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# Gabarito completo anton_calculo_8ed_caps_01_08

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### Gabarito completo anton_calculo_8ed_caps_01_08

1. 1. 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. 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
3. 3. 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. 4. 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
5. 5. 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. 6. 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
7. 7. 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. 8. 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
9. 9. 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. 10. 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
11. 11. 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. 12. 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
13. 13. 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. 14. 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
15. 15. 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. 16. 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.
17. 17. 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. 18. 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
19. 19. 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. 20. 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
21. 21. 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. 22. 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
23. 23. 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. 24. 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)
25. 25. 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. 26. 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 < +∞
27. 27. 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. 28. 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◦
29. 29. 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. 30. 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)
31. 31. 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. 32. 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
33. 33. 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. 34. 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.
35. 35. 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. 36. 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)
37. 37. 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. 38. 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.
39. 39. 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. 40. 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
41. 41. 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. 42. 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
43. 43. 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. 44. 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.
45. 45. 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. 46. 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
47. 47. 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. 48. 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)
49. 49. 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. 50. 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.
51. 51. 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. 52. 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
53. 53. 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. 54. 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.
55. 55. 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. 56. 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
57. 57. 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. 58. 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
59. 59. 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. 60. 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
61. 61. 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. 62. 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
63. 63. 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. 64. 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
65. 65. 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. 66. 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
67. 67. 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. 68. 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
69. 69. 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