 + tan2θ) sec2θ dθ =15tan5θ +17tan7θ + C33. sec4x(sec2x − 1) sec x tan x dx = (sec6x − sec4x) sec x tan x dx =17sec7x −15sec5x + C34. (sec2θ − 1)2sec θ tan θdθ = (sec4θ − 2 sec2θ + 1) sec θ tan θdθ =15sec5θ −23sec3θ + sec θ + C35. (sec2x − 1)2sec x dx = (sec5x − 2 sec3x + sec x)dx = sec5x dx − 2 sec3x dx + sec x dx=14sec3x tan x +34sec3x dx − 2 sec3x dx + ln | sec x + tan x|=14sec3x tan x −5412sec x tan x +12ln | sec x + tan x| + ln | sec x + tan x| + C=14sec3x tan x −58sec x tan x +38ln | sec x + tan x| + C36. [sec2x − 1] sec3x dx = [sec5x − sec3x]dx=14sec3x tan x +34sec3x dx − sec3x dx (equation (20))=14sec3x tan x −14sec3x dx=14sec3x tan x −18sec x tan x −18ln | sec x + tan x| + C (equation (20), (22))37. sec2t(sec t tan t)dt =13sec3t + C 38. sec4x(sec x tan x)dx =15sec5x + C39. sec4x dx = (1 + tan2x) sec2x dx = (sec2x + tan2x sec2x)dx = tan x +13tan3x + C40. Using equation (20),sec5x dx =14sec3x tan x +34sec3x dx=14sec3x tan x +38sec x tan x +38ln | sec x + tan x| + C
• 354 Chapter 841. u = 4x, use equation (19) to get14tan3u du =1412tan2u + ln | cos u| + C =18tan24x +14ln | cos 4x| + C42. Use equation (19) to get tan4x dx =13tan3x − tan x + x + C43.√tan x(1 + tan2x) sec2x dx =23tan3/2x +27tan7/2x + C44. sec1/2x(sec x tan x)dx =23sec3/2x + C45.π/80(sec22x − 1)dx =12tan 2x − xπ/80= 1/2 − π/846.π/60sec22θ(sec 2θ tan 2θ)dθ =16sec32θπ/60= (1/6)(2)3− (1/6)(1) = 7/647. u = x/2,2π/40tan5u du =12tan4u − tan2u − 2 ln | cos u|π/40= 1/2 − 1 − 2 ln(1/√2) = −1/2 + ln 248. u = πx,1ππ/40sec u tan u du =1πsec uπ/40= (√2 − 1)/π49. (csc2x − 1) csc2x(csc x cot x)dx = (csc4x − csc2x)(csc x cot x)dx = −15csc5x +13csc3x + C50.cos23tsin23t·1cos 3tdt = csc 3t cot 3t dt = −13csc 3t + C51. (csc2x − 1) cot x dx = csc x(csc x cot x)dx −cos xsin xdx = −12csc2x − ln | sin x| + C52. (cot2x + 1) csc2x dx = −13cot3x − cot x + C53. (a)2π0sin mx cos nx dx =122π0[sin(m + n)x + sin(m − n)x]dx= −cos(m + n)x2(m + n)−cos(m − n)x2(m − n)2π0but cos(m + n)x2π0= 0, cos(m − n)x2π0= 0.(b)2π0cos mx cos nx dx =122π0[cos(m + n)x + cos(m − n)x]dx;since m = n, evaluate sin at integer multiples of 2π to get 0.(c)2π0sin mx sin nx dx =122π0[cos(m − n)x − cos(m + n)x] dx;since m = n, evaluate sin at integer multiples of 2π to get 0.
• Exercise Set 8.3 35554. (a)2π0sin mx cos mx dx =122π0sin 2mx dx = −14mcos 2mx2π0= 0(b)2π0cos2mx dx =122π0(1 + cos 2mx) dx =12x +12msin 2mx2π0= π(c)2π0sin2mx dx =122π0(1 − cos 2mx) dx =12x −12msin 2mx2π0= π55. y = tan x, 1 + (y )2= 1 + tan2x = sec2x,L =π/40√sec2 x dx =π/40sec x dx = ln | sec x + tan x|π/40= ln(√2 + 1)56. V = ππ/40(1 − tan2x)dx = ππ/40(2 − sec2x)dx = π(2x − tan x)π/40=12π(π − 2)57. V = ππ/40(cos2x − sin2x)dx = ππ/40cos 2x dx =12π sin 2xπ/40= π/258. V = ππ0sin2x dx =π2π0(1 − cos 2x)dx =π2x −12sin 2xπ0= π2/259. With 0 < α < β, D = Dβ−Dα =L2πβαsec x dx =L2πln | sec x + tan x|βα=L2πlnsec β + tan βsec α + tan α60. (a) D =1002πln(sec 25◦+ tan 25◦) = 7.18 cm(b) D =1002πlnsec 50◦+ tan 50◦sec 30◦ + tan 30◦= 7.34 cm61. (a) csc x dx = sec(π/2 − x)dx = − ln | sec(π/2 − x) + tan(π/2 − x)| + C= − ln | csc x + cot x| + C(b) − ln | csc x + cot x| = ln1| csc x + cot x|= ln| csc x − cot x|| csc2 x − cot2x|= ln | csc x − cot x|,− ln | csc x + cot x| = − ln1sin x+cos xsin x= lnsin x1 + cos x= ln2 sin(x/2) cos(x/2)2 cos2(x/2)= ln | tan(x/2)|62. sin x + cos x =√2 (1/√2) sin x + (1/√2) cos x=√2 [sin x cos(π/4) + cos x sin(π/4)] =√2 sin(x + π/4),dxsin x + cos x=1√2csc(x + π/4)dx = −1√2ln | csc(x + π/4) + cot(x + π/4)| + C= −1√2ln√2 + cos x − sin xsin x + cos x+ C
• 356 Chapter 863. a sin x + b cos x = a2 + b2a√a2 + b2sin x +b√a2 + b2cos x = a2 + b2(sin x cos θ + cos x sin θ)where cos θ = a/√a2 + b2 and sin θ = b/√a2 + b2 so a sin x + b cos x =√a2 + b2 sin(x + θ)anddxa sin x + b cos x=1√a2 + b2csc(x + θ)dx = −1√a2 + b2ln | csc(x + θ) + cot(x + θ)| + C= −1√a2 + b2ln√a2 + b2 + a cos x − b sin xa sin x + b cos x+ C64. (a)π/20sinnx dx = −1nsinn−1x cos xπ/20+n − 1nπ/20sinn−2x dx =n − 1nπ/20sinn−2x dx(b) By repeated application of the formula in Part (a)π/20sinnx dx =n − 1nn − 3n − 2π/20sinn−4x dx=n − 1nn − 3n − 2n − 5n − 4· · ·12π/20dx, n evenn − 1nn − 3n − 2n − 5n − 4· · ·23π/20sin x dx, n odd=1 · 3 · 5 · · · (n − 1)2 · 4 · 6 · · · n·π2, n even2 · 4 · 6 · · · (n − 1)3 · 5 · 7 · · · n, n odd65. (a)π/20sin3x dx =23(b)π/20sin4x dx =1 · 32 · 4·π2= 3π/16(c)π/20sin5x dx =2 · 43 · 5= 8/15 (d)π/20sin6x dx =1 · 3 · 52 · 4 · 6·π2= 5π/3266. Similar to proof in Exercise 64.EXERCISE SET 8.41. x = 2 sin θ, dx = 2 cos θ dθ,4 cos2θ dθ = 2 (1 + cos 2θ)dθ = 2θ + sin 2θ + C= 2θ + 2 sin θ cos θ + C = 2 sin−1(x/2) +12x 4 − x2 + C2. x =12sin θ, dx =12cos θ dθ,12cos2θ dθ =14(1 + cos 2θ)dθ =14θ +18sin 2θ + C=14θ +14sin θ cos θ + C =14sin−12x +12x 1 − 4x2 + C
• Exercise Set 8.4 3573. x = 4 sin θ, dx = 4 cos θ dθ,16 sin2θ dθ = 8 (1 − cos 2θ)dθ = 8θ − 4 sin 2θ + C = 8θ − 8 sin θ cos θ + C= 8 sin−1(x/4) −12x 16 − x2 + C4. x = 3 sin θ, dx = 3 cos θ dθ,191sin2θdθ =19csc2θ dθ = −19cot θ + C = −√9 − x29x+ C5. x = 2 tan θ, dx = 2 sec2θ dθ,181sec2 θdθ =18cos2θ dθ =116(1 + cos 2θ)dθ =116θ +132sin 2θ + C=116θ +116sin θ cos θ + C =116tan−1 x2+x8(4 + x2)+ C6. x =√5 tan θ, dx =√5 sec2θ dθ,5 tan2θ sec θ dθ = 5 (sec3θ − sec θ)dθ = 512sec θ tan θ −12ln | sec θ + tan θ| + C1=12x 5 + x2 −52ln√5 + x2 + x√5+ C1 =12x 5 + x2 −52ln( 5 + x2 + x) + C7. x = 3 sec θ, dx = 3 sec θ tan θ dθ,3 tan2θ dθ = 3 (sec2θ − 1)dθ = 3 tan θ − 3θ + C = x2 − 9 − 3 sec−1 x3+ C8. x = 4 sec θ, dx = 4 sec θ tan θ dθ,1161sec θdθ =116cos θ dθ =116sin θ + C =√x2 − 1616x+ C9. x = sin θ, dx = cos θ dθ,3 sin3θ dθ = 3 1 − cos2θ sin θ dθ= 3 − cos θ + cos3θ + C = −3 1 − x2 + (1 − x2)3/2+ C10. x =√5 sin θ, dx =√5 cos θ dθ,25√5 sin3θ cos2θ dθ = 25√5 −13cos3θ +15cos5θ + C = −53(5 − x2)3/2+15(5 − x2)5/2+ C11. x =23sec θ, dx =23sec θ tan θ dθ,341sec θdθ =34cos θ dθ =34sin θ + C =14x9x2 − 4 + C12. t = tan θ, dt = sec2θ dθ,sec3θtan θdθ =tan2θ + 1tan θsec θ dθ = (sec θ tan θ + csc θ)dθ= sec θ + ln | csc θ − cot θ| + C = 1 + t2 + ln√1 + t2 − 1|t|+ C13. x = sin θ, dx = cos θ dθ,1cos2 θdθ = sec2θ dθ = tan θ + C = x/ 1 − x2 + C
• 358 Chapter 814. x = 5 tan θ, dx = 5 sec2θ dθ,125sec θtan2θdθ =125csc θ cot θ dθ = −125csc θ+C = −√x2 + 2525x+ C15. x = 3 sec θ, dx = 3 sec θ tan θ dθ, sec θ dθ = ln | sec θ + tan θ| + C = ln13x +13x2 − 9 + C16. 1 + 2x2+ x4= (1 + x2)2, x = tan θ, dx = sec2θ dθ,1sec2 θdθ = cos2θ dθ =12(1 + cos 2θ)dθ =12θ +14sin 2θ + C=12θ +12sin θ cos θ + C =12tan−1x +x2(1 + x2)+ C17. x =32sec θ, dx =32sec θ tan θ dθ,32sec θ tan θ dθ27 tan3θ=118cos θsin2θdθ = −1181sin θ+ C = −118csc θ + C = −x9√4x2 − 9+ C18. x = 5 sec θ, dx = 5 sec θ tan θ dθ,= 375 sec4θ dθ= 125 sec2θ tan θ + 250 sec2θ dθ= 125 sec2θ tan θ + 250 tan θ + C= x2x2 − 25 + 50 x2 − 25 + C19. ex= sin θ, exdx = cos θ dθ,cos2θ dθ =12(1 + cos 2θ)dθ =12θ +14sin 2θ + C =12sin−1(ex) +12ex1 − e2x + C20. u = sin θ,1√2 − u2du = sin−1 sin θ√2+ C21. x = sin θ, dx = cos θ dθ,510sin3θ cos2θ dθ = 5 −13cos3θ +15cos5θπ/20= 5(1/3 − 1/5) = 2/322. x = sin θ, dx = cos θ dθ,π/60sec3θ, dθ =12sec θ tan θ +12ln | sec θ + tan θ|π/60=122√31√3+12ln(2√3+1√3=13+14ln 323. x = sec θ, dx = sec θ tan θ dθ,π/3π/41sec θdθ =π/3π/4cos θ dθ = sin θπ/3π/4= (√3 −√2)/224. x =√2 sec θ, dx =√2 sec θ tan θ dθ, 2π/40tan2θ dθ = 2 tan θ − 2θπ/40= 2 − π/2
• Exercise Set 8.4 35925. x =√3 tan θ, dx =√3 sec2θ dθ,19π/3π/6sec θtan4θdθ =19π/3π/6cos3θsin4θdθ =19π/3π/61 − sin2θsin4θcos θ dθ =19√3/21/21 − u2u4du (u = sin θ)=19√3/21/2(u−4− u−2)du =19−13u3+1u√3/21/2=10√3 + 1824326. x =√3 tan θ, dx =√3 sec2θ dθ,√33π/30tan3θsec3 θdθ =√33π/30sin3θ dθ =√33π/301 − cos2θ sin θ dθ=√33− cos θ +13cos3θπ/30=√33−12+124− −1 +13= 5√3/7227. u = x2+ 4, du = 2x dx,121udu =12ln |u| + C =12ln(x2+ 4) + C; or x = 2 tan θ, dx = 2 sec2θ dθ,tan θ dθ = ln | sec θ| + C1 = ln√x2 + 42+ C1 = ln(x2+ 4)1/2− ln 2 + C1=12ln(x2+ 4) + C with C = C1 − ln 228. x = 2 tan θ, dx = 2 sec2θ dθ, 2 tan2θ dθ = 2 tan θ − 2θ + C = x − 2 tan−1 x2+ C; alternativelyx2x2 + 4dx = dx − 4dxx2 + 4= x − 2 tan−1 x2+ C29. y =1x, 1 + (y )2= 1 +1x2=x2+ 1x2,L =21x2 + 1x2dx; x = tan θ, dx = sec2θ dθ,L =tan−1(2)π/4sec3θtan θdθ =tan−1(2)π/4tan2θ + 1tan θsec θ dθ =tan−1(2)π/4(sec θ tan θ + csc θ)dθ= sec θ + ln | csc θ − cot θ|tan−1(2)π/4=√5 + ln√52−12−√2 + ln |√2 − 1|=√5 −√2 + ln2 + 2√21 +√530. y = 2x, 1 + (y )2= 1 + 4x2,L =101 + 4x2dx; x =12tan θ, dx =12sec2θ dθ,L =12tan−120sec3θ dθ =1212sec θ tan θ +12ln | sec θ + tan θ|tan−120=14(√5)(2) +14ln |√5 + 2| =12√5 +14ln(2 +√5)
• 360 Chapter 831. y = 2x, 1 + (y )2= 1 + 4x2,S = 2π10x21 + 4x2dx; x =12tan θ, dx =12sec2θ dθ,S =π4tan−120tan2θ sec3θ dθ =π4tan−120(sec2θ − 1) sec3θ dθ =π4tan−120(sec5θ − sec3θ)dθ=π414sec3θ tan θ −18sec θ tan θ −18ln | sec θ + tan θ|tan−120=π32[18√5 − ln(2 +√5)]32. V = π10y21 − y2dy; y = sin θ, dy = cos θ dθ,V = ππ/20sin2θ cos2θ dθ =π4π/20sin22θ dθ =π8π/20(1 − cos 4θ)dθ =π8θ −14sin 4θπ/20=π21633.1(x − 2)2 + 1dx = tan−1(x − 2) + C 34.11 − (x − 1)2dx = sin−1(x − 1) + C35.14 − (x − 1)2dx = sin−1 x − 12+ C36.116(x + 1/2)2 + 1dx =141(x + 2)2 + 1dx =14tan−1(4x + 2) + C37.1(x − 3)2 + 1dx = ln x − 3 + (x − 3)2 + 1 + C38.x(x + 1)2 + 1dx, let u = x + 1,u − 1u2 + 1du =uu2 + 1−1u2 + 1du =12ln(u2+ 1) − tan−1u + C=12ln(x2+ 2x + 2) − tan−1(x + 1) + C39. 4 − (x + 1)2 dx, let x + 1 = 2 sin θ,= 4 cos2θ dθ = 2(1 + cos 2θ) dθ= 2θ + sin 2θ + C = 2 sin−1 x + 12+12(x + 1) 3 − 2x − x2 + C40.ex(ex + 1/2)2 + 3/4dx, let u = ex+ 1/2,1u2 + 3/4du = sinh−1(2u/√3) + C = sinh−1 2ex+ 1√3+ CAlternate solution: let ex+ 1/2 =√32tan θ,sec θ dθ = ln | sec θ + tan θ| + C = ln2√e2x + ex + 1√3+2ex+ 1√3+ C1= ln(2 e2x + ex + 1 + 2ex+ 1) + C
• Exercise Set 8.5 36141.12(x + 1)2 + 5dx =121(x + 1)2 + 5/2dx =1√10tan−12/5(x + 1) + C42.2x + 34(x + 1/2)2 + 4dx, let u = x + 1/2,2u + 24u2 + 4du =12uu2 + 1+1u2 + 1du =14ln(u2+ 1) +12tan−1u + C=14ln(x2+ x + 5/4) +12tan−1(x + 1/2) + C43.211√4x − x2dx =2114 − (x − 2)2dx = sin−1 x − 2221= π/644.404x − x2dx =404 − (x − 2)2dx, let x − 2 = 2 sin θ,4π/2−π/2cos2θ dθ = 2θ + sin 2θπ/2−π/2= 2π45. u = sin2x, du = 2 sin x cos x dx;121 − u2 du =14u 1 − u2 + sin−1u + C =14sin2x 1 − sin4x + sin−1(sin2x) + C46. u = x sin x, du = (x cos x + sin x) dx;1 + u2 du =12u 1 + u2 +12sinh−1u + C =12x sin x 1 + x2 sin2x +12sinh−1(x sin x) + C47. (a) x = 3 sinh u, dx = 3 cosh u du, du = u + C = sinh−1(x/3) + C(b) x = 3 tan θ, dx = 3 sec2θ dθ,sec θ dθ = ln | sec θ + tan θ| + C = ln x2 + 9/3 + x/3 + Cbut sinh−1(x/3) = ln x/3 + x2/9 + 1 = ln x/3 +√x2 + 9/3 so the results agree.48. x = cosh u, dx = sinh u du,sinh2u du =12(cosh 2u − 1)du =14sinh 2u −12u + C=12sinh u cosh u −12u + C =12x x2 − 1 −12cosh−1x + Cbecause cosh u = x, and sinh u = cosh2u − 1 =√x2 − 1EXERCISE SET 8.51.A(x − 3)+B(x + 4)2.5x(x − 2)(x + 2)=Ax+Bx − 2+Cx + 23.2x − 3x2(x − 1)=Ax+Bx2+Cx − 14.Ax + 2+B(x + 2)2+C(x + 2)35.Ax+Bx2+Cx3+Dx + Ex2 + 26.Ax − 1+Bx + Cx2 + 6
• 362 Chapter 87.Ax + Bx2 + 5+Cx + D(x2 + 5)28.Ax − 2+Bx + Cx2 + 1+Dx + E(x2 + 1)29.1(x − 4)(x + 1)=Ax − 4+Bx + 1; A =15, B = −15so151x − 4dx −151x + 1dx =15ln |x − 4| −15ln |x + 1| + C =15lnx − 4x + 1+ C10.1(x + 1)(x − 7)=Ax + 1+Bx − 7; A = −18, B =18so−181x + 1dx +181x − 7dx = −18ln |x + 1| +18ln |x − 7| + C =18lnx − 7x + 1+ C11.11x + 17(2x − 1)(x + 4)=A2x − 1+Bx + 4; A = 5, B = 3 so512x − 1dx + 31x + 4dx =52ln |2x − 1| + 3 ln |x + 4| + C12.5x − 5(x − 3)(3x + 1)=Ax − 3+B3x + 1; A = 1, B = 2 so1x − 3dx + 213x + 1dx = ln |x − 3| +23ln |3x + 1| + C13.2x2− 9x − 9x(x + 3)(x − 3)=Ax+Bx + 3+Cx − 3; A = 1, B = 2, C = −1 so1xdx + 21x + 3dx −1x − 3dx = ln |x| + 2 ln |x + 3| − ln |x − 3| + C = lnx(x + 3)2x − 3+ CNote that the symbol C has been recycled; to save space this recycling is usually not mentioned.14.1x(x + 1)(x − 1)=Ax+Bx + 1+Cx − 1; A = −1, B =12, C =12so−1xdx +121x + 1dx +121x − 1dx = − ln |x| +12ln |x + 1| +12ln |x − 1| + C=12ln(x + 1)(x − 1)x2+ C =12ln|x2− 1|x2+ C15.x2− 8x + 3= x − 3 +1x + 3, x − 3 +1x + 3dx =12x2− 3x + ln |x + 3| + C16.x2+ 1x − 1= x + 1 +2x − 1, x + 1 +2x − 1dx =12x2+ x + 2 ln |x − 1| + C17.3x2− 10x2 − 4x + 4= 3 +12x − 22x2 − 4x + 4,12x − 22(x − 2)2=Ax − 2+B(x − 2)2; A = 12, B = 2 so3dx + 121x − 2dx + 21(x − 2)2dx = 3x + 12 ln |x − 2| − 2/(x − 2) + C18.x2x2 − 3x + 2= 1 +3x − 2x2 − 3x + 2,3x − 2(x − 1)(x − 2)=Ax − 1+Bx − 2; A = −1, B = 4 sodx −1x − 1dx + 41x − 2dx = x − ln |x − 1| + 4 ln |x − 2| + C
• Exercise Set 8.5 36319.x5+ x2+ 2x3 − x= x2+ 1 +x2+ x + 2x3 − x,x2+ x + 2x(x + 1)(x − 1)=Ax+Bx + 1+Cx − 1; A = −2, B = 1, C = 2 so(x2+ 1)dx −2xdx +1x + 1dx +2x − 1dx=13x3+ x − 2 ln |x| + ln |x + 1| + 2 ln |x − 1| + C =13x3+ x + ln(x + 1)(x − 1)2x2+ C20.x5− 4x3+ 1x3 − 4x= x2+1x3 − 4x,1x(x + 2)(x − 2)=Ax+Bx + 2+Cx − 2; A = −14, B = +18, C = +18sox2dx −141xdx +181x + 2dx +181x − 2dx=13x3−14ln |x| +18ln |x + 2| +18ln |x − 2| + C21.2x2+ 3x(x − 1)2=Ax+Bx − 1+C(x − 1)2; A = 3, B = −1, C = 5 so31xdx −1x − 1dx + 51(x − 1)2dx = 3 ln |x| − ln |x − 1| − 5/(x − 1) + C22.3x2− x + 1x2(x − 1)=Ax+Bx2+Cx − 1; A = 0, B = −1, C = 3 so−1x2dx + 31x − 1dx = 1/x + 3 ln |x − 1| + C23.2x2− 10x + 4(x + 1)(x − 3)2=Ax + 1+Bx − 3+C(x − 3)2; A = 1, B = 1, C = −2 so1x + 1dx +1x − 3dx −2(x − 3)2dx = ln |x + 1| + ln |x − 3| +2x − 3+ C124.2x2− 2x − 1x2(x − 1)=Ax+Bx2+Cx − 1; A = 3, B = 1, C = −1 so31xdx +1x2dx −1x − 1dx = 3 ln |x| −1x− ln |x − 1| + C25.x2(x + 1)3=Ax + 1+B(x + 1)2+C(x + 1)3; A = 1, B = −2, C = 1 so1x + 1dx −2(x + 1)2dx +1(x + 1)3dx = ln |x + 1| +2x + 1−12(x + 1)2+ C26.2x2+ 3x + 3(x + 1)3=Ax + 1+B(x + 1)2+C(x + 1)3; A = 2, B = −1, C = 2 so21x + 1dx −1(x + 1)2dx + 21(x + 1)3dx = 2 ln |x + 1| +1x + 1−1(x + 1)2+ C
• 364 Chapter 827.2x2− 1(4x − 1)(x2 + 1)=A4x − 1+Bx + Cx2 + 1; A = −14/17, B = 12/17, C = 3/17 so2x2− 1(4x − 1)(x2 + 1)dx = −734ln |4x − 1| +617ln(x2+ 1) +317tan−1x + C28.1x(x2 + 2)=Ax+Bx + Cx2 + 2; A =12, B = −12, C = 0 so1x3 + 2xdx =12ln |x| −14ln(x2+ 2) + C =14lnx2x2 + 2+ C29.x3+ 3x2+ x + 9(x2 + 1)(x2 + 3)=Ax + Bx2 + 1+Cx + Dx2 + 3; A = 0, B = 3, C = 1, D = 0 sox3+ 3x2+ x + 9(x2 + 1)(x2 + 3)dx = 3 tan−1x +12ln(x2+ 3) + C30.x3+ x2+ x + 2(x2 + 1)(x2 + 2)=Ax + Bx2 + 1+Cx + Dx2 + 2; A = D = 0, B = C = 1 sox3+ x2+ x + 2(x2 + 1)(x2 + 2)dx = tan−1x +12ln(x2+ 2) + C31.x3− 2x2+ 2x − 2x2 + 1= x − 2 +xx2 + 1,x3− 3x2+ 2x − 3x2 + 1dx =12x2− 2x +12ln(x2+ 1) + C32.x4+ 6x3+ 10x2+ xx2 + 6x + 10= x2+xx2 + 6x + 10,xx2 + 6x + 10dx =x(x + 3)2 + 1dx =u − 3u2 + 1du, u = x + 3=12ln(u2+ 1) − 3 tan−1u + C1sox4+ 6x3+ 10x2+ xx2 + 6x + 10dx =13x3+12ln(x2+ 6x + 10) − 3 tan−1(x + 3) + C33. Let x = sin θ to get1x2 + 4x − 5dx, and1(x + 5)(x − 1)=Ax + 5+Bx − 1; A = −1/6,B = 1/6 so we get −161x + 5dx +161x − 1dx =16lnx − 1x + 5+ C =16ln1 − sin θ5 + sin θ+ C.34. Let x = et; thenete2t − 4dt =1x2 − 4dx,1(x + 2)(x − 2)=Ax + 2+Bx − 2; A = −1/4, B = 1/4 so−141x + 2dx +141x − 2dx =14lnx − 2x + 2+ C =14lnet− 2et + 2+ C.
• Exercise Set 8.5 36535. V = π20x4(9 − x2)2dx,x4x4 − 18x2 + 81= 1 +18x2− 81x4 − 18x2 + 81,18x2− 81(9 − x2)2=18x2− 81(x + 3)2(x − 3)2=Ax + 3+B(x + 3)2+Cx − 3+D(x − 3)2;A = −94, B =94, C =94, D =94soV = π x −94ln |x + 3| −9/4x + 3+94ln |x − 3| −9/4x − 320= π195−94ln 536. Let u = exto getln 5− ln 5dx1 + ex=ln 5− ln 5exdxex(1 + ex)=51/5duu(1 + u),1u(1 + u)=Au+B1 + u; A = 1, B = −1;51/5duu(1 + u)= (ln u − ln(1 + u))51/5= ln 537.x2+ 1(x2 + 2x + 3)2=Ax + Bx2 + 2x + 3+Cx + D(x2 + 2x + 3)2; A = 0, B = 1, C = D = −2 sox2+ 1(x2 + 2x + 3)2dx =1(x + 1)2 + 2dx −2x + 2(x2 + 2x + 3)2dx=1√2tan−1 x + 1√2+ 1/(x2+ 2x + 3) + C38.x5+ x4+ 4x3+ 4x2+ 4x + 4(x2 + 2)3=Ax + Bx2 + 2+Cx + D(x2 + 2)2+Ex + F(x2 + 2)3;A = B = 1, C = D = E = F = 0 sox + 1x2 + 2dx =12ln(x2+ 2) +1√2tan−1(x/√2) + C39. x4− 3x3− 7x2+ 27x − 18 = (x − 1)(x − 2)(x − 3)(x + 3),1(x − 1)(x − 2)(x − 3)(x + 3)=Ax − 1+Bx − 2+Cx − 3+Dx + 3;A = 1/8, B = −1/5, C = 1/12, D = −1/120 sodxx4 − 3x3 − 7x2 + 27x − 18=18ln |x − 1| −15ln |x − 2| +112ln |x − 3| −1120ln |x + 3| + C40. 16x3− 4x2+ 4x − 1 = (4x − 1)(4x2+ 1),1(4x − 1)(4x2 + 1)=A4x − 1+Bx + C4x2 + 1; A = 4/5, B = −4/5, C = −1/5 sodx16x3 − 4x2 + 4x − 1=15ln |4x − 1| −110ln(4x2+ 1) −110tan−1(2x) + C41. Let u = x2, du = 2x dx,10xx4 + 1dx =121011 + u2du =12tan−1u10=12π4=π8.42.1a2 − x2=Aa − x+Ba + x; A =12a, B =12aso12a1a − x+1a + xdx =12a(− ln |a − x| + ln |a + x| ) + C =12alna + xa − x+ C
• 366 Chapter 843. If the polynomial has distinct roots r1, r2, r1 = r2, then the partial fraction decomposition willcontain terms of the formAx − r1,Bx − r2, and they will give logarithms and no inverse tangents. Ifthere are two roots not distinct, say x = r, then the termsAx − r,B(x − r)2will appear, and neitherwill give an inverse tangent term. The only other possibility is no real roots, and the integrandcan be written in the form1a x + b2a2+ c − b24a, which will yield an inverse tangent, speciﬁcallyof the form tan−1A x +b2afor some constant A.44. Since there are no inverse tangent terms, the roots are real. Since there are no logarithmic terms,there are no terms of the form1x − r, so the only terms that can arise in the partial fractiondecomposition form is one like1(x − a)2. Therefore the original quadratic had a multiple root.45. Yes, for instance the integrand1x2 + 1, whose integral is precisely tan−1x + C.EXERCISE SET 8.61. Formula (60):493x + ln |−1 + 3x| + C 2. Formula (62):12544 − 5x+ ln |4 − 5x| + C3. Formula (65):15lnx5 + 2x+ C 4. Formula (66): −1x− 5 ln1 − 5xx+ C5. Formula (102):15(x − 1)(2x + 3)3/2+ C 6. Formula (105):23(−x − 4)√2 − x + C7. Formula (108):12ln√4 − 3x − 2√4 − 3x + 2+ C 8. Formula (108): tan−1√3x − 42+ C9. Formula (69):18lnx + 4x − 4+ C 10. Formula (70):16lnx − 3x + 3+ C11. Formula (73):x2x2 − 3 −32ln x + x2 − 3 + C12. Formula (94): −√x2 − 5x+ ln(x + x2 − 5) + C13. Formula (95):x2x2 + 4 − 2 ln(x + x2 + 4) + C14. Formula (90): −√x2 − 22x+ C 15. Formula (74):x29 − x2 +92sin−1 x3+ C16. Formula (80): −√4 − x2x− sin−1 x2+ C
• Exercise Set 8.6 36717. Formula (79):√4 − x2 − 2 ln2 +√4 − x2x+ C18. Formula (117): −√6x − x23x+ C 19. Formula (38): −114sin(7x) +12sin x + C20. Formula (40): −114cos(7x) +16cos(3x) + C21. Formula (50):x416[4 ln x − 1] + C 22. Formula (50): −2ln x + 2√x23. Formula (42):e−2x13(−2 sin(3x) − 3 cos(3x)) + C24. Formula (43):ex5(cos(2x) + 2 sin(2x)) + C25. u = e2x, du = 2e2xdx, Formula (62):12u du(4 − 3u)2=11844 − 3e2x+ ln 4 − 3e2x+ C26. u = cos 2x, du = −2 sin 2xdx, Formula (65): −du2u(3 − u)= −16lncos 2x3 − cos 2x+ C27. u = 3√x, du =32√xdx, Formula (68):23duu2 + 4=13tan−1 3√x2+ C28. u = sin 4x, du = 4 cos 4xdx, Formula (68):14du9 + u2=112tan−1 sin 4x3+ C29. u = 2x, du = 2dx, Formula (76):12du√u2 − 9=12ln 2x + 4x2 − 9 + C30. u =√2x2, du = 2√2xdx, Formula (72):12√2u2 + 3 du =x242x4 + 3 +34√2ln√2x2+ 2x4 + 3 + C31. u = 2x2, du = 4xdx, u2du = 16x5dx, Formula (81):14u2du√2 − u2= −x242 − 4x4 +14sin−1(√2x2) + C32. u = 2x, du = 2dx, Formula (83): 2duu2√3 − u2= −13x3 − 4x2 + C33. u = ln x, du = dx/x, Formula (26): sin2u du =12ln x +14sin(2 ln x) + C34. u = e−2x, du = −2e−2x, Formula (27): −12cos2u du = −14e−2x−18sin 2e−2x+ C35. u = −2x, du = −2dx, Formula (51):14ueudu =14(−2x − 1)e−2x+ C
• 368 Chapter 836. u = 3x+1, du = 3 dx, Formula (11):13ln u du =13(u ln u−u)+C =13(3x+1)[ln(3x+1)−1]+C37. u = sin 3x, du = 3 cos 3x dx, Formula (67):13duu(u + 1)2=1311 + sin 3x+ lnsin 3x1 + sin 3x+ C38. u = ln x, du =1xdx, Formula (105):u du√4u − 1=112(2 ln x + 1)√4 ln x − 1 + C39. u = 4x2, du = 8xdx, Formula (70):18duu2 − 1=116ln4x2− 14x2 + 1+ C40. u = 2ex, du = 2exdx, Formula (69):12du3 − u2=14√3ln2ex+√32ex −√3+ C41. u = 2ex, du = 2exdx, Formula (74):123 − u2 du =14u 3 − u2 +34sin−1(u/√3) + C =12ex3 − 4e2x +34sin−1(2ex/√3) + C42. u = 3x, du = 3dx, Formula (80):3√4 − u2duu2= −3√4 − u2u− 3 sin−1(u/2) + C = −√4 − 9x2x− 3 sin−1(3x/2) + C43. u = 3x, du = 3dx, Formula (112):1353u − u2 du =16u −5653u − u2 +25216sin−1 u − 55+ C=18x − 5365x − 9x2 +25216sin−1 18x − 55+ C44. u =√5x, du =√5 dx, Formula (117):duu (u/√5) − u2= −(u/√5) − u2u/(2√5)+ C = −2√x − 5x2x+ C45. u = 2x, du = 2dx, Formula (44):u sin u du = (sin u − u cos u) + C = sin 2x − 2x cos 2x + C46. u =√x, u2= x, 2udu = dx, Formula (45): 2 u cos u du = 2 cos√x + 2√x sin√x + C47. u = −√x, u2= x, 2udu = dx, Formula (51): 2 ueudu = −2(√x + 1)e−√x+ C48. u = 2 + x2, du = 2x dx, Formula (11):12ln u du =12(u ln u − u) + C =12(2 + x2) ln(2 + x2) −12(2 + x2) + C49. x2+ 6x − 7 = (x + 3)2− 16; u = x + 3, du = dx, Formula (70):duu2 − 16=18lnu − 4u + 4+ C =18lnx − 1x + 7+ C
• Exercise Set 8.6 36950. x2+ 2x − 3 = (x + 1)2− 4, u = x + 1, du = dx, Formula (77):4 − u2 du =12u 4 − u2 + 2 sin−1(u/2) + C=12(x + 1) 3 − 2x − x2 + 2 sin−1((x + 1)/2) + C51. x2− 4x − 5 = (x − 2)2− 9, u = x − 2, du = dx, Formula (77):u + 2√9 − u2du =u du√9 − u2+ 2du√9 − u2= − 9 − u2 + 2 sin−1 u3+ C= − 5 + 4x − x2 + 2 sin−1 x − 23+ C52. x2+ 6x + 13 = (x + 3)2+ 4, u = x + 3, du = dx, Formula (71):(u − 3) duu2 + 4=12ln(u2+ 4) −32tan−1(u/2) + C =12ln(x2+ 6x + 13) −32tan−1((x + 3)/2) + C53. u =√x − 2, x = u2+ 2, dx = 2u du;2u2(u2+ 2)du = 2 (u4+ 2u2)du =25u5+43u3+ C =25(x − 2)5/2+43(x − 2)3/2+ C54. u =√x + 1, x = u2− 1, dx = 2u du;2 (u2− 1)du =23u3− 2u + C =23(x + 1)3/2− 2√x + 1 + C55. u =√x3 + 1, x3= u2− 1, 3x2dx = 2u du;23u2(u2− 1)du =23(u4− u2)du =215u5−29u3+ C =215(x3+ 1)5/2−29(x3+ 1)3/2+ C56. u =√x3 − 1, x3= u2+ 1, 3x2dx = 2u du;231u2 + 1du =23tan−1u + C =23tan−1x3 − 1 + C57. u = x1/3, x = u3, dx = 3u2du;3u2u3 − udu = 3uu2 − 1du = 312(u + 1)+12(u − 1)du=32ln |x1/3+ 1| +32ln |x1/3− 1| + C58. u = x1/6, x = u6, dx = 6u5du;6u5u3 + u2du = 6u3u + 1du = 6 u2− u + 1 −1u + 1du= 2x1/2− 3x1/3+ 6x1/6− 6 ln(x1/6+ 1) + C59. u = x1/4, x = u4, dx = 4u3du; 41u(1 − u)du = 41u+11 − udu = 4 lnx1/4|1 − x1/4|+ C60. u = x1/2, x = u2, dx = 2u du;2u2u2 + 1du = 2u2+ 1 − 1u2 + 1du = 2u − 2 tan−1u + C = 2√x − 2 tan−1√x + C
• 370 Chapter 861. u = x1/6, x = u6, dx = 6u5du;6u3u − 1du = 6 u2+ u + 1 +1u − 1du = 2x1/2+ 3x1/3+ 6x1/6+ 6 ln |x1/6− 1| + C62. u =√x, x = u2, dx = 2u du;−2u2+ uu − 1du = −2 u + 2 +2u − 1du = −x − 4√x − 4 ln |√x − 1| + C63. u =√1 + x2, x2= u2− 1, 2x dx = 2u du, x dx = u du;(u2− 1)du =13(1 + x2)3/2− (1 + x2)1/2+ C64. u = (x + 3)1/5, x = u5− 3, dx = 5u4du;5 (u8− 3u3)du =59(x + 3)9/5−154(x + 3)4/5+ C65.11 +2u1 + u2+1 − u21 + u221 + u2du =1u + 1du = ln | tan(x/2) + 1| + C66.12 +2u1 + u221 + u2du =1u2 + u + 1du=1(u + 1/2)2 + 3/4du =2√3tan−1 2 tan(x/2) + 1√3+ C67. u = tan(θ/2),dθ1 − cos θ=1u2du = −1u+ C = − cot(θ/2) + C68. u = tan(x/2),23u2 + 8u − 3du =231(u + 4/3)2 − 25/9du =231z2 − 25/9dz (z = u + 4/3)=15lnz − 5/3z + 5/3+ C =15lntan(x/2) − 1/3tan(x/2) + 3+ C69. u = tan(x/2),121 − u2udu =12(1/u − u)du =12ln | tan(x/2)| −14tan2(x/2) + C70. u = tan(x/2),1 − u21 + u2du = −u + 2 tan−1u + C = x − tan(x/2) + C71.x21t(4 − t)dt =14lnt4 − tx2(Formula (65), a = 4, b = −1)=14lnx4 − x− ln 1 =14lnx4 − x,14lnx4 − x= 0.5, lnx4 − x= 2,x4 − x= e2, x = 4e2− e2x, x(1 + e2) = 4e2, x = 4e2/(1 + e2) ≈ 3.523188312
• Exercise Set 8.6 37172.x11t√2t − 1dt = 2 tan−1√2t − 1x1(Formula (108), a = −1, b = 2)= 2 tan−1√2x − 1 − tan−11 = 2 tan−1√2x − 1 − π/4 ,2(tan−1√2x − 1 − π/4) = 1, tan−1√2x − 1 = 1/2 + π/4,√2x − 1 = tan(1/2 + π/4),x = [1 + tan2(1/2 + π/4)]/2 ≈ 6.30799351673. A =4025 − x2 dx =12x 25 − x2 +252sin−1 x540(Formula (74), a = 5)= 6 +252sin−1 45≈ 17.5911902374. A =22/39x2 − 4 dx; u = 3x,A =1362u2 − 4 du =1312u u2 − 4 − 2 ln u + u2 − 462(Formula (73), a2= 4)=133√32 − 2 ln(6 +√32) + 2 ln 2 = 4√2 −23ln(3 + 2√2) ≈ 4.48168946775. A =10125 − 16x2dx; u = 4x,A =1440125 − u2du =140lnu + 5u − 540=140ln 9 ≈ 0.054930614 (Formula (69), a = 5)76. A =41√x ln x dx =49x3/2 32ln x − 141(Formula (50), n = 1/2)=49(12 ln 4 − 7) ≈ 4.28245881577. V = 2ππ/20x cos x dx = 2π(cos x + x sin x)π/20= π(π − 2) ≈ 3.586419094 (Formula (45))78. V = 2π84x√x − 4 dx =4π15(3x + 8)(x − 4)3/284(Formula (102), a = −4, b = 1)=102415π ≈ 214.466058579. V = 2π30xe−xdx; u = −x,V = 2π−30ueudu = 2πeu(u − 1)−30= 2π(1 − 4e−3) ≈ 5.031899801 (Formula (51))80. V = 2π51x ln x dx =π2x2(2 ln x − 1)51= π(25 ln 5 − 12) ≈ 88.70584621 (Formula (50), n = 1)
• 372 Chapter 881. L =201 + 16x2 dx; u = 4x,L =14801 + u2 du =14u21 + u2 +12ln u + 1 + u280(Formula (72), a2= 1)=√65 +18ln(8 +√65) ≈ 8.40931678382. L =311 + 9/x2 dx =31√x2 + 9xdx = x2 + 9 − 3 ln3 +√x2 + 9x31= 3√2 −√10 + 3 ln3 +√101 +√2≈ 3.891581644 (Formula (89), a = 3)83. S = 2ππ0(sin x) 1 + cos2 x dx; u = cos x,S = −2π−111 + u2 du = 4π101 + u2 du = 4πu21 + u2 +12ln u + 1 + u210a2= 1= 2π√2 + ln(1 +√2) ≈ 14.42359945 (Formula (72))84. S = 2π411x1 + 1/x4 dx = 2π41√x4 + 1x3dx; u = x2,S = π161√u2 + 1u2du = π −√u2 + 1u+ ln u + u2 + 1161= π√2 −√25716+ ln16 +√2571 +√2≈ 9.417237485 (Formula (93), a2= 1)85. (a) s(t) = 2 +t020 cos6u sin3u du= −209sin2t cos7t −4063cos7t +16663(b)3 6 9 12 151234ts(t)86. (a) v(t) =t0a(u) du = −110e−tcos 2t +15e−tsin 2t +174e−tcos 6t −337e−tsin 6t +110−174s(t) = 10 +t0v(u) du= −350e−tcos 2t −225e−tsin 2t +352738e−tcos 6t +61369e−tsin 6t +16185t +34386634225
• Exercise Set 8.6 373(b)3 6 9 12 151234ts(t)87. (a) sec x dx =1cos xdx =21 − u2du = ln1 + u1 − u+ C = ln1 + tan(x/2)1 − tan(x/2)+ C= lncos(x/2) + sin(x/2)cos(x/2) − sin(x/2)cos(x/2) + sin(x/2)cos(x/2) + sin(x/2)+ C = ln1 + sin xcos x+ C= ln |sec x + tan x| + C(b) tanπ4+x2=tanπ4+ tanx21 − tanπ4tanx2=1 + tanx21 − tanx288. csc x dx =1sin xdx = 1/u du = ln | tan(x/2)| + C butln | tan(x/2)| =12lnsin2(x/2)cos2(x/2)=12ln(1 − cos x)/2(1 + cos x)/2=12ln1 − cos x1 + cos x; also,1 − cos x1 + cos x=1 − cos2x(1 + cos x)2=1(csc x + cot x)2so12ln1 − cos x1 + cos x= − ln | csc x + cot x|89. Let u = tanh(x/2) then cosh(x/2) = 1/ sech(x/2) = 1/ 1 − tanh2(x/2) = 1/√1 − u2,sinh(x/2) = tanh(x/2) cosh(x/2) = u/√1 − u2, so sinh x = 2 sinh(x/2) cosh(x/2) = 2u/(1 − u2),cosh x = cosh2(x/2) + sinh2(x/2) = (1 + u2)/(1 − u2), x = 2 tanh−1u, dx = [2/(1 − u2)]du;dx2 cosh x + sinh x=1u2 + u + 1du =2√3tan−1 2u + 1√3+ C =2√3tan−1 2 tanh(x/2) + 1√3+ C.90. Let u = x4to get141√1 − u2du =14sin−1u + C =14sin−1(x4) + C.91. (cos32x sin30x − cos30x sin32x)dx = cos30x sin30x(cos2x − sin2x)dx=1230sin302x cos 2x dx =sin312x31(231)+ C92. x − x2 − 4dx =1√2(√x + 2 −√x − 2)dx =√23[(x + 2)3/2− (x − 2)3/2] + C93.1x10(1 + x−9)dx = −191udu = −19ln |u| + C = −19ln |1 + x−9| + C
• 374 Chapter 894. (a) (x + 4)(x − 5)(x2+ 1)2;Ax + 4+Bx − 5+Cx + Dx2 + 1+Ex + F(x2 + 1)2(b) −3x + 4+2x − 5−x − 2x2 + 1−3(x2 + 1)2(c) −3 ln |x + 4| + 2 ln |x − 5| + 2 tan−1x −12ln(x2+ 1) −32xx2 + 1+ tan−1x + CEXERCISE SET 8.71. exact value = 14/3 ≈ 4.666666667(a) 4.667600663, |EM | ≈ 0.000933996(b) 4.664795679, |ET | ≈ 0.001870988(c) 4.666651630, |ES| ≈ 0.0000150372. exact value = 2(a) 1.998377048, |EM | ≈ 0.001622952(b) 2.003260982, |ET | ≈ 0.003260982(c) 2.000072698, |ES| ≈ 0.0000726983. exact value = 2(a) 2.008248408, |EM | ≈ 0.008248408(b) 1.983523538, |ET | ≈ 0.016476462(c) 2.000109517, |ES| ≈ 0.0001095174. exact value = sin(1) ≈ 0.841470985(a) 0.841821700, |EM | ≈ 0.000350715(b) 0.840769642, |ET | ≈ 0.000701343(c) 0.841471453, |ES| ≈ 0.0000004685. exact value = e−1− e−4≈ 0.3495638023(a) 0.3482563710, |EM | ≈ 0.0013074313(b) 0.3521816066, |ET | ≈ 0.0026178043(c) 0.3495793657, |ES| ≈ 0.00001556346. exact value =12ln 5 ≈ 0.804718956(a) 0.801605339, |EM | ≈ 0.003113617(b) 0.811019505, |ET | ≈ 0.006300549(c) 0.805041497, |ES| ≈ 0.0003225417. f(x) =√x − 1, f (x) = −14(x − 1)−3/2, f(4)(x) = −1516(x − 1)−7/2; K2 = 1/4, K4 = 15/16(a) |EM | ≤272400(1/4) = 0.002812500 (b) |ET | ≤271200(1/4) = 0.005625000(c) |ES| ≤243180 × 104(15/16) ≈ 0.0001265638. f(x) = 1/√x, f (x) =34x−5/2, f(4)(x) =10516x−9/2; K2 = 3/4, K4 = 105/16(a) |EM | ≤272400(3/4) = 0.008437500 (b) |ET | ≤271200(3/4) = 0.016875000(c) |ES| ≤243180 × 104(105/16) ≈ 0.0008859389. f(x) = sin x, f (x) = − sin x, f(4)(x) = sin x; K2 = K4 = 1(a) |EM | ≤π32400(1) ≈ 0.012919282 (b) |ET | ≤π31200(1) ≈ 0.025838564(c) |ES| ≤π5180 × 104(1) ≈ 0.00017001110. f(x) = cos x, f (x) = − cos x, f(4)(x) = cos x; K2 = K4 = 1(a) |EM | ≤12400(1) ≈ 0.000416667 (b) |ET | ≤11200(1) ≈ 0.000833333(c) |ES| ≤1180 × 104(1) ≈ 0.000000556
• Exercise Set 8.7 37511. f(x) = e−x, f (x) = f(4)(x) = e−x; K2 = K4 = e−1(a) |EM | ≤9800(e−1) ≈ 0.001226265 (b) |ET | ≤9400(e−1) ≈ 0.002452530(c) |ES| ≤272 × 105(e−1) ≈ 0.00004966412. f(x) = 1/(2x + 1), f (x) = 8(2x + 1)−3, f(4)(x) = 384(2x + 1)−5; K2 = 8, K4 = 384(a) |EM | ≤82400(8) ≈ 0.026666667 (b) |ET | ≤81200(8) ≈ 0.053333333(c) |ES| ≤32180 × 104(384) ≈ 0.00682666713. (a) n >(27)(1/4)(24)(5 × 10−4)1/2≈ 23.7; n = 24 (b) n >(27)(1/4)(12)(5 × 10−4)1/2≈ 33.5; n=34(c) n >(243)(15/16)(180)(5 × 10−4)1/4≈ 7.1; n = 814. (a) n >(27)(3/4)(24)(5 × 10−4)1/2≈ 41.1; n = 42 (b) n >(27)(3/4)(12)(5 × 10−4)1/2≈ 58.1; n=59(c) n >(243)(105/16)(180)(5 × 10−4)1/4≈ 11.5; n = 1215. (a) n >(π3)(1)(24)(10−3)1/2≈ 35.9; n = 36 (b) n >(π3)(1)(12)(10−3)1/2≈ 50.8; n = 51(c) n >(π5)(1)(180)(10−3)1/4≈ 6.4; n = 816. (a) n >(1)(1)(24)(10−3)1/2≈ 6.5; n = 7 (b) n >(1)(1)(12)(10−3)1/2≈ 9.1; n = 10(c) n >(1)(1)(180)(10−3)1/4≈ 1.5; n = 217. (a) n >(8)(e−1)(24)(10−6)1/2≈ 350.2; n = 351 (b) n >(8)(e−1)(12)(10−6)1/2≈ 495.2; n = 496(c) n >(32)(e−1)(180)(10−6)1/4≈ 15.99; n = 1618. (a) n >(8)(8)(24)(10−6)1/2≈ 1632.99; n = 1633 (b) n >(8)(8)(12)(10−6)1/2≈ 2309.4; n = 2310(c) n >(32)(384)(180)(10−6)1/4≈ 90.9; n = 91
• 376 Chapter 819. g(X0) = aX20 + bX0 + c = 4a + 2b + c = f(X0) = 1/X0 = 1/2; similarly9a + 3b + c = 1/3, 16a + 4b + c = 1/4. Three equations in three unknowns, with solutiona = 1/24, b = −3/8, c = 13/12, g(x) = x2/24 − 3x/8 + 13/12.42g(x) dx =42x224−3x8+1312dx =2536∆x3[f(X0) + 4f(X1) + f(X2)] =1312+43+14=253620. Suppose g(x) = ax2+ bx + c passes through the points (0, f(0)) = (0, 0), (m, f(m)) = (1/6, 1/4),and (1/3, f(2m)) = (1/3, 3/4). Then g(0) = c = 0, 1/4 = g(1/6) = a/36 + b/6, and3/4 = g(1/3) = a/9 + b/3, with solution a = 9/2, b = 3/4 or g(x) = 9x2/2 + 3x/4. Then(∆x/3)(Y0 + 4Y1O + Y2) = (1/18)(0 + 4(1/4) + 3/4) = 7/72, and1/30g(x) dx = (3/2)x3+ (3/8)x21/30= 1/18 + 1/24 = 7/7221. 0.746824948,0.74682413322. 1.236098366,1.23606797723. 1.511518747,1.51592714224. 2.418388347,2.41839915225. 0.805376152,0.80477648926. 1.536963087,1.54429477427. (a) 3.142425985, |EM | ≈ 0.000833331(b) 3.139925989, |ET | ≈ 0.001666665(c) 3.141592614, |ES| ≈ 0.00000004028. (a) 3.152411433, |EM | ≈ 0.010818779(b) 3.104518326, |ET | ≈ 0.037074328(c) 3.127008159, |ES| ≈ 0.01458449529. S14 = 0.693147984, |ES| ≈ 0.000000803 = 8.03 × 10−7; the method used in Example 6 resultsin a value of n which ensures that the magnitude of the error will be less than 10−6, this is notnecessarily the smallest value of n.30. (a) underestimates, because the graph of cos x2is concave down on the interval (0, 1)(b) overestimates, because the graph of cos x2is concave up on the interval (3/2, 2)31. f(x) = x sin x, f (x) = 2 cos x − x sin x, |f (x)| ≤ 2| cos x| + |x| | sin x| ≤ 2 + 2 = 4 so K2 ≤ 4,n >(8)(4)(24)(10−4)1/2≈ 115.5; n = 116 (a smaller n might suﬃce)32. f(x) = ecos x, f (x) = (sin2x)ecos x− (cos x)ecos x, |f (x)| ≤ ecos x(sin2x + | cos x|) ≤ 2e soK2 ≤ 2e, n >(1)(2e)(24)(10−4)1/2≈ 47.6; n = 48 (a smaller n might suﬃce)33. f(x) = x√x, f (x) =34√x, limx→0+|f (x)| = +∞34. f(x) = sin√x, f (x) = −√x sin√x + cos√x4x3/2, limx→0+|f (x)| = +∞35. L =π/2−π/21 + sin2x dx ≈ 3.820187623 36. L =311 + 1/x4 dx ≈ 2.146822803
• Exercise Set 8.7 37737.200v dt ≈15(3)(6)[0 + 4(35.2) + 2(60.1) + 4(79.2) + 2(90.9) + 4(104.1) + 114.4] ≈ 1078 ft38. 0010.0220.0830.2040.40ta50.6060.7070.608080a dt ≈8(3)(8)[0 + 4(0.02) + 2(0.08) + 4(0.20) + 2(0.40) + 4(0.60) + 2(0.70) + 4(0.60) + 0]≈ 2.7 cm/s39.1800v dt ≈180(3)(6)[0.00 + 4(0.03) + 2(0.08) + 4(0.16) + 2(0.27) + 4(0.42) + 0.65] = 37.9 mi40.18000(1/v)dx ≈1800(3)(6)13100+42908+22725+42549+22379+42216+12059≈ 0.71 s41. V =160πr2dy = π160r2dy ≈ π16(3)(4)[(8.5)2+ 4(11.5)2+ 2(13.8)2+ 4(15.4)2+ (16.8)2]≈ 9270 cm3≈ 9.3 L42. A =6000h dx ≈600(3)(6)[0 + 4(7) + 2(16) + 4(24) + 2(25) + 4(16) + 0] = 9000 ft2,V = 75A ≈ 75(9000) = 675, 000 ft343. (a) The maximum value of |f (x)| is approximately 3.8442(b) n = 18(c) 0.904740668444. (a) The maximum value of |f (x)| is approximately 1.46789(b) n = 8(c) 1.11283035045. (a) The maximum value of |f(4)(x)| is approximately 42.5518.(b) n = 4(c) 0.904524159446. (a) The maximum value of |f(4)(x)| is approximately 7.02267.(b) n = 4(c) 1.11144270247. (a) Left endpoint approximation ≈b − an[y0 + y1 + . . . + yn−2 + yn−1]Right endpoint approximation ≈b − an[y1 + y2 + . . . + yn−1 + yn]Average of the two =b − an12[y0 + 2y1 + 2y2 + . . . + 2yn−2 + 2yn−1 + yn]
• 378 Chapter 8(b) Area of trapezoid = (xk+1 − xk)yk + yk+12. If we sum from k = 0to k = n − 1 then we get the right hand side of (2).yx1x xkykk+1yk+12348. right endpoint, trapezoidal, midpoint, left endpoint49. Given g(x) = Ax2+ Bx + C, suppose ∆x = 1 and m = 0. Then set Y0 = g(−1), Y1 = g(0),Y2 = g(1). Also Y0 = g(−1) = A − B + C, Y1 = g(0) = C, Y2 = g(1) = A + B + C, with solutionC = Y1, B = 12 (Y2 − Y0), and A = 12 (Y0 + Y2) − Y1.Then1−1g(x) dx = 210(Ax2+ C) dx =23A + 2C =13(Y0 + Y2) −23Y1 + 2Y1 =13(Y0 + 4Y1 + Y2),which is exactly what one gets applying the Simpson’s Rule.The general case with the interval (m − ∆x, m + ∆x) and values Y0, Y1, Y2, can be converted bythe change of variables z =x − m∆x. Set g(x) = h(z) = h((x − m)/∆x) to get dx = ∆x dz and∆xm+∆xm−∆xh(z) dz =1−1g(x) dx. Finally, Y0 = g(m − ∆x) = h(−1),Y1 = g(m) = h(0), Y2 = g(m + ∆x) = h(1).50. From Exercise 49 we know, for i = 0, 1, . . . , n − 1 thatx2i+2x2igi(x) dx =13b − a2n[y2i + 4y2i+1 + y2i+2],becauseb − a2nis the width of the partition and acts as ∆x in Exercise 49.Summing over all the subintervals note that y0 and y2n are only listed once; sobaf(x) dx =n−1i=0x2i+2x2igi(x) dx=13b − a2n[y0 + 4y1 + 2y2 + 4y3 + . . . + 4y2n−2 + 2y2n−1 + y2n]EXERCISE SET 8.81. (a) improper; inﬁnite discontinuity at x = 3(b) continuous integrand, not improper(c) improper; inﬁnite discontinuity at x = 0(d) improper; inﬁnite interval of integration(e) improper; inﬁnite interval of integration and inﬁnite discontinuity at x = 1(f) continuous integrand, not improper2. (a) improper if p > 0 (b) improper if 1 ≤ p ≤ 2(c) integrand is continuous for all p, not improper
• Exercise Set 8.8 3793. lim→+∞−12e−2x0=12lim→+∞(−e−2+ 1) =124. lim→+∞12ln(1 + x2)−1= lim→+∞12[ln(1 + 2) − ln 2] = +∞, divergent5. lim→+∞−2 coth−1x3= lim→+∞2 coth−13 − 2 coth−1= 2 coth−136. lim→+∞−12e−x20= lim→+∞12−e− 2+ 1 = 1/27. lim→+∞−12 ln2x e= lim→+∞−12 ln2 +12=128. lim→+∞2√ln x2= lim→+∞(2√ln − 2√ln 2) = +∞, divergent9. lim→−∞−14(2x − 1)20= lim→−∞14[−1 + 1/(2 − 1)2] = −1/410. lim→−∞13tan−1 x33= lim→−∞13π4− tan−13=13[π/4 − (−π/2)] = π/411. lim→−∞13e3x0= lim→−∞13−13e3=1312. lim→−∞−12ln(3 − 2ex)0= lim→−∞12ln(3 − 2e ) =12ln 313.+∞−∞x dx converges if0−∞x dx and+∞0x dx both converge; it diverges if either (or both)diverges.+∞0x dx = lim→+∞12x20= lim→+∞122= +∞ so+∞−∞x dx is divergent.14.+∞0x√x2 + 2dx = lim→+∞x2 + 20= lim→+∞( 2 + 2 −√2) = +∞so∞−∞x√x2 + 2dx is divergent.15.+∞0x(x2 + 3)2dx = lim→+∞−12(x2 + 3) 0= lim→+∞12[−1/( 2+ 3) + 1/3] =16,similarly0−∞x(x2 + 3)2dx = −1/6 so∞−∞x(x2 + 3)2dx = 1/6 + (−1/6) = 0
• 380 Chapter 816.+∞0e−t1 + e−2tdt = lim→+∞− tan−1(e−t)0= lim→+∞− tan−1(e−) +π4=π4,0−∞e−t1 + e−2tdt = lim→−∞− tan−1(e−t)0= lim→−∞−π4+ tan−1(e−) =π4,+∞−∞e−t1 + e−2tdt =π4+π4=π217. lim→4−−1x − 4 0= lim→4−−1− 4−14= +∞, divergent18. lim→0+32x2/38= lim→0+32(4 − 2/3) = 619. lim→π/2−− ln(cos x)0= lim→π/2−− ln(cos ) = +∞, divergent20. lim→4−−2√4 − x0= lim→4−2(−√4 − + 2) = 421. lim→1−sin−1x0= lim→1−sin−1= π/222. lim→−3+− 9 − x21= lim→−3+(−√8 + 9 − 2) = −√823. lim→π/3+√1 − 2 cos xπ/2= lim→π/3+(1 −√1 − 2 cos ) = 124. lim→π/4−− ln(1 − tan x)0= lim→π/4−− ln(1 − tan ) = +∞, divergent25.20dxx − 2= lim→2−ln |x − 2|0= lim→2−(ln | − 2| − ln 2) = −∞, divergent26.20dxx2= lim→0+−1/x2= lim→0+(−1/2 + 1/ ) = +∞ so2−2dxx2is divergent27.80x−1/3dx = lim→0+32x2/38= lim→0+32(4 − 2/3) = 6,0−1x−1/3dx = lim→0−32x2/3−1= lim→0−32( 2/3− 1) = −3/2so8−1x−1/3dx = 6 + (−3/2) = 9/228.10dx(x − 1)2/3= lim→1−3(x − 1)1/30= lim→1−3[( − 1)1/3− (−1)1/3] = 3
• Exercise Set 8.8 38129. Deﬁne+∞01x2dx =a01x2dx ++∞a1x2dx where a > 0; take a = 1 for convenience,101x2dx = lim→0+(−1/x)1= lim→0+(1/ − 1) = +∞ so+∞01x2dx is divergent.30. Deﬁne+∞1dxx√x2 − 1=a1dxx√x2 − 1++∞adxx√x2 − 1where a > 1,take a = 2 for convenience to get21dxx√x2 − 1= lim→1+sec−1x2= lim→1+(π/3 − sec−1) = π/3,+∞2dxx√x2 − 1= lim→+∞sec−1x2= π/2 − π/3 so+∞1dxx√x2 − 1= π/2.31.+∞0e−√x√xdx = 2+∞0e−udu = 2 lim→+∞−e−u0= 2 lim→+∞1 − e−= 232.+∞12dx√x(x + 4)= 2+∞2√3duu2 + 4= 2 lim→+∞12tan−1 u2 2√3= lim→+∞tan−12− tan−1√3 =π633.+∞0e−x√1 − e−xdx =10du√u= lim→0+2√u1= lim→0+2(1 −√) = 234.+∞0e−x√1 − e−2xdx = −01du√1 − u2=10du√1 − u2= lim→1sin−1u0= lim→1sin−1=π235. lim→+∞ 0e−xcos x dx = lim→+∞12e−x(sin x − cos x)0= 1/236. A =+∞0xe−3xdx = lim→+∞−19(3x + 1)e−3x0= 1/337. (a) 2.726585 (b) 2.804364 (c) 0.219384 (d) 0.50406739. 1 +dydx2= 1 +4 − x2/3x2/3=4x2/3; the arc length is802x1/3dx = 3x2/380= 1240. 1 +dydx2= 1 +x24 − x2=44 − x2; the arc length is2044 − x2dx = lim→2−02√4 − x2dx = lim→2−2 sin−1 x2 0= 2 sin−11 = π41. ln x dx = x ln x − x + C,10ln x dx = lim→0+1ln x dx = lim→0+(x ln x − x)1= lim→0+(−1 − ln + ),but lim→0+ln = lim→0+ln1/= lim→0+(− ) = 0 so10ln x dx = −1
• 382 Chapter 842.ln xx2dx = −ln xx−1x+ C,+∞1ln xx2dx = lim→+∞ 1ln xx2dx = lim→+∞−ln xx−1x 1= lim→+∞−ln−1+ 1 ,but lim→+∞ln= lim→+∞1= 0 so+∞1ln xx2= 143.+∞0e−3xdx = lim→+∞ 0e−3xdx = lim→+∞−13e−3x0= lim→+∞−13e−3+19=1344. A =+∞48x2 − 4dx = lim→+∞2 lnx − 2x + 2 4= lim→+∞2 ln− 2+ 2− ln13= 2 ln 345. (a) V = π+∞0e−2xdx = −π2lim→+∞e−2x0= π/2(b) S = 2π+∞0e−x1 + e−2xdx, let u = e−xto getS = −2π011 + u2du = 2πu21 + u2 +12ln u + 1 + u210= π√2 + ln(1 +√2)47. (a) For x ≥ 1, x2≥ x, e−x2≤ e−x(b)+∞1e−xdx = lim→+∞ 1e−xdx = lim→+∞−e−x1= lim→+∞(e−1− e−) = 1/e(c) By Parts (a) and (b) and Exercise 46(b),+∞1e−x2dx is convergent and is ≤ 1/e.48. (a) If x ≥ 0 then ex≥ 1,12x + 1≤ex2x + 1(b) lim→+∞ 0dx2x + 1= lim→+∞12ln(2x + 1)0= +∞(c) By Parts (a) and (b) and Exercise 46(a),+∞0ex2x + 1dx is divergent.49. V = lim→+∞ 1(π/x2) dx = lim→+∞−(π/x)1= lim→+∞(π − π/ ) = πA = lim→+∞ 12π(1/x) 1 + 1/x4 dx;use Exercise 46(a) with f(x) = 2π/x, g(x) = (2π/x) 1 + 1/x4and a = 1 to see that the area is inﬁnite.50. (a) 1 ≤√x3 + 1xfor x ≥ 2,+∞21dx = +∞(b)+∞2xx5 + 1dx ≤+∞2dxx4= lim→+∞−13x32= 1/24(c)∞0xex2x + 1dx ≥+∞1xex2x + 1≥+∞1dx2x + 1= +∞
• Exercise Set 8.8 38351. The area under the curve y =11 + x2, above the x-axis, and to theright of the y-axis is given by∞011 + x2. Solving fory = 1−yy , the area is also given by the improper integral101 − yydy.0 42x52. (b) u =√x,+∞0cos√x√xdx = 2+∞0cos u du;+∞0cos u du diverges by Part (a).53. Let x = r tan θ to getdx(r2 + x2)3/2=1r2cos θ dθ =1r2sin θ + C =xr2√r2 + x2+ Cso u =2πNIrklim→+∞xr2√r2 + x2a=2πNIkrlim→+∞( / r2 + 2 − a/ r2 + a2)=2πNIkr(1 − a/ r2 + a2).54. Let a2=M2RTto get(a) ¯v =4√πM2RT3/212M2RT−2=2√π2RTM=8RTπM(b) v2rms =4√πM2RT3/23√π8M2RT−5/2=3RTMso vrms =3RTM55. (a) Satellite’s weight = w(x) = k/x2lb when x = distance from center of Earth; w(4000) = 6000so k = 9.6 × 1010and W =4000+40009.6 × 1010x−2dx mi·lb.(b)+∞40009.6 × 1010x−2dx = lim→+∞−9.6 × 1010/x4000= 2.4 × 107mi·lb56. (a) L{1} =+∞0e−stdt = lim→+∞−1se−st0=1s(b) L{e2t} =+∞0e−ste2tdt =+∞0e−(s−2)tdt = lim→+∞−1s − 2e−(s−2)t0=1s − 2(c) L{sin t} =+∞0e−stsin t dt = lim→+∞e−sts2 + 1(−s sin t − cos t)0=1s2 + 1(d) L{cos t} =+∞0e−stcos t dt = lim→+∞e−sts2 + 1(−s cos t + sin t)0=ss2 + 1
• 384 Chapter 857. (a) L{f(t)} =+∞0te−stdt = lim→+∞−(t/s + 1/s2)e−st0=1s2(b) L{f(t)} =+∞0t2e−stdt = lim→+∞−(t2/s + 2t/s2+ 2/s3)e−st0=2s3(c) L{f(t)} =+∞3e−stdt = lim→+∞−1se−st3=e−3ss58. 10 100 1000 10,0000.8862269 0.8862269 0.8862269 0.886226959. (a) u =√ax, du =√a dx, 2+∞0e−ax2dx =2√a+∞0e−u2du = π/a(b) x =√2σu, dx =√2σ du,2√2πσ+∞0e−x2/2σ2dx =2√π+∞0e−u2du = 160. (a)30e−x2dx ≈ 0.8862;√π/2 ≈ 0.8862(b)+∞0e−x2dx =30e−x2dx ++∞3e−x2dx so E =+∞3e−x2dx <+∞3xe−x2dx =12e−9< 7 × 10−561. (a)401x6 + 1dx ≈ 1.047; π/3 ≈ 1.047(b)+∞01x6 + 1dx =401x6 + 1dx ++∞41x6 + 1dx soE =+∞41x6 + 1dx <+∞41x6dx =15(4)5< 2 × 10−462. If p = 0, then+∞0(1)dx = lim→+∞x0= +∞,if p = 0, then+∞0epxdx = lim→+∞1pepx0= lim→+∞1p(ep− 1) =−1/p, p < 0+∞, p > 0.63. If p = 1, then10dxx= lim→0+ln x1= +∞;if p = 1, then10dxxp= lim→0+x1−p1 − p1= lim→0+[(1 − 1−p)/(1 − p)] =1/(1 − p), p < 1+∞, p > 1.64. u =√1 − x, u2= 1 − x, 2u du = −dx;−2012 − u2 du = 2102 − u2 du = u 2 − u2 + 2 sin−1(u/√2)10=√2 + π/265. 210cos(u2)du ≈ 1.80966. −201sin(1 − u2)du = 210sin(1 − u2)du ≈ 1.187
• Review Exercise Set 38567. (a) Γ(1) =+∞0e−tdt = lim→+∞−e−t0= lim→+∞(−e−+ 1) = 1(b) Γ(x + 1) =+∞0txe−tdt; let u = tx, dv = e−tdt to getΓ(x + 1) = −txe−t+∞0+ x+∞0tx−1e−tdt = −txe−t+∞0+ xΓ(x)limt→+∞txe−t= limt→+∞txet= 0 (by multiple applications of L’Hˆopital’s rule)so Γ(x + 1) = xΓ(x)(c) Γ(2) = (1)Γ(1) = (1)(1) = 1, Γ(3) = 2Γ(2) = (2)(1) = 2, Γ(4) = 3Γ(3) = (3)(2) = 6It appears that Γ(n) = (n − 1)! if n is a positive integer.(d) Γ12=+∞0t−1/2e−tdt = 2+∞0e−u2du (with u =√t) = 2(√π/2) =√π(e) Γ32=12Γ12=12√π, Γ52=32Γ32=34√π68. (a) t = − ln x, x = e−t, dx = −e−tdt,10(ln x)ndx = −0+∞(−t)ne−tdt = (−1)n+∞0tne−tdt = (−1)nΓ(n + 1)(b) t = xn, x = t1/n, dx = (1/n)t1/n−1dt,+∞0e−xndx = (1/n)+∞0t1/n−1e−tdt = (1/n)Γ(1/n) = Γ(1/n + 1)69. (a) cos θ − cos θ0 = 2 sin2(θ0/2) − sin2(θ/2) = 2(k2 − k2 sin2φ) = 2k2 cos2 φ=√2 k cos φ; k sin φ = sin(θ/2) so k cos φ dφ =12cos(θ/2) dθ =121 − sin2(θ/2) dθ=121 − k2 sin2φ dθ, thus dθ =2k cos φ1 − k2 sin2φdφ and henceT =8Lgπ/201√2k cos φ·2k cos φ1 − k2 sin2φdφ = 4Lgπ/2011 − k2 sin2φdφ(b) If L = 1.5 ft and θ0 = (π/180)(20) = π/9, thenT =√32π/20dφ1 − sin2(π/18) sin2φ≈ 1.37 s.REVIEW EXERCISE SET1. u = 4 + 9x, du = 9 dx,19u1/2du =227(4 + 9x)3/2+ C2. u = πx, du = π dx,1πcos u du =1πsin u + C =1πsin πx + C
• 386 Chapter 83. u = cos θ, − u1/2du = −23cos3/2θ + C4. u = ln x, du =dxx,duu= ln |u| + C = ln |ln x| + C5. u = tan(x2),12u2du =16tan3(x2) + C6. u =√x, x = u2, dx = 2u du,230u2u2 + 9du = 2301 −9u2 + 9du = 2u − 6 tan−1 u330= 6 −32π7. (a) With u =√x:1√x√2 − xdx = 21√2 − u2du = 2 sin−1(u/√2) + C = 2 sin−1( x/2) + C;with u =√2 − x :1√x√2 − xdx = −21√2 − u2du = −2 sin−1(u/√2) + C = −2 sin−1(√2 − x/√2) + C1;completing the square:11 − (x − 1)2dx = sin−1(x − 1) + C.(b) In the three results in Part (a) the antiderivatives diﬀer by a constant, in particular2 sin−1( x/2) = π − 2 sin−1(√2 − x/√2) = π/2 + sin−1(x − 1).8. (a) u = x2, dv =x√x2 + 1dx, du = 2x dx, v =√x2 + 1;10x3√x2 + 1dx = x2x2 + 110− 210x(x2+ 1)1/2dx=√2 −23(x2+ 1)3/210=√2 −23[2√2 − 1] = (2 −√2)/3(b) u2= x2+ 1, x2= u2− 1, 2x dx = 2u du, x dx = u du;10x3√x2 + 1dx =10x2√x2 + 1x dx =√21u2− 1uu du=√21(u2− 1)du =13u3− u√21= (2 −√2)/39. u = x, dv = e−xdx, du = dx, v = −e−x;xe−xdx = −xe−x+ e−xdx = −xe−x− e−x+ C10. u = x, dv = sin 2x dx, du = dx, v = −12cos 2x;x sin 2x dx = −12x cos 2x +12cos 2x dx = −12x cos 2x +14sin 2x + C
• Review Exercise Set 38711. u = ln(2x + 3), dv = dx, du =22x + 3dx, v = x; ln(2x + 3)dx = x ln(2x + 3) −2x2x + 3dxbut2x2x + 3dx = 1 −32x + 3dx = x −32ln(2x + 3) + C1 soln(2x + 3)dx = x ln(2x + 3) − x +32ln(2x + 3) + C12. u = tan−1(2x), dv = dx, du =21 + 4x2dx, v = x;tan−1(2x)dx = x tan−1(2x) −2x1 + 4x2dx = x tan−1(2x) −14ln(1 + 4x2) + C13. Repeated RepeatedDiﬀerentiation Antidiﬀerentiation8x4cos 2x+32x3 12sin 2x−96x2−14cos 2x+192x −18sin 2x−192116cos 2x+0132sin 2x8x4cos 2x dx = (4x4− 12x2+ 6) sin 2x + (8x3− 12x) cos 2x + C14. distance =50t2e−tdt; u = t2, dv = e−tdt, du = 2tdt, v = −e−t,distance = −t2e−t50+ 250te−tdt; u = 2t, dv = e−tdt, du = 2dt, v = −e−t,distance = −25e−5− 2te−t50+ 250e−tdt = −25e−5− 10e−5− 2e−t50= −25e−5− 10e−5− 2e−5+ 2 = −37e−5+ 215. sin25θ dθ =12(1 − cos 10θ)dθ =12θ −120sin 10θ + C16. sin32x cos22x dx = (1 − cos22x) cos22x sin 2x dx= (cos22x − cos42x) sin 2x dx = −16cos32x +110cos52x + C17. sin x cos 2x dx =12(sin 3x − sin x)dx = −16cos 3x +12cos x + C
• 388 Chapter 818.π/60sin 2x cos 4x dx =12π/60(sin 6x − sin 2x)dx = −112cos 6x +14cos 2xπ/60= [(−1/12)(−1) + (1/4)(1/2)] − [−1/12 + 1/4] = 1/2419. u = 2x,sin42x dx =12sin4u du =12−14sin3u cos u +34sin2u du= −18sin3u cos u +38−12sin u cos u +12du= −18sin3u cos u −316sin u cos u +316u + C= −18sin32x cos 2x −316sin 2x cos 2x +38x + C20. u = x2,x cos5(x2)dx =12cos5u du =12(cos u)(1 − sin2u)2du=12cos u du − cos u sin2u du +12cos u sin4u du=12sin u −13sin3u +110sin5u + C=12sin(x2) −13sin3(x2) +110sin5(x2) + C21. x = 3 sin θ, dx = 3 cos θ dθ,9 sin2θ dθ =92(1 − cos 2θ)dθ =92θ −94sin 2θ + C =92θ −92sin θ cos θ + C=92sin−1(x/3) −12x 9 − x2 + C22. x = 4 sin θ, dx = 4 cos θ dθ,1161sin2θdθ =116csc2θ dθ = −116cot θ + C = −√16 − x216x+ C23. x = sec θ, dx = sec θ tan θ dθ, sec θ dθ = ln | sec θ + tan θ| + C = ln x + x2 − 1 + C24. x = 5 sec θ, dx = 5 sec θ tan θ dθ,25 sec3θ dθ =252sec θ tan θ +252ln | sec θ + tan θ| + C1=12x x2 − 25 +252ln |x + x2 − 25| + C
• Review Exercise Set 38925. x = 3 tan θ, dx = 3 sec2θ dθ,9 tan2θ sec θ dθ = 9 sec3θ dθ − 9 sec θ dθ=92sec θ tan θ −92ln | sec θ + tan θ| + C=12x 9 + x2 −92ln |139 + x2 +13x| + C26. 2x = tan θ, 2 dx = sec2θ dθ,sec2θ csc θ dθ = (sec θ tan θ + csc θ) dθ= sec θ − ln | csc θ + cot θ| + C= 1 + 4x2 − ln√1 + 4x22x+12x+ C27.1(x + 4)(x − 1)=Ax + 4+Bx − 1; A = −15, B =15so−151x + 4dx +151x − 1dx = −15ln |x + 4| +15ln |x − 1| + C =15lnx − 1x + 4+ C28.1(x + 1)(x + 7)=Ax + 1+Bx + 7; A =16, B = −16so161x + 1dx −161x + 7dx =16ln |x + 1| −16ln |x + 7| + C =16lnx + 1x + 7+ C29.x2+ 2x + 2= x − 2 +6x + 2, x − 2 +6x + 2dx =12x2− 2x + 6 ln |x + 2| + C30.x2+ x − 16(x + 1)(x − 3)2=Ax + 1+Bx − 3+C(x − 3)2; A = −1, B = 2, C = −1 so−1x + 1dx + 21x − 3dx −1(x − 3)2dx= − ln |x + 1| + 2 ln |x − 3| +1x − 3+ C = ln(x − 3)2|x + 1|+1x − 3+ C31.x2(x + 2)3=Ax + 2+B(x + 2)2+C(x + 2)3; A = 1, B = −4, C = 4 so1x + 2dx − 41(x + 2)2dx + 41(x + 2)3dx = ln |x + 2| +4x + 2−2(x + 2)2+ C32.1x(x2 + 1)=Ax+Bx + Cx2 + 1; A = 1, B = −1, C = 0 so1x3 + xdx = ln |x| −12ln(x2+ 1) + C =12lnx2x2 + 1+ C
• 390 Chapter 833. (a) With x = sec θ:1x3 − xdx = cot θ dθ = ln | sin θ| + C = ln√x2 − 1|x|+ C; valid for |x| > 1.(b) With x = sin θ:1x3 − xdx = −1sin θ cos θdθ = − 2 csc 2θ dθ= − ln | csc 2θ − cot 2θ| + C = ln | cot θ| + C = ln√1 − x2|x|+ C, 0 < |x| < 1.34. A =213 − xx3 + x2dx,3 − xx2(x + 1)=Ax+Bx2+Cx + 1; A = −4, B = 3, C = 4A = −4 ln |x| −3x+ 4 ln |x + 1|21= (−4 ln 2 −32+ 4 ln 3) − (−4 ln 1 − 3 + 4 ln 2) =32− 8 ln 2 + 4 ln 3 =32+ 4 ln3435. #40 36. #52 37. #11338. #108 39. #28 40. #7141. exact value = 14/3 ≈ 4.666666667(a) 4.667600663, |EM | ≈ 0.000933996(b) 4.664795679, |ET | ≈ 0.001870988(c) 4.666651630, |ES| ≈ 0.00001503742. exact value =12ln 5 ≈ 0.804718956(a) 0.801605339, |EM | ≈ 0.003113617(b) 0.811019505, |ET | ≈ 0.006300549(c) 0.805041497, |ES| ≈ 0.00032254143. f(x) =√x + 1, f (x) = −14(x + 1)−3/2, f(4)(x) = −1516(x + 1)−7/2; K2 = 1/4, K4 = 15/16(a) |EM | ≤272400(1/4) = 0.002812500 (b) |ET | ≤271200(1/4) = 0.005625000(c) |ES| ≤243180 × 104(15/16) ≈ 0.00012656344. f(x) = 1/(2x + 3), f (x) = 8(2x + 3)−3, f(4)(x) = 384(2x + 3)−5; K2 = 8, K4 = 384(a) |EM | ≤82400(8) ≈ 0.026666667 (b) |ET | ≤81200(8) ≈ 0.053333333(c) |ES| ≤32180 × 104(384) ≈ 0.00682666745. (a) n >(27)(1/4)(24)(5 × 10−4)1/2≈ 23.7; n = 24 (b) n >(27)(1/4)(12)(5 × 10−4)1/2≈ 33.5; n=34(c) n >(243)(15/16)(180)(5 × 10−4)1/4≈ 7.1; n = 8
• Review Exercise Set 39146. (a) n >(8)(8)(24)(10−6)1/2≈ 1632.99; n = 1633 (b) n >(8)(8)(12)(10−6)1/2≈ 2309.4; n = 2310(c) n >(32)(384)(180)(10−6)1/4≈ 90.9; n = 9247. lim→+∞(−e−x)0= lim→+∞(−e−+ 1) = 148. lim→−∞12tan−1 x22= lim→−∞12π4− tan−12=12[π/4 − (−π/2)] = 3π/849. lim→9−−2√9 − x0= lim→9−2(−√9 − + 3) = 650.1012x − 1dx =1/2012x − 1dx +11/212x − 1dx = lim→1/2−12ln(2x − 1) + lim→1/2+12ln(2x − 1) + Cneither limit exists hence the integral diverges51. A =+∞eln x − 1x2dx = lim→+∞c −ln xx e= 1/e52. V = 2π+∞0xe−xdx = 2π lim→+∞−e−x(x + 1)0= 2π lim→+∞1 − e−( + 1)but lim→+∞e−( + 1) = lim→+∞+ 1e= lim→+∞1e= 0 so V = 2π53.+∞0dxx2 + a2= lim→+∞1atan−1(x/a)0= lim→+∞1atan−1( /a) =π2a= 1, a = π/254. (a) integration by parts, u = x, dv = sin x dx (b) u-substitution: u = sin x(c) reduction formula (d) u-substitution: u = tan x(e) u-substitution: u = x3+ 1 (f) u-substitution: u = x + 1(g) integration by parts: dv = dx, u = tan−1x (h) trigonometric substitution: x = 2 sin θ(i) u-substitution: u = 4 − x255. x =√3 tan θ, dx =√3 sec2θ dθ,131sec θdθ =13cos θ dθ =13sin θ + C =x3√3 + x2+ C56. u = x, dv = cos 3x dx, du = dx, v =13sin 3x;x cos 3x dx =13x sin 3x −13sin 3x dx =13x sin 3x +19cos 3x + C57. Use Endpaper Formula (31) to get tan7θ dθ =16tan6θ −14tan4θ +12tan2θ + ln | cos θ| + C.58.cos θ(sin θ − 3)2 + 3dθ, let u = sin θ − 3,1u2 + 3du =1√3f tan−1[(sin θ − 3)/√3] + C
• 392 Chapter 859. sin22t cos32t dt = sin22t(1 − sin22t) cos 2t dt = (sin22t − sin42t) cos 2t dt=16sin32t −110sin52t + C60.301(x − 3)2dx = lim→3−01(x − 3)2dx = lim→3−−1x − 3 0which is clearly divergent,so the integral diverges.61. u = e2x, dv = cos 3x dx, du = 2e2xdx, v =13sin 3x;e2xcos 3x dx =13e2xsin 3x −23e2xsin 3x dx. Use u = e2x, dv = sin 3x dx to gete2xsin 3x dx = −13e2xcos 3x +23e2xcos 3x dx soe2xcos 3x dx =13e2xsin 3x +29e2xcos 3x −49e2xcos 3x dx,139e2xcos 3x dx =19e2x(3 sin 3x+2 cos 3x)+C1, e2xcos 3x dx =113e2x(3 sin 3x+2 cos 3x)+C62. x = (1/√2) sin θ, dx = (1/√2) cos θ dθ,1√2π/2−π/2cos4θ dθ =1√214cos3θ sin θπ/2−π/2+34π/2−π/2cos2θ dθ=34√212cos θ sin θπ/2−π/2+12π/2−π/2dθ =34√212π =3π8√263.1(x − 1)(x + 2)(x − 3)=Ax − 1+Bx + 2+Cx − 3; A = −16, B =115, C =110so−161x − 1dx +1151x + 2dx +1101x − 3dx= −16ln |x − 1| +115ln |x + 2| +110ln |x − 3| + C64. x =23sin θ, dx =23cos θ dθ,124π/601cos3 θdθ =124π/60sec3θ dθ =148sec θ tan θ +148ln | sec θ + tan θ|π/60=148[(2/√3)(1/√3) + ln |2/√3 + 1/√3|] =14823+12ln 365. u =√x − 4, x = u2+ 4, dx = 2u du,202u2u2 + 4du = 2201 −4u2 + 4du = 2u − 4 tan−1(u/2)20= 4 − π
• Review Exercise Set 39366. u =√ex − 1, ex= u2+ 1, x = ln(u2+ 1), dx =2uu2 + 1du,102u2u2 + 1du = 2101 −1u2 + 1du = (2u − 2 tan−1u)10= 2 −π267. u =√ex + 1, ex= u2− 1, x = ln(u2− 1), dx =2uu2 − 1du,2u2 − 1du =1u − 1−1u + 1du = ln |u − 1| − ln |u + 1| + C = ln√ex + 1 − 1√ex + 1 + 1+ C68.1x(x2 + x + 1)=Ax+Bx + Cx2 + x + 1; A = 1, B = C = −1 so−x − 1x2 + x + 1dx = −x + 1(x + 1/2)2 + 3/4dx = −u + 1/2u2 + 3/4du, u = x + 1/2= −12ln(u2+ 3/4) −1√3tan−1(2u/√3) + C1sodxx(x2 + x + 1)= ln |x| −12ln(x2+ x + 1) −1√3tan−1 2x + 1√3+ C69. u = sin−1x, dv = dx, du =1√1 − x2dx, v = x;1/20sin−1x dx = x sin−1x1/20−1/20x√1 − x2dx =12sin−1 12+ 1 − x21/20=12π6+34− 1 =π12+√32− 170. tan34x(1 + tan24x) sec24x dx = (tan34x + tan54x) sec24x dx =116tan44x +124tan64x + C71.x + 3(x + 1)2 + 1dx, let u = x + 1,u + 2√u2 + 1du = u(u2+ 1)−1/2+2√u2 + 1du = u2 + 1 + 2 sinh−1u + C= x2 + 2x + 2 + 2 sinh−1(x + 1) + CAlternate solution: let x + 1 = tan θ,(tan θ + 2) sec θ dθ = sec θ tan θ dθ + 2 sec θ dθ = sec θ + 2 ln | sec θ + tan θ| + C= x2 + 2x + 2 + 2 ln( x2 + 2x + 2 + x + 1) + C.
• 394 Chapter 872. Let x = tan θ to get1x3 − x2dx.1x2(x − 1)=Ax+Bx2+Cx − 1; A = −1, B = −1, C = 1 so−1xdx −1x2dx +1x − 1dx = − ln |x| +1x+ ln |x − 1| + C=1x+ lnx − 1x+ C = cot θ + lntan θ − 1tan θ+ C = cot θ + ln |1 − cot θ| + C73. lim→+∞−12(x2 + 1) a= lim→+∞−12( 2 + 1)+12(a2 + 1)=12(a2 + 1)74. lim→+∞1abtan−1 bxa 0= lim→+∞1abtan−1 ba=π2ab