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# solucionario de purcell 3

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capitulo 3

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## solucionario de purcell 3Document Transcript

• Applications of the CHAPTER 3 Derivative 3 3.1 Concepts Review 7. Ψ ′( x) = 2 x + 3; 2x + 3 = 0 when x = – . 2 1. continuous; closed and bounded 3 Critical points: –2, – , 1 2 2. extreme ⎛ 3⎞ 9 Ψ (–2) = –2, Ψ ⎜ – ⎟ = – , Ψ (1) = 4 ⎝ 2⎠ 4 3. endpoints; stationary points; singular points 9 Maximum value = 4, minimum value = – 4. f ′(c) = 0; f ′(c) does not exist 4 1 6 8. G ′( x) = (6 x 2 + 6 x –12) = ( x 2 + x – 2); Problem Set 3.1 5 5 1. Endpoints: −2 , 4 x 2 + x – 2 = 0 when x = –2, 1 Singular points: none Critical points: –3, –2, 1, 3 Stationary points: 0, 2 9 7 G (–3) = , G (–2) = 4, G (1) = – , G (3) = 9 Critical points: −2, 0, 2, 4 5 5 Maximum value = 9, 2. Endpoints: −2 , 4 7 Singular points: 2 minimum value = – Stationary points: 0 5 Critical points: −2, 0, 2, 4 9. f ′( x) = 3 x 2 – 3; 3x 2 – 3 = 0 when x = –1, 1. 3. Endpoints: −2 , 4 Critical points: –1, 1 Singular points: none f(–1) = 3, f(1) = –1 Stationary points: −1, 0,1, 2,3 No maximum value, minimum value = –1 Critical points: −2, −1, 0,1, 2,3, 4 (See graph.) 4. Endpoints: −2 , 4 Singular points: none Stationary points: none Critical points: −2, 4 5. f ′( x) = 2 x + 4; 2 x + 4 = 0 when x = –2. Critical points: –4, –2, 0 f(–4) = 4, f(–2) = 0, f(0) = 4 Maximum value = 4, minimum value = 0 10. f ′( x) = 3 x 2 – 3; 3x 2 – 3 = 0 when x = –1, 1. 1 3 6. h′( x) = 2 x + 1; 2 x + 1 = 0 when x = – . Critical points: – , –1, 1, 3 2 2 1 ⎛ 3 ⎞ 17 Critical points: –2, – , 2 f ⎜ – ⎟ = , f (–1) = 3, f (1) = –1, f (3) = 19 2 ⎝ 2⎠ 8 ⎛ 1⎞ 1 Maximum value = 19, minimum value = –1 h(–2) = 2, h ⎜ – ⎟ = – , h(2) = 6 ⎝ 2⎠ 4 1 Maximum value = 6, minimum value = – 4 154 Section 3.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 1 2x 2x 11. h′(r ) = − ; h′(r ) is never 0; h′(r ) is not defined 15. g ′( x) = − ; − = 0 when x = 0. 2 2 r 2 (1 + x ) (1 + x 2 ) 2 when r = 0, but r = 0 is not in the domain on Critical point: 0 [–1, 3] since h(0) is not defined. g(0) = 1 Critical points: –1, 3 As x → ∞, g ( x) → 0+ ; as x → −∞, g ( x) → 0+. Note that lim h(r ) = −∞ and lim h( x) = ∞. x → 0− x → 0+ Maximum value = 1, no minimum value No maximum value, no minimum value. (See graph.) 2x 2x 12. g ′( x) = − ; − = 0 when x = 0 2 2 (1 + x ) (1 + x 2 ) 2 Critical points: –3, 0, 1 1 1 g(–3) = , g(0) = 1, g(1) = 10 2 1 Maximum value = 1, minimum value = 10 1 − x2 16. f ′( x) = ; 13. f '( x) = 4x − 4x 3 (1 + x 2 )2 ( = 4 x x2 − 1 ) 1 − x2 = 0 when x = –1, 1 = 4 x ( x − 1)( x + 1) (1 + x 2 )2 4 x ( x − 1)( x + 1) = 0 when x = 0,1, −1 . Critical points: –1, 1, 4 Critical points: −2, −1, 0,1, 2 1 1 4 f (−1) = − , f (1) = , f (4) = 2 2 17 f ( −2 ) = 10 ; f ( −1) = 1 ; f ( 0 ) = 2 ; f (1) = 1 ; 1 f ( 2 ) = 10 Maximum value = , 2 Maximum value: 10 1 Minimum value: 1 minimum value = – 2 14. f ' ( x ) = 5 x 4 − 25 x 2 + 20 π 17. r ′(θ ) = cos θ ; cos θ = 0 when θ = + kπ ( 4 = 5 x − 5x + 4 )2 2 π π = 5 ( x 2 − 4 )( x 2 − 1) Critical points: – , 4 6 = 5 ( x − 2 )( x + 2 )( x − 1)( x + 1) ⎛ π⎞ 1 ⎛ π⎞ 1 r⎜− ⎟ = − , r⎜ ⎟ = 5 ( x − 2 )( x + 2 )( x − 1)( x + 1) = 0 when ⎝ 4⎠ 2 ⎝6⎠ 2 x = −2, −1,1, 2 1 1 Maximum value = , minimum value = – Critical points: −3, −2, −1,1, 2 2 2 19 41 f ( −3) = −79 ; f ( −2 ) = − ; f ( −1) = − ; 18. s ′(t ) = cos t + sin t ; cos t + sin t = 0 when 3 3 35 13 π f (1) = ; f ( 2) = tan t = –1 or t = – + k π. 3 3 4 35 3π Maximum value: Critical points: 0, ,π 3 4 Minimum value: −79 ⎛ 3π ⎞ s(0) = –1, s ⎜ ⎟ = 2, s (π ) = 1 . ⎝ 4 ⎠ Maximum value = 2, minimum value = –1 Instructor’s Resource Manual Section 3.1 155 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• x –1 25. g ' (θ ) = θ 2 ( sec θ tan θ ) + 2θ sec θ 19. a ′( x) = ; a ′( x) does not exist when x = 1. x –1 = θ sec θ (θ tan θ + 2 ) Critical points: 0, 1, 3 θ sec θ (θ tan θ + 2 ) = 0 when θ = 0 . a(0) = 1, a(1) = 0, a(3) = 2 Maximum value = 2, minimum value = 0 Consider the graph: y 3(3s – 2) 20. f ′( s ) = ; f ′( s ) does not exist when s = 2 . 3s – 2 3 Critical points: −1, 2 , 4 1 3 f(–1) = 5, f 3 ( ) 2 = 0, f(4) = 10 −π π x Maximum value = 10, minimum value = 0 4 4 −1 1 21. g ′( x) = ; f ′( x) does not exist when x = 0. 3x2 / 3 Critical points: –1, 0, 27 π π Critical points: − , 0, g(–1) = –1, g(0) = 0, g(27) = 3 4 4 Maximum value = 3, minimum value = –1 ⎛ π⎞ π 2 ⎛π ⎞ π 2 2 2 g⎜− ⎟ = ; g ( 0) = 0 ; g ⎜ ⎟ = 2 ⎝ 4⎠ 16 ⎝4⎠ 16 22. s ′(t ) = ; s ′(t ) does not exist when t = 0. 3/ 5 5t π2 2 Critical points: –1, 0, 32 Maximum value: ; Minimum value: 0 16 s(–1) = 1, s(0) = 0, s(32) = 4 Maximum value = 4, minimum value = 0 ( 2 + t ) ⎛ t 2 / 3 ⎞ − t 5/ 3 (1) 5 ⎜ ⎟ 23. H ' ( t ) = − sin t 26. h ' ( t ) = ⎝3 ⎠ − sin t = 0 when ( 2 + t )2 t = 0, π , 2π ,3π , 4π ,5π , 6π , 7π ,8π ⎛5 ⎞ ⎛ 10 2 ⎞ t2/3 ⎜ (2 + t ) − t ⎟ t2/3 ⎜ + t ⎟ Critical points: 0, π , 2π ,3π , 4π ,5π , 6π , 7π ,8π ⎝3 ⎠= ⎝ 3 3 ⎠ = H ( 0 ) = 1 ; H (π ) = −1 ; H ( 2π ) = 1 ; ( 2 + t )2 ( 2 + t )2 H ( 3π ) = −1 ; H ( 4π ) = 1 ; H ( 5π ) = −1 ; 2t 2 / 3 ( t + 5 ) = H ( 6π ) = 1 ; H ( 7π ) = −1 ; H ( 8π ) = 1 3( 2 + t ) 2 Maximum value: 1 h ' ( t ) is undefined when t = −2 and h ' ( t ) = 0 Minimum value: −1 when t = 0 or t = −5 . Since −5 is not in the 24. g ' ( x ) = 1 − 2 cos x interval of interest, it is not a critical point. Critical points: −1, 0,8 1 1 − 2 cos x = 0 → cos x = when h ( −1) = −1 ; h ( 0 ) = 0 ; h ( 8 ) = 16 2 5 5π π π 5π Maximum value: 16 ; Minimum value: −1 x=− ,− , , 5 3 3 3 3 5π π π 5π f ′( x) = 3 x 2 –12 x + 1;3x 2 –12 x + 1 = 0 Critical points: −2π , − , − , , , 2π 27. a. 3 3 3 3 33 33 ⎛ 5π ⎞ −5π when x = 2 – and x = 2 + . g ( −2π ) = −2π ; g ⎜ − ⎟= − 3; 3 3 ⎝ 3 ⎠ 3 33 33 ⎛ π⎞ π ⎛π ⎞ π Critical points: –1, 2 – ,2+ ,5 g⎜− ⎟ = − + 3 ; g⎜ ⎟ = − 3 ; 3 3 ⎝ 3⎠ 3 ⎝3⎠ 3 ⎛ 33 ⎞ ⎛ 5π ⎞ 5π f(–1) = –6, f ⎜ 2 – ⎟ ≈ 2.04, g⎜ ⎟= + 3 ; g ( 2π ) = 2π ⎜ 3 ⎟ ⎝ 3 ⎠ 3 ⎝ ⎠ 5π ⎛ 33 ⎞ Maximum value: + 3 f ⎜2+ ⎜ ⎟ ≈ –26.04, f(5) = –18 3 ⎝ 3 ⎟⎠ 5π Maximum value ≈ 2.04; Minimum value: − − 3 3 minimum value ≈ −26.04 156 Section 3.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• ( x3 – 6 x 2 + x + 2)(3x 2 – 12 x + 1) 29. Answers will vary. One possibility: b. g ′( x) = ; y 3 2 x – 6x + x + 2 5 33 g '( x) = 0 when x = 2 – and 3 33 x = 2+ . g ′( x) does not exist when 3 −5 5 x f(x) = 0; on [–1, 5], f(x) = 0 when x ≈ –0.4836 and x ≈ 0.7172 33 −5 Critical points: –1, –0.4836, 2 – , 3 30. Answers will vary. One possibility: 33 y 0.7172, 2 + , 5 3 g(–1) = 6, g(–0.4836) = 0, 5 ⎛ 33 ⎞ g⎜2 – ⎜ ⎟ ≈ 2.04, g(0.7172) = 0, ⎝ 3 ⎟⎠ ⎛ 33 ⎞ g⎜2+ ⎜ ⎟ ≈ 26.04, g(5) = 18 5 x ⎝ 3 ⎟⎠ Maximum value ≈ 26.04, minimum value = 0 −5 28. a. f ′( x) = x cos x; on [–1, 5], x cos x = 0 when 31. Answers will vary. One possibility: π 3π y x = 0, x = , x = 2 2 5 π 3π Critical points: –1, 0, , , 5 2 2 ⎛π⎞ f(–1) ≈3.38, f(0) = 3, f ⎜ ⎟ ≈ 3.57, ⎝2⎠ x 5 ⎛ 3π ⎞ f ⎜ ⎟ ≈ –2.71, f(5) ≈ −2.51 ⎝ 2 ⎠ Maximum value ≈ 3.57, −5 minimum value ≈–2.71 32. Answers will vary. One possibility: (cos x + x sin x + 2)( x cos x) y b. g ′( x) = ; cos x + x sin x + 2 5 π 3π g ′( x ) = 0 when x = 0, x = , x= 2 2 g ′( x) does not exist when f(x) = 0; on [–1, 5], f(x) = 0 when x ≈ 3.45 5 x π 3π Critical points: –1, 0, , 3.45, , 5 2 2 ⎛π⎞ −5 g(–1) ≈ 3.38, g(0) = 3, g ⎜ ⎟ ≈ 3.57, ⎝2⎠ ⎛ 3π ⎞ g(3.45) = 0, g ⎜ ⎟ ≈ 2.71, g(5) ≈ 2.51 ⎝ 2 ⎠ Maximum value ≈ 3.57; minimum value = 0 Instructor’s Resource Manual Section 3.1 157 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 33. Answers will vary. One possibility: 3.2 Concepts Review y 1. Increasing; concave up 5 2. f ′( x) > 0; f ′′( x) < 0 3. An inflection point 5 x 4. f ′′(c) = 0; f ′′(c) does not exist. −5 Problem Set 3.2 34. Answers will vary. One possibility: 1. f ′( x) = 3; 3 > 0 for all x. f(x) is increasing y for all x. 5 1 2. g ′( x) = 2 x – 1; 2x – 1 > 0 when x > . g(x) is 2 ⎡1 ⎞ increasing on ⎢ , ∞ ⎟ and decreasing on ⎣2 ⎠ 5 x ⎛ 1⎤ ⎜ – ∞, ⎥ . ⎝ 2⎦ −5 3. h′(t ) = 2t + 2; 2t + 2 > 0 when t > –1. h(t) is 35. Answers will vary. One possibility: increasing on [–1, ∞ ) and decreasing on y ( −∞ , –1]. 5 4. f ′( x) = 3x 2 ; 3 x 2 > 0 for x ≠ 0 . f(x) is increasing for all x. 5. G ′( x ) = 6 x 2 – 18 x + 12 = 6( x – 2)( x – 1) 5 x Split the x-axis into the intervals (– ∞ , 1), (1, 2), (2, ∞ ). 3 ⎛3⎞ 3 Test points: x = 0, , 3; G ′(0) = 12, G ′ ⎜ ⎟ = – , −5 2 ⎝2⎠ 2 G ′(3) = 12 36. Answers will vary. One possibility: G(x) is increasing on (– ∞ , 1] ∪ [2, ∞ ) and y decreasing on [1, 2]. 5 6. f ′(t ) = 3t 2 + 6t = 3t (t + 2) Split the x-axis into the intervals (– ∞ , –2), (–2, 0), (0, ∞ ). Test points: t = –3, –1, 1; f ′(–3) = 9, 5 x f ′(–1) = –3, f ′(1) = 9 f(t) is increasing on (– ∞ , –2] ∪ [0, ∞ ) and −5 decreasing on [–2, 0]. 7. h′( z ) = z 3 – 2 z 2 = z 2 ( z – 2) Split the x-axis into the intervals (– ∞ , 0), (0, 2), (2, ∞ ). Test points: z = –1, 1, 3; h′(–1) = –3, h′(1) = –1, h′(3) = 9 h(z) is increasing on [2, ∞ ) and decreasing on (– ∞ , 2]. 158 Section 3.2 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 8. f ′( x) = 2– x 16. f ′′( x) = 12 x 2 + 48 x = 12 x( x + 4); f ′′( x) > 0 when 3 x x < –4 and x > 0. Split the x-axis into the intervals (– ∞ , 0), (0, 2), f(x) is concave up on (– ∞ , –4) ∪ (0, ∞ ) and (2, ∞ ). concave down on (–4, 0); inflection points are Test points: –1, 1, 3; f ′(–1) = –3, f ′(1) = 1, (–4, –258) and (0, –2). 1 f ′(3) = – 17. F ′′( x) = 2sin 2 x – 2 cos 2 x + 4 = 6 – 4 cos 2 x; 27 f(x) is increasing on (0, 2] and decreasing on 6 – 4 cos 2 x > 0 for all x since 0 ≤ cos 2 x ≤ 1. (– ∞ , 0) ∪ [2, ∞ ). F(x) is concave up for all x; no inflection points. π 18. G ′′( x) = 48 + 24 cos 2 x – 24sin 2 x 9. H ′(t ) = cos t ; H ′(t ) > 0 when 0 ≤ t < and 2 3π = 24 + 48cos 2 x; 24 + 48cos 2 x > 0 for all x. < t ≤ 2π. G(x) is concave up for all x; no inflection points. 2 ⎡ π ⎤ ⎡ 3π ⎤ H(t) is increasing on ⎢ 0, ⎥ ∪ ⎢ , 2π ⎥ and 19. f ′( x) = 3 x 2 – 12; 3 x 2 – 12 > 0 when ⎣ 2⎦ ⎣ 2 ⎦ x < –2 or x > 2. ⎡ π 3π ⎤ f(x) is increasing on (– ∞ , –2] ∪ [2, ∞ ) and decreasing on ⎢ , ⎥ . ⎣2 2 ⎦ decreasing on [–2, 2]. f ′′( x) = 6 x; 6x > 0 when x > 0. f(x) is concave up π on (0, ∞ ) and concave down on (– ∞ , 0). 10. R ′(θ ) = –2 cos θ sin θ ; R ′(θ ) > 0 when <θ < π 2 3π and < θ < 2π. 2 ⎡ π ⎤ ⎡ 3π ⎤ R( θ ) is increasing on ⎢ , π ⎥ ∪ ⎢ , 2π ⎥ and ⎣2 ⎦ ⎣ 2 ⎦ ⎡ π ⎤ ⎡ 3π ⎤ decreasing on ⎢ 0, ⎥ ∪ ⎢ π, ⎥ . ⎣ 2⎦ ⎣ 2 ⎦ 11. f ′′( x) = 2; 2 > 0 for all x. f(x) is concave up for all x; no inflection points. 20. g ′( x) = 12 x 2 – 6 x – 6 = 6(2 x + 1)( x – 1); g ′( x) > 0 12. G ′′( w) = 2; 2 > 0 for all w. G(w) is concave up for 1 when x < – or x > 1. g(x) is increasing on all w; no inflection points. 2 ⎛ 1⎤ ⎡ 1 ⎤ 13. T ′′(t ) = 18t ; 18t > 0 when t > 0. T(t) is concave up ⎜ – ∞, – ⎥ ∪ [1, ∞) and decreasing on ⎢ – , 1⎥ . ⎝ 2⎦ ⎣ 2 ⎦ on (0, ∞ ) and concave down on (– ∞ , 0); g ′′( x) = 24 x – 6 = 6(4 x – 1); g ′′( x) > 0 when (0, 0) is the only inflection point. 1 x> . 6 2 4 14. f ′′( z ) = 2 – = ( z 4 – 3); z 4 – 3 > 0 for ⎛1 ⎞ 4 z z4 g(x) is concave up on ⎜ , ∞ ⎟ and concave down ⎝4 ⎠ z < – 4 3 and z > 4 3. ⎛ 1⎞ f(z) is concave up on (– ∞, – 4 3) ∪ ( 4 3, ∞) and on ⎜ – ∞, ⎟ . ⎝ 4⎠ concave down on (– 4 3, 0) ∪ (0, 4 3); inflection ⎛ 1 ⎞ ⎛4 1 ⎞ points are ⎜ – 4 3, 3 – ⎟ and ⎜ 3, 3 – ⎟. ⎝ 3⎠ ⎝ 3⎠ 15. q ′′( x ) = 12 x 2 – 36 x – 48; q ′′( x) > 0 when x < –1 and x > 4. q(x) is concave up on (– ∞ , –1) ∪ (4, ∞ ) and concave down on (–1, 4); inflection points are (–1, –19) and (4, –499). Instructor’s Resource Manual Section 3.2 159 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 21. g ′( x) = 12 x3 – 12 x 2 = 12 x 2 ( x – 1); g ′( x) > 0 23. G ′( x ) = 15 x 4 – 15 x 2 = 15 x 2 ( x 2 – 1); G ′( x) > 0 when x > 1. g(x) is increasing on [1, ∞ ) and when x < –1 or x > 1. G(x) is increasing on decreasing on (−∞,1]. (– ∞ , –1] ∪ [1, ∞ ) and decreasing on [–1, 1]. g ′′( x) = 36 x 2 – 24 x = 12 x(3 x – 2); g ′′( x) > 0 G ′′( x) = 60 x3 – 30 x = 30 x(2 x 2 – 1); 2 ⎛ 1 ⎞ when x < 0 or x > . g(x) is concave up on Split the x-axis into the intervals ⎜ −∞, − ⎟, 3 ⎝ 2⎠ ⎛2 ⎞ ⎛ 2⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ (– ∞, 0) ∪ ⎜ , ∞ ⎟ and concave down on ⎜ 0, ⎟ . ⎜− , 0 ⎟ , ⎜ 0, ⎟, ⎜ , ∞ ⎟. ⎝ 3 ⎠ ⎝ 3⎠ ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 2 ⎠ 1 1 Test points: x = –1, – , , 1; G ′′(–1) = –30, 2 2 ⎛ 1 ⎞ 15 ′′ ⎛ 1 ⎞ 15 G ′′ ⎜ – ⎟ = , G ⎜ ⎟ = – , G ′′(1) = 30. ⎝ 2⎠ 2 ⎝2⎠ 2 ⎛ 1 ⎞ ⎛ 1 ⎞ G(x) is concave up on ⎜ – , 0⎟ ∪ ⎜ , ∞ ⎟ and ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ concave down on ⎜ – ∞, – ⎟ ∪ ⎜ 0, ⎟. ⎝ 2⎠ ⎝ 2⎠ 22. F ′( x) = 6 x5 – 12 x3 = 6 x3 ( x 2 – 2) Split the x-axis into the intervals (– ∞ , − 2) , (− 2, 0), (0, 2), ( 2, ∞) . Test points: x = –2, –1, 1, 2; F ′(–2) = –96, F ′(–1) = 6, F ′(1) = –6, F ′(2) = 96 F(x) is increasing on [– 2, 0] ∪ [ 2, ∞) and decreasing on (– ∞, – 2] ∪ [0, 2] 2x 4 2 2 2 2 24. H ′( x ) = ; H ′( x) > 0 when x > 0. F ′′( x) = 30 x – 36 x = 6 x (5 x – 6); 5 x – 6 > 0 ( x + 1)2 2 6 6 H(x) is increasing on [0, ∞ ) and decreasing on when x < – or x > . (– ∞ , 0]. 5 5 ⎛ 2(1 – 3 x 2 ) 6⎞ ⎛ 6 ⎞ H ′′( x) = ; H ′′( x) > 0 when F(x) is concave up on ⎜ – ∞, – ⎜ ⎟∪⎜ , ∞ ⎟ and ( x 2 + 1)3 ⎝ 5⎟ ⎜ 5 ⎟ ⎠ ⎝ ⎠ 1 1 ⎛ 6 6⎞ – <x< . concave down on ⎜ – , ⎜ 5 5 ⎟. ⎟ 3 3 ⎝ ⎠ ⎛ 1 1 ⎞ H(x) is concave up on ⎜ – , ⎟ and concave ⎝ 3 3⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ down on ⎜ – ∞, – ⎟∪⎜ , ∞ ⎟. ⎝ 3⎠ ⎝ 3 ⎠ 160 Section 3.2 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• cos x π –2(5 x + 1) 25. f ′( x) = ; f ′( x) > 0 when 0 < x < . f(x) f ′′( x) = ; –2(5x + 1) > 0 when 2 sin x 2 9 x4 / 3 ⎡ π⎤ 1 is increasing on ⎢ 0, ⎥ and decreasing on x < – , f ′′( x) does not exist at x = 0. ⎣ 2⎦ 5 ⎡π ⎤ 1 Test points: –1, – , 1; f ′′(–1) = , 8 ⎢ 2 , π⎥ . 10 9 ⎣ ⎦ – cos 2 x – 2sin 2 x ⎛ 1⎞ 104 / 3 4 f ′′( x) = ; f ′′( x) < 0 for all x in f ′′ ⎜ – ⎟ = – , f (1) = – . 4sin 3 / 2 x ⎝ 10 ⎠ 9 3 (0, ∞ ). f(x) is concave down on (0, π ). ⎛ 1⎞ f(x) is concave up on ⎜ – ∞, – ⎟ and concave ⎝ 5⎠ ⎛ 1 ⎞ down on ⎜ – , 0 ⎟ ∪ (0, ∞). ⎝ 5 ⎠ 3x – 4 4 26. g ′( x) = ; 3x – 4 > 0 when x > . 2 x–2 3 g(x) is increasing on [2, ∞ ). 3x – 8 8 g ′′( x) = ; 3x – 8 > 0 when x > . 3/ 2 3 4( x – 2) 4( x + 2) 28. g ′( x) = ; x + 2 > 0 when x > –2, g ′( x) ⎛8 ⎞ 3x 2 / 3 g(x) is concave up on ⎜ , ∞ ⎟ and concave down ⎝3 ⎠ does not exist at x = 0. ⎛ 8⎞ Split the x-axis into the intervals ( −∞, −2 ) , on ⎜ 2, ⎟ . ⎝ 3⎠ (–2, 0), (0, ∞ ). 4 Test points: –3, –1, 1; g ′(–3) = – , 5/3 3 4 g ′(–1) = , g ′(1) = 4. 3 g(x) is increasing on [–2, ∞ ) and decreasing on (– ∞ , –2]. 4( x – 4) g ′′( x ) = ; x – 4 > 0 when x > 4, g ′′( x) 9 x5 / 3 does not exist at x = 0. 2 – 5x 2 20 27. f ′( x) = ; 2 – 5x > 0 when x < , f ′( x ) Test points: –1, 1, 5; g ′′(–1) = , 3x 1/ 3 5 9 does not exist at x = 0. 4 4 g ′′(1) = – , g ′′(5) = . Split the x-axis into the intervals ( − ∞, 0), 3 9(5)5 / 3 ⎛ 2⎞ ⎛2 ⎞ g(x) is concave up on (– ∞ , 0) ∪ (4, ∞ ) and ⎜ 0, ⎟ , ⎜ , ∞ ⎟ . concave down on (0, 4). ⎝ 5⎠ ⎝5 ⎠ 1 7 Test points: –1, , 1; f ′(−1) = – , 5 3 ⎛1⎞ 35 ′ f ′⎜ ⎟ = , f (1) = –1. ⎝5⎠ 3 ⎡ 2⎤ f(x) is increasing on ⎢ 0, ⎥ and decreasing on ⎣ 5⎦ ⎡2 ⎞ (– ∞, 0] ∪ ⎢ , ∞ ⎟ . ⎣5 ⎠ Instructor’s Resource Manual Section 3.2 161 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 29. 35. f ( x) = ax 2 + bx + c; f ′( x) = 2ax + b; f ′′( x) = 2a An inflection point would occur where f ′′( x) = 0 , or 2a = 0. This would only occur when a = 0, but if a = 0, the equation is not quadratic. Thus, quadratic functions have no points of inflection. 36. f ( x) = ax3 + bx 2 + cx + d ; 30. f ′( x) = 3ax 2 + 2bx + c; f ′′( x) = 6ax + 2b An inflection point occurs where f ′′( x) = 0 , or 6ax + 2b = 0. The function will have an inflection point at b x = – , a ≠ 0. 3a 31. 37. Suppose that there are points x1 and x2 in I where f ′( x1 ) > 0 and f ′( x2 ) < 0. Since f ′ is continuous on I, the Intermediate Value Theorem says that there is some number c between x1 and x2 such that f ′(c) = 0, which is a contradiction. Thus, either f ′( x) > 0 for all x in I and f is increasing throughout I or f ′( x) < 0 for all x in I and f is decreasing throughout I. 32. 38. Since x 2 + 1 = 0 has no real solutions, f ′( x ) exists and is continuous everywhere. x 2 – x + 1 = 0 has no real solutions. x 2 – x + 1 > 0 and x 2 + 1 > 0 for all x, so f ′( x) > 0 for all x. Thus f is increasing everywhere. 39. a. Let f ( x) = x 2 and let I = [ 0, a ] , a > y . 33. f ′( x) = 2 x > 0 on I. Therefore, f(x) is increasing on I, so f(x) < f(y) for x < y. b. Let f ( x) = x and let I = [ 0, a ] , a > y . 1 f ′( x) = > 0 on I. Therefore, f(x) is 2 x increasing on I, so f(x) < f(y) for x < y. 1 c. Let f ( x) = and let I = [0, a], a > y. 34. x 1 f ′( x) = − < 0 on I. Therefore f(x) is x2 decreasing on I, so f(x) > f(y) for x < y. 40. f ′( x) = 3ax 2 + 2bx + c In order for f(x) to always be increasing, a, b, and c must meet the condition 3ax 2 + 2bx + c > 0 for all x. More specifically, a > 0 and b 2 − 3ac < 0. 162 Section 3.2 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• f ′′( x) = − 2x ⎛ 2x ⎞ 2 ⎛ 2x ⎞ cos ⎜ ⎟ − sin ⎜ ⎟ d 3s d 2s e. c. < 0, >0 9 ⎝ 3 ⎠ 3 ⎝ 3 ⎠ dt 3 dt 2 s 47. f ′( x) > 0 on (–0.598, 0.680) f is increasing on [–0.598, 0.680]. 48. f ′′( x) < 0 when x > 1.63 in [–2, 3] f is concave down on (1.63, 3). t Concave up. ds 49. Let s be the distance traveled. Then is the d 2s dt d. = 10 mph/min speed of the car. dt 2 s ds a. = ks, k a constant dt s t Concave up. ds d 2s t e. and are approaching zero. Concave up. dt dt 2 d 2s s b. >0 dt 2 s t Concave down. t Concave up. 164 Section 3.2 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• ds is constant. dV dh d 2h f. c. = k, > 0, <0 dt dt dt dt 2 s Concave down. h ( t) t t Neither concave up nor down. dI d 2 I 50. a. dV = k <0, V is the volume of water in the d. I ( t) = k now, but , > 0 in the future dt dt dt 2 tank, k is a constant. where I is inflation. I ( t) Neither concave up nor down. v(t) t t dp d2 p dp e. < 0, but > 0 and at t = 2: >0. dV 1 1 dt dt 2 dt b. = 3 – = 2 gal/min dt 2 2 where p is the price of oil. Neither concave up nor down. Concave up. v(t) P ( t) t 2 t Instructor’s Resource Manual Section 3.2 165 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• dT d 2T dP d 2P f. > 0, < 0 , where T is David’s c. > 0, < 0 , where P is world dt dt 2 dt dt 2 temperature. population. Concave down. Concave down. T ( t) P ( t) t t dC d 2C dθ d 2θ 51. a. > 0, > 0 , where C is the car’s cost. d. > 0, > 0 , where θ is the angle that dt dt 2 dt dt 2 Concave up. the tower makes with the vertical. C ( t) Concave up. θ( t) t t b. f(t) is oil consumption at time t. df d2 f e. P = f(t) is profit at time t. < 0, >0 dt dt 2 dP d 2P > 0, <0 Concave up. dt dt 2 f( t) Concave down. P ( t) t t 166 Section 3.2 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• f. R is revenue at time t. dP P < 0, >0 dt Could be either concave up or down. P 54. The height is always increasing so h '(t ) > 0 . The rate of change of the height decreases for the first 50 minutes and then increases over the next 50 minutes. Thus h ''(t ) < 0 for 0 ≤ t ≤ 50 and t h ''(t ) > 0 for 50 < t ≤ 100 . P t 55. V = 3t , 0 ≤ t ≤ 8 . The height is always increasing, so h '(t ) > 0. The rate of change of the height decreases from time t = 0 until time t1 when the water reaches the middle of the rounded bottom 52. a. R(t) ≈ 0.28, t < 1981 part. The rate of change then increases until time dR d 2R t2 when the water reaches the middle of the neck. b. On [1981, 1983], > 0, >0, dt dt 2 Then the rate of change decreases until t = 8 and R(1983) ≈ 0.36 the vase is full. Thus, h ''(t ) > 0 for t1 < t < t2 and h ''(t ) < 0 for t2 < t < 8 . dV 53. = 2 in 3 / sec h ( t) dt The cup is a portion of a cone with the bottom cut off. If we let x represent the height of the missing 24 cone, we can use similar triangles to show that x x+5 = 3 3.5 3.5 x = 3 x + 15 0.5 x = 15 x = 30 Similar triangles can be used again to show that, at t2 t1 8 t any given time, the radius of the cone at water level is h + 30 r= 20 Therefore, the volume of water can be expressed as π (h + 30) 3 45π V= − . 1200 2 We also know that V = 2t from above. Setting the two volume equations equal to each other and 2400 solving for h gives h = 3 t + 27000 − 30 . π Instructor’s Resource Manual Section 3.2 167 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 56. V = 20 − .1t , 0 ≤ t ≤ 200 . The height of the water 3.3 Concepts Review is always decreasing so h '(t ) < 0 . The rate of change in the height increases (the rate is negative, 1. maximum and its absolute value decreases) for the first 100 days and then decreases for the remaining time. 2. maximum; minimum Therefore we have h ''(t ) > 0 for 0 < t < 100 , and h ''(t ) < 0 for 100 < t < 200 . 3. maximum 4. local maximum, local minimum, 0 Problem Set 3.3 1. f ′( x) = 3 x 2 –12 x = 3 x( x – 4) Critical points: 0, 4 57. a. The cross-sectional area of the vase is f ′( x) > 0 on (– ∞ , 0), f ′( x) < 0 on (0, 4), approximately equal to ΔV and the f ′( x) > 0 on (4, ∞ ) corresponding radius is r = ΔV / π . The table below gives the approximate values for r. f ′′( x) = 6 x –12; f ′′(0) = –12, f ′′(4) = 12. The vase becomes slightly narrower as you Local minimum at x = 4; move above the base, and then gets wider as local maximum at x = 0 you near the top. 2. f ′( x) = 3 x 2 –12 = 3( x 2 – 4) Depth V A ≈ ΔV r = ΔV / π Critical points: –2, 2 f ′( x) > 0 on (– ∞ , –2), f ′( x) < 0 on (–2, 2), 1 4 4 1.13 f ′( x) > 0 on (2, ∞ ) 2 8 4 1.13 f ′′( x) = 6 x; f ′′(–2) = –12, f ′′(2) = 12 Local minimum at x = 2; 3 11 3 0.98 local maximum at x = –2 4 14 3 0.98 ⎛ π⎞ 3. f ′(θ ) = 2 cos 2θ ; 2 cos 2θ ≠ 0 on ⎜ 0, ⎟ 5 20 6 1.38 ⎝ 4⎠ No critical points; no local maxima or minima on 6 28 8 1.60 ⎛ π⎞ ⎜ 0, ⎟ . ⎝ 4⎠ b. Near the base, this vase is like the one in part (a), but just above the base it becomes larger. 1 1 1 Near the middle of the vase it becomes very 4. f ′( x) = + cos x; + cos x = 0 when cos x = – . 2 2 2 narrow. The top of the vase is similar to the 2π 4π one in part (a). Critical points: , 3 3 Depth V A ≈ ΔV r = ΔV / π ⎛ 2π ⎞ ⎛ 2π 4π ⎞ f ′( x) > 0 on ⎜ 0, ⎟ , f ′( x) < 0 on ⎜ , ⎟, ⎝ 3 ⎠ ⎝ 3 3 ⎠ 1 4 4 1.13 ⎛ 4π ⎞ f ′( x) > 0 on ⎜ , 2π ⎟ 2 9 5 1.26 ⎝ 3 ⎠ ⎛ 2π ⎞ 3 ⎛ 4π ⎞ 3 3 12 3 0.98 f ′′( x) = – sin x; f ′′ ⎜ ⎟ = – , f ′′ ⎜ ⎟ = ⎝ 3 ⎠ 2 ⎝ 3 ⎠ 2 4 14 2 0.80 4π Local minimum at x = ; local maximum at 3 5 20 6 1.38 2π x= . 3 6 28 8 1.60 168 Section 3.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 5. Ψ ′(θ ) = 2sin θ cosθ 1 9. h ' ( y ) = 2 y + π π y2 − <θ < 2 2 3 4 Critical point: 0 Critical point: − 2 ⎛ π ⎞ ⎛ π⎞ ⎛ Ψ ′(θ ) < 0 on ⎜ − , 0 ⎟ , Ψ ′(θ ) > 0 on ⎜ 0, ⎟ , 3 4⎞ ⎝ 2 ⎠ ⎝ 2⎠ h ' ( y ) < 0 on ⎜ −∞, − ⎟ ⎝ 2 ⎠ Ψ ′′(θ ) = 2 cos 2 θ – 2sin 2 θ ; Ψ ′′(0) = 2 ⎛ 34 ⎞ Local minimum at x = 0 h ' ( y ) > 0 on ⎜ − , 0 ⎟ and ( 0, ∞ ) ⎝ 2 ⎠ 6. r ′( z ) = 4 z 3 2 h '' ( y ) = 2 − 3 Critical point: 0 y r ′( z ) < 0 on (−∞, 0); ⎛ 34⎞ 2 16 r ′( z ) > 0 on (0, ∞) h⎜ − ⎟ = 2− 3 = 2+ =6 ( ) 3 ⎝ 2 ⎠ − 24 4 r ′′( x ) = 12 x 2 ; r ′′(0) = 0; the Second Derivative 3 Test fails. 4 Local minima at − Local minimum at z = 0; no local maxima 2 f '( x) = ( x 2 + 4 ) ⋅1 − x ( 2 x ) = 4 − x 2 (x 2 ) + 1 ( 3) − ( 3 x + 1)( 2 x ) 3 − 2 x − 3x 2 7. f '( x) = = ( x2 + 4) ( x2 + 4) 2 2 10. (x ) (x ) 2 2 2 2 +1 +1 Critical points: −2, 2 The only critical points are stationary points. Find f ' ( x ) < 0 on ( −∞, −2 ) and ( 2, ∞ ) ; these by setting the numerator equal to 0 and f ' ( x ) > 0 on ( −2, 2 ) solving. 3 − 2 x − 3x2 = 0 f '' ( x ) = ( 2 x x 2 − 12 ) a = −3, b = −2, c = 3 ( x2 + 4) 3 2± ( −2 )2 − 4 ( −3)( 3) 2 ± 40 −1 ± 10 x= = = 1 f '' ( −2 ) = ; f '' ( 2 ) = − 1 2 ( −3) −6 3 16 16 −1 − 10 −1 + 10 Local minima at x = −2 ; Local maxima at x = 2 Critical points: and 3 3 g '( z ) = (1 + z 2 ) ( 2 z ) − z 2 ( 2 z ) = 2 z ⎛ f ' ( x ) < 0 on ⎜ −∞, −1 − 10 ⎞ ⎟ and 8. ⎜ ⎟ (1 + z 2 ) (1 + z 2 ) 3 2 2 ⎝ ⎠ ⎛ −1 + 10 ⎞ Critical point: z = 0 ⎜ ,∞⎟ . ⎜ 3 ⎟ g ' ( z ) < 0 on ( −∞, 0 ) ⎝ ⎠ g ' ( z ) > 0 on ( 0, ∞ ) ⎛ −1 − 10 −1 + 10 ⎞ f ' ( 0 ) > 0 on ⎜ ⎜ , ⎟ ⎟ 3 3 ( −2 3 z 2 − 1 ) ⎝ ⎠ g '' ( z ) = ( 2 3x3 + 3x 2 − 9 x − 1 ) ( ) 3 2 z +1 f '' ( x ) = (x ) 2 3 +1 g '' ( 0 ) = 2 Local minima at z = 0 . ⎛ −1 − 10 ⎞ f '' ⎜ ⎜ ⎟ ≈ 0.739 ⎟ ⎝ 3 ⎠ ⎛ −1 + 10 ⎞ f '' ⎜ ⎜ ⎟ ≈ −2.739 ⎟ ⎝ 3 ⎠ −1 − 10 Local minima at x = ; 3 −1 + 10 Local maxima at x = 3 Instructor’s Resource Manual Section 3.3 169 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 11. f ′( x) = 3 x 2 – 3 = 3( x 2 – 1) 6 ⎛ ⎛ 2 ⎞5 / 3 ⎞ 6 ⎛ 15 ⎞ 8/3 r ′′( s ) = – ; r ′′⎜ – ⎜ ⎟ ⎟ = – ⎜ ⎟ Critical points: –1, 1 25s8 / 5 ⎜ ⎝ 15 ⎠ ⎟ 25 ⎝ 2 ⎠ ⎝ ⎠ f ′′( x) = 6 x; f ′′(–1) = –6, f ′′(1) = 6 ⎛ ⎛ 2 ⎞5 3 ⎞ Local minimum value f(1) = –2; r ′( s ) < 0 on ⎜ − ⎜ ⎟ , 0 ⎟ , r ′( s ) > 0 on (0, ∞) local maximum value f(–1) = 2 ⎜ ⎝ 15 ⎠ ⎟ ⎝ ⎠ 12. g ′( x) = 4 x3 + 2 x = 2 x(2 x 2 + 1) Local minimum value r(0) = 0; local maximum Critical point: 0 value g ′′( x) = 12 x 2 + 2; g ′′(0) = 2 ⎛ ⎛ 2 ⎞5 / 3 ⎞ ⎛ 2⎞ 5/3 ⎛ 2⎞ 2/3 3⎛ 2 ⎞ 2/3 r ⎜ – ⎜ ⎟ ⎟ = –3 ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ Local minimum value g(0) = 3; no local maximum ⎜ ⎝ 15 ⎠ ⎟ ⎝ 15 ⎠ ⎝ 15 ⎠ 5 ⎝ 15 ⎠ values ⎝ ⎠ 1 13. H ′( x ) = 4 x3 – 6 x 2 = 2 x 2 (2 x – 3) 17. f ′(t ) = 1 + t2 3 Critical points: 0, No critical points 2 No local minimum or maximum values H ′′( x) = 12 x 2 – 12 x = 12 x( x – 1); H ′′(0) = 0, ⎛3⎞ x( x 2 + 8) H ′′ ⎜ ⎟ = 9 18. f ′( x) = ⎝2⎠ ( x 2 + 4)3 / 2 ⎛ 3⎞ Critical point: 0 H ′( x) < 0 on (−∞, 0), H ′( x) < 0 on ⎜ 0, ⎟ f ′( x) < 0 on (−∞, 0), f ′( x) > 0 on (0, ∞) ⎝ 2⎠ Local minimum value f(0) = 0, no local maximum ⎛3⎞ 27 Local minimum value H ⎜ ⎟ = – ; no local values ⎝ 2⎠ 16 maximum values (x = 0 is neither a local 1 minimum nor maximum) 19. Λ ′(θ ) = – ; Λ ′(θ ) does not exist at 1 + sin θ 3π 14. f ′( x) = 5( x – 2) 4 θ= , but Λ (θ ) does not exist at that point 2 Critical point: 2 either. f ′′( x) = 20( x – 2)3 ; f ′′(2) = 0 No critical points f ′( x) > 0 on (−∞, 2), f ′( x) > 0 on (2, ∞) No local minimum or maximum values No local minimum or maximum values sin θ cos θ π 3π 20. g ′(θ ) = ; g ′(θ ) = 0 when θ = , ; 2 sin θ 2 2 15. g ′(t ) = – ; g ′(t ) does not exist at t = 2. 3(t – 2)1/ 3 g ′(θ ) does not exist at x = π . Critical point: 2 ⎛ π⎞ Split the x -axis into the intervals ⎜ 0, ⎟ , 2 2 ⎝ 2⎠ g ′(1) = , g ′(3) = – 3 3 ⎛ π ⎞ ⎛ 3π ⎞ ⎛ 3π ⎞ No local minimum values; local maximum value ⎜ , π ⎟ , ⎜ π, ⎟ , ⎜ , 2π ⎟ . ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ g(2) = π . π 3π 5π 7 π ⎛π⎞ 1 Test points: , , , ; g ′ ⎜ ⎟ = , 4 4 4 4 ⎝4⎠ 2 2 15s3 / 5 + 2 16. r ′( s ) = 3 + = ; r ′( s ) = 0 when ⎛ 3π ⎞ 1 ⎛ 5π ⎞ 1 ⎛ 7π ⎞ 1 5s 3 / 5 5s 3 / 5 g′⎜ ⎟ = – , g′⎜ ⎟ = , g′⎜ ⎟ = – 5/3 ⎝ 4 ⎠ 2 ⎝ 4 ⎠ 2 ⎝ 4 ⎠ 2 ⎛ 2⎞ Local minimum value g( π ) = 0; local maximum s = –⎜ ⎟ , r ′( s ) does not exist at s = 0. ⎝ 15 ⎠ ⎛π⎞ ⎛ 3π ⎞ 5/3 values g ⎜ ⎟ = 1 and g ⎜ ⎟ = 1 ⎛ 2⎞ ⎝2⎠ ⎝ 2 ⎠ Critical points: – ⎜ ⎟ ,0 ⎝ 15 ⎠ 170 Section 3.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 21. f ' ( x ) = 4 ( sin 2 x )( cos 2 x ) 3 3 9 25. F ′( x) = – 4; – 4 = 0 when x = ( 2k − 1) π x x 16 4 ( sin 2 x )( cos 2 x ) = 0 when x = or 9 4 Critical points: 0, , 4 kπ 16 x= where k is an integer. 2 ⎛9⎞ 9 F(0) = 0, F ⎜ ⎟ = , F(4) = –4 Critical points: 0, π , π , 2 4 2 ⎝ 16 ⎠ 4 Minimum value F(4) = –4; maximum value ⎛π ⎞ ⎛π ⎞ f (0) = 0 ; f ⎜ ⎟ = 1 ; f ⎜ ⎟ = 0 ; ⎛9⎞ 9 ⎝4⎠ ⎝2⎠ F⎜ ⎟= ⎝ 16 ⎠ 4 f ( 2 ) ≈ 0.5728 ⎛π ⎞ 9 Minimum value: f ( 0 ) = f ⎜ ⎟ = 0 26. From Problem 25, the critical points are 0 and . ⎝2⎠ 16 ⎛ π⎞ ⎛ 9⎞ ⎛9 ⎞ Maximum value: f ⎜ ⎟ = 1 F ′( x) > 0 on ⎜ 0, ⎟ , F ′( x) < 0 on ⎜ , ∞ ⎟ ⎝4⎠ ⎝ 16 ⎠ ⎝ 16 ⎠ ⎛ 9 ⎞ F decreases without bound on ⎜ , ∞ ⎟ . No 22. f '( x) = ( −2 x 2 − 4 ) ⎝ 16 ⎠ (x ) ⎛9⎞ 9 2 2 +4 minimum values; maximum value F ⎜ ⎟ = ⎝ 16 ⎠ 4 f ' ( x ) = 0 when x = 2 or x = −2 . (there are no singular points) 27. f ′( x) = 64(−1)(sin x)−2 cos x Critical points: 0, 2 (note: −2 is not in the given domain) +27(−1)(cos x)−2 (− sin x) 1 64 cos x 27 sin x f ( 0 ) = 0 ; f ( 2 ) = ; f ( x ) → 0 as x → ∞ . =− + 2 2 sin x cos 2 x Minimum value: f ( 0 ) = 0 2 2 (3sin x − 4 cos x)(9 sin x + 12 cos x sin x + 16 cos x ) = 1 2 2 Maximum value: f ( 2 ) = sin x cos x 2 ⎛ π⎞ On ⎜ 0, ⎟ , f ′( x) = 0 only where 3sin x = 4cos x; ⎝ 2⎠ 23. g ' ( x ) = ( − x x3 − 64 ) 4 tan x = ; ( x3 + 32 ) 2 3 g ' ( x ) = 0 when x = 0 or x = 4 . 4 x = tan −1 ≈ 0.9273 Critical points: 0, 4 3 Critical point: 0.9273 1 g ( 0) = 0 ; g ( 4) = For 0 < x < 0.9273, f ′( x) < 0, while for 6 As x approaches ∞ , the value of g approaches 0 π 0.9273 < x < , f '( x) > 0 but never actually gets there. 2 1 ⎛ 4 ⎞ 64 27 Maximum value: g ( 4 ) = Minimum value f ⎜ tan −1 ⎟ = + = 125; 6 ⎝ 3⎠ 4 3 5 5 Minimum value: g ( 0 ) = 0 no maximum value −2 x 24. h ' ( x ) = ( x2 + 4) 2 h ' ( x ) = 0 when x = 0 . (there are no singular points) Critical points: 0 Since h ' ( x ) < 0 for x > 0 , the function is always decreasing. Thus, there is no minimum value. 1 Maximum value: h ( 0 ) = 4 Instructor’s Resource Manual Section 3.3 171 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• (8 − x)2 (32 x) − (16 x 2 )2(8 − x)(−1) 32. f '( x) = 0 at x = 1, 2,3, 4 ; f '( x) is negative on 28. g ′( x) = 2 x + (8 − x) 4 (−∞,1) ∪ (2,3) ∪ (4, ∞) and positive on (1, 2) ∪ (3, 4). Thus, the function has a local 256 x 2 x[(8 − x)3 + 128] = 2x + = minimum at x = 1,3 and a local maximum at (8 − x )3 (8 − x)3 x = 2, 4 . For x > 8, g ′( x) = 0 when (8 − x)3 + 128 = 0; 33. f '( x) = 0 at x = 1, 2,3, 4 ; f '( x) is negative on (8 − x)3 = −128; 8 − x = − 3 128 ; (3, 4) and positive on x = 8 + 4 3 2 ≈ 13.04 (−∞,1) ∪ (1, 2) ∪ (2,3) ∪ (4, ∞) Thus, the function g ′( x) < 0 on (8, 8 + 4 3 2), has a local minimum at x = 4 and a local g ′( x) > 0 on (8 + 4 3 2, ∞) maximum at x = 3 . g(13.04) ≈ 277 is the minimum value 34. Since f ' ( x ) ≥ 0 for all x, the function is always 29. H ' ( x ) = ( 2 x x2 − 1 ) increasing. Therefore, there are no local extrema. x2 − 1 35. Since f ' ( x ) ≥ 0 for all x, the function is always H ' ( x ) = 0 when x = 0 . increasing. Therefore, there are no local extrema. H ' ( x ) is undefined when x = −1 or x = 1 36. f ' ( x ) = 0 at x = 0, A, and B . Critical points: −2 , −1 , 0, 1, 2 f ' ( x ) is negative on ( −∞, 0 ) and ( A, B ) H ( −2 ) = 3 ; H ( −1) = 0 ; H ( 0 ) = 1 ; H (1) = 0 ; f ' ( x ) is positive on ( 0, A ) and ( B, ∞ ) H ( 2) = 3 Therefore, the function has a local minimum at Minimum value: H ( −1) = H (1) = 0 x = 0 and x = B , and a local maximum at x = A . Maximum value: H ( −2 ) = H ( 2 ) = 3 37. Answers will vary. One possibility: y 30. h ' ( t ) = 2t cos t 2 5 2π 6π h ' ( t ) = 0 when t = 0 , t = , t= , and 2 2 10π t= 2 3 6 x 2 π 3π 2 5π (Consider t = , t = , and t 2 = ) 2 2 2 2π 6π 10π −5 Critical points: 0, , , ,π 2 2 2 ⎛ 2π ⎞ ⎛ 6π ⎞ 38. Answers will vary. One possibility: h ( 0) = 0 ; h ⎜ ⎟ = 1; h⎜ ⎟ = −1 ; y ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 10π ⎞ 5 h⎜ ⎟ = 1 ; h (π ) ≈ −0.4303 ⎝ 2 ⎠ ⎛ 6π ⎞ Minimum value: h ⎜ ⎟ = −1 ⎝ 2 ⎠ 3 6 x ⎛ 2π ⎞ ⎛ 10π ⎞ Maximum value: h ⎜ ⎟ = h⎜ ⎟ =1 ⎝ 2 ⎠ ⎝ 2 ⎠ −5 31. f '( x) = 0 when x = 0 and x = 1 . On the interval (−∞, 0) we get f '( x) < 0 . On (0, ∞) , we get f '( x) > 0 . Thus there is a local min at x = 0 but no local max. 172 Section 3.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• ⎛ − B − B 2 − 3 AC ⎞ Problem Set 3.4 f '' ⎜ ⎟ ⎜ 3A ⎟ 1. Let x be one number, y be the other, and Q be the ⎝ ⎠ sum of the squares. ⎛ − B − B 2 − 3 AC ⎞ xy = –16 = 6A⎜ ⎟ + 2B 16 ⎜ 3A ⎟ y=– ⎝ ⎠ x = −2 B − 2 B 2 − 3 AC + 2 B The possible values for x are in (– ∞ , 0) or (0, ∞) . = −2 B 2 − 3 AC 256 Q = x2 + y 2 = x2 + and x2 ⎛ − B + B 2 − 3 AC ⎞ dQ 512 f '' ⎜ ⎟ = 2x – ⎜ 3A ⎟ dx x3 ⎝ ⎠ 512 ⎛ − B + B 2 − 3 AC ⎞ 2x – =0 = 6A⎜ ⎟ + 2B x3 ⎜ 3A ⎟ ⎝ ⎠ x 4 = 256 = −2 B + 2 B 2 − 3 AC + 2 B x = ±4 The critical points are –4, 4. = 2 B 2 − 3 AC dQ dQ < 0 on (– ∞ , –4) and (0, 4). > 0 on If B 2 − 3 AC > 0 , then −2 B 2 − 3 AC exists and dx dx (–4, 0) and (4, ∞ ). is negative, and 2 B 2 − 3 AC exists and is When x = –4, y = 4 and when x = 4, y = –4. The positive. two numbers are –4 and 4. Thus, from the Second Derivative Test, 2. Let x be the number. − B − B 2 − 3 AC would yield a local maximum Q = x – 8x 3A x will be in the interval (0, ∞ ). − B + B 2 − 3 AC dQ 1 –1/ 2 and would yield a local = x –8 3A dx 2 minimum. 1 –1/ 2 x –8 = 0 45. f ′′′(c) > 0 implies that f ′′ is increasing at c, so f 2 is concave up to the right of c (since f ′′( x) > 0 to x –1/ 2 = 16 the right of c) and concave down to the left of c 1 x= (since f ′′( x) < 0 to the left of c). Therefore f has a 256 point of inflection at c. dQ ⎛ 1 ⎞ dQ ⎛ 1 ⎞ > 0 on ⎜ 0, ⎟ and < 0 on ⎜ , ∞ ⎟. dx ⎝ 256 ⎠ dx ⎝ 256 ⎠ 1 3.4 Concepts Review Q attains its maximum value at x = . 256 1. 0 < x < ∞ 3. Let x be the number. Q = 4 x – 2x 200 2. 2x + x will be in the interval (0, ∞ ). x dQ 1 –3 / 4 = x –2 n dx 4 3. S = ∑ ( yi − bxi ) 2 1 –3 / 4 i =1 x –2=0 4 x –3 / 4 = 8 4. marginal revenue; marginal cost 1 x= 16 dQ ⎛ 1⎞ dQ ⎛1 ⎞ > 0 on ⎜ 0, ⎟ and < 0 on ⎜ , ∞ ⎟ dx ⎝ 16 ⎠ dx ⎝ 16 ⎠ 1 Q attains its maximum value at x = . 16 174 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 4. Let x be one number, y be the other, and Q be the 6. Let Q be the square of the distance between (x, y) sum of the squares. and (10, 0). xy = –12 Q = ( x – 10)2 + ( y – 0) 2 = (2 y 2 – 10) 2 + y 2 12 y=– = 4 y 4 – 39 y 2 + 100 x The possible values for x are in (– ∞ , 0) or (0, ∞) . dQ = 16 y 3 – 78 y 144 dy Q = x2 + y2 = x2 + x2 16 y 3 – 78 y = 0 dQ 288 2 y (8 y 2 – 39) = 0 = 2x – dx x3 39 288 y = 0, ± 2x – =0 2 2 x3 dQ ⎛ 39 ⎞ ⎛ 39 ⎞ x 4 = 144 < 0 on ⎜ – ∞, – ⎜ ⎟ and ⎜ 0, ⎟ ⎜ 2 2 ⎟. ⎟ dy ⎝ 2 2⎠ ⎝ ⎠ x = ±2 3 dQ ⎛ 39 ⎞ ⎛ 39 ⎞ The critical points are –2 3, 2 3 > 0 on ⎜ – ⎜ 2 2 , 0 ⎟ and ⎜ 2 2 , ∞ ⎟ . ⎟ ⎜ ⎟ dy ⎝ ⎠ ⎝ ⎠ dQ < 0 on (– ∞, – 2 3) and (0, 2 3). dx 39 39 When y = – ,x= and when dQ 2 2 4 > 0 on (–2 3, 0) and (2 3, ∞). dx 39 39 y= ,x= . When x = –2 3, y = 2 3 and when 2 2 4 x = 2 3, y = –2 3. ⎛ 39 39 ⎞ ⎛ 39 39 ⎞ The points are ⎜ , – ⎜ 4 ⎟ and ⎜ , ⎟ ⎜ 4 2 2 ⎟. ⎟ The two numbers are –2 3 and 2 3. ⎝ 2 2⎠ ⎝ ⎠ 5. Let Q be the square of the distance between (x, y) 7. x ≥ x 2 if 0 ≤ x ≤ 1 and (0, 5). f ( x) = x − x 2 ; f ′( x ) = 1 − 2 x; Q = ( x – 0)2 + ( y – 5)2 = x 2 + ( x 2 – 5)2 1 = x 4 – 9 x 2 + 25 f ′( x) = 0 when x = 2 dQ = 4 x3 – 18 x 1 Critical points: 0, , 1 dx 2 4 x3 – 18 x = 0 ⎛1⎞ 1 1 f(0) = 0, f(1) = 0, f ⎜ ⎟ = ; therefore, 2 x(2 x 2 – 9) = 0 ⎝ 2⎠ 4 2 3 exceeds its square by the maximum amount. x = 0, ± 2 8. For a rectangle with perimeter K and width x, the dQ ⎛ 3 ⎞ ⎛ 3 ⎞ K < 0 on ⎜ – ∞, – ⎟ and⎜ 0, ⎟. length is − x . Then the area is dx ⎝ 2⎠ ⎝ 2⎠ 2 ⎛ 3 ⎞ ⎛3 ⎞ ⎛K ⎞ Kx dQ > 0 on ⎜ – , 0 ⎟ and ⎜ , ∞ ⎟. A = x⎜ − x⎟ = − x2 . dx ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝2 ⎠ 2 3 9 3 dA K dA K When x = – , y = and when x = , = − 2 x; = 0 when x = 2 2 2 dx 2 dx 4 9 K K y= . Critical points: 0, , 2 4 2 ⎛ 3 9⎞ ⎛ 3 9⎞ K K K2 The points are ⎜ – , ⎟ and ⎜ , ⎟. At x = 0 or , A = 0; at x = , A = . 2 2⎠ 2 4 16 ⎝ ⎝ 2 2⎠ The area is maximized when the width is one fourth of the perimeter, so the rectangle is a square. Instructor’s Resource Manual Section 3.4 175 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 9. Let x be the width of the square to be cut out and V dQ the volume of the resulting open box. < 0 on (0, 15 3) and dx V = x (24 − 2 x)2 = 4 x3 − 96 x 2 + 576 x dQ > 0 on (15 3, ∞). dV dx = 12 x 2 − 192 x + 576 = 12( x − 12)( x − 4); dx 900 When x = 15 3, y = = 20 3. 12(x – 12)(x – 4) = 0; x = 12 or x = 4. 15 3 Critical points: 0, 4, 12 At x = 0 or 12, V = 0; at x = 4, V = 1024. Q has a minimum when x = 15 3 ≈ 25.98 ft and The volume of the largest box is 1024 in. 3 y = 20 3 ≈ 34.64 ft. 10. Let A be the area of the pen. 300 14. xy = 300; y = dA x A = x(80 − 2 x) = 80 x − 2 x 2 ; = 80 − 4 x; dx The possible values for x are in (0, ∞ ). 80 − 4 x = 0; x = 20 1200 Q = 6x + 4 y = 6x + Critical points: 0, 20, 40. x At x = 0 or 40, A = 0; at x = 20, A = 800. dQ 1200 The dimensions are 20 ft by 80 – 2(20) = 40 ft, =6– with the length along the barn being 40 ft. dx x2 1200 6– =0 11. Let x be the width of each pen, then the length x2 along the barn is 80 – 4x. x 2 = 200 dA A = x(80 − 4 x) = 80 x − 4 x 2 ; = 80 − 8 x; x = ±10 2 dx dA x = 10 2 is the only critical point in (0, ∞ ). = 0 when x = 10. dQ dQ dx < 0 on (0, 10 2) and > 0 on (10 2, ∞) Critical points: 0, 10, 20 dx dx At x = 0 or 20, A = 0; at x = 10, A = 400. 300 When x = 10 2, y = = 15 2. The area is largest with width 10 ft 10 2 and length 40 ft. Q has a minimum when x = 10 2 ≈ 14.14 ft and 12. Let A be the area of the pen. The perimeter is y = 15 2 ≈ 21.21 ft. 100 + 180 = 280 ft. y + y – 100 + 2x = 180; y = 140 – x 300 dA 15. xy = 300; y = A = x(140 − x ) = 140 x − x 2 ; = 140 − 2 x; x dx The possible values for x are in (0, ∞). 140 − 2 x = 0; x = 70 3000 Since 0 ≤ x ≤ 40 , the critical points are 0 and 40. Q = 3(6x + 2y) + 2(2y) = 18x + 10y = 18x + x When x = 0, A = 0. When x = 40, A = 4000. The dQ 3000 dimensions are 40 ft by 100 ft. = 18 – dx x2 900 3000 13. xy = 900; y = 18 – =0 x x2 The possible values for x are in (0, ∞ ). 500 ⎛ 900 ⎞ 2700 x2 = Q = 4x + 3 y = 4x + 3⎜ 3 ⎟ = 4x + ⎝ x ⎠ x 10 5 dQ 2700 x=± = 4− 3 dx x2 10 5 2700 x= is the only critical point in (0, ∞). 4– =0 3 x2 dQ ⎛ 10 5 ⎞ x 2 = 675 < 0 on ⎜ 0, ⎜ ⎟ and dx ⎝ 3 ⎟ ⎠ x = ±15 3 dQ ⎛ 10 5 ⎞ x = 15 3 is the only critical point in (0, ∞ ). > 0 on ⎜ dx ⎜ 3 , ∞ ⎟. ⎟ ⎝ ⎠ 176 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 10 5 300 2A A 3A When x = ,y= = 6 15 When x = ,y= = 3 10 5 3 2A 2 3 3 10 5 2A Q has a minimum when x = ≈ 12.91 ft and x 3 2 = = 3 y 3A 3 2 y = 6 15 ≈ 23.24 ft. 17. Let D be the square of the distance. 900 2 16. xy = 900; y = ⎛ x2 ⎞ 2 2 2 x D = ( x − 0) + ( y − 4) = x + ⎜ − 4⎟ The possible values for x are in (0, ∞ ). ⎜ 4 ⎟ ⎝ ⎠ 3600 x4 Q = 6x + 4 y = 6x + = − x 2 + 16 x 16 dQ 3600 =6– dD x3 x3 dx x2 = − 2 x; − 2 x = 0; x( x 2 − 8) = 0 dx 4 4 3600 6– =0 x = 0, x = ± 2 2 x2 Critical points: 0, 2 2, 2 3 x 2 = 600 Since D is continuous and we are considering a x = ±10 6 closed interval for x, there is a maximum and x = 10 6 is the only critical point in (0, ∞ ). minimum value of D on the interval. These dQ dQ extrema must occur at one of the critical points. < 0 on (0, 10 6) and > 0 on (10 6, ∞). At x = 0, y = 0, and D = 16. At x = 2 2, y = 2, dx dx 900 and D = 12. At x = 2 3 , y = 3, and D = 13. When x = 10 6, y = = 15 6 10 6 x2 Therefore, the point on y = closest to ( 0, 4 ) is Q has a minimum when x = 10 6 ≈ 24.49 ft and 4 y = 15 6 ≈ 36.74. ( ) P 2 2, 2 and the point farthest from ( 0, 4 ) is It appears that x 2 = . Q ( 0, 0 ) . y 3 Suppose that each pen has area A. 18. Let r1 and h1 be the radius and altitude of the A outer cone; r2 and h2 the radius and altitude of xy = A; y = x the inner cone. The possible values for x are in (0, ∞ ). 1 3V1 4A V1 = πr12 h1 is fixed. r1 = Q = 6x + 4 y = 6x + 3 πh1 x 4A h1 – h2 r2 dQ =6– By similar triangles = (see figure). h1 r1 dx x2 4A 6– =0 x2 2A x2 = 3 2A x=± 3 2A x= is the only critical point on (0, ∞ ). 3 dQ ⎛ 2A ⎞ < 0 on ⎜ 0, ⎜ ⎟ and dx ⎝ 3 ⎟⎠ ⎛ h ⎞ 3V1 ⎛ h2 ⎞ dQ ⎛ 2A ⎞ r2 = r1 ⎜ 1 – 2 ⎟= ⎜1 – ⎟ > 0 on ⎜ ⎝ ⎠ πh1 ⎝ h1 ⎠ ⎜ 3 , ∞ ⎟. h1 dx ⎟ ⎝ ⎠ Instructor’s Resource Manual Section 3.4 177 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 2 1 1 ⎡ 3V1 ⎛ h2 ⎞ ⎤ x 2 + 4 10 – x V2 = πr22 h2 = π ⎢ ⎜ 1 – ⎟ ⎥ h2 20. T ( x) = + , 0 ≤ x ≤ 10. 3 3 ⎢ πh1 ⎝ ⎣ h1 ⎠ ⎥ ⎦ 3 50 x 1 π 3V1h2 ⎛ h2 ⎞ 2 h2 ⎛ h2 ⎞ 2 T ′( x) = – ; T ′( x) = 0 when = ⋅ ⎜ 1 – ⎟ = V1 ⎜ 1 – ⎟ 3 x 2 – 4 50 3 πh1 ⎝ h1 ⎠ h1 ⎝ h1 ⎠ 6 h x= Let k = 2 , the ratio of the altitudes of the cones, 2491 h1 13 ⎛ 6 ⎞ then V2 = V1k (1 – k ) 2 . T (0) = ≈ 0.867 hr; T ⎜ ⎟ ≈ 0.865 hr; 15 ⎝ 2491 ⎠ dV2 = V1 (1 – k ) 2 – 2kV1 (1 – k ) = V1 (1 – k )(1 – 3k ) T (10) ≈ 3.399 hr dk 6 dV2 1 She should land the boat  ≈ 0.12 mi down 0 < k < 1 so = 0 when k = . 2491 dk 3 the shore from P. d 2V2 d 2V2 1 = V1 (6k − 4); < 0 when k = dk 2 dk 2 3 x 2 + 4 10 – x 21. T ( x) = + , 0 ≤ x ≤ 10. 1 20 4 The altitude of the inner cone must be the 3 x 1 T ′( x ) = – ; T ′( x) = 0 has no solution. altitude of the outer cone. 20 x + 4 4 2 19. Let x be the distance from P to where the woman 2 10 13 T (0) = + = hr = 2 hr, 36 min lands the boat. She must row a distance of 20 4 5 x 2 + 4 miles and walk 10 – x miles. This will T (10) = 104 ≈ 0.5 hr 20 x 2 + 4 10 – x take her T ( x) = + hours; She should take the boat all the way to town. 3 4 x 1 22. Let x be the length of cable on land, 0 ≤ x ≤ L. 0 ≤ x ≤ 10. T ′( x) = – ; T ′( x) = 0 Let C be the cost. 3 x2 + 4 4 6 C = a ( L − x) 2 + w2 + bx when x = . 7 dC a( L − x) =− +b T (0) = 19 hr = 3 hr 10 min ≈ 3.17 hr , dx ( L − x ) 2 + w2 6 a( L − x) − + b = 0 when ⎛ 6 ⎞ 15 + 7 T⎜ ⎟= ≈ 2.94 hr, ( L − x ) 2 + w2 ⎝ 7⎠ 6 b 2 [( L − x) 2 + w2 ] = a 2 ( L − x)2 104 T (10) = ≈ 3.40 hr (a 2 − b 2 )( L − x) 2 = b 2 w2 3 6 bw She should land the boat ≈ 2.27 mi down the x = L− ft on land; 7 a – b2 2 shore from P. aw ft under water a 2 – b2 d 2C aw2 = > 0 for all x, so this dx 2 [( L − x)2 + w2 ]3 2 minimizes the cost. 178 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 23. Let the coordinates of the first ship at 7:00 a.m. be a 3b (0, 0). Thus, the coordinates of the second ship at = (2 x 2 – a 2 ) 2 2 2 3/ 2 7:00 a.m. are (–60, 0). Let t be the time in hours 2 x (a − x ) since 7:00 a.m. The coordinates of the first and a 3b second ships at t are (–20t, 0) and (2 x 2 − a 2 ) = 0 when 2 2 2 3/ 2 2 x (a − x ) ( −60 + 15 ) 2t , − 15 2t respectively. Let D be the 2 a b 2 ⎛ a ⎞ b square of the distances at t. x= ;y= a −⎜ ⎟ = 2 ⎝ 2⎠ 2 ( ) + ( 0 + 15 2t ) 2 2 a D = −20t + 60 − 15 2t ⎛ a ⎞ = (1300 + 600 2 ) t 2 − ( 2400 + 1800 2 ) t + 3600 b⎜ ⎟ y′ = − ⎝ 2⎠ =− b 2 = 2 (1300 + 600 2 ) t − ( 2400 + 1800 2 ) a dD ⎛ a ⎞ a a2 − ⎜ ⎟ dt ⎝ 2⎠ 2 (1300 + 600 2 ) t − ( 2400 + 1800 2 ) = 0 when dA ⎛ a ⎞ Note that < 0 on ⎜ 0, ⎟ and 12 + 9 2 dx ⎝ 2⎠ t= ≈ 1.15 hrs or 1 hr, 9 min 13 + 6 2 dA ⎛ a ⎞ > 0 on ⎜ , a ⎟ , so A is a minimum at 2 ⎝ 2 ⎠ 12 + 9 2 d D dx D is the minimum at t = since >0 13 + 6 2 dt 2 x= a . Then the equation of the tangent line is for all t. 2 The ships are closest at 8:09 A.M. b⎛ a ⎞ b y = − ⎜x− ⎟+ or bx + ay − ab 2 = 0 . b 2 a⎝ 2⎠ 2 24. Write y in terms of x: y = a − x 2 (positive a 25. Let x be the radius of the base of the cylinder and square root since the point is in the first quadrant). h the height. Compute the slope of the tangent line: 2 2 ⎛h⎞ h y′ = − bx . V = πx 2 h; r 2 = x 2 + ⎜ ⎟ ; x 2 = r 2 − ⎝2⎠ 4 a a2 − x2 ⎛ h 2⎞ πh 3 Find the y-intercept, y0 , of the tangent line V = π ⎜ r 2 − ⎟ h = πhr 2 − ⎜ 4 ⎟ 4 through the point (x, y): ⎝ ⎠ y0 − y bx 3πh 2 2 3r =− dV = πr 2 − ; V ′ = 0 when h = ± 0− x a a2 − x2 dh 4 3 bx 2 bx 2 b 2 d 2V 3πh y0 = +y= + a − x2 Since 2 =− , the volume is maximized 2 a a −x 2 2 a a −x 2 a dh 2 ab 2 3r = when h = . 3 a2 − x2 ( r) 3 Find the x-intercept, x0 , of the tangent line 2 3 ⎛2 3 ⎞ 2 π 3 through the point (x, y): V = π⎜ ⎜ 3 r⎟r − ⎟ y–0 bx ⎝ ⎠ 4 =– x – x0 a a2 – x2 2 π 3 3 2 π 3 3 4π 3 3 = r − r = r 3 9 9 ay a 2 – x 2 a2 – x2 a2 x0 = +x= +x= bx x x 26. Let r be the radius of the circle, x the length of the Compute the Area A of the resulting triangle and rectangle, and y the width of the rectangle. maximize: 2 2 ⎛ x⎞ ⎛ y⎞ x2 y2 −1 P = 2x + 2y; r 2 = ⎜ ⎟ + ⎜ ⎟ ; r 2 = + ; 1 a 3b a 3b ⎛ 2 2⎞ ⎝2⎠ ⎝ 2⎠ 4 4 A = x0 y0 = = ⎜x a −x ⎟ 2 2 x a2 − x2 2 ⎝ ⎠ y = 4r 2 − x 2 ; P = 2 x + 2 4r 2 − x 2 −2 ⎛ ⎞ dA a 3b ⎛ 2 2⎞ 2 2 x2 dP 2x =− ⎜x a −x ⎟ ⎜ a −x − ⎟ = 2− ; dx 2 ⎝ ⎠ ⎜ a2 − x2 ⎟ ⎝ ⎠ dx 4r 2 − x 2 Instructor’s Resource Manual Section 3.4 179 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 2x 300 3 400 2− = 0; 2 4r 2 − x 2 = 2 x; A ' ( x ) = 0 when x = − ≈ 10.874 . 2 2 11 11 4r − x Critical points: x = 0, 10.874, 25 16r − 4 x 2 = 4 x 2 ; x = ± 2r 2 At x = 0, A ≈ 481; at x = 10.874, A ≈ 272; at d 2P 8r 2 x = 25, A = 625. =− < 0 when x = 2r ; dx 2 (4r 2 − x 2 )3 2 a. For minimum area, the cut should be y = 4 r − 2r = 2 r2 2 approximately 4(10.874) ≈ 43.50 cm from one end and the shorter length should be bent The rectangle with maximum perimeter is a square to form the square. with side length 2r . b. For maximum area, the wire should not be 27. Let x be the radius of the cylinder, r the radius of cut; it should be bent to form a square. the sphere, and h the height of the cylinder. 30. Let x be the length of the sides of the base, y be h2 h2 A = 2π xh ; r 2 = x 2 + ; x = r2 − the height of the box, and k be the cost per square 4 4 inch of the material in the sides of the box. h2 h4 V = x 2 y; A = 2π r 2 − h = 2π h 2 r 2 − 4 4 The cost is C = 1.2kx 2 + 1.5kx 2 + 4kxy dA π 2r h − h = ( 2 3 ; A′ = 0 when h = 0, ± 2r ) ⎛V ⎞ = 2.7 kx 2 + 4kx ⎜ ⎟ = 2.7 kx 2 + 4kV 2 dh 2 2 h4 ⎝x ⎠ x h r − 4 dC 4kV dC dA dA = 5.4kx − ; = 0 when x ≈ 0.9053 V > 0 on (0, 2r ) and < 0 on ( 2r , 2r ), dx x 2 dx dh dh V so A is a maximum when h = 2r. y≈ ≈ 1.22 3 V (0.9053 V )2 r The dimensions are h = 2r , x = . 2 31. Let r be the radius of the cylinder and h the height of the cylinder. 28. Let x be the distance from I1. 2 V − 2 πr 3 V 2 kI kI 2 V = πr 2 h + πr 3 ; h = 3 = − r Q= 1+ 3 πr 2 πr 2 3 2 x ( s − x)2 Let k be the cost per square foot of the cylindrical dQ −2kI1 2kI 2 wall. The cost is = + 3 dx x ( s − x )3 C = k (2πrh) + 2k (2πr 2 ) 2kI1 2kI 2 x3 I ⎛ ⎛ V 2 ⎞ ⎞ ⎛ 2V 8πr 2 ⎞ − + = 0; = 1; = k ⎜ 2πr ⎜ − r ⎟ + 4πr 2 ⎟ = k ⎜ + ⎟ 3 3 3 ⎜ ⎟ x ( s − x) ( s − x) I2 ⎝ ⎝ πr 2 3 ⎠ ⎠ ⎝ r 3 ⎠ s 3 I1 dC ⎛ 2V 16πr ⎞ ⎛ 2V 16πr ⎞ x= = k ⎜− + ⎟; k ⎜ − + ⎟=0 3 I1 + 3 I 2 dr ⎝ r2 3 ⎠ ⎝ r2 3 ⎠ 1/ 3 d 2Q 6kI1 6kI 2 3V 1 ⎛ 3V ⎞ = + > 0, so this point when r 3 = ,r = ⎜ ⎟ dx 2 x 4 ( s − x) 4 8π 2⎝ π ⎠ minimizes the sum. 1/ 3 13 4V 1 ⎛ 3V ⎞ ⎛ 3V ⎞ h= − =⎜ 2/3 3⎜ π ⎟ ⎟ 29. Let x be the length of a side of the square, so π 3π V ( ) ⎝ ⎠ ⎝ π ⎠ 100 − 4 x For a given volume V, the height of the cylinder is is the side of the triangle, 0 ≤ x ≤ 25 3 1/ 3 1/ 3 ⎛ 3V ⎞ 1 ⎛ 3V ⎞ 1 ⎛ 100 − 4 x ⎞ 3 ⎛ 100 − 4 x ⎞ ⎜ ⎟ and the radius is ⎜ ⎟ . A = x2 + ⎜ ⎟ ⎜ ⎟ ⎝ π ⎠ 2⎝ π ⎠ 2⎝ 3 ⎠ 2 ⎝ 3 ⎠ 3 ⎛ 10, 000 − 800 x + 16 x 2 ⎞ = x2 + ⎜ ⎟ 4 ⎜⎝ 9 ⎟ ⎠ dA 200 3 8 3 = 2x − + x dx 9 9 180 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• dx 3 3 = 2 cos 2t − 2 3 sin 2t ; h h 32. tan θ = ; sin θ = , dt 3 2/3 w h + w2 / 3 dx 1 = 0 when tan2t= ; 3 w dt 3 cos θ = 2/3 π h + w2 / 3 2t = + πn for any integer n 6 ⎛ h 2 / 3 + w2 / 3 ⎞ ⎛ 2 / 3 + w2 / 3 ⎞ x = h⎜ ⎟ + w⎜ h ⎟ π π ⎜ 3 ⎟ ⎜ 3 ⎟ t= + n ⎝ h ⎠ ⎝ w ⎠ 12 2 π π = ( h 2 / 3 + w 2 / 3 )3 / 2 When t = + n , 12 2 35. x is limited by 0 ≤ x ≤ 12 . ⎛π ⎞ ⎛π ⎞ dA x = sin ⎜ + πn ⎟ + 3 cos ⎜ + πn ⎟ A = 2 x(12 − x 2 ) = 24 x − 2 x3 ; = 24 − 6 x 2 ; ⎝6 ⎠ ⎝6 ⎠ dx π π 24 − 6 x 2 = 0; x = −2, 2 = sin cos πn + cos sin πn 6 6 Critical points: 0, 2, 12. π π When x = 0 or 12, A = 0. + 3(cos cos πn − sin sin πn) 6 6 When x = 2, y = 12 − (2)2 = 8. 1 3 The dimensions are 2x = 2(2) = 4 by 8. = (−1)n + (−1)n = 2. 2 2 The farthest the weight gets from the origin is 2 36. Let the x-axis lie on the diameter of the semicircle units. and the y-axis pass through the middle. 2 Then the equation y = r 2 − x 2 describes the r θ 2A 33. A = ; θ= semicircle. Let (x, y) be the upper-right corner of 2 r2 the rectangle. x is limited by 0 ≤ x ≤ r . The perimeter is A = 2 xy = 2 x r 2 − x 2 2 Ar 2A Q = 2r + rθ = 2r + = 2r + 2 dA 2x2 2 r r = 2 r 2 − x2 − = (r 2 − 2 x 2 ) dx 2 2 2 2 dQ 2A r −x r −x = 2− ; Q′ = 0 when r = A dr r2 2 r (r 2 − 2 x 2 ) = 0; x = 2A 2 r −x 2 2 θ= =2 ( A )2 r Critical points: 0, ,r d 2Q 4A 2 = > 0, so this minimizes the perimeter. 2 dr r3 When x = 0 or r, A = 0. When x = r , A = r2. 2 34. The distance from the fence to the base of the 2 ⎛ r ⎞ r ladder is h . y = r2 − ⎜ ⎟ = tan θ ⎝ 2⎠ 2 The length of the ladder is x. r 2r h +w The dimensions are by . h 2 2 cos θ = tan θ ; x cos θ = + w; x tan θ h w x= + sin θ cos θ dx h cosθ w sin θ w sin 3 θ − h cos3 θ =− + ; =0 dθ sin 2 θ cos 2 θ sin 2 θ cos 2 θ h when tan 3 θ = w h θ = tan −1 3 w Instructor’s Resource Manual Section 3.4 181 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 37. If the end of the cylinder has radius r and h is the ⎛ d ⎞ height of the cylinder, the surface area is A′( w) < 0 on ⎜ , d ⎟ . Maximum area is for a ⎝ 2 ⎠ A A = 2πr 2 + 2πrh so h = – r. square. 2πr The volume is h ⎛ A ⎞ Ar 40. Note that cos t = , so h = r cos t, V = πr 2 h = πr 2 ⎜ –r⎟ = – πr 3 . r ⎝ 2πr ⎠ 2 1 2 sin t = r – h 2 , and r 2 – h 2 = r sin t A A r V ′(r ) = – 3πr 2 ; V ′(r ) = 0 when r = , 2 6π Area of submerged region = tr 2 – h r 2 – h 2 V ′′(r ) = −6πr , so the volume is maximum when = tr 2 – (r cos t )(r sin t ) = r 2 (t – cos t sin t ) A A = area of exposed wetted region r= . 6π = πr 2 – πh 2 – r 2 (t – cos t sin t ) A A h= –r =2 = 2r = r 2 (π – π cos 2 t – t + cos t sin t ) 2πr 6π dA = r 2 (2π cos t sin t –1 + cos 2 t – sin 2 t ) 38. The ellipse has equation dt b2 x2 b 2 = r 2 (2π cos t sin t – 2sin 2 t ) y = ± b2 – =± a – x2 a2 a = 2r 2 sin t (π cos t – sin t ) ⎛ b 2 ⎞ Let ( x, y ) = ⎜ x, a – x 2 ⎟ be the upper right- dA ⎝ a ⎠ Since 0 < t < π , = 0 only when hand corner of the rectangle (use a and b positive). dt Then the dimensions of the rectangle are 2x by π cos t = sin t or tan t = π . In terms of r and h, 2b 2 1 r 2 – h2 a – x 2 and the area is r a this is r = π or h = . h 4bx 2 r 1 + π2 A( x) = a – x2 . a 41. The carrying capacity of the gutter is maximized 4b 2 4bx 2 4b(a 2 – 2 x 2 ) when the area of the vertical end of the gutter is A′( x) = a – x2 – = ; a a a2 – x2 a a2 – x2 maximized. The height of the gutter is 3sin θ . The area is a A′( x) = 0 when x = , so the corner is at ⎛1⎞ 2 A = 3(3sin θ ) + 2 ⎜ ⎟ (3cos θ )(3sin θ ) ⎝2⎠ ⎛ a b ⎞ ⎜ , ⎟ . The corners of the rectangle are at = 9sin θ + 9 cos θ sin θ . ⎝ 2 2⎠ dA = 9 cos θ + 9(− sin θ ) sin θ + 9 cosθ cosθ ⎛ a b ⎞ ⎛ a b ⎞ ⎛ a b ⎞ dθ ⎜ , ⎟, ⎜ – , ⎟, ⎜ – ,– ⎟, ⎝ 2 2⎠ ⎝ 2 2⎠ ⎝ 2 2⎠ = 9(cos θ − sin 2 θ + cos 2 θ ) ⎛ a b ⎞ ⎜ ,− ⎟. = 9(2 cos 2 θ + cosθ − 1) ⎝ 2 2⎠ 1 π The dimensions are a 2 and b 2 . 2 cos 2 θ + cos θ − 1 = 0; cos θ = −1, ; θ = π, 2 3 π 39. If the rectangle has length l and width w, the Since 0 ≤ θ ≤ , the critical points are 2 diagonal is d = l 2 + w2 , so l = d 2 – w2 . The π π 0, , and . area is A = lw = w d 2 – w2 . 3 2 w2 d 2 – 2 w2 When θ = 0 , A = 0. A′( w) = d 2 – w2 – = ; π 27 3 d 2 – w2 d 2 – w2 When θ = ,A= ≈ 11.7. 3 4 d A′( w) = 0 when w = and so π 2 When θ = , A = 9. 2 d2 d ⎛ d ⎞ π l = d2 – = . A′( w) > 0 on ⎜ 0, ⎟ and The carrying capacity is maximized when θ = . 2 2 ⎝ 2⎠ 3 182 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 42. The circumference of the top of the tank is the 44. Let x be the length of the edges of the cube. The circumference of the circular sheet minus the arc 1 length of the sector, surface area of the cube is 6x 2 so 0 ≤ x ≤ . 6 20π − 10θ meters. The radius of the top of the 20π − 10θ 5 The surface area of the sphere is 4πr 2 , so tank is r = = (2π − θ ) meters. The 2π π 1 – 6x2 slant height of the tank is 10 meters, so the height 6 x 2 + 4πr 2 = 1, r = 4π of the tank is 4 1 2 V = x3 + πr 3 = x3 + (1 – 6 x 2 )3 / 2 ⎛ 5θ ⎞ 5 3 6 π h = 102 − ⎜ 10 − ⎟ = 4πθ − θ 2 meters. ⎝ π ⎠ π ⎛ ⎞ dV 3 1 – 6x2 2 = 3x2 – x 1 – 6 x 2 = 3x ⎜ x – ⎟ 1 1 ⎡5 ⎤ ⎡5 ⎤ π ⎜ π ⎟ V = πr 2 h = π ⎢ (2π − θ ) ⎥ ⎢ 4πθ − θ 2 ⎥ dx ⎝ ⎠ 3 3 ⎣π ⎦ ⎣π ⎦ dV 1 125 = 0 when x = 0, = (2π − θ )2 4πθ − θ 2 dx 6+π 3π2 1 dV 125 ⎛ 2 V (0) = ≈ 0.094 m3 . = ⎜ 2(2π − θ )(−1) 4πθ − θ 6 π dθ 3π2 ⎝ 3/ 2 ⎛ 1 ⎞ 1 ⎛ 6 ⎞ + (2π − θ )2 1 (4π − 2θ ) ⎞ 2 ⎟ ( ) V⎜ ⎝ 6+π ⎠ ⎟ = (6 + π) –3 / 2 + 6 π⎝ ⎜1 – ⎟ 6+π⎠ 4πθ − θ 2 ⎟ ⎠ ⎛ π⎞ 1 = ⎜ 1 + ⎟ (6 + π) –3 / 2 = ≈ 0.055 m3 125(2π − θ ) ⎝ 6⎠ 6 6+π = (3θ 2 − 12πθ + 4π2 ) ; For maximum volume: no cube, a sphere of radius 2 2 3π 4πθ − θ 1 ≈ 0.282 meters. 125(2π − θ ) 2 π (3θ 2 − 12πθ + 4π2 ) = 0 For minimum volume: cube with sides of length 2 2 3π 4πθ − θ 1 ≈ 0.331 meters, 2π − θ = 0 or 3θ 2 − 12πθ + 4π2 = 0 6+π 2 6 2 6 1 θ = 2π, θ = 2π − π, θ = 2π + π sphere of radius ≈ 0.165 meters 3 3 2 6+π Since 0 < θ < 2π, the only critical point is 45. Consider the figure below. 2 6 2π − π . A graph shows that this maximizes 3 the volume. 43. Let V be the volume. y = 4 – x and z = 5 – 2x. x is limited by 0 ≤ x ≤ 2.5 . V = x(4 − x)(5 − 2 x) = 20 x − 13 x 2 + 2 x3 dV = 20 − 26 x + 6 x 2 ; 2(3x 2 − 13 x + 10) = 0; dx a. y = x 2 − (a − x )2 = 2ax − a 2 2(3 x − 10)( x − 1) = 0; 1 Area of A = A = (a − x) y 10 2 x = 1, 3 1 = (a − x) 2ax − a 2 Critical points: 0, 1, 2.5 2 At x = 0 or 2.5, V = 0. At x = 1, V = 9. Maximum volume when x = 1, y = 4 – 1 = 3, and dA =− 1 1 (a − x) 1 (2a ) 2ax − a + 2 2 2 ( ) z = 5 – 2(1) = 3. dx 2 2ax − a 2 a 2 − 3 ax 2 = 2ax − a 2 Instructor’s Resource Manual Section 3.4 183 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• a 2 − 3 ax 2a a2 x2 2 = 0 when x = . c. z = x 2 + w2 = x 2 + 2ax − a 2 3 2ax – a 2 dA ⎛ a 2a ⎞ dA ⎛ 2a ⎞ 2ax3 > 0 on ⎜ , ⎟ and < 0 on ⎜ , a ⎟ , = dx ⎝2 3 ⎠ dx ⎝ 3 ⎠ 2ax – a 2 2a so x = maximizes the area of triangle A. dz 1 2ax − a 2 ⎛ 6ax 2 (2ax − a 2 ) − 2ax3 (2a) ⎞ 3 = ⎜ ⎟ dx 2 2ax3 ⎜ (2ax − a 2 ) 2 ⎟ ⎝ ⎠ b. Triangle A is similar to triangle C, so 4a 2 x3 − 3a3 x 2 ax ax = w= = 2ax3 (2ax − a 2 )3 y 2ax − a 2 dz 3a 3a 1 ax 2 = 0 when x = 0, → x= Area of B = B = xw = dx 4 4 2 2 2ax − a 2 dz ⎛ a 3a ⎞ dz ⎛ 3a ⎞ < 0 on ⎜ , ⎟ and > 0 on ⎜ , a ⎟ , ⎛ 2 x 2ax − a 2 − x 2 a ⎞ dx ⎝ 2 4 ⎠ dx ⎝ 4 ⎠ dB a ⎜ 2 ax − a 2 ⎟ 3a = ⎜ ⎟ so x = minimizes length z. dx 2 ⎜ 2ax − a 2 ⎟ 4 ⎜ ⎟ ⎝ ⎠ a ⎛ 2 x(2ax − a 2 ) − ax 2 ⎞ a ⎛ 3ax 2 − 2 xa 2 ⎞ =⎜ ⎟= ⎜ ⎟ 2 ⎜ (2ax − a 2 )3 / 2 ⎟ 2 ⎜ (2ax − a 2 )3 / 2 ⎟ ⎝ ⎠ ⎝ ⎠ 2⎛ a 3x − 2 xa ⎞ 2 2a ⎜ ⎟ = 0 when x = 0, ⎜ (2ax − a 2 )3 / 2 ⎟ 2 ⎝ 3 ⎠ 2a Since x = 0 is not possible, x = . 3 dB ⎛ a 2a ⎞ dB ⎛ 2a ⎞ < 0 on ⎜ , ⎟ and > 0 on ⎜ , a ⎟ , dx ⎝2 3 ⎠ dx ⎝ 3 ⎠ 2a so x = minimizes the area of triangle B. 3 46. Let 2x be the length of a bar and 2y be the width of a bar. ⎛π θ ⎞ ⎛ 1 θ 1 θ⎞ a ⎛ θ θ⎞ x = a cos ⎜ − ⎟ = a ⎜ cos + sin ⎟ = ⎜ cos + sin ⎟ ⎝4 2⎠ ⎝ 2 2 2 2⎠ 2⎝ 2 2⎠ ⎛π θ ⎞ ⎛ 1 θ 1 θ⎞ a ⎛ θ θ⎞ y = a sin ⎜ − ⎟ = a ⎜ cos − sin ⎟ = ⎜ cos − sin ⎟ ⎝ 4 2⎠ ⎝ 2 2 2 2⎠ 2⎝ 2 2⎠ Compute the area A of the cross and maximize. 2 ⎡ a ⎛ θ θ ⎞⎤ ⎡ a ⎛ θ θ ⎞⎤ ⎡ a ⎛ θ θ ⎞⎤ A = 2(2 x)(2 y ) − (2 y ) 2 = 8 ⎢ ⎜ cos + sin ⎟ ⎥ ⎢ ⎜ cos − sin ⎟ ⎥ − 4 ⎢ ⎜ cos − sin ⎟ ⎥ ⎣ 2⎝ 2 2 ⎠⎦ ⎣ 2 ⎝ 2 2 ⎠⎦ ⎣ 2⎝ 2 2 ⎠⎦ ⎛ θ θ⎞ ⎛ θ θ⎞ = 4a 2 ⎜ cos 2 − sin 2 ⎟ − 2a 2 ⎜1 − 2 cos sin ⎟ = 4a 2 cosθ − 2a 2 (1 − sin θ ) ⎝ 2 2⎠ ⎝ 2 2⎠ dA 1 = −4a 2 sin θ + 2a 2 cosθ ; −4a 2 sin θ + 2a 2 cos θ = 0 when tan θ = ; dθ 2 1 2 sin θ = , cos θ = 5 5 d2A 1 < 0 when tan θ = , so this maximizes the area. 2 dθ 2 ⎛ 2 ⎞ ⎛ 1 ⎞ 10a 2 A = 4a 2 ⎜ ⎟ – 2a 2 ⎜ 1 – ⎟= – 2a 2 = 2a 2 ( 5 – 1) ⎝ 5⎠ ⎝ 5⎠ 5 184 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 47. a. L′(θ ) = 15(9 + 25 − 30 cosθ )−1/ 2 sin θ = 15(34 − 30 cosθ )−1/ 2 sin θ 15 L′′(θ ) = − (34 − 30 cosθ )−3 / 2 (30sin θ ) sin θ + 15(34 − 30 cosθ )−1/ 2 cosθ 2 = −225(34 − 30 cos θ ) −3 / 2 sin 2 θ + 15(34 − 30 cos θ ) −1/ 2 cosθ = 15(34 − 30 cos θ )−3 / 2 [−15sin 2 θ + (34 − 30 cos θ ) cosθ ] = 15(34 − 30 cosθ ) −3 / 2 [−15sin 2 θ + 34 cos θ − 30 cos 2 θ ] = 15(34 − 30 cos θ ) −3 / 2 [−15 + 34 cos θ − 15cos 2 θ ] = −15(34 − 30 cos θ )−3 / 2 [15cos 2 θ − 34 cosθ + 15] 34 ± (34)2 − 4(15)(15) 5 3 L′′ = 0 when cos θ = = , 2(15) 3 5 ⎛3⎞ θ = cos −1 ⎜ ⎟ 5 ⎝ ⎠ −1/ 2 ⎛ ⎛ 3 ⎞⎞ ⎛ ⎛ 3 ⎞⎞ ⎛4⎞ L′ ⎜ cos −1 ⎜ ⎟ ⎟ = 15 ⎜ 9 + 25 − 30 ⎜ ⎟ ⎟ ⎜ ⎟=3 ⎝ ⎝ 5 ⎠⎠ ⎝ ⎝ 5 ⎠⎠ ⎝5⎠ 1/ 2 ⎛ ⎛ 3 ⎞⎞ ⎛ ⎛ 3 ⎞⎞ L ⎜ cos−1 ⎜ ⎟ ⎟ = ⎜ 9 + 25 − 30 ⎜ ⎟ ⎟ =4 ⎝ ⎝ 5 ⎠⎠ ⎝ ⎝ 5 ⎠⎠ φ = 90° since the resulting triangle is a 3-4-5 right triangle. b. L′(θ ) = 65(25 + 169 − 130 cos θ )−1/ 2 sin θ = 65(194 − 130 cosθ ) −1/ 2 sin θ 65 L′′(θ ) = − (194 − 130 cos θ )−3 / 2 (130sin θ ) sin θ + 65(194 − 130 cos θ ) −1/ 2 cosθ 2 = −4225(194 − 130 cos θ )−3 / 2 sin 2 θ + 65(194 − 130 cosθ ) −1/ 2 cosθ = 65(194 − 130 cosθ )−3 / 2 [−65sin 2 θ + (194 − 130 cos θ ) cosθ ] = 65(194 − 130 cos θ )−3 / 2 [−65sin 2 θ + 194 cos θ − 130 cos 2 θ ] = 65(194 − 130 cos θ )−3 / 2 [−65cos 2 θ + 194 cosθ − 65] = −65(194 − 130 cosθ ) −3 / 2 [65cos 2 θ − 194 cos θ + 65] 194 ± (194)2 − 4(65)(65) 13 5 L′′ = 0 when cos θ = = , 2(65) 5 13 ⎛5⎞ θ = cos −1 ⎜ ⎟ ⎝ 13 ⎠ 1/ 2 ⎛ ⎛ 5 ⎞⎞ ⎛ ⎛ 5 ⎞⎞ ⎛ 12 ⎞ L′ ⎜ cos −1 ⎜ ⎟ ⎟ = 65 ⎜ 25 + 169 − 130 ⎜ ⎟ ⎟ ⎜ ⎟=5 ⎝ ⎝ 13 ⎠ ⎠ ⎝ ⎝ 13 ⎠ ⎠ ⎝ 13 ⎠ 1/ 2 ⎛ ⎛ 5 ⎞⎞ ⎛ ⎛ 5 ⎞⎞ L ⎜ cos−1 ⎜ ⎟ ⎟ = ⎜ 25 + 169 − 130 ⎜ ⎟ ⎟ = 12 ⎝ ⎝ 13 ⎠ ⎠ ⎝ ⎝ 13 ⎠ ⎠ φ = 90° since the resulting triangle is a 5-12-13 right triangle. c. When the tips are separating most rapidly, φ = 90°, L = m 2 − h 2 , L′ = h Instructor’s Resource Manual Section 3.4 185 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• d. L′(θ ) = hm(h 2 + m2 − 2hm cos θ )−1/ 2 sin θ L′′(θ ) = −h 2 m2 (h 2 + m 2 − 2hm cos θ ) −3 / 2 sin 2 θ + hm(h 2 + m 2 − 2hm cosθ )−1/ 2 cos θ = hm(h 2 + m 2 − 2hm cos θ )−3 / 2 [− hm sin 2 θ + (h 2 + m 2 ) cos θ − 2hm cos 2 θ ] = hm(h 2 + m 2 − 2hm cosθ ) −3 / 2 [− hm cos 2 θ + (h 2 + m 2 ) cos θ − hm] = −hm(h 2 + m2 − 2hm cos θ )−3 / 2 [hm cos 2 θ − (h 2 + m2 ) cosθ + hm] L′′ = 0 when hm cos 2 θ − (h 2 + m2 ) cos θ + hm = 0 (h cos θ − m)(m cos θ − h) = 0 m h cos θ = , h m h ⎛h⎞ Since h < m, cos θ = so θ = cos −1 ⎜ ⎟ . m ⎝m⎠ −1/ 2 ⎛ ⎛ h ⎞⎞ ⎛ ⎛ h ⎞⎞ m2 − h2 m2 − h2 L′ ⎜ cos −1 ⎜ ⎟ ⎟ = hm ⎜ h 2 + m2 − 2hm ⎜ ⎟ ⎟ = hm(m2 − h 2 ) −1/ 2 =h ⎝ ⎝ m ⎠⎠ ⎝ ⎝ m ⎠⎠ m m 1/ 2 ⎛ ⎛ h ⎞⎞ ⎛ ⎛ h ⎞⎞ L ⎜ cos −1 ⎜ ⎟ ⎟ = ⎜ h 2 + m2 − 2hm ⎜ ⎟ ⎟ = m2 − h2 ⎝ ⎝ m ⎠⎠ ⎝ ⎝ m ⎠⎠ Since h 2 + L2 = m 2 , φ = 90°. 48. We are interested in finding the global extrema for 49. Here we are interested in minimizing the distance the distance of the object from the observer. We between the earth and the asteroid. Using the will obtain this result by considering the squared coordinates P and Q for the two bodies, we can distance instead. The squared distance can be use the distance formula to obtain a suitable expressed as equation. However, for simplicity, we will 2 minimize the squared distance to find the critical ⎛ 1 ⎞ D( x) = ( x − 2)2 + ⎜ 100 + x − x 2 ⎟ points. The squared distance between the objects ⎝ 10 ⎠ is given by The first and second derivatives are given by D(t ) = (93cos(2π t ) − 60 cos[2π (1.51t − 1)]) 2 1 3 3 2 D '( x) = x − x − 36 x + 196 and 25 5 + (93sin(2π t ) − 120sin[2π (1.51t − 1)])2 ( ) 3 2 The first derivative is D ''( x) = x − 10 x − 300 25 D '(t ) ≈ −34359 [ cos(2π t )][ sin(9.48761t ) ] Using a computer package, we can solve the + [ cos(9.48761t ) ][(204932sin(9.48761t ) equation D '( x) = 0 to find the critical points. The −141643sin(2π t ))] critical points are x ≈ 5.1538,36.148 . Using the Plotting the function and its derivative reveal a second derivative we see that periodic relationship due to the orbiting of the D ''(5.1538) ≈ −38.9972 (max) and objects. Careful examination of the graphs reveals D ''(36.148) ≈ 77.4237 (min) that there is indeed a minimum squared distance Therefore, the position of the object closest to the (and hence a minimum distance) that occurs only observer is ≈ ( 36.148,5.48 ) while the position of once. The critical value for this occurrence is t ≈ 13.82790355 . This value gives a squared the object farthest from the person is distance between the objects of ≈ 0.0022743 ≈ (5.1538,102.5) . million miles. The actual distance is ≈ 0.047851 (Remember to go back to the original equation for million miles ≈ 47,851 miles. the path of the object once you find the critical points.) 186 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 50. Let x be the width and y the height of the flyer. 51. Consider the following sketch. 2 inches 1 inch 1 inch 2 inches We wish to minimize the area of the flyer, A = xy . As it stands, A is expressed in terms of two x t By similar triangles, = . variables so we need to write one in terms of the 2 2 27 − t + 64 t + 64 other. 27t The printed area of the flyer has an area of 50 x= −t 2 square inches. The equation for this area is t + 64 ( x − 2 )( y − 4 ) = 50 27 t 2 + 64 − 27t 2 We can solve this equation for y to obtain dx t 2 + 64 1728 = −1 = −1 2 50 dt t + 64 (t + 64)3 / 2 2 y= +4 x−2 1728 Substituting this expression for y in our equation − 1 = 0 when t = 4 5 (t + 64)3 / 2 2 for A, we get A in terms of a single variable, x. A = xy d2x −5184t d2x = ; <0 ⎛ 50 ⎞ 50 x dt 2 (t 2 + 64)5 / 2 dt 2 t = 4 = x⎜ + 4⎟ = + 4x 5 ⎝ x−2 ⎠ x−2 Therefore The allowable values for x are 2 < x < ∞ ; we want to minimize A on the open interval ( 2, ∞ ) . x= ( 27 4 5 ) − 4 5 = 5 5 ≈ 11.18 ft is the (4 5) 2 dA ( x − 2 ) 50 − 50 x −100 + 64 = +4= +4 dx ( x − 2) 2 ( x − 2 )2 maximum horizontal overhang. 4 x 2 − 16 x − 84 4 ( x − 7 )( x + 3) = = 52. a. ( x − 2) 2 ( x − 2 )2 The only critical points are obtained by solving dA = 0 ; this yields x = 7 and x = −3 . We reject dx x = −3 because it is not in the feasible domain dA dA ( 2, ∞ ) . Since < 0 for x in ( 2, 7 ) and > 0 dx dx b. There are only a few data points, but they do for x in ( 7, ∞ ) , we conclude that A attains its seem fairly linear. minimum value at x = 7 . This value of x makes y = 14 . So, the dimensions for the flyer that will c. The data values can be entered into most use the least amount of paper are 7 inches by 14 scientific calculators to utilize the Least inches. Squares Regression feature. Alternately one could use the formulas for the slope and intercept provided in the text. The resulting line should be y = 0.56852 + 2.6074 x Instructor’s Resource Manual Section 3.4 187 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• d. Using the result from c., the predicted number 54. C(x) = 7000 + 100x of surface imperfections on a sheet with area 2.0 square feet is 250 – p (n) n 55. n = 100 + 10 so p(n) = 300 – y = 0.56852 + 2.6074(2.0) = 5.7833 ≈ 6 5 2 since we can't have partial imperfections n2 R (n) = np (n) = 300n – 2 dS d n ∑ [ yi − (5 + bxi )] 2 53. a. = db db i =1 56. P (n) = R (n) – C (n) n n2 d = ∑ [ yi − (5 + bxi ) ] 2 = 300n – – (7000 + 100n) db 2 i =1 n2 n = −7000 + 200n – = ∑ 2( yi − 5 − bxi )(− xi ) 2 i =1 ⎡n ⎤ 57. ( = 2 ⎢ ∑ − xi yi + 5 xi + bxi 2 ⎢ i =1 ⎣ )⎥⎥⎦ n n n = −2∑ xi yi + 10∑ xi + 2b∑ xi 2 i =1 i =1 i =1 dS Setting = 0 gives db n n n Estimate n ≈ 200 0 = −2∑ xi yi + 10∑ xi + 2b∑ xi 2 P ′(n) = 200 – n; 200 – n = 0 when n = 200. i =1 i =1 i =1 P ′′(n) = –1, so profit is maximum at n = 200. n n n 0 = −∑ xi yi + 5∑ xi + b∑ xi 2 C ( x) 100 i =1 i =1 i =1 58. = + 3.002 – 0.0001x n n n x x b∑ xi 2 =∑ xi yi − 5∑ xi C ( x) i =1 i =1 i =1 When x = 1600, = 2.9045 or \$2.90 per unit. x n n ∑ xi yi − 5∑ xi dC = 3.002 − 0.0002 x dx b = i =1 i =1 n C ′(1600) = 2.682 or \$2.68 ∑ xi 2 i =1 C (n) 1000 n You should check that this is indeed the value 59. = + of b that minimizes the sum. Taking the n n 1200 second derivative yields C ( n) When n = 800, ≈ 1.9167 or \$1.92 per unit. d 2S n n = 2∑ xi 2 db 2 i =1 dC = n which is always positive (unless all the x dn 600 values are zero). Therefore, the value for b C ′(800) ≈ 1.333 or \$1.33 above does minimize the sum as required. dC b. Using the formula from a., we get that 60. a. = 33 − 18 x + 3 x 2 dx (2037) − 5(52) b= ≈ 3.0119 d 2C d 2C 590 = −18 + 6 x; = 0 when x = 3 dx 2 dx 2 c. The Least Squares Regression line is d 2C d 2C y = 5 + 3.0119 x < 0 on (0, 3), > 0 on (3, ∞) dx 2 dx 2 Using this line, the predicted total number of Thus, the marginal cost is a minimum when labor hours to produce a lot of 15 brass x = 3 or 300 units. bookcases is y = 5 + 3.0119(15) ≈ 50.179 hours b. 33 − 18(3) + 3(3) 2 = 6 188 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• x3 65. The revenue function would be 61. a. R ( x) = xp( x) = 20 x + 4 x 2 − R ( x ) = x ⋅ p ( x ) = 200 x − 0.15 x 2 . This, together 3 dR with the cost function yields the following profit = 20 + 8 x − x 2 = (10 − x )( x + 2 ) function: dx ⎧−5000 + 194 x − 0.148 x 2 if 0 ≤ x ≤ 500 ⎪ P ( x) = ⎨ dR 2 b. Increasing when >0 ⎪−9000 + 194 x − 0.148 x if 500 < x ≤ 750 ⎩ dx 20 + 8 x − x 2 > 0 on [0, 10) a. The only difference in the two pieces of the Total revenue is increasing if 0 ≤ x ≤ 10. profit function is the constant. Since the derivative of a constant is 0, we can say that on the interval 0 < x < 750 , d 2R d 2R c. = 8 – 2 x; = 0 when x = 4 dP dx 2 dx 2 = 194 − 0.296 x dx d 3R dR There are no singular points in the given = −2; is maximum at x = 4. 3 dx interval. To find stationary points, we solve dx dP 1/ 2 =0 ⎛ x ⎞ dx 62. R ( x) = x ⎜ 182 − ⎟ ⎝ 36 ⎠ 194 − 0.296 x = 0 dR 1⎛ x ⎞ −1/ 2 ⎛ 1 ⎞ ⎛ 1/ 2 x ⎞ −0.296 x = −194 = x ⎜182 − ⎟ ⎜ − ⎟ + ⎜182 − ⎟ dx 2⎝ 36 ⎠ ⎝ 36 ⎠ ⎝ 36 ⎠ x ≈ 655 −1 2 Thus, the critical points are 0, 500, 655, and ⎛ x ⎞ ⎛ x ⎞ 750. = ⎜ 182 − ⎟ ⎜ 182 − ⎟ ⎝ 36 ⎠ ⎝ 24 ⎠ P ( 0 ) = −5000 ; P ( 500 ) = 55, 000 ; dR = 0 when x = 4368 P ( 655 ) = 54,574.30 ; P ( 750 ) = 53, 250 dx The profit is maximized if the company x1 = 4368; R(4368) ≈ 34, 021.83 produces 500 chairs. The current machine can dR handle this work, so they should not buy the At x1 , =0. new machine. dx b. Without the new machine, a production level 800 x of 500 chairs would yield a maximum profit 63. R( x) = − 3x x+3 of \$55,000. dR ( x + 3)(800) − 800 x 2400 = −3 = − 3; 66. The revenue function would be 2 dx ( x + 3) ( x + 3)2 R ( x ) = x ⋅ p ( x ) = 200 x − 0.15 x 2 . This, together dR = 0 when x = 20 2 − 3 ≈ 25 with the cost function yields the following profit dx function: x1 = 25; R (25) ≈ 639.29 ⎧−5000 + 194 x − 0.148 x 2 if 0 ≤ x ≤ 500 ⎪ dR P ( x) = ⎨ 2 At x1 , =0. ⎪−8000 + 194 x − 0.148 x if 500 < x ≤ 750 ⎩ dx a. The only difference in the two pieces of the ( x − 400) 64. p( x) = 12 − (0.20) = 20 − 0.02 x profit function is the constant. Since the 10 derivative of a constant is 0, we can say that R ( x) = 20 x − 0.02 x 2 on the interval 0 < x < 750 , dP dR = 20 − 0.04 x; dR = 0 when x = 500 = 194 − 0.296 x dx dx dx Total revenue is maximized at x1 = 500 . There are no singular points in the given interval. To find stationary points, we solve dP =0 dx 194 − 0.296 x = 0 −0.296 x = −194 x ≈ 655 Instructor’s Resource Manual Section 3.4 189 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• Thus, the critical points are 0, 500, 655, and 750. a 2 + 2ab + b 2 P ( 0 ) = −5000 ; P ( 500 ) = 55, 000 ; b. F (b) = 4b P ( 655 ) = 55,574.30 ; P ( 750 ) = 54, 250 As b → 0+ , a 2 + 2ab + b 2 → a 2 while The profit is maximized if the company produces 4b → 0+ , thus lim F (b) = ∞ which is not 655 chairs. The current machine cannot handle b → 0+ this work, so they should buy the new machine. close to a. 2 a + 2a + b b. With the new machine, a production level of a 2 + 2ab + b 2 655 chairs would yield a maximum profit of lim = lim b =∞ , b →∞ 4b b →∞ 4 \$55,574.30. so when b is very large, F(b) is not close to a. 67. R ( x) = 10 x − 0.001x 2 ; 0 ≤ x ≤ 300 2(a + b)(4b) – 4(a + b)2 F ′(b) = 16b 2 P ( x) = (10 x – 0.001x 2 ) – (200 + 4 x – 0.01x 2 ) 4b 2 – 4a 2 b 2 – a 2 = –200 + 6 x + 0.009 x 2 = = ; 16b 2 4b 2 dP dP = 6 + 0.018 x; = 0 when x ≈ −333 F ′(b) = 0 when b 2 = a 2 or b = a since a and dx dx Critical numbers: x = 0, 300; P(0) = –200; b are both positive. P(300) = 2410; Maximum profit is \$2410 ( a + a ) 2 4a 2 F (a) = = =a at x = 300. 4a 4a ( a + b) 2 ⎧200 + 4 x − 0.01x 2 if 0 ≤ x ≤ 300 ⎪ Thus a ≤ for all b > 0 or 68. C ( x) = ⎨ 4b 2 ⎪800 + 3 x − 0.01x ⎩ if 300 < x ≤ 450 ( a + b) 2 a+b ab ≤ which leads to ab ≤ . ⎧−200 + 6 x + 0.009 x 2 if 0 ≤ x ≤ 300 ⎪ 4 2 P( x) = ⎨ 2 ⎪−800 + 7 x + 0.009 x ⎩ if 300 < x ≤ 450 1⎛ a+b+c⎞ 3 ( a + b + c )3 There are no stationary points on the interval c. Let F (b) = ⎜ ⎟ = b⎝ 3 ⎠ 27b [0, 300]. On [300, 450]: dP dP 3(a + b + c)2 (27b) – 27(a + b + c)3 = 7 + 0.018 x; = 0 when x ≈ −389 F ′(b) = dx dx 27 2 b 2 The critical numbers are 0, 300, 450. (a + b + c) 2 [3b – (a + b + c)] P(0) = –200, P(300) = 2410, P(450) = 4172.5 = 27b 2 Monthly profit is maximized at x = 450, P(450) = 4172.50 (a + b + c)2 (2b – a – c) = ; 27b 2 a+c F ′(b) = 0 when b = . 2 3 ⎛a+c⎞ 2 ⎛a+c a+c⎞ F⎜ ⎟= ⋅⎜ + ⎟ ⎝ 2 ⎠ a+c ⎝ 3 6 ⎠ 3 3 2 2 ⎛ 3(a + c) ⎞ 2 ⎛a+c⎞ ⎛a+c⎞ = ⎜ ⎟ = ⎜ ⎟ =⎜ ⎟ a+c⎝ 6 ⎠ a+c⎝ 2 ⎠ ⎝ 2 ⎠ 2 3 2 a 2 + 2ab + b 2 ⎛a+c⎞ 1⎛ a+b+c⎞ ⎛ a+b⎞ Thus ⎜ ⎟ ≤ ⎜ ⎟ for all b > 0. 69. a. ab ≤ ⎜ ⎟ = ⎝ 2 ⎠ b⎝ 3 ⎠ ⎝ 2 ⎠ 4 2 a2 1 b2 ⎛a+c⎞ = + ab + From (b), ac ≤ ⎜ ⎟ , thus 4 2 4 ⎝ 2 ⎠ This is true if 3 3 1⎛ a+b+c⎞ ⎛ a+b+c⎞ 2 2 ac ≤ ⎜ ⎟ or abc ≤ ⎜ ⎟ a2 1 b2 ⎛ a b ⎞ ⎛a–b⎞ b⎝ 3 ⎠ ⎝ 3 ⎠ 0≤ – ab + =⎜ – ⎟ =⎜ ⎟ 4 2 4 ⎝ 2 2⎠ ⎝ 2 ⎠ which gives the desired result Since a square can never be negative, this is a+b+c (abc)1/ 3 ≤ . always true. 3 190 Section 3.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 70. Let a = lw, b = lh, and c = hw, then 1 1 Critical points: – , S = 2(a + b + c) while V 2 = abc. By problem 2 2 a+b+c 1 1 69(c), (abc)1/ 3 ≤ so f ′( x) > 0 when x < – or x > 3 2 2 2(a + b + c) S ⎛ 1 ⎤ ⎡ 1 ⎞ (V 2 )1/ 3 ≤ = . f(x) is increasing on ⎜ – ∞, – 2⋅3 6 ⎥∪⎢ , ∞ ⎟ and ⎝ 2⎦ ⎣ 2 ⎠ In problem 1c, the minimum occurs, hence a+c ⎡ 1 1 ⎤ equality holds, when b = . In the result used decreasing on ⎢ – , ⎥. 2 ⎣ 2 2⎦ from Problem 69(b), equality holds when c = a, ⎛ 1 ⎞ a+a Local minimum f ⎜ ⎟ = – 2 –10 ≈ –11.4 thus b = = a, so a = b = c. For the boxes, ⎝ 2⎠ 2 this means l = w = h, so the box is a cube. ⎛ 1 ⎞ Local maximum f ⎜ – ⎟ = 2 –10 ≈ –8.6 ⎝ 2⎠ f ′′( x) = 12 x; f ′′( x) > 0 when x > 0. f(x) is concave 3.5 Concepts Review up on (0, ∞ ) and concave down on 1. f(x); –f(x) (– ∞ , 0); inflection point (0, –10). 2. decreasing; concave up 3. x = –1, x = 2, x = 3; y = 1 4. polynomial; rational. Problem Set 3.5 1. Domain: (−∞, ∞) ; range: (−∞, ∞) Neither an even nor an odd function. y-intercept: 5; x-intercept: ≈ –2.3 3. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) f ′( x) = 3x 2 – 3; 3 x 2 – 3 = 0 when x = –1, 1 Neither an even nor an odd function. Critical points: –1, 1 y-intercept: 3; x-intercepts: ≈ –2.0, 0.2, 3.2 f ′( x) > 0 when x < –1 or x > 1 f ′( x) = 6 x 2 – 6 x –12 = 6( x – 2)( x + 1); f(x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) and f ′( x) = 0 when x = –1, 2 decreasing on [–1, 1]. Local minimum f(1) = 3; Critical points: –1, 2 local maximum f(–1) = 7 f ′( x) > 0 when x < –1 or x > 2 f ′′( x) = 6 x; f ′′( x) > 0 when x > 0. f(x) is increasing on (– ∞ , –1] ∪ [2, ∞ ) and f(x) is concave up on (0, ∞ ) and concave down decreasing on [–1, 2]. on (– ∞ , 0); inflection point (0, 5). Local minimum f(2) = –17; local maximum f(–1) = 10 1 f ′′( x) = 12 x − 6 = 6(2 x − 1); f ′′( x) > 0 when x > . 2 ⎛1 ⎞ f(x) is concave up on ⎜ , ∞ ⎟ and concave down ⎝2 ⎠ ⎛ 1⎞ ⎛1 7⎞ on ⎜ – ∞, ⎟ ; inflection point: ⎜ , – ⎟ ⎝ 2⎠ ⎝2 2⎠ 2. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) Neither an even nor an odd function. y-intercept: –10; x-intercept: 2 f ′( x) = 6 x 2 – 3 = 3(2 x 2 –1); 2 x 2 –1 = 0 when 1 1 x=– , 2 2 Instructor’s Resource Manual Section 3.5 191 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 4. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) 1 1 Neither an even nor an odd function H ′(t ) > 0 for – < t < 0 or < t. 2 2 y-intercept: –1; x-intercept: 1 ⎡ 1 ⎤ ⎡ 1 ⎞ f ′( x) = 3( x –1)2 ; f ′( x) = 0 when x = 1 H(t) is increasing on ⎢ – , 0⎥ ∪ ⎢ , ∞ ⎟ and ⎣ 2 ⎦ ⎣ 2 ⎠ Critical point: 1 f ′( x) > 0 for all x ≠ 1 ⎛ 1 ⎤ ⎡ 1 ⎤ decreasing on ⎜ – ∞, – ⎥ ∪ ⎢ 0, ⎥ f(x) is increasing on (– ∞ , ∞ ) ⎝ 2⎦ ⎣ 2⎦ No local minima or maxima ⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 f ′′( x) = 6( x –1); f ′′( x) > 0 when x > 1. Global minima f ⎜ – ⎟=– , f ⎜ ⎟=– ; ⎝ 2⎠ 4 ⎝ 2⎠ 4 f(x) is concave up on (1, ∞ ) and concave down Local maximum f(0) = 0 on (– ∞ , 1); inflection point (1, 0) H ′′(t ) = 12t 2 – 2 = 2(6t 2 –1); H ′′ > 0 when 1 1 t<– or t > 6 6 ⎛ 1 ⎞ ⎛ 1 ⎞ H(t) is concave up on ⎜ – ∞, –⎟∪⎜ , ∞⎟ ⎝ 6⎠ ⎝ 6 ⎠ ⎛ 1 1 ⎞ and concave down on ⎜ – ⎟ ; inflection , ⎝ 66⎠ ⎛ 1 5 ⎞ ⎛ 1 5 ⎞ points H ⎜ – , − ⎟ and H ⎜ , ⎟ 5. Domain: (– ∞ , ∞ ); range: [0, ∞ ) ⎝ 6 36 ⎠ ⎝ 6 36 ⎠ Neither an even nor an odd function. y-intercept: 1; x-intercept: 1 G ′( x ) = 4( x – 1)3 ; G ′( x) = 0 when x = 1 Critical point: 1 G ′( x) > 0 for x > 1 G(x) is increasing on [1, ∞ ) and decreasing on (– ∞ , 1]. Global minimum f(1) = 0; no local maxima G ′′( x) = 12( x –1) 2 ; G ′′( x) > 0 for all x ≠ 1 G(x) is concave up on (– ∞ , 1) ∪ (1, ∞ ); no 7. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) inflection points Neither an even nor an odd function. y-intercept: 10; x-intercept: 1 –111/ 3 ≈ –1.2 f ′( x) = 3 x 2 – 6 x + 3 = 3( x –1) 2 ; f ′( x) = 0 when x = 1. Critical point: 1 f ′( x) > 0 for all x ≠ 1. f(x) is increasing on (– ∞ , ∞ ) and decreasing nowhere. No local maxima or minima f ′′( x) = 6 x – 6 = 6( x –1); f ′′( x) > 0 when x > 1. f(x) is concave up on (1, ∞ ) and concave down ⎡ 1 ⎞ on (– ∞ , 1); inflection point (1, 11) 6. Domain: (– ∞ , ∞ ); range: ⎢ – , ∞ ⎟ ⎣ 4 ⎠ H (–t ) = (– t )2 [(– t )2 – 1] = t 2 (t 2 – 1) = H (t ); even function; symmetric with respect to the y-axis. y-intercept: 0; t-intercepts: –1, 0, 1 H ′(t ) = 4t 3 – 2t = 2t (2t 2 – 1); H ′(t ) = 0 when 1 1 t=– , 0, 2 2 1 1 Critical points: – , 0, 2 2 192 Section 3.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• ⎡ 16 ⎞ down on (–1, ∞ ); no inflection points (–1 is not 8. Domain: (– ∞ , ∞ ); range: ⎢ – , ∞ ⎟ in the domain of g). ⎣ 3 ⎠ x 1 4(– s )4 – 8(– s ) 2 –12 4s 4 – 8s 2 –12 lim = lim = 1; F (– s ) = = x →∞ x + 1 x →∞ 1 + 1 3 3 x = F ( s ); even function; symmetric with respect x 1 lim = lim = 1; to the y-axis x→ – ∞ x + 1 x→ – ∞ 1 + 1 x y-intercept: –4; s-intercepts: – 3, 3 horizontal asymptote: y = 1 16 16 16 x F ′( s ) = s3 – s = s ( s 2 –1); F ′( s ) = 0 As x → –1– , x + 1 → 0 – so lim = ∞; 3 3 3 x → –1– x +1 when s = –1, 0, 1. x Critical points: –1, 0, 1 as x → –1+ , x + 1 → 0+ so lim = – ∞; x → –1+ x +1 F ′( s ) > 0 when –1 < x < 0 or x > 1. vertical asymptote: x = –1 F(s) is increasing on [–1, 0] ∪ [1, ∞ ) and decreasing on (– ∞ , –1] ∪ [0, 1] 16 16 Global minima F (−1) = − , F (1) = − ; local 3 3 maximum F(0) = –4 16 ⎛ 1⎞ F ′′( s ) = 16 s 2 − = 16 ⎜ s 2 − ⎟ ; F ′′( s ) > 0 3 ⎝ 3⎠ 1 1 when s < – or s > 3 3 ⎛ 1 ⎞ ⎛ 1 ⎞ 10. Domain: (– ∞ , 0) ∪ (0, ∞ ); F(s) is concave up on ⎜ −∞, − ⎟∪⎜ , ∞⎟ ⎝ 3⎠ ⎝ 3 ⎠ range: (– ∞ , –4 π ] ∪ [0, ∞ ) ⎛ 1 1 ⎞ and concave down on ⎜ – , ⎟; Neither an even nor an odd function ⎝ 3 3⎠ No y-intercept; s-intercept: π inflection points ⎛ 1 ,− 128 ⎞ ⎛ 1 ,− 128 ⎞ s 2 – π2 F⎜– ⎟, F ⎜ ⎟ g ′( s ) = ; g ′( s ) = 0 when s = – π , π ⎝ 3 27 ⎠ ⎝ 3 27 ⎠ s2 Critical points: −π , π g ′( s ) > 0 when s < – π or s > π g(s) is increasing on (−∞, −π ] ∪ [π , ∞) and decreasing on [– π , 0) ∪ (0, π ]. Local minimum g( π ) = 0; local maximum g(– π ) = –4 π 9. Domain: (– ∞ , –1) ∪ (–1, ∞ ); 2π2 range: (– ∞ , 1) ∪ (1, ∞ ) g ′′( s) = ; g ′′( s ) > 0 when s > 0 Neither an even nor an odd function s3 y-intercept: 0; x-intercept: 0 g(s) is concave up on (0, ∞ ) and concave down 1 g ′( x) = ; g ′( x) is never 0. on (– ∞ , 0); no inflection points (0 is not in the ( x + 1) 2 No critical points domain of g(s)). g ′( x) > 0 for all x ≠ –1. π2 g(x) is increasing on (– ∞ , –1) ∪ (–1, ∞ ). g ( s ) = s – 2π + ; y = s – 2π is an oblique s No local minima or maxima asymptote. 2 g ′′( x) = – ; g ′′( x) > 0 when x < –1. As s → 0 – , ( s – π) 2 → π2 , so lim g ( s ) = – ∞; ( x + 1)3 s →0 – g(x) is concave up on (– ∞ , –1) and concave Instructor’s Resource Manual Section 3.5 193 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• as s → 0+ , ( s – π)2 → π2 , so lim g ( s ) = ∞; No vertical asymptotes s →0 + s = 0 is a vertical asymptote. 12. Domain: (– ∞ , ∞ ); range: [0, 1) (–θ ) 2 θ2 Λ (–θ ) = = = Λ (θ ); even (–θ )2 + 1 θ 2 + 1 ⎡ 1 1⎤ function; symmetric with respect to the y-axis. 11. Domain: (– ∞ , ∞ ); range: ⎢ – , ⎥ ⎣ 4 4⎦ y-intercept: 0; θ -intercept: 0 –x 2θ f (– x) = =– x = – f ( x); odd Λ ′(θ ) = ; Λ ′(θ ) = 0 when θ = 0 (– x) 2 + 4 x2 + 4 (θ + 1)2 2 function; symmetric with respect to the origin. Critical point: 0 y-intercept: 0; x-intercept: 0 Λ ′(θ ) > 0 when θ > 0 4 – x2 f ′( x) = ; f ′( x) = 0 when x = –2, 2 Λ(θ) is increasing on [0, ∞ ) and ( x 2 + 4)2 decreasing on (– ∞ , 0]. Critical points: –2, 2 Global minimum Λ(0) = 0; no local maxima f ′( x) > 0 for –2 < x < 2 2(1 – 3θ 2 ) f(x) is increasing on [–2, 2] and Λ ′′(θ ) = ; Λ ′′(θ ) > 0 when (θ 2 + 1)3 decreasing on (– ∞ , –2] ∪ [2, ∞ ). 1 1 1 Global minimum f (–2) = – ; global maximum – <θ < 4 3 3 1 ⎛ 1 1 ⎞ f (2) = Λ(θ) is concave up on ⎜ – , ⎟ and 4 ⎝ 3 3⎠ 2 x( x 2 – 12) ⎛ 1 ⎞ ⎛ 1 ⎞ f ′′( x) = ; f ′′( x) > 0 when concave down on ⎜ – ∞, – ⎟∪⎜ , ∞ ⎟; ( x 2 + 4)3 ⎝ 3⎠ ⎝ 3 ⎠ –2 3 < x < 0 or x > 2 3 ⎛ 1 1⎞ ⎛ 1 1⎞ inflection points ⎜ – , ⎟, ⎜ , ⎟ f(x) is concave up on (–2 3, 0) ∪ (2 3, ∞) and ⎝ 3 4⎠ ⎝ 3 4⎠ concave down on (– ∞, – 2 3) ∪ (0, 2 3); θ2 1 lim = lim = 1; ⎛ 3⎞ θ →∞ θ 2 + 1 θ →∞ 1 + 1 inflection points ⎜ −2 3, − ⎟ , (0, 0) , θ2 ⎜ 8 ⎟ ⎝ ⎠ θ2 1 ⎛ lim = lim = 1; 3⎞ θ →–∞ θ + 1 θ →–∞ 1 + 1 2 ⎜ 2 3, ⎜ ⎟ 2 ⎝ 8 ⎟ ⎠ θ y = 1 is a horizontal asymptote. No vertical 1 x asymptotes lim = lim x = 0; x →∞ x 2 +4 x →∞ 1 + 4 x2 1 x lim = lim x = 0; x→ – ∞ x 2 +4 x→ – ∞ 1 + 4 x2 y = 0 is a horizontal asymptote. 194 Section 3.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 13. Domain: (– ∞ , 1) ∪ (1, ∞ ); (−∞, 0] and decreasing on [0, ∞) . Global range (– ∞ , 1) ∪ (1, ∞ ) maximum P (0) = 1 ; no local minima. Neither an even nor an odd function y-intercept: 0; x-intercept: 0 6x2 − 2 P ''( x) = 1 ( x 2 + 1)3 h( x ) = − ; h′( x) is never 0. ( x − 1)2 P ''( x) > 0 on (−∞, −1/ 3) ∪ (1/ 3, ∞) (concave No critical points up) and P ''( x) < 0 on (−1/ 3,1/ 3) (concave h′( x) < 0 for all x ≠ 1. down). h(x) is increasing nowhere and ⎛ 1 3⎞ Inflection points: ⎜ ± , ⎟ decreasing on (– ∞ , 1) ∪ (1, ∞ ). ⎝ 3 4⎠ No local maxima or minima No vertical asymptotes. 2 lim P ( x) = 0; lim P ( x) = 0 h′′( x) = ; h′′( x) > 0 when x > 1 x →∞ x →−∞ ( x – 1)3 y = 0 is a horizontal asymptote. h( x) is concave up on (1, ∞ ) and concave down on (– ∞ , 1); no inflection points (1 is not in the domain of h( x) ) x 1 lim = lim = 1; x →∞ x – 1 x→∞ 1 – 1 x x 1 lim = lim = 1; 15. Domain: (– ∞ , –1) ∪ (–1, 2) ∪ (2, ∞ ); x →−∞ x − 1 x →−∞ 1 − 1 x range: (– ∞ , ∞ ) y = 1 is a horizontal asymptote. Neither an even nor an odd function x 3 As x → 1– , x – 1 → 0 – so lim = – ∞; y-intercept: – ; x-intercepts: 1, 3 x →1– x –1 2 2 x 3x – 10 x + 11 as x → 1+ , x – 1 → 0+ so lim = ∞; f ′( x) = ; f ′( x) is never 0. x →1+ x –1 ( x + 1) 2 ( x – 2) 2 x = 1 is a vertical asymptote. No critical points f ′( x) > 0 for all x ≠ –1, 2 f(x) is increasing on (– ∞ , –1) ∪ (–1, 2) ∪ ( 2, ∞ ) . No local minima or maxima –6 x3 + 30 x 2 – 66 x + 42 f ′′( x) = ; f ′′( x) > 0 when ( x + 1)3 ( x – 2)3 x < –1 or 1 < x < 2 f(x) is concave up on (– ∞ , –1) ∪ (1, 2) and concave down on (–1, 1) ∪ (2, ∞ ); 14. Domain: ( −∞, ∞ ) inflection point f(1) = 0 Range: ( 0,1] ( x – 1)( x – 3) x2 – 4 x + 3 lim = lim Even function since x →∞ ( x + 1)( x – 2) x →∞ x 2 – x – 2 1 1 1– 4 + 3 P(− x) = = = P( x) 2 2 x x2 (− x) + 1 x + 1 = lim = 1; x →∞ 1 – 1 – 2 so the function is symmetric with respect to the x x2 y-axis. 3 1– 4 + y-intercept: y = 1 ( x – 1)( x – 3) x x2 lim = lim = 1; x-intercept: none x → – ∞ ( x + 1)( x – 2) x → – ∞ 1 – 1– 2 x x2 −2 x P '( x) = ; P '( x) is 0 when x = 0 . y =1 is a horizontal asymptote. ( x + 1) 2 2 As x → –1– , x – 1 → –2, x – 3 → –4, critical point: x = 0 P '( x) > 0 when x < 0 so P ( x) is increasing on x – 2 → –3, and x + 1 → 0 – so lim f ( x) = ∞; x → –1– Instructor’s Resource Manual Section 3.5 195 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• as x → –1+ , x – 1 → –2, x – 3 → –4, 17. Domain: ( −∞,1) ∪ (1, ∞ ) x – 2 → –3, and x + 1 → 0+ , so Range: ( −∞, ∞ ) lim f ( x) = – ∞ Neither even nor odd function. x → –1+ y-intercept: y = 6 ; x-intercept: x = −3, 2 As x → 2 – , x – 1 → 1, x – 3 → –1, x + 1 → 3, and x2 − 2 x + 5 g '( x) = ; g '( x) is never zero. No x – 2 → 0 – , so lim f ( x) = ∞; as ( x − 1) 2 x→2– critical points. x → 2+ , x – 1 → 1, x – 3 → –1, x + 1 → 3, and g '( x) > 0 over the entire domain so the function x – 2 → 0+ , so lim f ( x) = – ∞ is always increasing. No local extrema. x → 2+ −8 x = –1 and x = 2 are vertical asymptotes. f ''( x) = ; f ''( x) > 0 when ( x − 1)3 x < 1 (concave up) and f ''( x) < 0 when x > 1 (concave down); no inflection points. No horizontal asymptote; x = 1 is a vertical asymptote; the line y = x + 2 is an oblique (or slant) asymptote. 16. Domain: ( −∞, 0 ) ∪ ( 0, ∞ ) Range: (−∞, −2] ∪ [2, ∞) Odd function since 18. Domain: (– ∞ , ∞ ); range [0, ∞ ) (− z )2 + 1 z2 +1 3 3 w(− z ) = =− = − w( z ) ; symmetric f (– x) = – x = x = f ( x); even function; −z z with respect to the origin. symmetric with respect to the y-axis. y-intercept: none y-intercept: 0; x-intercept: 0 x-intercept: none 2⎛ x ⎞ 1 f ′( x) = 3 x ⎜ ⎟ = 3x x ; f ′( x) = 0 when x = 0 ⎜ x⎟ w '( z ) = 1 − ; w '( z ) = 0 when z = ±1 . ⎝ ⎠ z2 Critical point: 0 critical points: z = ±1 . w '( z ) > 0 on f ′( x) > 0 when x > 0 (−∞, −1) ∪ (1, ∞) so the function is increasing on f(x) is increasing on [0, ∞ ) and decreasing on (−∞, −1] ∪ [1, ∞) . The function is decreasing on (– ∞ , 0]. [−1, 0) ∪ (0,1) . Global minimum f(0) = 0; no local maxima local minimum w(1) = 2 and local maximum 3x 2 2 f ′′( x) = 3 x + = 6 x as x 2 = x ; w(−1) = −2 . No global extrema. x 2 f ′′( x) > 0 when x ≠ 0 w ''( z ) = > 0 when z > 0 . Concave up on f(x) is concave up on (– ∞ , 0) ∪ (0, ∞ ); no z3 inflection points (0, ∞) and concave down on ( −∞, 0 ) . No horizontal asymptote; x = 0 is a vertical asymptote; the line y = z is an oblique (or slant) asymptote. No inflection points. 196 Section 3.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 19. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) 21. Domain: (– ∞ , ∞ ); range: [0, ∞ ) R (– z ) = – z – z = – z z = – R ( z ); odd function; Neither an even nor an odd function. symmetric with respect to the origin. Note that for x ≤ 0, x = – x so x + x = 0, while y-intercept: 0; z-intercept: 0 x +x for x > 0, x = x so = x. z2 2 2 R ′( z ) = z + = 2 z since z 2 = z for all z; z ⎧0 ⎪ if x ≤ 0 g ( x) = ⎨ 2 R ′( z ) = 0 when z = 0 ⎪3x + 2 x if x > 0 ⎩ Critical point: 0 y-intercept: 0; x-intercepts: ( − ∞, 0] R ′( z ) > 0 when z ≠ 0 ⎧0 if x ≤ 0 R(z) is increasing on (– ∞ , ∞ ) and decreasing g ′( x) = ⎨ nowhere. ⎩ 6 x + 2 if x > 0 No local minima or maxima No critical points for x > 0. 2z g(x) is increasing on [0, ∞ ) and decreasing R′′( z ) = ; R′′( z ) > 0 when z > 0. nowhere. z ⎧0 if x ≤ 0 R(z) is concave up on (0, ∞ ) and concave down g ′′( x) = ⎨ on (– ∞ , 0); inflection point (0, 0). ⎩6 if x > 0 g(x) is concave up on (0, ∞ ); no inflection points 20. Domain: (– ∞ , ∞ ); range: [0, ∞ ) H (– q) = (– q )2 – q = q 2 q = H (q); even 22. Domain: (– ∞ , ∞ ); range: [0, ∞ ) Neither an even nor an odd function. Note that function; symmetric with respect to the y-axis. y-intercept: 0; q-intercept: 0 x –x for x < 0, x = – x so = – x , while for q3 3q3 2 2 H ′(q) = 2q q + = = 3q q as q = q 2 x −x q q x ≥ 0, x = x so = 0. 2 for all q; H ′(q ) = 0 when q = 0 ⎧− x3 + x 2 − 6 x if x < 0 ⎪ Critical point: 0 h( x ) = ⎨ H ′(q ) > 0 when q > 0 ⎪0 ⎩ if x ≥ 0 H(q) is increasing on [0, ∞ ) and y-intercept: 0; x-intercepts: [0, ∞ ) decreasing on (– ∞ , 0]. ⎧−3 x 2 + 2 x − 6 if x < 0 ⎪ Global minimum H(0) = 0; no local maxima h′( x) = ⎨ ⎪0 ⎩ if x ≥ 0 3q 2 H ′′(q) = 3 q + = 6 q ; H ′′(q ) > 0 when No critical points for x < 0 q h(x) is increasing nowhere and decreasing on q ≠ 0. (– ∞ , 0]. H(q) is concave up on (– ∞ , 0) ∪ (0, ∞ ); no ⎧−6 x + 2 if x < 0 inflection points. h′′( x) = ⎨ ⎩0 if x ≥ 0 h(x) is concave up on (– ∞ , 0); no inflection Instructor’s Resource Manual Section 3.5 197 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• points 24. Domain: [2k π , (2k + 1) π ] where k is any integer; range: [0, 1] Neither an even nor an odd function y-intercept: 0; x-intercepts: k π , where k is any integer. cos x π f ′( x) = ; f ′( x) = 0 when x = 2k π + 2 sin x 2 while f ′( x) does not exist when x = k π , k any integer. π Critical points: k π, 2k π + where k is any 23. Domain: (– ∞ , ∞ ); range: [0, 1] 2 f (− x) = sin(− x) = − sin x = sin x = f ( x); even integer π function; symmetric with respect to the y-axis. f ′( x) > 0 when 2k π < x < 2k π + y-intercept: 0; x-intercepts: k π where k is any 2 integer. ⎡ π⎤ f(x) is increasing on ⎢ 2k π, 2k π + ⎥ and sin x π ⎣ 2⎦ f ′( x) = cos x; f ′( x) = 0 when x = + k π sin x 2 ⎡ π ⎤ decreasing on ⎢ 2k π + , (2k + 1)π ⎥ , k any and f ′( x) does not exist when x = k π , where k ⎣ 2 ⎦ is any integer. integer. kπ π Global minima f(k π ) = 0; global maxima Critical points: and kπ + , where k is any ⎛ π⎞ 2 2 f ⎜ 2k π + ⎟ = 1, k any integer integer; f ′( x) > 0 when sin x and cos x are either ⎝ 2⎠ both positive or both negative. – cos 2 x – 2sin 2 x –1 – sin 2 x f ′′( x) = = ⎡ π⎤ 4sin 3 / 2 x 4sin 3 / 2 x f(x) is increasing on ⎢ k π, k π + ⎥ and decreasing ⎣ 2⎦ 1 + sin 2 x =– ; ⎡ π ⎤ 4sin 3 / 2 x on ⎢ k π + , (k + 1)π ⎥ where k is any integer. ⎣ 2 ⎦ f ′′( x) < 0 for all x. Global minima f(k π ) = 0; global maxima f(x) is concave down on (2k π , (2k + 1) π ); ⎛ π⎞ no inflection points f ⎜ k π + ⎟ = 1, where k is any integer. ⎝ 2⎠ cos 2 x sin 2 x f ′′( x) = − sin x sin x ⎛ 1 ⎞ ⎛ sin x ⎞ + sin x cos x ⎜ − ⎟⎜ ⎟ (cos x) ⎜ sin x 2 ⎟ ⎜ sin x ⎟ ⎝ ⎠⎝ ⎠ cos 2 x sin 2 x cos 2 x sin 2 x = − − =− = − sin x sin x sin x sin x sin x f ′′( x) < 0 when x ≠ k π , k any integer 25. Domain: (−∞, ∞) f(x) is never concave up and concave down on Range: [0,1] (k π , (k + 1) π ) where k is any integer. Even function since No inflection points h(−t ) = cos 2 (−t ) = cos 2 t = h(t ) so the function is symmetric with respect to the y-axis. π y-intercept: y = 1 ; t-intercepts: x = + kπ 2 where k is any integer. kπ h '(t ) = −2 cos t sin t ; h '(t ) = 0 at t = . 2 kπ Critical points: t = 2 198 Section 3.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• π cos4 t + sin t (3) cos 2 t sin t h '(t ) > 0 when kπ + < t < (k + 1)π . The g '(t ) = 2 2 cos6 t function is increasing on the intervals cos 2 t + 3sin 2 t [ kπ + (π / 2), (k + 1)π ] and decreasing on the =2 cos 4 t intervals [ kπ , kπ + (π / 2) ] . 1 + 2sin 2 t Global maxima h ( kπ ) = 1 =2 >0 cos 4 t ⎛π ⎞ over the entire domain. Thus the function is Global minima h ⎜ + kπ ⎟ = 0 ⎝2 ⎠ ⎛ π π⎞ concave up on ⎜ kπ − , kπ + ⎟ ; no inflection h ''(t ) = 2sin 2 t − 2 cos 2 t = −2(cos 2t ) ⎝ 2 2⎠ ⎛ π π⎞ points. h ''(t ) < 0 on ⎜ kπ − , kπ + ⎟ so h is concave ⎝ 4 4⎠ π No horizontal asymptotes; t = + kπ are ⎛ π 3π ⎞ 2 down, and h ''(t ) > 0 on ⎜ kπ + , kπ + ⎟ so h vertical asymptotes. ⎝ 4 4 ⎠ is concave up. ⎛ kπ π 1 ⎞ Inflection points: ⎜ + , ⎟ ⎝ 2 4 2⎠ No vertical asymptotes; no horizontal asymptotes. 27. Domain: ≈ (– ∞ , 0.44) ∪ (0.44, ∞ ); range: (– ∞ , ∞ ) Neither an even nor an odd function π 26. Domain: all reals except t = + kπ y-intercept: 0; x-intercepts: 0, ≈ 0.24 2 Range: [0, ∞) 74.6092 x3 – 58.2013 x 2 + 7.82109 x y-intercepts: y = 0 ; t-intercepts: t = kπ where k f ′( x) = ; is any integer. (7.126 x – 3.141) 2 Even function since f ′( x) = 0 when x = 0, ≈ 0.17, ≈ 0.61 g (−t ) = tan 2 (−t ) = (− tan t )2 = tan 2 t so the function is symmetric with respect to the y-axis. Critical points: 0, ≈ 0.17, ≈ 0.61 2sin t g '(t ) = 2sec 2 t tan t = ; g '(t ) = 0 when f ′( x) > 0 when 0 < x < 0.17 or 0.61 < x cos3 t t = kπ . f(x) is increasing on ≈ [0, 0.17] ∪ [0.61, ∞ ) Critical points: kπ ⎡ π⎞ and decreasing on g ( t ) is increasing on ⎢ kπ , kπ + ⎟ and ⎣ 2⎠ π (– ∞ , 0] ∪ [0.17, 0.44) ∪ (0.44, 0.61] ⎛ ⎤ decreasing on ⎜ kπ − , kπ ⎥ . ⎝ 2 ⎦ Local minima f(0) = 0, f(0.61) ≈ 0.60; local Global minima g (kπ ) = 0 ; no local maxima maximum f(0.17) ≈ 0.01 531.665 x3 – 703.043 x 2 + 309.887 x – 24.566 f ′′( x) = ; 3 (7.126 x – 3.141) f ′′( x) > 0 when x < 0.10 or x > 0.44 Instructor’s Resource Manual Section 3.5 199 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• f(x) is concave up on (– ∞ , 0.10) ∪ (0.44, ∞ ) 30. and concave down on (0.10, 0.44); inflection point ≈ (0.10, 0.003) 5.235 x3 − 1.245 x 2 5.235 x 2 − 1.245 x lim = lim =∞ x →∞ 7.126 x − 3.141 x →∞ 7.126 − 3.141 x so f(x) does not have a horizontal asymptote. 31. As x → 0.44 – , 5.235 x3 – 1.245 x 2 → 0.20 while 7.126 x – 3.141 → 0 – , so lim f ( x) = – ∞; x →0.44 – as x → 0.44+ , 5.235 x3 – 1.245 x 2 → 0.20 while 7.126 x – 3.141 → 0+ , so lim f ( x) = ∞; x →0.44+ 32. x ≈ 0.44 is a vertical asymptote of f(x). 33. 28. 34. 29. 200 Section 3.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 35. 40. Let f ( x) = ax 2 + bx + c, then f ′( x) = 2ax + b and f ′′( x) = 2a. An inflection point occurs where f ′′( x) changes from positive to negative, but 2a is either always positive or always negative, so f(x) does not have any inflection points. ( f ′′( x) = 0 only when a = 0, but then f(x) is not a quadratic curve.) 36. y ′ = 5( x – 1) 4 ; y ′′ = 20( x – 1)3 ; y ′′( x) > 0 41. Let f ( x ) = ax3 + bx 2 + cx + d , then when x > 1; inflection point (1, 3) f ′( x) = 3ax 2 + 2bx + c and f ′′( x) = 6ax + 2b. As At x = 1, y ′ = 0, so the linear approximation is a long as a ≠ 0 , f ′′( x) will be positive on one horizontal line. b side of x = and negative on the other side. 3a b x= is the only inflection point. 3a 3 42. Let f ( x) = ax 4 + bx3 + cx 2 + dx + c, then f ′( x) = 4ax3 + 3bx 2 + 2cx + d and f ′′( x) = 12ax 2 + 6bx + 2c = 2(6ax 2 + 3bx + c) 37. Inflection points can only occur when f ′′( x) changes sign from positive to negative and f ′′( x) = 0. f ′′( x) has at most 2 zeros, thus f(x) has at most 2 inflection points. 43. Since the c term is squared, the only difference occurs when c = 0. When c = 0, 3 y = x2 x2 = x which has domain (– ∞ , ∞ ) 38. and range [0, ∞ ). When c ≠ 0, y = x 2 x 2 – c 2 has domain (– ∞ , –|c|] ∪ [|c|, ∞ ) and range [0, ∞ ). 39. Instructor’s Resource Manual Section 3.5 201 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• The only extremum points are ± c . For c = 0 , 1 45. f ( x) = , then there is one minimum, for c ≠ 0 there are two. (cx – 4)2 + cx 2 2 No maxima, independent of c. No inflection 2cx(7 − 2cx 2 ) points, independent of c. f ′( x) = ; [(cx 2 – 4) 2 + cx 2 ]2 f ( x) = cx = cx 7 44. 2 If c > 0, f ′( x) = 0 when x = 0, ± . 4 + (cx) 4 + c2 x2 2c c(4 – c 2 x 2 ) 2 If c < 0, f ′( x) = 0 when x = 0. f ′( x) = ; f ′( x) = 0 when x = ± (4 + c 2 x 2 )2 c 1 Note that f(x) = (a horizontal line) if c = 0. unless c = 0, in which case f(x) = 0 and 16 f ′( x) = 0. 7 If c > 0, f ′( x) > 0 when x < − and ⎡ 2 2⎤ 2c If c > 0, f(x) is increasing on ⎢ – , ⎥ and ⎣ c c⎦ 7 0< x< , so f(x) is increasing on ⎛ 2⎤ ⎡2 ⎞ 2c decreasing on ⎜ – ∞, – ⎥ ∪ ⎢ , ∞ ⎟ , thus, f(x) has ⎝ c⎦ ⎣c ⎠ ⎛ 7 ⎤ ⎡ 7 ⎤ ⎜ −∞, − ⎜ ⎥ ∪ ⎢ 0, ⎥ and decreasing on ⎛ 2⎞ 1 ⎝ 2c ⎦ ⎣ 2c ⎦ a global minimum at f ⎜ – ⎟ = – and a global ⎝ c⎠ 4 ⎡ 7 ⎤ ⎡ 7 ⎞ ⎛ 2⎞ 1 ⎢− , 0⎥ ∪ ⎢ , ∞ ⎟ . Thus, f(x) has local ⎟ maximum of f ⎜ ⎟ = . ⎣ 2c ⎦ ⎣ 2 c ⎠ ⎝c⎠ 4 ⎛ 7 ⎞ 4 ⎛ 7 ⎞ 4 ⎛ 2⎤ ⎡ 2 ⎞ maxima f ⎜ − ⎜ 2c ⎟ = 15 , f ⎜ 2c ⎟ = 15 and ⎟ ⎜ ⎟ If c < 0, f(x) is increasing on ⎜ – ∞, ⎥ ∪ ⎢ – , ∞ ⎟ ⎝ c⎦ ⎣ c ⎠ ⎝ ⎠ ⎝ ⎠ ⎡2 2⎤ 1 and decreasing on ⎢ , – ⎥ . Thus, f(x) has a local minimum f (0) = . If c < 0, f ′( x) > 0 ⎣c c⎦ 16 when x < 0, so f(x) is increasing on (– ∞ , 0] and ⎛ 2⎞ 1 decreasing on [0, ∞ ). Thus, f(x) has a local global minimum at f ⎜ – ⎟ = – and a global ⎝ c⎠ 4 1 maximum f (0) = . Note that f(x) > 0 and has ⎛ 2⎞ 1 16 maximum at f ⎜ ⎟ = . ⎝c⎠ 4 horizontal asymptote y = 0. 2c3 x(c 2 x 2 – 12) f ′′( x) = , so f(x) has inflection (4 + c 2 x 2 )3 2 3 points at x = 0, ± , c ≠ 0 c 202 Section 3.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 1 If c < 0 : 46. f ( x) = . By the quadratic formula, 2 x + 4x + c ⎡ (4k + 1)π ( 4k − 1) π ⎤ f ( x ) is decreasing on ⎢ , ⎥ x 2 + 4 x + c = 0 when x = –2 ± 4 – c . Thus f(x) ⎣ 2c 2c ⎦ has vertical asymptote(s) at x = –2 ± 4 – c ⎡ ( 4k − 1) π ( 4k − 3) π ⎤ f ( x ) is increasing on ⎢ , ⎥ –2 x – 4 ⎣ 2c 2c ⎦ when c ≤ 4. f ′( x) = ; f ′( x) = 0 ( x + 4 x + c) 2 2 ( 4k − 1) f ( x ) has local minima at x = π and local when x = –2, unless c = 4 since then x = –2 is a 2c vertical asymptote. ( 4k − 3 ) π For c ≠ 4, f ′( x) > 0 when x < –2, so f(x) is maxima at x = where k is an integer. 2c increasing on (– ∞ , –2] and decreasing on [–2, ∞ ) (with the asymptotes excluded). Thus If c = 0 , f ( x ) = 0 and there are no extrema. 1 f(x) has a local maximum at f (–2) = . For c–4 If c > 0 : c = 4, f ′( x) = – 2 ⎡ ( 4k − 3) π ( 4k − 1) π ⎤ so f(x) is increasing on f ( x ) is decreasing on ⎢ , ⎥ ( x + 2)3 ⎣ 2c 2c ⎦ (– ∞ , –2) and decreasing on (–2, ∞ ). ⎡ ( 4k − 1) π (4k + 1)π ⎤ f ( x ) is increasing on ⎢ , ⎣ 2c 2c ⎥ ⎦ ( 4k − 1) f ( x ) has local minima at x = π and 2c ( 4k − 3 ) π local maxima at x = where k is an 2c integer. y −3π −2π −π π 2π 3π x c = −2 c y 47. f ( x ) = c + sin cx . Since c is constant for all x and sin cx is continuous everywhere, the function f ( x ) is continuous everywhere. f ' ( x ) = c ⋅ cos cx −3π −2π −π π 2π 3π x ( )π ( ) c f ' ( x ) = 0 when cx = k + 1 2 or x = k + 1 π 2 c c = −1 where k is an integer. f '' ( x ) = −c 2 ⋅ sin cx (( ) ) ( ( ) ) f '' k + 1 π = −c 2 ⋅ sin c ⋅ k + 1 π = −c 2 ⋅ ( −1) 2 c 2 c k In general, the graph of f will resemble the graph of y = sin x . The period will decrease as c increases and the graph will shift up or down depending on whether c is positive or negative. If c = 0 , then f ( x ) = 0 . Instructor’s Resource Manual Section 3.5 203 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• y Justification: f (1) = g (1) = 1 f (− x) = g ((− x)4 ) = g ( x 4 ) = f ( x) c=0 f is an even function; symmetric with respect to the y-axis. −3π −2π −π π 2π 3π x f '( x) = g '( x 4 )4 x3 f '( x) > 0 for x on (0,1) ∪ (1, ∞) f '( x) < 0 for x on (−∞, −1) ∪ (−1, 0) f '( x) = 0 for x = −1, 0,1 since f ' is continuous. y f ''( x) = g ''( x 4 )16 x6 + g '( x)12 x 2 f ''( x) = 0 for x = −1, 0,1 c=1 f ''( x) > 0 for x on (0, x0 ) ∪ (1, ∞) c f ''( x) < 0 for x on ( x0 ,1) −3π −2π −π π 2π 3π x Where x0 is a root of f ''( x) = 0 (assume that there is only one root on (0, 1)). y 50. Suppose H ′′′(1) < 0, then H ′′( x) is decreasing in a neighborhood around x = 1. Thus, H ′′( x) > 0 c c=2 to the left of 1 and H ′′( x) < 0 to the right of 1, so H(x) is concave up to the left of 1 and concave down to the right of 1. Suppose H ′′′(1) > 0, then −3π −2π −π π 2π 3π x H ′′( x) is increasing in a neighborhood around x = 1. Thus, H ′′( x) < 0 to the left of 1 and H ′′( x) > 0 to the right of 1, so H(x) is concave up to the right of 1 and concave down to the left of 1. In either case, H(x) has a point of inflection 48. Since we have f ''' ( c ) > 0 , we know that f ' ( x ) at x = 1 and not a local max or min. is concave up in a neighborhood around x = c . Since f ' ( c ) = 0 , we then know that the graph of 51. a. Not possible; F ′( x) > 0 means that F(x) is increasing. F ′′( x) > 0 means that the rate at f ' ( x ) must be positive in that neighborhood. which F(x) is increasing never slows down. This means that the graph of f must be increasing Thus the values of F must eventually to the left of c and increasing to the right of c. become positive. Therefore, there is a point of inflection at c. b. Not possible; If F(x) is concave down for all 49. x, then F(x) cannot always be positive. c. 204 Section 3.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 52. a. 53. a. No global extrema; inflection point at (0, 0) f ′( x) = 2 cos x – 2 cos x sin x b. = 2 cos x(1 – sin x); π π f ′( x) = 0 when x = – , 2 2 f ′′( x) = –2sin x – 2 cos 2 x + 2sin 2 x = 4sin 2 x – 2sin x – 2; f ′′( x) = 0 when 1 No global maximum; global minimum at sin x = – or sin x = 1 which occur when 2 (0, 0); no inflection points π 5π π x= – ,– , c. 6 6 2 ⎛ π⎞ Global minimum f ⎜ – ⎟ = –2; global ⎝ 2⎠ ⎛π⎞ maximum f ⎜ ⎟ = 2; inflection points ⎝2⎠ ⎛ π⎞ 1 ⎛ 5π ⎞ 1 f ⎜– ⎟ = – , f ⎜– ⎟ = – Global minimum ⎝ 6⎠ 4 ⎝ 6 ⎠ 4 f(– π ) = –2 π + sin (– π ) = –2 π ≈ –6.3; b. global maximum f (π ) = 2π + sin π = 2π ≈ 6.3 ; inflection point at (0, 0) d. f ′( x) = 2 cos x + 2sin x cos x = 2 cos x(1 + sin x); f ′( x) = 0 when π π x=– , 2 2 f ′′( x) = –2sin x + 2 cos 2 x – 2sin 2 x Global minimum = –4sin 2 x – 2sin x + 2; f ′′( x) = 0 when sin(– π) 1 f (– π) = – π – = – π ≈ 3.1; global sin x = –1 or sin x = which occur when 2 2 sin π π π 5π maximum f (π) = π + = π ≈ 3.1; x=– , , 2 2 6 6 inflection point at (0, 0). ⎛ π⎞ Global minimum f ⎜ – ⎟ = –1; global ⎝ 2⎠ ⎛π⎞ maximum f ⎜ ⎟ = 3; inflection points ⎝2⎠ ⎛ π ⎞ 5 ⎛ 5π ⎞ 5 f ⎜ ⎟= , f ⎜ ⎟= . ⎝6⎠ 4 ⎝ 6 ⎠ 4 Instructor’s Resource Manual Section 3.5 205 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• c. f ′( x) = 2 cos 2 x + 3sin 3 x Using the graphs, f(x) has a global minimum at f(2.17) ≈ –1.9 and a global maximum at f(0.97) ≈ 1.9 f ′′( x) = –4sin 2 x + 9 cos 3 x; f ′′( x) = 0 when π π x = – , and when f ′( x) = –2sin 2 x + 2sin x 2 2 = –4sin x cos x + 2sin x = 2sin x(1 – 2 cos x); x ≈ –2.469, –0.673, 0.413, 2.729. π π ⎛ π ⎞ ⎛π ⎞ Inflection points: ⎜ − , 0 ⎟ , ⎜ , 0 ⎟ , f ′( x) = 0 when x = – π, – , 0, , π ⎝ 2 ⎠ ⎝2 ⎠ 3 3 f ′′( x) = –4 cos 2 x + 2 cos x; f ′′( x) = 0 when ≈ ( −2.469, 0.542 ) , ( −0.673, −0.542 ) , x ≈ –2.206, –0.568, 0.568, 2.206 ( 0.413, 0.408 ) , ( 2.729, −0.408) ⎛ π⎞ ⎛π⎞ Global minimum f ⎜ – ⎟ = f ⎜ ⎟ = –1.5; ⎝ 3⎠ ⎝3⎠ 54. Global maximum f(– π ) = f( π ) = 3; Inflection points: ≈ ( −2.206, 0.890 ) , ( −0.568, −1.265 ) , ( 0.568, −1.265) , ( 2.206, 0.890 ) d. y 55. 5 f ′( x) = 3cos 3 x – cos x; f ′( x) = 0 when 3cos 3x = cos x which occurs when −5 5 π π x x = – , and when 2 2 x ≈ –2.7, –0.4, 0.4, 2.7 −5 f ′′( x) = –9sin 3x + sin x which occurs when x = – π , 0, π and when x ≈ –2.126, –1.016, 1.016, 2.126 a. f is increasing on the intervals ( −∞, −3] ⎛π⎞ Global minimum f ⎜ ⎟ = –2; and [ −1, 0] . ⎝2⎠ f is decreasing on the intervals [ −3, −1] ⎛ π⎞ global maximum f ⎜ – ⎟ = 2; and [ 0, ∞ ) . ⎝ 2⎠ Inflection points: ≈ ( −2.126, 0.755 ) , b. f is concave down on the intervals ( −1.016, 0.755 ) , ( 0, 0 ) , (1.016, −0.755 ) , ( −∞, −2 ) and ( 2, ∞ ) . ( 2.126, −0.755) f is concave up on the intervals ( −2, 0 ) and ( 0, 2 ) . e. c. f attains a local maximum at x = −3 and x = 0. f attains a local minimum at x = −1 . d. f has a point of inflection at x = −2 and x = 2. 206 Section 3.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• y 56. 59. a. 5 −5 5 x 2 x 2 – 9 x + 40 f ′( x) = ; f ′( x) is never 0, −5 x 2 – 6 x + 40 and always positive, so f(x) is increasing for a. f is increasing on the interval [ −1, ∞ ) . all x. Thus, on [–1, 7], the global minimum is f is decreasing on the interval ( −∞, −1] f(–1) ≈ –6.9 and the global maximum if f(7) ≈ 48.0. 2 x3 − 18 x 2 + 147 x – 240 b. f is concave up on the intervals ( −2, 0 ) f ′′( x) = ; f ′′( x) = 0 ( x 2 – 6 x + 40)3 / 2 and ( 2, ∞ ) . when x ≈ 2.02; inflection point f is concave down on the interval ( 0, 2 ) . f(2.02) ≈ 11.4 c. f does not have any local maxima. b. f attains a local minimum at x = −1 . d. f has inflection points at x = 0 and x = 2. 57. Global minimum f(0) = 0; global maximum f(7) ≈ 124.4; inflection point at x ≈ 2.34, f(2.34) ≈ 48.09 c. 58. No global minimum or maximum; no inflection points d. Global minimum f(3) ≈ –0.9; global maximum f(–1) ≈ 1.0 or f(7) ≈ 1.0; Inflection points at x ≈ 0.05 and x ≈ 5.9, f(0.05) ≈ 0.3, f(5.9) ≈ 0.3. Instructor’s Resource Manual Section 3.5 207 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 60. a. 3.6 Concepts Review 1. continuous; (a, b); f (b) – f (a) = f ′(c)(b – a) 2. f ′(0) does not exist. 3. F(x) = G(x) + C f ′( x) = 3 x 2 –16 x + 5; f ′( x) = 0 when 1 4. x 4 + C x = , 5. 3 Global minimum f(5) = –46; ⎛1⎞ Problem Set 3.6 global maximum f ⎜ ⎟ ≈ 4.8 ⎝3⎠ x 8 1. f ′( x) = f ′′( x) = 6 x –16; f ′′( x) = 0 when x = ; x 3 f (2) – f (1) 2 –1 inflection point: ( 8 , −20.6 3 ) 2 –1 = 1 =1 c b. = 1 for all c > 0, hence for all c in (1, 2) c Global minimum when x ≈ –0.5 and x ≈ 1.2, f(–0.5) ≈ 0, f(1.2) ≈ 0; global maximum f(5) = 46 Inflection point: ( −0.5, 0 ) , (1.2, 0 ) , ( 8 , 20.6) 3 2. The Mean Value Theorem does not apply because g ′(0) does not exist. c. No global minimum or maximum; inflection point at x ≈ −0.26, f (−0.26) ≈ −1.7 d. 3. f ′( x) = 2 x + 1 f (2) – f (–2) 6 – 2 = =1 2 – (–2) 4 2c + 1 = 1 when c = 0 No global minimum, global maximum when x ≈ 0.26, f(0.26) ≈ 4.7 Inflection points when x ≈ 0.75 and x ≈ 3.15, f(0.75) ≈ 2.6, f(3.15) ≈ –0.88 208 Section 3.6 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 4. g ′( x) = 3( x + 1)2 7. 1 f ′( z ) = (3 z 2 + 1) = z 2 + 1 g (1) – g (–1) 8 – 0 3 3 = =4 f (2) – f (–1) 2 – (–2) 4 1 – (–1) 2 = = 2 – (–1) 3 3 2 3(c + 1)2 = 4 when c = –1 + ≈ 0.15 1 4 3 c2 + = when c = –1, 1, but −1 is not in 3 3 (−1, 2) so c = 1 is the only solution. 5. H ′( s ) = 2s + 3 H (1) – H (–3) 3 – (–1) 8. The Mean Value Theorem does not apply = =1 1 – (–3) 1 – (–3) because F(t) is not continuous at t = 1. 2c + 3 = 1 when c = –1 3 9. h′( x) = – 6. F ′( x) = x 2 ( x – 3) 2 8 8 F (2) – F (–2) 3 – – 3 = ( ) = 4 h(2) – h(0) –2 – 0 2–0 = 2 = –1 2 – (–2) 4 3 3 – = –1 when c = 3 ± 3, 4 2 (c – 3) 2 c2 = when c = ± ≈ ±1.15 3 3 c = 3 – 3 ≈ 1.27 (3 + 3 is not in (0, 2).) Instructor’s Resource Manual Section 3.6 209 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 10. The Mean Value Theorem does not apply 5 2/3 because f(x) is not continuous at x = 3. 14. g ′( x) = x 3 g (1) – g (–1) 1 – (–1) = =1 1 – (–1) 2 3/ 2 5 2/3 ⎛3⎞ c = 1 when c = ± ⎜ ⎟ ≈ ±0.46 3 ⎝5⎠ 2 11. h′(t ) = 1/ 3 3t h(2) – h(0) 22 / 3 − 0 = = 2 –1/ 3 2–0 2 2 16 = 2 –1/ 3 when c = ≈ 0.59 1/ 3 27 3c 15. S ′(θ ) = cos θ S (π) – S (– π) 0 – 0 = =0 π – (– π) 2π π cos c = 0 when c = ± . 2 12. The Mean Value Theorem does not apply because h′(0) does not exist. 16. The Mean Value Theorem does not apply because C (θ ) is not continuous at θ = −π , 0, π . 5 2/3 13. g ′( x) = x 3 g (1) – g (0) 1 – 0 = =1 1– 0 1 3/ 2 5 2/3 ⎛3⎞ 17. The Mean Value Theorem does not apply c = 1 when c = ± ⎜ ⎟ , 3 ⎝5⎠ π because T (θ ) is not continuous at θ = . 3/ 2 ⎛ ⎛ 3 ⎞3 / 2 ⎞ 2 ⎛3⎞ c=⎜ ⎟ ≈ 0.46, ⎜ – ⎜ ⎟ is not in (0, 1). ⎟ ⎝5⎠ ⎜ ⎝5⎠ ⎟ ⎝ ⎠ 210 Section 3.6 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 18. The Mean Value Theorem does not apply 22. By the Mean Value Theorem because f(x) is not continuous at x = 0. f (b) − f (a) = f ′(c) for some c in (a, b). b−a 0 Since f(b) = f(a), = f ′(c); f ′(c) = 0 . b−a f (8) − f (0) 1 23. =− 8−0 4 There are three values for c such that 1 f ′(c) = − . 4 1 They are approximately 1.5, 3.75, and 7. 19. f ′( x) = 1 – x2 5 24. f ′( x) = 2α x + β f (2) – f (1) 2 – 2 1 f (b) – f (a ) 1 = = = [α (b 2 – a 2 ) + β (b – a )] 2 –1 1 2 b–a b–a 1 1 1– = when c = ± 2, c = 2 ≈ 1.41 = α ( a + b) + β c2 2 a+b (c = – 2 is not in (1, 2).) 2α c + β = α (a + b) + β when c = which is 2 the midpoint of [a, b]. 25. By the Monotonicity Theorem, f is increasing on the intervals (a, x0 ) and ( x0 , b) . To show that f ( x0 ) > f ( x) for x in (a, x0 ) , consider f on the interval (a, x0 ] . f satisfies the conditions of the Mean Value Theorem on the interval [ x, x0 ] for x in (a, x0 ) . So for some c in ( x, x0 ), 20. The Mean Value Theorem does not apply f ( x0 ) − f ( x) = f ′(c)( x0 − x) . because f(x) is not continuous at x = 2. Because f ′(c) > 0 and x0 − x > 0, f ( x0 ) − f ( x ) > 0, so f ( x0 ) > f ( x ) . Similar reasoning shows that f ( x) > f ( x0 ) for x in ( x0 , b) . Therefore, f is increasing on (a, b). 26. a. f ′( x) = 3 x 2 > 0 except at x = 0 in (– ∞ , ∞ ). f ( x) = x3 is increasing on (– ∞ , ∞ ) by Problem 25. 21. The Mean Value Theorem does not apply because f is not differentiable at x = 0 . b. f ′( x) = 5 x 4 > 0 except at x = 0 in (– ∞ , ∞ ). f ( x) = x5 is increasing on (– ∞ , ∞ ) by Problem 25. ⎧3 x 2 x ≤ 0 ⎪ c. f ′( x) = ⎨ > 0 except at x = 0 in ⎪1 ⎩ x>0 (– ∞ , ∞ ). ⎧ x3 x ≤ 0 ⎪ f ( x) = ⎨ is increasing on ⎪x ⎩ x>0 (– ∞ , ∞ ) by Problem 25. Instructor’s Resource Manual Section 3.6 211 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 27. s(t) is defined in any interval not containing t = 0. 34. f ′( x) = 6 x 2 – 18 x = 6 x( x – 3); f ′( x) = 0 when 1 s ′(c) = – < 0 for all c ≠ 0. For any a, b with x = 0 or x = 3. c2 f(–1) = –10, f(0) = 1 so, by Problem 33, f(x) = 0 a < b and both either positive or negative, the has exactly one solution on (–1, 0). Mean Value Theorem says f(0) = 1, f(1) = –6 so, by Problem 33, f(x) = 0 has s (b) – s (a) = s ′(c)(b – a ) for some c in (a, b). exactly one solution on (0, 1). Since a < b, b – a > 0 while s ′(c) < 0, hence f(4) = –15, f(5) = 26 so, by Problem 33, f(x) = 0 has exactly one solution on (4, 5). s(b) – s(a) < 0, or s(b) < s(a). Thus, s(t) is decreasing on any interval not 35. Suppose there is more than one zero between containing t = 0. successive distinct zeros of f ′ . That is, there are 2 a and b such that f(a) = f(b) = 0 with a and b 28. s ′(c) = – < 0 for all c > 0. If 0 < a < b, the between successive distinct zeros of f ′ . Then by c3 Rolle’s Theorem, there is a c between a and b Mean Value Theorem says such that f ′(c) = 0 . This contradicts the s (b) – s (a) = s ′(c)(b – a ) for some c in (a, b). supposition that a and b lie between successive Since a < b, b – a > 0 while s ′(c) < 0, hence distinct zeros. s(b) – s(a) < 0, or s(b) < s(a). Thus, s(t) is decreasing on any interval to the right of the 36. Let x1 , x2 , and x3 be the three values such that origin. g ( x1 ) = g ( x2 ) = g ( x3 ) = 0 and 29. F ′( x) = 0 and G ( x ) = 0; G ′( x) = 0 . a ≤ x1 < x2 < x3 ≤ b . By applying Rolle’s By Theorem B, Theorem (see Problem 22) there is at least one F(x) = G(x) + C, so F(x) = 0 + C = C. number x4 in ( x1 , x2 ) and one number x5 in ( x2 , x3 ) such that g ′( x4 ) = g ′( x5 ) = 0 . Then by 30. F ( x ) = cos 2 x + sin 2 x; F (0) = 12 + 02 = 1 applying Rolle’s Theorem to g ′( x) , there is at F ′( x) = 2 cos x (− sin x) + 2sin x(cos x) = 0 least one number x6 in ( x4 , x5 ) such that By Problem 29, F(x) = C for all x. g ′′( x6 ) = 0 . Since F(0) = 1, C = 1, so sin 2 x + cos 2 x = 1 for all x. 37. f(x) is a polynomial function so it is continuous on [0, 4] and f ′′( x) exists for all x on (0, 4). 31. Let G ( x) = Dx; F ′( x) = D and G ′( x) = D . f(1) = f(2) = f(3) = 0, so by Problem 36, there are By Theorem B, F(x) = G(x) + C; F(x) = Dx + C. at least two values of x in [0, 4] where f ′( x) = 0 32. F ′( x) = 5; F (0) = 4 and at least one value of x in [0, 4] where f ′′( x) = 0. F(x) = 5x + C by Problem 31. F(0) = 4 so C = 4. 38. By applying the Mean Value Theorem and taking F(x) = 5x + 4 the absolute value of both sides, 33. Since f(a) and f(b) have opposite signs, 0 is f ( x2 ) − f ( x1 ) = f ′(c) , for some c in ( x1 , x2 ) . between f(a) and f(b). f(x) is continuous on [a, b], x2 − x1 since it has a derivative. Thus, by the Intermediate Value Theorem, there is at least one Since f ′( x) ≤ M for all x in (a, b), point c, f ( x2 ) − f ( x1 ) a < c < b with f(c) = 0. ≤ M ; f ( x2 ) − f ( x1 ) ≤ M x2 − x1 . x2 − x1 Suppose there are two points, c and c ′, c < c ′ in (a, b) with f (c) = f (c′) = 0. Then by Rolle’s 39. f ′( x) = 2 cos 2 x; f ′( x) ≤ 2 Theorem, there is at least one number d in (c, c ′) f ( x2 ) − f ( x1 ) f ( x2 ) − f ( x1 ) with f ′(d ) = 0. This contradicts the given = f ′( x) ; ≤2 x2 − x1 x2 − x1 information that f ′( x) ≠ 0 for all x in [a, b], thus f ( x2 ) − f ( x1 ) ≤ 2 x2 − x1 ; there cannot be more than one x in [a, b] where f(x) = 0. sin 2 x2 − sin 2 x1 ≤ 2 x2 − x1 212 Section 3.6 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 40. a. 1 44. Let f ( x) = x so f ′( x) = . Apply the Mean 2 x Value Theorem to f on the interval [x, x + 2] for x > 0. 1 1 Thus x + 2 − x = (2) = for some c in 2 c c 1 1 1 (x, x + 2). Observe < < . x+2 c x 1 b. Thus as x → ∞, → 0. c ( Therefore lim x + 2 − x = lim x →∞ x →∞ c )1 =0. 45. Let f(x) = sin x. f ′( x) = cos x, so f ′( x) = cos x ≤ 1 for all x. By the Mean Value Theorem, f ( x) − f ( y ) = f ′(c) for some c in (x, y). x− y 41. Suppose f ′( x) ≥ 0 . Let a and b lie in the interior f ( x) − f ( y ) of I such that b > a. By the Mean Value Theorem, Thus, = f ′(c) ≤ 1; there is a point c between a and b such that x− y f (b) − f (a ) f (b) − f (a) sin x − sin y ≤ x − y . f ′(c) = ; ≥0. b−a b−a Since a < b, f(b) ≥ f(a), so f is nondecreasing. 46. Let d be the difference in distance between horse Suppose f ′( x) ≤ 0. Let a and b lie in the interior A and horse B as a function of time t. of I such that b > a. By the Mean Value Theorem, Then d ′ is the difference in speeds. there is a point c between a and b such that Let t0 and t1 and be the start and finish times of f (b) − f (a ) f (b) − f (a) the race. f ′(c) = ; ≤ 0 . Since b−a b−a d (t0 ) = d (t1 ) = 0 a < b, f(a) ≥ f(b), so f is nonincreasing. By the Mean Value Theorem, d (t1 ) − d (t0 ) = d ′(c) for some c in (t0 , t1 ) . 42. [ f 2 ( x)]′ = 2 f ( x) f ′( x) t1 − t0 Because f(x) ≥ 0 and f ′( x) ≥ 0 on I , [ f 2 ( x)]′ ≥ 0 Therefore d ′(c) = 0 for some c in (t0 , t1 ) . on I. As a consequence of the Mean Value Theorem, 47. Let s be the difference in speeds between horse A and horse B as function of time t. f 2 ( x2 ) − f 2 ( x1 ) ≥ 0 for all x2 > x1 on I. Then s ′ is the difference in accelerations. Therefore f 2 is nondecreasing. Let t2 be the time in Problem 46 at which the horses had the same speeds and let t1 be the 43. Let f(x) = h(x) – g(x). finish time of the race. f ′( x) = h′( x) − g ′( x); f ′( x) ≥ 0 for all x in s (t2 ) = s (t1 ) = 0 (a, b) since g ′( x) ≤ h′( x ) for all x in (a, b), so f is By the Mean Value Theorem, nondecreasing on (a, b) by Problem 41. Thus s (t1 ) − s (t2 ) x1 < x2 ⇒ f ( x1 ) ≤ f ( x2 ); = s ′(c) for some c in (t2 , t1 ) . t1 − t2 h( x1 ) − g ( x1 ) ≤ h( x2 ) − g ( x2 ); Therefore s ′(c) = 0 for some c in (t2 , t1 ) . g ( x2 ) − g ( x1 ) ≤ h( x2 ) − h( x1 ) for all x1 and x2 in (a, b). Instructor’s Resource Manual Section 3.6 213 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 48. Suppose x > c. Then by the Mean Value 52. s is differentiable with s (0) = 0 and s (18) = 20 so Theorem, we can apply the Mean Value Theorem. There f ( x) − f (c ) = f ′(a )( x − c) for some a in (c, x) . exists a c in the interval ( 0,18 ) such that Since f is concave up, f ′′ > 0 and by the (20 − 0) Monotonicity Theorem f ′ is increasing. v(c) = s '(c) = ≈ 1.11 miles per minute (18 − 0 ) Therefore f ′(a ) > f ′(c) and ≈ 66.67 miles per hour f ( x) − f (c ) = f ′(a )( x − c) > f ′(c)( x − c) f ( x) > f (c ) + f ′(c )( x − c), x > c Suppose x < c. Then by the Mean Value Theorem, f (c) − f ( x) = f ′(a )(c − x) for some a in ( x, c) . Since f is concave up, f ′′ > 0 , and by the Monotonicity Theorem f ′ is increasing. Therefore, f ′(c) > f ′(a) and f (c) − f ( x ) = f ′(a )(c − x) < f ′(c)(c − x) . − f ( x) < − f (c ) + f ′(c)(c − x) f ( x) > f (c) − f ′(c)(c − x) f ( x) > f (c ) + f ′(c )( x − c), x < c Therefore f ( x) > f (c ) + f ′(c )( x − c), x ≠ c . 49. Fix an arbitrary x. f ( y) − f ( x) f ' ( x ) = lim = 0 , since y→x y−x 53. Since the car is stationary at t = 0 , and since v is f ( y) − f ( x) ≤M y−x . 1 y−x continuous, there exists a δ such that v(t ) < 2 So, f ' ≡ 0 → f = constant . for all t in the interval [0, δ ] . v(t ) is therefore 1 1 δ less than and s (δ ) < δ ⋅ = . By the Mean 50. 1/ 3 f ( x) = x on [0, a] or [–a, 0] where a is any 2 2 2 Value Theorem, there exists a c in the interval positive number. f ′(0) does not exist, but f(x) (δ , 20) such that has a vertical tangent line at x = 0. ⎛ δ⎞ ⎜ 20 − ⎟ 51. Let f(t) be the distance traveled at time t. v(c) = s '(c) = ⎝ 2⎠ f (2) − f (0) 112 − 0 (20 − δ ) = = 56 20 − δ 2−0 2 > By the Mean Value Theorem, there is a time c 20 − δ such that f ′(c) = 56. = 1 mile per minute At some time during the trip, Johnny must have = 60 miles per hour gone 56 miles per hour. 214 Section 3.6 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 54. Given the position function s ( t ) = at 2 + bt + c , the car’s instantaneous velocity is given by the Problem Set 3.7 function s ' ( t ) = 2at + b . 1. Let f ( x) = x3 + 2 x – 6. A+ B The midpoint of the interval [ A, B ] is . f(1) = –3, f(2) = 6 2 Thus, the car’s instantaneous velocity at the n hn mn f (mn ) midpoint of the interval is given by ⎛ A+ B ⎞ ⎛ A+ B ⎞ 1 0.5 1.5 0.375 s '⎜ ⎟ = 2a ⎜ ⎟+b ⎝ 2 ⎠ ⎝ 2 ⎠ 2 0.25 1.25 –1.546875 = a ( A + B) + b 3 0.125 1.375 –0.650391 The car’s average velocity will be its change in 4 0.0625 1.4375 –0.154541 position divided by the length of the interval. 5 0.03125 1.46875 0.105927 That is, 6 0.015625 1.45312 –0.0253716 s ( B ) − s ( A ) ( aB + bB + c ) − ( aA + bA + c ) 2 2 7 0.0078125 1.46094 0.04001 = B− A B− A 8 0.00390625 1.45703 0.00725670 2 2 aB − aA + bB − bA 9 0.00195312 1.45508 –0.00907617 = B− A r ≈ 1.46 = ( ) a B 2 − A2 + b ( B − A ) B− A 2. Let f ( x) = x 4 + 5 x3 + 1. a ( B − A )( B + A ) + b ( B − A ) f(–1) = –3, f(0) = 1 = B− A n f (mn ) = a ( B + A) + b hn mn = a ( A + B) + b 1 0.5 –0.5 0.4375 This is the same result as the instantaneous 2 0.25 –0.75 –0.792969 velocity at the midpoint. 3 0.125 –0.625 –0.0681152 4 0.0625 –0.5625 0.21022 5 0.03125 –0.59375 0.0776834 6 0.015625 –0.609375 0.00647169 3.7 Concepts Review 7 0.0078125 –0.617187 –0.0303962 8 0.00390625 –0.613281 –0.011854 1. slowness of convergence 9 0.00195312 –0.611328 –0.00266589 2. root; Intermediate Value r ≈ –0.61 3. algorithms 3. Let f ( x ) = 2 cos x − sin x . 4. fixed point f (1) ≈ 0.23913 ; f ( 2 ) ≈ −1.74159 n hn mn f ( mn ) 1 0.5 1.5 −0.856021 2 0.25 1.25 −0.318340 3 0.125 1.125 −0.039915 4 0.0625 1.0625 0.998044 5 0.03125 1.09375 0.029960 6 0.01563 1.109375 −0.004978 r ≈ 1.11 Instructor’s Resource Manual Section 3.7 215 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 4. Let f ( x ) = x − 2 + 2 cos x n xn f (1) = 1 − 2 + 2 cos (1) ≈ 0.080605 1 1 f ( 2 ) = 2 − 2 + 2 cos ( 2 ) ≈ −0.832294 2 0.8636364 n hn mn f ( mn ) 3 0.8412670 1 0.5 1.5 −0.358526 4 0.8406998 2 0.25 1.25 −0.119355 5 0.8406994 3 0.125 1.125 −0.012647 6 0.8406994 4 0.0625 1.0625 0.035879 r ≈ 0.84070 5 0.03125 1.09375 0.012065 7. Let f ( x ) = x − 2 + 2 cos x . 6 0.01563 1.109375 −0.000183 y r ≈ 1.11 5 5. Let f ( x) = x3 + 6 x 2 + 9 x + 1 = 0 . −5 5 x −5 f ' ( x ) = 1 − 2sin x n xn f ′( x) = 3x 2 + 12 x + 9 1 4 2 3.724415 n xn 3 3.698429 1 0 4 3.698154 2 –0.1111111 5 3.698154 3 –0.1205484 r ≈ 3.69815 4 –0.1206148 8. Let f ( x ) = 2 cos x − sin x . 5 –0.1206148 y r ≈ –0.12061 5 6. Let f ( x) = 7 x3 + x – 5 −5 5 x −5 f ' ( x ) = −2sin x − cos x n xn f ′( x) = 21x 2 + 1 1 0.5 2 1.1946833 3 1.1069244 4 1.1071487 5 1.1071487 r ≈ 1.10715 216 Section 3.7 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 9. Let f(x) = cos x – 2x. f ′( x) = 4 x3 – 24 x 2 + 44 x – 24 Note that f(2) = 0. n xn 1 0.5 2 0.575 3 0.585586 4 0.585786 f ′( x) = – sin x – 2 n xn 1 3.5 n xn 2 3.425 1 0.5 3 3.414414 2 0.4506267 4 3.414214 3 0.4501836 5 3.414214 4 0.4501836 r = 2, r ≈ 0.58579, r ≈ 3.41421 r ≈ 0.45018 12. Let f ( x) = x 4 + 6 x3 + 2 x 2 + 24 x – 8. 10. Let f ( x ) = 2 x − sin x − 1 . y 5 −5 5 x f ′( x) = 4 x3 + 18 x 2 + 4 x + 24 −5 f ' ( x ) = 2 − cos x n xn n xn 1 –6.5 1 1 2 –6.3299632 2 0.891396 3 –6.3167022 3 0.887866 4 –6.3166248 4 0.887862 5 –6.3166248 5 0.887862 r ≈ 0.88786 n xn 11. Let f ( x) = x 4 – 8 x3 + 22 x 2 – 24 x + 8. 1 0.5 2 0.3286290 3 0.3166694 4 0.3166248 5 0.3166248 r ≈ –6.31662, r ≈ 0.31662 Instructor’s Resource Manual Section 3.7 217 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 13. Let f ( x ) = 2 x 2 − sin x . 16. Let f ( x) = x 4 – 47 . y f ′( x) = 4 x3 2 n xn 1 2.5 2 2.627 −2 2 x 3 2.618373 4 2.618330 −2 5 2.618330 4 f ' ( x ) = 4 x − cos x 47 ≈ 2.61833 n xn 17. f ( x ) = x 4 + x3 + x 2 + x is continuous on the 1 0.5 given interval. 2 0.481670 3 0.480947 4 0.480946 r ≈ 0.48095 14. Let f ( x ) = 2 cot x − x . From the graph of f, we see that the maximum y value of the function on the interval occurs at the right endpoint. The minimum occurs at a stationary point within the interval. To find 2 where the minimum occurs, we solve f ' ( x ) = 0 on the interval [ −1,1] . f ' ( x ) = 4 x3 + 3x 2 + 2 x + 1 = g ( x ) −2 2 x Using Newton’s Method to solve g ( x ) = 0 , we −2 get: n xn f ' ( x ) = −2 csc 2 x − 1 1 0 2 −0.5 n xn 3 −0.625 1 1 4 −0.60638 2 1.074305 5 −0.60583 3 1.076871 6 −0.60583 4 1.076874 Minimum: f ( −0.60583) ≈ −0.32645 r ≈ 1.07687 Maximum: f (1) = 4 15. Let f ( x) = x3 – 6. f ′( x) = 3 x 2 n xn 1 1.5 2 1.888889 3 1.819813 4 1.817125 5 1.817121 6 1.817121 3 6 ≈ 1.81712 218 Section 3.7 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• x3 + 1 n xn n xn 18. f ( x) = is continuous on the given x4 + 1 1 4.712389 1 7.853982 interval. 2 4.479179 2 7.722391 3 4.793365 3 7.725251 4 4.493409 4 7.725252 5 4.493409 5 7.725252 Minimum: f ( 4.493409 ) ≈ −0.21723 From the graph of f, we see that the maximum Maximum: f ( 7.725252 ) ≈ 0.128375 and minimum will both occur at stationary points within the interval. The minimum appears to x 20. f ( x ) = x 2 sin is continuous on the given occur at about x = −1.5 while the maximum 2 appears to occur at about x = 0.8 . To find the interval. stationary points, we solve f ' ( x ) = 0 . f '( x) = ( − x2 x4 + 4 x − 3 ) = g ( x) (x ) 4 2 +1 Using Newton’s method to solve g ( x ) = 0 on From the graph of f, we see that the minimum value and maximum value on the interval will the interval, we use the starting values of −1.5 occur at stationary points within the interval. To and 0.8 . find these points, we need to solve f ' ( x ) = 0 on n xn n xn the interval. 1 −1.5 1 0.8 x x x 2 cos + 4 x sin 2 −1.680734 2 0.694908 f '( x) = 2 2 = g ( x) 3 −1.766642 3 0.692512 2 4 −1.783766 4 0.692505 Using Newton’s method to solve g ( x ) = 0 on 5 −1.784357 5 0.692505 3π the interval, we use the starting values of and 6 −1.784358 2 13π 7 −1.784358 . 4 Maximum: f ( 0.692505 ) ≈ 1.08302 n xn n xn Minimum: f ( −1.78436 ) ≈ −0.42032 1 4.712389 1 10.210176 sin x 2 4.583037 2 10.174197 19. f ( x) = is continuous on the given interval. x 3 4.577868 3 10.173970 4 4.577859 4 10.173970 5 4.577859 Minimum: f (10.173970 ) ≈ −96.331841 Maximum: f ( 4.577859 ) ≈ 15.78121 From the graph of f, we see that the minimum value and maximum value on the interval will occur at stationary points within the interval. To find these points, we need to solve f ' ( x ) = 0 on the interval. x cos x − sin x f '( x) = = g ( x) x2 Using Newton’s method to solve g ( x ) = 0 on 3π the interval, we use the starting values of and 2 5π . 2 Instructor’s Resource Manual Section 3.7 219 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 21. Graph y = x and y = 0.8 + 0.2 sin x. b. Let f (i ) = 20i (1 + i ) 24 − (1 + i )24 + 1 = (1 + i )24 (20i − 1) + 1 . Then f ′(i ) = 20(1 + i ) 24 + 480i (1 + i )23 − 24(1 + i )23 = (1 + i )23 (500i − 4), so f (in ) (1 + in ) 24 (20in − 1) + 1 in +1 = in − = in − f ′(in ) (1 + in )23 (500in − 4) xn +1 = 0.8 + 0.2sin xn Let x1 = 1. ⎡ 20i 2 + 19in − 1 + (1 + in ) −23 ⎤ = in − ⎢ n ⎥. ⎢ ⎣ 500in − 4 ⎥ ⎦ n xn 1 1 c. n in 2 0.96829 1 0.012 3 0.96478 2 0.0165297 4 0.96439 5 0.96434 3 0.0152651 6 0.96433 4 0.0151323 7 0.96433 5 0.0151308 x ≈ 0.9643 6 0.0151308 22. i = 0.0151308 r = 18.157% f ( xn ) 24. From Newton’s algorithm, xn +1 – xn = − . f ′( xn ) lim ( xn +1 – xn ) = lim xn +1 – lim xn xn → x xn → x xn → x =x–x =0 f ( xn ) lim exists if f and f ′ are continuous at xn → x f ′( xn ) f ( xn ) x 1/ 3 xn +1 = xn – = xn – n x and f ′( x ) ≠ 0. f ′( xn ) 1 x –2 / 3 3 n f ( xn ) f (x ) = xn – 3 xn = –2 xn Thus, lim = = 0, so f ( x ) = 0. xn → x f ′( xn ) f ′( x ) Thus, every iteration of Newton’s Method gets x is a solution of f(x) = 0. further from zero. Note that xn +1 = (–2) n +1 x0 . Newton’s Method is based on approximating f by xn + 1.5cos xn its tangent line near the root. This function has a 25. xn +1 = 2 vertical tangent at the root. n xn n xn 23. a. For Tom’s car, P = 2000, R = 100, and 1 1 5 0.914864 k = 24, thus 2 0.905227 6 0.914856 100 ⎡ 1 ⎤ 3 0.915744 7 0.914857 2000 = ⎢1 − ⎥ or i ⎢ (1 + i ) 24 ⎥ ⎣ ⎦ 4 0.914773 1 x ≈ 0.91486 20i = 1 − , which is equivalent to (1 + i )24 20i (1 + i ) 24 − (1 + i )24 + 1 = 0 . 220 Section 3.7 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 26. xn +1 = 2 − sin x 29. a. n xn n xn n xn 1 2 5 1.10746 9 1.10603 2 1.09070 6 1.10543 10 1.10607 3 1.11305 7 1.10634 11 1.10606 4 1.10295 8 1.10612 12 1.10606 x ≈ 1.10606 27. xn +1 = 2.7 + xn x ≈ 0.5 2 n xn b. xn +1 = 2( xn – xn ) 1 1 n xn 2 1.923538 1 0.7 3 2.150241 2 0.42 4 2.202326 3 0.4872 5 2.214120 4 0.4996723 6 2.216781 5 0.4999998 7 2.217382 6 0.5 8 2.217517 7 0.5 9 2.217548 c. x = 2( x – x 2 ) 10 2.217554 2 x2 – x = 0 11 2.217556 x(2x – 1) = 0 12 2.217556 1 x = 0, x = x ≈ 2.21756 2 28. xn +1 = 3.2 + xn 30. a. n xn 1 47 2 7.085196 3 3.207054 4 2.531216 5 2.393996 x ≈ 0.8 2 6 2.365163 b. xn +1 = 5( xn – xn ) 7 2.359060 n xn 8 2.357766 1 0.7 9 2.357491 2 1.05 10 2.357433 3 –0.2625 11 2.357421 4 –1.657031 12 2.357418 5 –22.01392 13 2.357418 6 –2533.133 x ≈ 2.35742 Instructor’s Resource Manual Section 3.7 221 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• c. x = 5( x – x 2 ) 33. a. x1 = 1 1 5x2 – 4 x = 0 x2 = 1 + = 2 x(5x – 4) = 0 1 4 1 3 x = 0, x = x3 = 1 + = = 1.5 5 1+ 1 2 1 x1 = 0 1 5 31. a. x4 = 1 + = ≈ 1.6666667 1 3 x2 = 1 = 1 1+ 1 1+ 1 x3 = 1 + 1 = 2 ≈ 1.4142136 1 8 x5 = 1 + = = 1.6 1 5 x4 = 1 + 1 + 1 ≈ 1.553774 1+ 1 1+ x5 = 1 + 1 + 1 + 1 ≈ 1.5980532 1+ 1 1 b. x = 1+ x 1 b. x = 1+ x x2 = 1 + x x2 = x + 1 x2 − x − 1 = 0 x2 − x − 1 = 0 1 ± 1 + 4 ⋅1⋅1 1 ± 5 x= = 1 ± 1 + 4 ⋅1⋅1 1 ± 5 2 2 x= = Taking the minus sign gives a negative 2 2 solution for x, violating the requirement that Taking the minus sign gives a negative solution for x, violating the requirement that 1+ 5 x ≥ 0 . Hence, x = ≈ 1.618034 . 1+ 5 2 x ≥ 0 . Hence, x = ≈ 1.618034 . 2 c. Let x = 1 + 1 + 1 +… . Then x satisfies c. Let the equation x = 1 + x . From part (b) we 1 x = 1+ . know that x must equal 1 1+ (1 + 5 ) / 2 ≈ 1.618034 . 1+ 1 32. a. x1 = 0 Then x satisfies the equation x = 1 + . x x2 = 5 ≈ 2.236068 From part (b) we know that x must equal x3 = 5 + 5 ≈ 2.689994 (1 + 5 ) / 2 ≈ 1.618034 . f (r ) x4 = 5 + 5 + 5 ≈ 2.7730839 34. a. Suppose r is a root. Then r = r – . f ′(r ) x5 = 5 + 5 + 5 + 5 ≈ 2.7880251 f (r ) = 0, so f(r) = 0. b. x = 5 + x , and x must satisfy x ≥ 0 f ′(r ) x2 = 5 + x f (r ) Suppose f(r) = 0. Then r – = r – 0 = r, f ′(r ) x2 − x − 5 = 0 f ( x) 1 ± 1 + 4 ⋅1 ⋅ 5 1 ± 21 so r is a root of x = x – . x= = f ′( x) 2 2 Taking the minus sign gives a negative solution for x, violating the requirement that x ≥ 0 . Hence, ( ) x = 1 + 21 / 2 ≈ 2.7912878 c. Let x = 5 + 5 + 5 + … . Then x satisfies the equation x = 5 + x . From part (b) we know that x must equal (1 + ) 21 / 2 ≈ 2.7912878 222 Section 3.7 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• b. If we want to solve f(x) = 0 and f ′( x) ≠ 0 in ⎛ π⎞ On the interval ⎜ 0, ⎟ , there is only one f ( x) ⎝ 2⎠ [a, b], then = 0 or f ′( x) stationary point (check graphically). We will use Newton’s Method to find the stationary point, f ( x) x= x– = g ( x) . π f ′( x) starting with x = ≈ 0.785398 . 4 f ′( x) f ( x) g ′( x) = 1 – + f ′′( x) n xn f ′( x) [ f ′( x)]2 π ≈ 0.785398 1 4 f ( x) f ′′( x) = 2 0.862443 [ f ′( x)]2 3 0.860335 f (r ) f ′′(r ) and g '(r ) = = 0. 4 0.860334 [ f ′(r )]2 5 0.860334 35. a. The algorithm computes the root of x ≈ 0.860334 will maximize the area of the 1 1 rectangle in quadrant I, and subsequently the – a = 0 for x1 close to . larger rectangle as well. x a y = cos x = cos ( 0.860334 ) ≈ 0.652184 1 The maximum area of the larger rectangle is b. Let f ( x) = – a. AL = ( 2 x ) y ≈ 2 ( 0.860334 )( 0.652184 ) x ≈ 1.122192 square units 1 f ′( x) = – x2 f ( x) = – x + ax 2 f ′( x) The recursion formula is f ( xn ) 37. The rod that barely fits around the corner will xn +1 = xn – = 2 xn – axn 2 . f ′( xn ) touch the outside walls as well as the inside corner. E F C 36. We can start by drawing a diagram: y θ 6.2 feet b 2 D θ B (x, y) a A π −π 2 0 x 2 x 8.6 feet From symmetry, maximizing the area of the As suggested in the diagram, let a and b represent entire rectangle is equivalent to maximizing the the lengths of the segments AB and BC, and let θ area of the rectangle in quadrant I. The area of denote the angles ∠DBA and ∠FCB . Consider the rectangle in quadrant I is given by the two similar triangles ΔADB and ΔBFC ; A = xy these have hypotenuses a and b respectively. A = x cos x little trigonometry applied to these angles gives To find the maximum area, we first need the 8.6 6.2 ⎛ π⎞ a= = 8.6sec θ and b = = 6.2 csc θ stationary points on the interval ⎜ 0, ⎟ . cos θ sin θ ⎝ 2⎠ Note that the angle θ determines the position of A ' ( x ) = cos x − x sin x the rod. The total length of the rod is then Therefore, we need to solve L = a + b = 8.6sec θ + 6.2 csc θ A'( x) = 0 ⎛ π⎞ The domain for θ is the open interval ⎜ 0, ⎟ . cos x − x sin x = 0 ⎝ 2⎠ Instructor’s Resource Manual Section 3.7 223 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• The derivative of L is As suggested in the diagram, let a and b represent 8.6sin 3 θ − 6.2 cos3 θ the lengths of the segments AB and BC, and let θ L ' (θ ) = denote the angle ∠ABD . Consider the two right sin 2 θ ⋅ cos 2 θ triangles ΔADB and ΔCEB ; these have Thus, L ' (θ ) = 0 provided hypotenuses a and b respectively. A little 8.6sin 3 θ − 6.2 cos3 θ = 0 trigonometry applied to these angles gives 8 8.6sin 3 θ = 6.2 cos3 θ a= = 8csc θ and sin θ sin 3 θ 6.2 8 = b= = 8csc ( 75 − θ ) cos3 θ 8.6 sin ( 75 − θ ) 6.2 tan 3 θ = Note that the angle θ determines the position of 8.6 the rod. The total length of the rod is then tan θ = 3 6.2 L = a + b = 8csc θ + 8csc ( 75 − θ ) 8.6 The domain for θ is the open interval ( 0, 75 ) . ⎛ π⎞ On the interval ⎜ 0, ⎟ , there will only be one A graph of of L indicates there is only one ⎝ 2⎠ extremum (a minimum) on the interval. solution to this equation. We will use Newton’s 6.2 method to solve tan θ − 3 = 0 starting with 8.6 π θ1 = . 4 The derivative of L is n θn π ≈ 0.78540 1 4 L ' (θ ) = ( 8 sin 2 θ ⋅ cos (θ − 75 ) − cos θ ⋅ sin 2 (θ − 75 ) ) 2 0.73373 sin θ ⋅ sin (θ − 75 ) 2 2 3 0.73098 We will use Newton’s method to solve 4 0.73097 L ' (θ ) = 0 starting with θ1 = 40 . 5 0.73097 n θn Note that θ ≈ 0.73097 minimizes the length of 1 40 the rod that does not fit around the corner, which 2 37.54338 in turn maximizes the length of the rod that will fit around the corner (verify by using the Second 3 37.50000 Derivative Test). 4 37.5 L ( 0.73097 ) = 8.6sec ( 0.73097 ) + 6.2 csc ( 0.73097 ) Note that θ = 37.5° minimizes the length of the ≈ 20.84 rod that does not fit around the corner, which in Thus, the length of the longest rod that will fit turn maximizes the length of the rod that will fit around the corner is about 20.84 feet. around the corner (verify by using the Second Derivative Test). 38. The rod that barely fits around the corner will L ( 37.5 ) = 8csc ( 37.5 ) + 8csc ( 75 − 37.5 ) touch the outside walls as well as the inside corner. = 16 csc ( 37.5 ) C ≈ 26.28 Thus, the length of the longest rod that will fit 8 feet around the corner is about 26.28 feet. b E 75 − θ B o 105 θ a A D 8 feet 224 Section 3.7 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 2x2 3.8 Concepts Review 39. We can solve the equation − + x + 42 = 0 to 25 x r +1 find the value for x when the object hits the 1. rx r −1 ; + C , r ≠ −1 ground. We want the value to be positive, so we r +1 use the quadratic formula, keeping only the r −1 2. r [ f ( x) ] f ′( x); [ f ( x) ] f ′( x) positive solution. r −1 − 12 − 4 ( −0.08 )( 42 ) x= = 30 2 ( −0.08 ) 3. u = x 4 + 3 x 2 + 1, du = (4 x3 + 6 x)dx 4 We are interested in the global extrema for the ∫ (x + 3x 2 + 1)8 (4 x3 + 6 x)dx = ∫ u8 du distance of the object from the observer. We obtain the same extrema by considering the u9 ( x 4 + 3 x 2 + 1)9 = +C = +C squared distance 9 9 D( x) = ( x − 3)2 + (42 + x − .08 x 2 ) 2 A graph of D will help us identify a starting point 4. c1 ∫ f ( x)dx + c2 ∫ g ( x)dx for our numeric approach. Problem Set 3.8 1. ∫ 5dx = 5 x + C From the graph, it appears that D (and thus the 2. ∫ ( x − 4)dx = ∫ xdx − 4∫ 1dx distance from the observer) is maximized at x2 about x = 7 feet and minimized just before the = − 4x + C 2 object hits the ground at about x = 28 feet. The first derivative is given by 2 2 x3 D '( x) = 16 3 12 2 236 x − x − x + 78 . 3. ∫ ( x + π)dx = ∫ x dx + π∫ 1dx = 3 + πx + C 625 25 25 ∫ ( 3x ) a. We will use Newton’s method to find the 4. 2 + 3 dx = 3∫ x 2 dx + 3 ∫ 1dx stationary point that yields the minimum distance, starting with x1 = 28 . x 3 =3 + 3 x + C = x3 + 3 x + C n xn 3 1 28 5/ 4 x9 / 4 4 9/4 2 28.0280 5. ∫ x dx = 9 +C = 9 x +C 3 28.0279 4 4 28.0279 x ≈ 28.0279; y ≈ 7.1828 ⎛ x5 / 3 ⎞ 6. ∫ 3x 2 / 3 dx = 3∫ x 2 / 3 dx = 3 ⎜ + C1 ⎟ The object is closest to the observer when it ⎜ 5 ⎟ ⎝ 3 ⎠ is at the point ( 28.0279, 7.1828 ) . 9 = x5 / 3 + C b. We will use Newton’s method to find the 5 stationary point that yields the maximum distance, starting with x1 = 7 . 1 7. ∫3 dx = ∫ x −2 / 3 dx = 3 x1/ 3 + C = 33 x + C n xn 2 x 1 7 −3 / 4 2 6.7726 8. ∫ 7x dx = 7 ∫ x −3 / 4 dx = 7(4 x1/ 4 + C1 ) 3 6.7728 = 28 x1/ 4 + C 4 6.7728 x ≈ 6.7728; y ≈ 45.1031 2 x3 x 2 9. ∫ (x − x)dx = ∫ x 2 dx − ∫ x dx = − +C The object is closest to the observer when it 3 2 is at the point ( 6.7728, 45.1031) . Instructor’s Resource Manual Section 3.8 225 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 10. ∫ (3x 2 − πx)dx = 3∫ x 2 dx − π∫ x dx 4 x6 + 3x 4 17. ∫ dx = ∫ (4 x3 + 3 x) dx 3 ⎛ x3 ⎞ ⎛ x2 ⎞ x = 3 ⎜ + C1 ⎟ − π ⎜ + C2 ⎟ = 4 ∫ x3 dx + 3∫ x dx ⎜ 3 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ πx 2 3x 2 =x − 3 +C = x4 + +C 2 2 5 − x3 )dx = 4 ∫ x5 dx − ∫ x3 dx x6 − x 11. ∫ (4 x 18. ∫ 3 dx = ∫ ( x3 − x −2 ) dx x ⎛ x6 ⎞ ⎛ x4 ⎞ = 4 ⎜ + C1 ⎟ − ⎜ + C2 ⎟ x 4 x −1 ⎜ 6 ⎟ ⎜ 4 ⎟ = ∫ x3 dx − ∫ x −2 dx = − +C ⎝ ⎠ ⎝ ⎠ 4 −1 2 x6 x 4 x4 1 = − +C = + +C 3 4 4 x 100 12. ∫ (x + x99 )dx = ∫ x100 dx + ∫ x99 dx 2 2 x3 x 2 19. ∫ ( x + x) dx = ∫ x dx + ∫ x dx = 3 + 2 +C x101 x100 = + +C ∫(x ) 101 100 3 20. + x dx = ∫ x3 dx + ∫ x1/ 2 dx 7 5 3 13. ∫ (27 x + 3x − 45 x + 2 x)dx x 4 x3 / 2 x 4 2 x3 = + +C = + +C = 27 ∫ x7 dx + 3∫ x5 dx − 45∫ x3 dx + 2 ∫ x dx 4 3 4 3 2 8 6 4 2 27 x x 45 x 2x = + − + +C 21. Let u = x + 1; then du = dx. 8 2 4 2 2 2 u3 ( x + 1)3 ∫ ( x + 1) dx = ∫ u du = 3 + C = 3 + C 14. ∫ ⎡x ⎢ ⎣ ( x3 + 5x2 − 3x + 3 )⎤⎥⎦ dx 2 = ∫ ( x5 + 5 x 4 − 3 x3 + 3 x 2 ) dx ∫(z + ) ( ) 2 2 22. 2z dz = ∫ ⎡ 1 + 2 z ⎤ dz ⎣ ⎦ = ∫ x5 dx + 5∫ x 4 dx −3∫ x3 dx + 3 ∫ x 2 dx (1 + 2 ) z3 + C 2 ( )∫ 2 2 x6 3x4 3 x3 = 1+ 2 z dz = = + x5 − + +C 3 6 4 3 ( z 2 + 1)2 z4 + 2z2 + 1 ⎛ 3 2 ⎞ −2 −3 23. ∫ dz = ∫ dz 15. ∫ ⎜ x2 − x3 ⎟ dx = ∫ (3x − 2 x ) dx ⎝ ⎠ z z = ∫ z 7 / 2 dz + 2∫ z 3 / 2 + ∫ z −1/ 2 dz −2 −3 = 3∫ x dx − 2 ∫ x dx 2 9/ 2 4 5/ 2 −1 −2 = z + z + 2 z1/ 2 + C 3x 2x 9 5 = − +C −1 −2 3 1 s( s + 1)2 s3 + 2 s 2 + s =− + x x2 +C 24. ∫ s ds = ∫ s ds = ∫ s5 / 2 ds + 2∫ s3 / 2 ds + ∫ s1/ 2 ds ⎛ 2x 3 ⎞ 16. ∫ ⎜ ⎜ x + x5 ⎟ dx = ∫ ⎝ ⎟ ⎠ ( ) 2 x −1/ 2 + 3 x −5 dx = 2s 7 / 2 4s5 / 2 2s3 / 2 + + +C 7 5 3 2 x1/ 2 3 x −4 = + +C 1 −4 2 25. ∫ (sin θ − cosθ )dθ = ∫ sin θ dθ − ∫ cosθ dθ 3 = − cos θ − sin θ + C = 2 2x − +C 4 4x 226 Section 3.8 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 26. ∫ (t 2 − 2 cos t )dt = ∫ t 2 dt − 2∫ cos t dt 32. Let u = 2 y 2 + 5; then du = 4 y dy t3 3y 3 = − 2sin t + C ∫ dy = ∫ (4 y )(2 y 2 + 5) −1/ 2 dy 3 2y + 5y 2 4 3 −1/ 2 3 1/ 2 27. Let g ( x) = 2 x + 1 ; then g ′( x) = 2 . = 4 ∫ u du = 4 (2u + C1 ) ∫( ) 3 2 dx = ∫ [ g ( x) ] g ′( x)dx 3 3 2 x +1 = 2 y2 + 5 + C 2 [ g ( x)]4 + C = ( ) 4 2 x +1 = +C 33. Let u = x3 + 4 ; then du = 3 x 2 dx . 4 4 2 3 1 2 3 ∫ x x + 4 dx = ∫ 3 3x x + 4 dx 28. Let g ( x) = πx3 + 1 ; then g ′( x) = 3πx 2 . 1 1 3 + 1)4 3πx 2 dx = ∫ [ g ( x) ] g ′( x) dx 4 = ∫ u du = ∫ u1/ 2 du ∫ (πx 3 3 1 ⎛ 2 3/ 2 ⎞ = [ g ( x)]5 + C = (πx3 + 1)5 + C = ⎜ u + C1 ⎟ 3⎝ 3 ⎠ 5 5 2 ( ) 3/ 2 = x3 + 4 +C 29. Let u = 5 x3 + 3x − 8 ; then du = (15 x 2 + 3) dx . 9 2 ∫ (5 x + 1)(5 x3 + 3x − 8)6 dx 34. Let u = x 4 + 2 x 2 ; then 1 = ∫ (15 x 2 + 3)(5 x3 + 3 x − 8)6 dx ( ) ( du = 4 x3 + 4 x dx = 4 x 3 + x dx . ) 3 1 1 ⎛ u7 ⎞ ∫ ( x3 + x ) x4 + 2 x2 dx = ∫ u 6 du = ⎜ + C1 ⎟ 3 3⎜ 7 ⎝ ⎟ ⎠ = ∫ ⋅ 4 ( x3 + x ) x 4 + 2 x 2 dx 1 4 (5 x3 + 3x − 8)7 1 1 = +C = 1/ 2 21 4 ∫ u du = 4 ∫ u du 1⎛2 ⎞ 30. Let u = 5 x3 + 3x − 2; then du = (15 x 2 + 3)dx . = ⎜ u 3/ 2 + C1 ⎟ 4⎝3 ⎠ 2 ∫ (5x + 1) 5 x3 + 3 x − 2 dx 1 4 ( ) 3/ 2 = x + 2 x2 +C 1 6 = ∫ (15 x 2 + 3) 5 x3 + 3 x − 2 dx 3 35. Let u = 1 + cos x ; then du = − sin x dx . 1 1/ 2 1⎛ 2 ⎞ = ∫ u du = ⎜ u 3 / 2 + C1 ⎟ ∫ sin x (1 + cos x ) dx = − ∫ − sin x (1 + cos x ) dx 4 4 3 3⎝ 3 ⎠ 2 ⎛1 ⎞ = (5 x3 + 3x − 2)3 / 2 + C = − ∫ u 4 du = − ⎜ u 5 + C1 ⎟ 9 ⎝5 ⎠ 2 1 = (5 x3 + 3x − 2)3 + C = − (1 + cos x ) + C 5 9 5 31. Let u = 2t 2 − 11; then du = 4t dt . 36. Let u = 1 + sin 2 x ; then du = 2sin x cos x dx . 3 3 2 ∫ sin x cos x 1 + sin x dx ∫ 3t2t 2 − 11 dt = ∫ (4t )(2t 2 − 11)1/ 3 dt 4 1 3 1/ 3 3 ⎛ 3 4/3 ⎞ =∫ ⋅ 2sin x cos x 1 + sin 2 x dx = ∫ u du = ⎜ u + C1 ⎟ 2 4 4⎝4 ⎠ 1 1 9 = ∫ u du = ∫ u1/ 2 du = (2t 2 − 11) 4 / 3 + C 2 2 16 1 ⎛ 2 3/ 2 ⎞ 9 = ⎜ u + C1 ⎟ = 3 (2t 2 − 11) 4 + C 2⎝3 ⎠ 16 1 ( ) 3/ 2 = 1 + sin 2 x +C 3 Instructor’s Resource Manual Section 3.8 227 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 3 2 44. The Quotient Rule for derivatives says 37. f ′( x) = ∫ (3 x + 1)dx = x + x + C1 2 d ⎡ f ( x) ⎤ g ( x) f ′( x) − f ( x) g ′( x) ⎢ g ( x) + C ⎥ = . ⎛3 ⎞ f ( x) = ∫ ⎜ x 2 + x + C1 ⎟ dx dx ⎣ ⎦ g 2 ( x) ⎝ 2 ⎠ g ( x) f ′( x) − f ( x) g ′( x) f ( x) 1 3 1 2 Thus, ∫ 2 g ( x) dx = g ( x) +C . = x + x + C1 x + C2 2 2 45. Let f ( x) = x 2 , g ( x) = x − 1 . 38. f ′( x) = ∫ (−2 x + 3) dx = − x 2 + 3x + C1 1 f ′( x) = 2 x, g ′( x) = f ( x) = ∫ (− x 2 + 3 x + C1 )dx 2 x −1 1 3 ⎡ x2 ⎤ = − x3 + x 2 + C1 x + C2 3 2 ∫ ⎢ 2 x − 1 + 2 x x − 1⎥ dx ⎢ ⎥ ⎣ ⎦ 2 3/ 2 = ∫ [ f ( x ) g ′( x) + f ′( x) g ( x) ] dx = f ( x) g ( x) + C 39. f ′( x) = ∫ x1/ 2 dx = x + C1 3 = x2 x − 1 + C ⎛2 ⎞ f ( x) = ∫ ⎜ x3 / 2 + C1 ⎟ dx ⎝ 3 ⎠ 46. Let f ( x) = x3 , g ( x) = (2 x + 5)−1/ 2 . 4 5/ 2 f ′( x) = 3 x 2 , g ′( x) = −(2 x + 5)−3 / 2 = x + C1 x + C2 15 1 =− 3 7/3 (2 x + 5)3 / 2 40. f ′( x) = ∫ x 4 / 3 dx =x + C1 7 ⎡ − x3 3x2 ⎤ ⎛3 ⎞ 9 10 / 3 ∫ ⎢ (2 x + 5)3 / 2 2 x + 5 ⎥ dx ⎢ + ⎥ f ( x) = ∫ ⎜ x7 / 3 + C1 ⎟ dx = x + C1 x + C2 ⎣ ⎦ ⎝7 ⎠ 70 = ∫ [ f ( x) g ′( x) + g ( x) f ′( x) ] dx 41. f ′′( x) = x + x −3 = f ( x) g ( x) + C = x3 (2 x + 5)−1/ 2 + C x 2 x −2 x3 f ′( x) = ∫ ( x + x −3 )dx =− + C1 = +C 2 2 2x + 5 ⎛1 1 ⎞ f ( x) = ∫ ⎜ x 2 − x −2 + C1 ⎟ dx ⎝2 2 ⎠ d 47. ∫ f ′′( x)dx = ∫ dx f ′( x)dx = f ′( x) + C 1 3 1 −1 = x + x + C1 x + C2 6 2 3 x3 5 x3 + 2 f ′( x) = x3 + 1 + = so 1 3 1 = x + + C1 x + C2 2 x3 + 1 2 x3 + 1 6 2x 5 x3 + 2 ∫ f ′′( x)dx = +C . 3 2 x3 + 1 42. f ′( x) = 2∫ ( x + 1)1/ 3 dx = ( x + 1)4 / 3 + C1 2 ⎡3 ⎤ d ⎛ f ( x) ⎞ f ( x) = ∫ ⎢ ( x + 1)4 / 3 + C1 ⎥ dx 48. ⎜ +C⎟ ⎜ g ( x) ⎟ ⎣2 ⎦ dx ⎝ ⎠ 9 g ( x) f ′( x) − f ( x) 1 [ g ( x)]−1/ 2 g ′( x) = ( x + 1)7 / 3 + C1 x + C2 2 14 = g ( x) 43. The Product Rule for derivatives says 2 g ( x) f ′( x) − f ( x) g ′( x) = d [ f ( x) g ( x) + C ] = f ( x) g ′( x) + f ′( x) g ( x) . 2[ g ( x)]3 / 2 dx Thus, Thus, 2 g ( x) f '( x) − f ( x) g '( x) f ( x) ∫ [ f ( x) g ′( x) + f ′( x) g ( x)]dx = f ( x) g ( x) + C . ∫ 2 [ g ( x)] 3/ 2 = g ( x) +C 228 Section 3.8 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 49. The Product Rule for derivatives says that 54. a. F1 ( x) = ∫ ( x sin x)dx = sin x − x cos x + C1 d m [ f ( x) g n ( x) + C ] F2 ( x) = ∫ (sin x − x cos x + C1 )dx dx = f m ( x )[ g n ( x)]′ + [ f m ( x)]′ g n ( x) = −2 cos x − x sin x + C1 x + C2 F3 ( x) = ∫ (−2 cos x − x sin x + C1 x + C2 )dx = f m ( x)[ng n −1 ( x) g ′( x)] + [mf m −1 ( x) f ′( x)]g n ( x) 1 = f m −1 ( x) g n −1 ( x)[nf ( x) g ′( x) + mg ( x) f ′( x)] . = x cos x − 3sin x + C1 x 2 + C2 x + C3 2 Thus, 1 m −1 n −1 F4 ( x) = ∫ ( x cos x − 3sin x + C1 x 2 + C2 x + C3 )dx ∫ f ( x) g ( x)[nf ( x) g ′( x) + mg ( x) f ′( x)]dx 2 = f m ( x) g n ( x) + C . 1 1 = x sin x + 4 cos x + C1 x3 + C2 x 2 + C3 x + C4 6 2 50. Let u = sin[( x 2 + 1)4 ]; 16 Cn x16− n b. F16 ( x) = x sin x + 16 cos x + ∑ then du = cos ⎡( x 2 + 1)4 ⎤ 4( x 2 + 1)3 (2 x)dx . n =1 (16 − n)! ⎣ ⎦ du = 8 x cos ⎡ ( x 2 + 1)4 ⎤ ( x 2 + 1)3 dx ⎣ ⎦ 3⎡ ∫ sin ( x 2 + 1)4 ⎤ cos ⎡ ( x 2 + 1) 4 ⎤ ( x 2 + 1)3 x dx 3.9 Concepts Review ⎣ ⎦ ⎣ ⎦ 1 1 1 ⎛ u4 ⎞ 1. differential equation = ∫ u 3 ⋅ du = ∫ u 3 du = ⎜ + C1 ⎟ 8 8 8⎜ 4 ⎝ ⎟ ⎠ 2. function sin 4 ⎡ ( x 2 + 1)4 ⎤ 3. separate variables = ⎣ ⎦ +C 32 4. −32t + v0 ; − 16t 2 + v0t + s0 1 2 51. If x ≥ 0, then x = x and ∫ x dx = 2 x +C . Problem Set 3.9 1 2 dy −2 x −x If x < 0, then x = − x and ∫ x dx = − 2 x +C . 1. = dx 2 1 − x 2 = 1 − x2 ⎧1 2 −x ⎪2 x + C if x ≥ 0 dy x x ⎪ + = + =0 ∫ x dx = ⎨ dx y 1 − x2 1 − x2 ⎪− 1 x 2 + C if x < 0 ⎪ 2 ⎩ dy 2. =C u 1 − cos u dx 52. Using sin 2 = , dy 2 2 − x + y = −Cx + Cx = 0 2 1 − cos 2 x 1 1 dx ∫ sin x dx =∫ 2 dx = 2 x − 4 sin 2 x + C . dy 3. = C1 cos x − C2 sin x; 53. Different software may produce different, but dx equivalent answers. These answers were d2y produced by Mathematica. = −C1 sin x − C2 cos x dx 2 a. ∫ 6sin ( 3( x − 2) ) dx = −2 cos ( 3( x − 2) ) + C d2y +y dx 2 3⎛ x⎞ 1 ⎛x⎞ 9 ⎛ x⎞ = (−C1 sin x − C2 cos x) + (C1 sin x + C2 cos x) = 0 b. ∫ sin ⎜ ⎟ dx = cos ⎜ ⎟ − cos ⎜ ⎟ + C ⎝ 6⎠ 2 ⎝ 2⎠ 2 ⎝6⎠ 2 x 2 sin 2 x c. ∫ (x cos 2 x + x sin 2 x)dx = 2 +C Instructor’s Resource Manual Section 3.9 229 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• dy dy x 4. For y = sin(x + C), = cos( x + C ) 7. = dx dx y 2 ⎛ dy ⎞ 2 2 2 ⎜ ⎟ + y = cos ( x + C ) + sin ( x + C ) = 1 ∫ y dy = ∫ x dx ⎝ dx ⎠ y2 x2 dy + C1 = + C2 For y = ±1, = 0. 2 2 dx 2 y 2 = x2 + C ⎛ dy ⎞ 2 2 2 ⎜ ⎟ + y = 0 + (±1) = 1 y = ± x2 + C ⎝ dx ⎠ At x = 1, y = 1: dy 1 = ± 1 + C ; C = 0 and the square root is 5. = x2 + 1 dx positive. dy = ( x 2 + 1) dx y = x 2 or y = x 2 ∫ dy = ∫ ( x + 1) dx dy x x 3 8. = y + C1 = + x + C2 dx y 3 x3 ∫ y dy = ∫ x dx y= + x+C 2 3/ 2 2 3 y + C1 = x3 / 2 + C2 At x = 1, y = 1: 3 3 1 1 y 3 / 2 = x3 / 2 + C 1 = + 1 + C; C = − 3 3 y = ( x3 / 2 + C ) 2 / 3 3 x 1 At x = 1, y = 4: y= +x− 3 3 4 = (1 + C )2 / 3 ; C = 7 dy y = ( x3 / 2 + 7)2 / 3 6. = x −3 + 2 dx dz 2 2 dy = ( x −3 + 2) dx 9. =t z dt −3 ∫ dy = ∫ ( x + 2) dx ∫z −2 dz = ∫ t 2 dt x −2 t3 y + C1 = − + 2 x + C2 − z −1 + C1 = + C2 2 3 1 y=− + 2x + C 1 t3 C − t3 2x 2 = − + C3 = z 3 3 At x = 1, y = 3: 3 1 3 z= 3 = − + 2 + C; C = C − t3 2 2 1 3 1 y=− + 2x + At t = 1, z = : 3 2 x2 2 1 3 = ; C − 1 = 9; C = 10 3 C −1 3 z= 10 − t 3 230 Section 3.9 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• dy dy 10. = y4 13. = (2 x + 1) 4 dt dx −4 ∫y dy = ∫ dt 1 y = ∫ (2 x + 1)4 dx = 4 ∫ (2 x + 1) 2 dx 2 1 − + C1 = t + C2 1 (2 x + 1)5 (2 x + 1)5 3 y3 = +C = +C 2 5 10 1 At x = 0, y = 6: y=− 3 3t + C 1 59 At t = 0, y = 1: 6 = + C; C = 10 10 C = –1 (2 x + 1)5 59 (2 x + 1)5 + 59 1 y= + = y=− 10 10 10 3 3t − 1 dy ds 14. = − y 2 x( x 2 + 2)4 11. = 16t 2 + 4t − 1 dx dt 1 2 − ∫ y −2 dy = ∫ 2 x( x 2 + 2)4 dx ∫ ds = ∫ (16t + 4t − 1) dt 2 s + C1 = 16 3 t + 2t 2 − t + C2 1 1 ( x 2 + 2)5 + C1 = + C2 3 y 2 5 16 3 s= t + 2t 2 − t + C 1 ( x 2 + 2)5 + C 3 = y 10 At t = 0, s = 100:C = 100 16 10 s = t 3 + 2t 2 − t + 100 y= 3 ( x + 2)5 + C 2 At x = 0, y = 1: du 10 12. = u 3 (t 3 − t ) 1= ; C = 10 − 32 = −22 dt 32 + C −3 ∫u du = ∫ (t 3 − t ) dt y= 10 ( x + 2)5 − 22 2 1 t4 t2 − + C1 = − + C2 2u 2 4 2 dy 15. = 3x −2 t42 dx u = t − +C 2 3 2 −1/ 2 y = ∫ 3 x dx = x +C ⎛ t4 ⎞ 2 u = ⎜t2 − + C ⎟ At (1, 2): ⎜ 2 ⎟ ⎝ ⎠ 3 At t = 0, u = 4: 2 = +C 2 1 4 = C −1/ 2 ; C = 1 16 C= 2 −1/ 2 ⎛ t4 1 ⎞ 3 2 1 3x 2 + 1 u = ⎜t2 − + ⎟ y= x + = ⎜ 2 16 ⎟ 2 2 2 ⎝ ⎠ Instructor’s Resource Manual Section 3.9 231 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• dy 1 = 3y2 19. v = ∫ (2t + 1)1/ 3 dt = (2t + 1)1/ 3 2dt 2∫ 16. dx −2 3 ∫y dy = 3∫ dx = (2t + 1)4 / 3 + C1 8 1 − + C1 = 3 x + C2 3 3 y v0 = 0 : 0 = + C1 ; C1 = − 8 8 1 3 3 = −3x + C y v = (2t + 1) 4 / 3 − 8 8 1 3 3 y= s = ∫ (2t + 1) 4 / 3 dt − ∫ 1dt C − 3x 8 8 At (1, 2): 3 3 1 = ∫ (2t + 1)4 3 2dt − ∫ 1dt 2= 16 8 C −3 9 3 7 = (2t + 1)7 3 − t + C2 C= 112 8 2 9 1111 1 2 s0 = 10 :10 = + C2 ; C2 = y= = 112 112 7 − 3x 7 − 6x 2 9 3 1111 s= (2t + 1)7 3 − t + 112 8 112 t2 17. v = ∫ t dt = + v0 3 3 At t = 2: v = (5) 4 3 − ≈ 2.83 2 8 8 t2 9 6 1111 v= +3 s= (5)7 3 − + ≈ 12.6 2 112 8 112 ⎛ t2 ⎞ t3 s = ∫ ⎜ + 3 ⎟ dt = + 3t + s0 ⎜2 ⎟ 6 1 20. v = ∫ (3t + 1) −3 dt = (3t + 1) −3 3dt 3∫ ⎝ ⎠ t3 t3 s= + 3t + 0 = + 3t 1 6 6 = − (3t + 1)−2 + C1 6 At t = 2: v = 5 cm/s 1 25 v0 = 4 : 4 = − + C1; C1 = 22 6 6 s= cm 1 25 3 v = − (3t + 1)−2 + 6 6 1 1 25 18. v = ∫ (1 + t )−4 dt = − +C s = − ∫ (3t + 1)−2 dt + ∫ dt 3(1 + t )3 6 6 1 1 1 −2 25 v0 = 0 : 0 = − + C; C = = − ∫ (3t + 1) 3dt + ∫ dt 3(1 + 0) 3 3 18 6 1 25 1 1 = (3t + 1) −1 + t + C2 v=− + 18 6 3 3 3(1 + t ) 1 1 ⎛ s0 = 0 : 0 = + C2 ; C2 = − 1 1⎞ 1 1 18 18 s = ∫⎜− + ⎟ dt = + t+C ⎜ 3(1 + t )3 3 ⎟ 6(1 + t ) 2 3 1 25 1 ⎝ ⎠ s = (3t + 1)−1 + t − 1 1 59 18 6 18 s0 = 10 :10 = + (0) + C ; C = 6(1 + 0) 2 3 6 1 −2 25 At t = 2: v = − (7) + ≈ 4.16 1 1 59 6 6 s= + t+ 1 25 1 6(1 + t ) 3 26 s = (7) −1 + − ≈ 8.29 18 3 18 At t = 2: 1 1 26 v=− + = cm/s 81 3 81 1 2 59 284 s= + + = cm 54 3 6 27 232 Section 3.9 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 21. v = –32t + 96, 27. vesc = 2 gR 2 2 s = −16t + 96t + s0 = −16t + 96t For the Moon, vesc ≈ 2(0.165)(32)(1080 ⋅ 5280) v = 0 at t = 3 ≈ 7760 ft/s ≈ 1.470 mi/s. At t = 3, s = −16(32 ) + 96(3) = 144 ft For Venus, vesc ≈ 2(0.85)(32)(3800 ⋅ 5280) ≈ 33,038 ft/s ≈ 6.257 mi/s. dv For Jupiter, vesc ≈ 194,369 ft/s ≈ 36.812 mi/s. 22. a = =k dt For the Sun, vesc ≈ 2,021,752 ft/s ds v = ∫ k dt = kt + v0 = ; ≈ 382.908 mi/s. dt k k s = ∫ (kt + v0 )dt = t 2 + v0t + s0 = t 2 + v0t 28. v0 = 60 mi/h = 88 ft/s 2 2 v = 0 = –11t + 88; t = 8 sec v0 11 v = 0 when t = − . Then s ( t ) = − t 2 + 88t k 2 2 k ⎛ v0 ⎞ ⎛ v0 ⎞ 2 2 v0 11 2 s ( 8 ) = − ( 8 ) + 88 ( 8 ) = 352 feet s= ⎜ − k ⎟ + ⎜ − k ⎟ = − 2k . 2⎝ ⎠ ⎜ ⎝ ⎟ ⎠ 2 The shortest distance in which the car can be braked to a halt is 352 feet. dv 23. = −5.28 dt dv Δv 60 − 45 29. a = = = = 1.5 mi/h/s = 2.2 ft/s2 ∫ dv = −∫ 5.28dt dt Δt 10 ds v= = −5.28t + v0 = –5.28t + 56 8 dt 30. 75 = (3.75) 2 + v0 (3.75) + 0; v0 = 5 ft/s 2 ∫ ds = ∫ (−5.28t + 56)dt dv s = −2.64t 2 + 56t + s0 = −2.64t 2 + 56t + 1000 31. For the first 10 s, a = = 6t , v = 3t 2 , and dt When t = 4.5, v = 32.24 ft/s and s = 1198.54 ft s = t 3 . So v(10) = 300 and s(10) = 1000. After −56 10 s, a = dv = −10 , v = –10(t – 10) + 300, and 24. v = 0 when t = ≈ 10.6061 . Then −5.28 dt s ≈ −2.64(10.6061) 2 + 56(10.6061) + 1000 s = −5(t − 10)2 + 300(t − 10) + 1000. v = 0 at ≈ 1296.97 ft t = 40, at which time s = 5500 m. dV 32. a. After accelerating for 8 seconds, the velocity 25. = −kS is 8 · 3 = 24 m/s. dt 4 3 b. Since acceleration and deceleration are Since V = πr and S = 4πr 2 , 3 constant, the average velocity during those dr dr times is 4πr 2 = −k 4πr 2 so = −k . 24 dt dt = 12 m/s . Solve 0 = –4t + 24 to get the 2 ∫ dr = − ∫ k dt 24 r = –kt + C time spent decelerating. t = = 6 s; 4 2 = –k(0) + C and 0.5 = –k(10) + C, so d = (12)(8) + (24)(100) + (12)(6) = 2568 m. 3 3 C = 2 and k = . Then, r = − t + 2 . 20 20 17 26. Solving v = –136 = –32t yields t = . 4 2 ⎛ 17 ⎞ ⎛ 17 ⎞ Then s = 0 = −16 ⎜ ⎟ + (0) ⎜ ⎟ + s0 , so ⎝ 4⎠ ⎝ 4⎠ s0 = 289 ft. Instructor’s Resource Manual Section 3.9 233 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 3.10 Chapter Review 1+ 1 x2 + 1 x2 15. True: lim = lim Concepts Test x →∞ 1 – x 2 x →∞ 1 –1 x2 1. True: Max-Min Existence Theorem 1 = = –1 and –1 2. True: Since c is an interior point and f is 1+ 1 differentiable ( f ′(c) exists), by the x2 + 1 2 lim = lim x Critical Point Theorem, c is a x→ – ∞ 1 – x 2 x→ – ∞ 1 –1 x2 stationary point ( f ′(c) = 0). 1 = = –1. 3. True: For example, let f(x) = sin x. –1 4. False: f ( x) = x1/ 3 is continuous and 3x 2 + 2 x + sin x sin x 16. True: – (3 x + 2) = ; increasing for all x, but f ' ( x ) does x x sin x sin x not exist at x = 0. lim = 0 and lim = 0. x →∞ x x→ – ∞ x 5. True: f ′( x) = 18 x5 + 16 x3 + 4 x; 17. True: The function is differentiable on f ′′( x) = 90 x 4 + 48 x 2 + 4 , which is (0, 2). greater than zero for all x. x 18. False: f ′( x) = so f ′(0) does not exist. 6. False: For example, f ( x) = x is increasing 3 x on [–1, 1] but f ′(0) = 0. 3 3 19. False: There are two points: x = − , . 7. True: When f ′( x) > 0, f ( x) is increasing. 3 3 8. False: If f ′′(c) = 0 , c is a candidate, but not 20. True: Let g(x) = D where D is any number. Then g ′( x) = 0 and so, by Theorem B necessarily an inflection point. For of Section 3.6, example, if f ( x) = x 4 , P ′′(0) = 0 but f(x) = g(x) + C = D + C, which is a x = 0 is not an inflection point. constant, for all x in (a, b). 9. True: f ( x) = ax 2 + bx + c; 21. False: For example if f ( x) = x 4 , f ′( x) = 2ax + b; f ′′( x) = 2a f ′(0) = f ′′(0) = 0 but f has a minimum at x = 0. 10. True: If f(x) is increasing for all x in [a, b], the maximum occurs at b. dy d2y 22. True: = cos x; = − sin x; –sin x = 0 11. False: tan 2 x has a minimum value of 0. dx dx 2 has infinitely many solutions. This occurs whenever x = kπ where k is an integer. 23. False: The rectangle will have minimum perimeter if it is a square. 12. True: lim (2 x3 + x) = ∞ while K x →∞ A = xy = K; y = x lim (2 x3 + x) = −∞ x →−∞ 2 K dP 2K d 2 P 4K P = 2x + ; = 2− ; = x dx x 2 dx 2 x3 13. True: lim (2 x3 + x + tan x) = ∞ while − dP d 2P x→ π = 0 and >0 2 dx dx 2 lim (2 x3 + x + tan x) = −∞. when x = K , y = K . + x →− π 2 24. True: By the Mean Value Theorem, the 14. False: At x = 3 there is a removable derivative must be zero between each discontinuity. pair of distinct x-intercepts. Instructor's Resource Manual Section 3.10 235 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 25. True: If f ( x1 ) < f ( x2 ) and g ( x1 ) < g ( x2 ) is an inflection point while if for x1 < x2 , b 2 – 3ac < 0 there are no critical f ( x1 ) + g ( x1 ) < f ( x2 ) + g ( x2 ), so points.) On an open interval, no local maxima f + g is increasing. can come from endpoints, so there can be at most one local maximum in an 26. False: Let f(x) = g(x) = 2x, f ′( x) > 0 and open interval. g ′( x) > 0 for all x, but f ( x) g ( x) = 4 x 2 is decreasing on 34. True: f ′( x) = a ≠ 0 so f(x) has no local (– ∞ , 0). minima or maxima. On an open interval, no local minima or maxima 27. True: Since f ′′( x) > 0, f ′( x) is increasing can come from endpoints, so f(x) has no local minima. for x ≥ 0. Therefore, f ′( x) > 0 for x in [0, ∞ ), so f(x) is increasing. 35. True: Intermediate Value Theorem 28. False: If f(3) = 4, the Mean Value Theorem 36. False: The Bisection Method can be very requires that at some point c in [0, 3], slow to converge. f (3) – f (0) 4 –1 f ′(c) = = = 1 which f ( xn ) 3–0 3–0 37. False: xn +1 = xn – = –2 xn . does not contradict that f ′( x) ≤ 2 for f ′( xn ) all x in [0, 3]. 38. False: Newton’s method can fail to exist for several reasons (e.g. if f’(x) is 0 at or 29. True: If the function is nondecreasing, near r). It may be possible to achieve f ′( x) must be greater than or equal to convergence by selecting a different zero, and if f ′( x) ≥ 0, f is starting value. nondecreasing. This can be seen using the Mean Value Theorem. 39. True: From the Fixed-point Theorem, if g is continuous on [ a, b ] and 30. True: However, if the constant is 0, the a ≤ g ( x ) ≤ b whenever a ≤ x ≤ b , functions are the same. then there is at least one fixed point 31. False: For example, let f ( x) = e . x on [ a, b ] . The given conditions satisfy these criteria. lim e x = 0, so y = 0 is a horizontal x →−∞ asymptote. 40. True: The Bisection Method always converges as long as the function is 32. True: If f(c) is a global maximum then f(c) continuous and the values of the is the maximum value of f on function at the endpoints are of (a, b) ↔ S where (a, b) is any interval opposite sign. containing c and S is the domain of f. Hence, f(c) is a local maximum value. 41. True: Theorem 3.8.C 42. True: Obtained by integrating both sides of 33. True: f ′( x) = 3ax 2 + 2bx + c; f ′( x) = 0 the Product Rule –b ± b 2 – 3ac when x = by the 43. True: (− sin x) 2 = sin 2 x = 1 − cos 2 x 3a Quadratic Formula. f ′′( x) = 6ax + 2b 44. True: If F ( x) = ∫ f ( x) dx, f ( x) is a so ⎛ – b ± b 2 – 3ac ⎞ derivative of F(x). f ′′ ⎜ ⎟ = ±2 b 2 – 3ac . ⎜ 3a ⎟ 45. False: f ( x) = x 2 + 2 x + 1 and ⎝ ⎠ Thus, if b 2 – 3ac > 0, one critical g ( x) = x 2 + 7 x − 5 are a counter- point is a local maximum and the example. other is a local minimum. (If b2 – 3ac = 0 the only critical point 236 Section 3.10 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 46. False: The two sides will in general differ by x a constant term. 5. f ′( x) = ; f ′( x) does not exist at x = 0. x 47. True: At any given height, speed on the 1 Critical points: – , 0, 1 downward trip is the negative of 2 speed on the upward. ⎛ 1⎞ 1 f ⎜ – ⎟ = , f (0) = 0, f (1) = 1 ⎝ 2⎠ 2 Global minimum f(0) = 0; Sample Test Problems global maximum f(1) = 1 1. f ′( x) = 2 x – 2; 2x – 2 = 0 when x = 1. s 6. f ′( s) = 1 + ; f ′( s ) does not exist when s = 0. Critical points: 0, 1, 4 s f(0) = 0, f(1) = –1, f(4) = 8 For s < 0, s = – s so f(s) = s – s = 0 and Global minimum f(1) = –1; global maximum f(4) = 8 f ′( s ) = 1 – 1 = 0. Critical points: 1 and all s in [–1, 0] 1 1 f(1) = 2, f(s) = 0 for s in [–1, 0] 2. f ′(t ) = – is never 0. ;– t 2 t2 Global minimum f(s) = 0, –1 ≤ s ≤ 0; Critical points: 1, 4 global maximum f(1) = 2. 1 f(1) = 1, f (4) = 7. f ′( x) = 12 x3 – 12 x 2 = 12 x 2 ( x – 1); f ′( x) = 0 4 1 when x = 0, 1 Global minimum f (4) = ; Critical points: –2, 0, 1, 3 4 f(–2) = 80, f(0) = 0, f(1) = –1, f(3) = 135 global maximum f(1) = 1. Global minimum f(1) = –1; global maximum f(3) = 135 2 2 3. f ′( z ) = – ;– is never 0. 3 z z3 u (7u – 12) 12 8. f ′(u ) = ; f ′(u ) = 0 when u = 0, 1 2/3 7 Critical points: –2, – 3(u – 2) 2 f ′(2) does not exist. 1 ⎛ 1⎞ f (–2) = , f ⎜ – ⎟ = 4 12 4 ⎝ 2⎠ Critical points: –1, 0, , 2, 3 7 1 Global minimum f (–2) = ; f (–1) = 3 –3 ≈ –1.44, f (0) = 0, 4 ⎛ 1⎞ ⎛ 12 ⎞ 144 3 2 global maximum f ⎜ – ⎟ = 4. f ⎜ ⎟= – ≈ –1.94, f(2) = 0, f(3) = 9 ⎝ 2⎠ ⎝ 7 ⎠ 49 7 ⎛ 12 ⎞ Global minimum f ⎜ ⎟ ≈ –1.94; 2 2 ⎝7⎠ 4. f ′( x) = – is never 0. ;– 3 x x3 global maximum f(3) = 9 Critical point: –2 1 9. f ′( x) = 10 x 4 – 20 x3 = 10 x3 ( x – 2); f (–2) = 4 f ′( x) = 0 when x = 0, 2 f ′( x) > 0 for x < 0, so f is increasing. Critical points: –1, 0, 2, 3 1 f(–1) = 0, f(0) = 7, f(2) = –9, f(3) = 88 Global minimum f (–2) = ; no global Global minimum f(2) = –9; 4 maximum. global maximum f(3) = 88 Instructor's Resource Manual Section 3.10 237 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 10. f ′( x) = 3( x – 1)2 ( x + 2)2 + 2( x – 1)3 ( x + 2) 14. f ′( x) = 9 x8 ; f ′( x) > 0 for all x ≠ 0. = ( x – 1)2 ( x + 2)(5 x + 4); f ′( x) = 0 when f ′′( x) = 72 x 7 ; f ′′( x) < 0 when x < 0. 4 f(x) is increasing on (– ∞ , ∞ ) and concave down x = –2, – , 1 on (– ∞ , 0). 5 4 Critical points: –2, – , 1 , 2 15. f ′( x) = 3 x 2 – 3 = 3( x 2 – 1); f ′( x) > 0 when 5 x < –1 or x > 1. ⎛ 4⎞ 26, 244 f ′′( x) = 6 x; f ′′( x) < 0 when x < 0. f(–2) = 0, f ⎜ – ⎟ = – ≈ –8.40, ⎝ 5⎠ 3125 f(x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) and f(1) = 0, f(2) = 16 concave down on (– ∞ , 0). ⎛ 4⎞ Global minimum f ⎜ – ⎟ ≈ –8.40; 16. f ′( x) = −6 x 2 − 6 x + 12 = –6(x + 2)(x – 1); ⎝ 5⎠ global maximum f(2) = 16 f ′( x) > 0 when –2 < x < 1. f ′′( x) = −12 x − 6 = –6(2x + 1); f ′′( x) < 0 when π 11. f ′(θ ) = cos θ ; f ′(θ ) = 0 when θ = in x>− . 1 2 2 ⎡ π 4π ⎤ f(x) is increasing on [–2, 1] and concave down on ⎢4 , 3 ⎥ ⎣ ⎦ ⎛ 1 ⎞ π π 4π ⎜− , ∞⎟ . Critical points: , , ⎝ 2 ⎠ 4 2 3 17. f ′( x) = 4 x3 – 20 x 4 = 4 x3 (1 – 5 x); f ′( x) > 0 ⎛π⎞ 1 ⎛π⎞ f ⎜ ⎟= ≈ 0.71, f ⎜ ⎟ = 1, 1 ⎝ 4⎠ 2 ⎝2⎠ when 0 < x < . 5 ⎛ 4π ⎞ 3 f ⎜ ⎟=– ≈ –0.87 f ′′( x) = 12 x 2 – 80 x3 = 4 x 2 (3 – 20 x); f ′′( x) < 0 ⎝ 3 ⎠ 2 ⎛ 4π ⎞ when x > 3 Global minimum f ⎜ ⎟ ≈ –0.87; 20 . ⎝ 3 ⎠ ⎛π⎞ f(x) is increasing on ⎡ 0, 1 ⎤ and concave down on global maximum f ⎜ ⎟ = 1 ⎣ 5⎦ ⎝2⎠ ⎛ 3 ⎞ ⎜ , ∞ ⎟. 12. f ′(θ ) = 2sin θ cosθ – cosθ = cosθ (2sin θ – 1); ⎝ 20 ⎠ π π 5π f ′(θ ) = 0 when θ = , , in [0, π ] 18. f ′( x) = 3 x 2 – 6 x 4 = 3 x 2 (1 – 2 x 2 ); f ′( x) > 0 6 2 6 1 1 π π 5π when – < x < 0 and 0 < x < . Critical points: 0, , , , π 2 2 6 2 6 ⎛π⎞ 1 ⎛π⎞ f ′′( x) = 6 x – 24 x3 = 6 x(1 – 4 x 2 ); f ′′( x) < 0 when f(0) = 0, f ⎜ ⎟ = – , f ⎜ ⎟ = 0, ⎝6⎠ 4 ⎝2⎠ – 1 < x < 0 or x > . 1 ⎛ 5π ⎞ 1 2 2 f ⎜ ⎟ = – , f( π ) = 0 ⎝ 6 ⎠ 4 ⎡ 1 1 ⎤ f(x) is increasing on ⎢ – , ⎥ and concave ⎛π⎞ 1 ⎛ 5π ⎞ 1 ⎣ 2 2⎦ Global minimum f ⎜ ⎟ = – or f ⎜ ⎟ = – ; ⎝6⎠ 4 ⎝ 6 ⎠ 4 ⎛ 1 ⎞ ⎛1 ⎞ down on ⎜ – , 0 ⎟ ∪ ⎜ , ∞ ⎟ . ⎛π⎞ ⎝ 2 ⎠ ⎝2 ⎠ global maximum f(0) = 0, f ⎜ ⎟ = 0, or ⎝2⎠ f( π ) = 0 3 13. f ′( x) = 3 – 2 x; f ′( x ) > 0 when x < . 2 f ′′( x) = –2; f ′′( x) is always negative. ⎛ 3⎤ f(x) is increasing on ⎜ – ∞, ⎥ and concave down ⎝ 2⎦ on (– ∞ , ∞ ). 238 Section 3.10 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 19. f ′( x) = 3 x 2 – 4 x3 = x 2 (3 – 4 x); f ′( x) > 0 when 4 f ′′( x) = 6 x – 8; f ′′( x) > 0 when x > . 3 3 x< . ⎛4 ⎞ 4 f(x) is concave up on ⎜ , ∞ ⎟ and concave down f ′′( x) = 6 x – 12 x 2 = 6 x(1 – 2 x); f ′′( x) < 0 when ⎝3 ⎠ ⎛ 4⎞ ⎛ 4 128 ⎞ 1 on ⎜ – ∞, ⎟ ; inflection point ⎜ , − ⎟ x < 0 or x > . ⎝ 3⎠ ⎝3 27 ⎠ 2 ⎛ 3⎤ f(x) is increasing on ⎜ – ∞, ⎥ and concave down ⎝ 4⎦ ⎛1 ⎞ on (– ∞, 0) ∪ ⎜ , ∞ ⎟ . ⎝2 ⎠ 1 1 20. g ′(t ) = 3t 2 – ; g ′(t ) > 0 when 3t 2 > or 2 t t2 1 1 1 t 4 > , so t < – or t > . 3 1/ 4 1/ 4 8x 3 3 22. f ′( x) = – ; f ′( x) = 0 when x = 0. ⎛ 1 ⎤ ⎡ 1 ⎞ ( x + 1)2 2 g ′(t ) is increasing on ⎜ – ∞, – ∪ , ∞⎟ 1/ 4 ⎥ ⎢ 1/ 4 8(3x 2 – 1) ⎝ 3 ⎦ ⎣3 ⎠ f ′′( x) = ; f ′′(0) = –8, so f(0) = 6 is a ⎡ 1 ⎞ ⎛ 1 ⎤ ( x 2 + 1)3 and decreasing on ⎢ – , 0 ⎟ ∪ ⎜ 0, . 1/ 4 1/ 4 ⎥ local maximum. f ′( x) > 0 for x < 0 and ⎣ 3 ⎠ ⎝ 3 ⎦ ⎛ 1 ⎞ 1 f ′( x) < 0 for x > 0 so Local minimum g ⎜ = + 31/ 4 ≈ 1.75; 1/ 4 ⎟ 3/ 4 f(0) = 6 is a global maximum value. f(x) has no ⎝3 ⎠ 3 local maximum minimum value. ⎛ 1 ⎞ 1 g⎜– =– – 31/ 4 ≈ –1.75 1 1/ 4 ⎟ 3/ 4 23. f ′( x) = 4 x3 – 2; f ′( x) = 0 when x = . ⎝ 3 ⎠ 3 3 2 2 g ′′(t ) = 6t + ; g ′′(t ) > 0 when t > 0. g(t) has no f ′′( x) = 12 x 2 ; f ′′( x) = 0 when x = 0. t3 ⎛ 1 ⎞ 12 inflection point since g(0) does not exist. f ′′ ⎜ = > 0, so 3 ⎟ 2/3 ⎝ 2⎠ 2 ⎛ 1 ⎞ 1 2 3 f⎜ = =– 3 ⎟ – is a global 4/3 1/ 3 4/3 ⎝ 2⎠ 2 2 2 minimum. f ′′( x) > 0 for all x ≠ 0; no inflection points No horizontal or vertical asymptotes 21. f ′( x) = 2 x ( x – 4) + x 2 = 3 x 2 – 8 x = x (3 x – 8); 8 f ′( x) > 0 when x < 0 or x > 3 ⎡8 ⎞ f(x) is increasing on (– ∞, 0] ∪ ⎢ , ∞ ⎟ and ⎣3 ⎠ ⎡ 8⎤ decreasing on ⎢ 0, ⎥ ⎣ 3⎦ ⎛8⎞ 256 Local minimum f ⎜ ⎟ = – ≈ –9.48; ⎝3⎠ 27 local maximum f(0) = 0 Instructor's Resource Manual Section 3.10 239 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 24. f ′( x) = 2( x 2 – 1)(2 x) = 4 x( x 2 – 1) = 4 x3 – 4 x; Vertical asymptote x = 3 f ′( x) = 0 when x = –1, 0, 1. f ′′( x) = 12 x 2 – 4 = 4(3 x 2 – 1); f ′′( x ) = 0 when 1 x=± . 3 f ′′(–1) = 8, f ′′(0) = –4, f ′′(1) = 8 Global minima f(–1) = 0, f(1) = 0; local maximum f(0) = 1 ⎛ 1 4⎞ Inflection points ⎜ ± , ⎟ 27. f ′( x) = 12 x3 – 12 x 2 = 12 x 2 ( x – 1); f ′( x) = 0 ⎝ 3 9⎠ when x = 0, 1. No horizontal or vertical asymptotes f ′′( x) = 36 x 2 – 24 x = 12 x(3 x – 2); f ′′( x) = 0 2 when x = 0, . 3 f ′′(1) = 12, so f(1) = –1 is a minimum. Global minimum f(1) = –1; no local maxima ⎛ 2 16 ⎞ Inflection points (0, 0), ⎜ , − ⎟ ⎝ 3 27 ⎠ No horizontal or vertical asymptotes. 3x – 6 25. f ′( x) = ; f ′( x) = 0 when x = 2, but x = 2 2 x–3 is not in the domain of f(x). f ′( x ) does not exist when x = 3. 3( x – 4) f ′′( x) = ; f ′′( x) = 0 when x = 4. 4( x – 3)3 / 2 Global minimum f(3) = 0; no local maxima Inflection point (4, 4) 1 No horizontal or vertical asymptotes. 28. f ′( x) = 1 + ; f ′( x) > 0 for all x ≠ 0. x2 2 f ′′( x) = – ; f ′′( x) > 0 when x < 0 and x3 f ′′( x) < 0 when x > 0. No local minima or maxima No inflection points 1 f ( x) = x – , so x ⎛ 1⎞ lim [ f ( x) − x] = lim ⎜ − ⎟ = 0 and y = x is an x →∞ x →∞ ⎝ x ⎠ 1 26. f ′( x) = – ; f ′( x) < 0 for all x ≠ 3. oblique asymptote. ( x – 3) 2 Vertical asymptote x = 0 2 f ′′( x) = ; f ′′( x) > 0 when x > 3. ( x – 3)3 No local minima or maxima No inflection points x–2 1– 2 lim = lim x =1 x →∞ x – 3 x →∞ 1 – 3 x Horizontal asymptote y = 1 240 Section 3.10 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 1 31. f ′( x) = – sin x – cos x; f ′( x) = 0 when 29. f ′( x) = 3 + ; f ′( x ) > 0 for all x ≠ 0. x 2 π 3π x=– , . 2 4 4 f ′′( x) = – ; f ′′( x) > 0 when x < 0 and x3 f ′′( x) = – cos x + sin x; f ′′( x) = 0 when f ′′( x) < 0 when x > 0 3π π x=– , . No local minima or maxima 4 4 No inflection points ⎛ π⎞ ⎛ 3π ⎞ f ′′ ⎜ – ⎟ = – 2, f ′′ ⎜ ⎟ = 2 1 ⎝ 4⎠ ⎝ 4 ⎠ f ( x) = 3x – , so x ⎛ 3π ⎞ Global minimum f ⎜ ⎟ = – 2; ⎛ 1⎞ ⎝ 4 ⎠ lim [ f ( x) − 3 x] = lim ⎜ − ⎟ = 0 and y = 3x is an x →∞ x →∞ ⎝ x ⎠ ⎛ π⎞ oblique asymptote. global maximum f ⎜ – ⎟ = 2 ⎝ 4⎠ Vertical asymptote x = 0 ⎛ 3π ⎞ ⎛ π ⎞ Inflection points ⎜ − , 0 ⎟ , ⎜ , 0 ⎟ ⎝ 4 ⎠ ⎝4 ⎠ 4 30. f ′( x) = – ; f ′( x ) > 0 when x < −1 and ( x + 1)3 f ′( x) < 0 when x > –1. 32. f ′( x) = cos x – sec 2 x; f ′( x ) = 0 when x = 0 12 f ′′( x) = – sin x – 2sec2 x tan x f ′′( x) = ; f ′′( x) > 0 for all x ≠ −1 . ( x + 1)4 = – sin x(1 + 2sec3 x ) No local minima or maxima f ′′( x) = 0 when x = 0 No inflection points No local minima or maxima lim f ( x) = 0, lim f ( x) = 0, so y = 0 is a Inflection point f(0) = 0 x →∞ x→ – ∞ π π horizontal asymptote. Vertical asymptotes x = – , Vertical asymptote x = –1 2 2 Instructor's Resource Manual Section 3.10 241 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 33. f ′( x) = x sec 2 x + tan x; f ′( x ) = 0 when x = 0 f ′′( x) = 2sec 2 x(1 + x tan x); f ′′( x) is never 0 on ⎛ π π⎞ ⎜– , ⎟. ⎝ 2 2⎠ f ′′(0) = 2 Global minimum f(0) = 0 36. f ′( x) = –2sin x – 2 cos x; f ′( x) = 0 when π 3π x=– , . 4 4 f ′′( x) = –2 cos x + 2sin x; f ′′( x) = 0 when 3π π x=– , . 4 4 34. f ′( x) = 2 + csc2 x; f ′( x ) > 0 on (0, π ) ⎛ π⎞ ⎛ 3π ⎞ f ′′ ⎜ – ⎟ = –2 2, f ′′ ⎜ ⎟ = 2 2 f ′′( x) = –2 cot x csc2 x; f ′′( x) = 0 when ⎝ 4⎠ ⎝ 4 ⎠ π ⎛π ⎞ ⎛ 3π ⎞ x= ; f ′′( x) > 0 on ⎜ , π ⎟ Global minimum f ⎜ ⎟ = –2 2; 2 ⎝2 ⎠ ⎝ 4 ⎠ ⎛π ⎞ ⎛ π⎞ Inflection point ⎜ , π ⎟ global maximum f ⎜ – ⎟ = 2 2 ⎝2 ⎠ ⎝ 4⎠ ⎛ 3π ⎞ ⎛ π ⎞ Inflection points ⎜ − , 0 ⎟ , ⎜ , 0 ⎟ ⎝ 4 ⎠ ⎝4 ⎠ 35. f ′( x) = cos x – 2 cos x sin x = cos x(1 – 2sin x); π π π 5π f ′( x) = 0 when x = – , , , 2 6 2 6 f ′′( x) = – sin x + 2sin 2 x – 2 cos 2 x; f ′′( x ) = 0 when x ≈ –2.51, –0.63, 1.00, 2.14 37. ⎛ π⎞ ⎛π⎞ 3 ⎛π⎞ f ′′ ⎜ – ⎟ = 3, f ′′ ⎜ ⎟ = – , f ′′ ⎜ ⎟ = 1, ⎝ 2⎠ ⎝ 6⎠ 2 ⎝2⎠ ⎛ 5π ⎞ 3 f ′′ ⎜ ⎟ = – ⎝ 6 ⎠ 2 ⎛ π⎞ Global minimum f ⎜ – ⎟ = –2, ⎝ 2⎠ ⎛ π⎞ local minimum f ⎜ ⎟ = 0; ⎝2⎠ ⎛ π ⎞ 1 ⎛ 5π ⎞ 1 global maxima f ⎜ ⎟ = , f ⎜ ⎟ = ⎝6⎠ 4 ⎝ 6 ⎠ 4 Inflection points (–2.51, –0.94), (–0.63, –0.94), (1.00, 0.13), (2.14, 0.13) 242 Section 3.10 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 38. 1 ⎛ ⎛ 1⎞ 3⎞ = ⎜ –( x + 64) + ⎜1 + ⎟ x ⎟ 2 x2 x 2 + 64 ⎝ ⎝ x⎠ ⎠ x3 − 64 = x 2 x 2 + 64 x3 − 64 = 0; x = 4 x 2 x 2 + 64 dp dp < 0 if x < 4, > 0 if x > 4 39. dx dx ⎛ 1⎞ When x = 4, p = ⎜ 1 + ⎟ 16 + 64 ≈ 11.18 ft. ⎝ 4⎠ 42. Let x be the width and y the height of a page. A = xy. Because of the margins, 27 (y – 4)(x – 3) = 27 or y = +4 x−3 27 x A= + 4 x; x −3 40. Let x be the length of a turned up side and let l be dA ( x − 3)(27) − 27 x 81 the (fixed) length of the sheet of metal. = +4= − +4 dx ( x − 3) 2 ( x − 3) 2 V = x (16 − 2 x )l = 16 xl − 2 x 2 l dA 3 15 dV = 0 when x = − , = 16l − 4 xl ; V ′ = 0 when x = 4 dx 2 2 dx d2A 162 d2A 15 d 2V = ; > 0 when x = = −4l ; 4 inches should be turned up for dx 2 ( x − 3) 3 dx 2 2 dx 2 each side. 15 x= ; y = 10 2 41. Let p be the length of the plank and let x be the distance from the fence to where the plank 1 2 touches the ground. 43. πr h = 128π 2 See the figure below. 256 h= r2 Let S be the surface area of the trough. 256π S = πr 2 + πrh = πr 2 + r dS 256π = 2πr − dr r2 256π 2πr − = 0; r 3 = 128, r = 4 3 2 2 r d 2S Since > 0 when r = 4 3 2 , r = 4 3 2 2 dr minimizes S. By properties of similar triangles, 256 h= = 83 2 x 2 + 64 p ( ) 2 = 43 2 x +1 x ⎛ 1⎞ p = ⎜ 1 + ⎟ x 2 + 64 ⎝ x⎠ Minimize p: dp 1 ⎛ 1⎞ x =− x 2 + 64 + ⎜ 1 + ⎟ dx x 2 ⎝ x ⎠ x + 64 2 Instructor's Resource Manual Section 3.10 243 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• ⎧x 3 ( x − 1) − ( x + 1) −2 ⎪2 + 2 if − 2 < x < 0 c. g ′( x) = = ⎪ ( x − 1) 2 ( x − 1) 2 44. f ′( x) = ⎨ ⎪− x + 2 if 0 < x < 2 g (3) − g (2) 2 − 3 ⎪ ⎩ 3 = = −1 3− 2 1 x 3 −2 + = 0; x = −3 , which is not in the domain. = –1; c = 1 ± 2 2 2 (c − 1)2 x+2 − = 0; x = −2, which is not in the domain. Only c = 1 + 2 is in the interval (2, 3). 3 Critical points: x = –2, 0, 2 f(–2) = 0, f(0) = 2, f(2) = 0 Minima f(–2) = 0, f(2) = 0, maximum f(0) = 2. ⎧1 ⎪ if − 2 < x < 0 ⎪ f ′′( x) = ⎨ 2 ⎪− 1 if 0 < x < 2 ⎪ 3 ⎩ Concave up on (–2, 0), concave down on (0, 2) dy 46. = 4 x3 − 18 x 2 + 24 x − 3 dx d2y = 12 x 2 − 36 x + 24; 12( x 2 − 3x + 2) = 0 when 2 dx x = 1, 2 Inflection points: x = 1, y = 5 and x = 2, y = 11 dy Slope at x = 1: =7 dx x =1 45. a. f ′( x) = x 2 Tangent line: y – 5 = 7(x – 1); y = 7x – 2 f (3) − f (−3) 9 + 9 dy = =3 Slope at x = 2: =5 3 − (−3) 6 dx x = 2 c 2 = 3; c = − 3, 3 Tangent line: y – 11 = 5(x – 2); y = 5x + 1 47. b. The Mean Value Theorem does not apply because F ′(0) does not exist. 48. 244 Section 3.10 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 49. Let f ( x ) = 3 x − cos 2 x ; a1 = 0 , b1 = 1 . cos 2 xn 51. xn +1 = f ( 0 ) = −1 ; f (1) ≈ 3.4161468 3 n hn mn f ( mn ) n xn 1 0.5 0.5 0.9596977 1 0.5 2 0.25 0.25 −0.1275826 2 0.18010 3 0.125 0.375 0.3933111 3 0.311942 4 0.0625 0.3125 0.1265369 4 0.270539 5 0.03125 0.28125 −0.0021745 5 0.285718 6 0.015625 0.296875 0.0617765 6 0.280375 7 0.0078125 0.2890625 0.0296988 7 0.282285 8 0.0039063 0.2851563 0.0137364 8 0.281606 9 0.0019532 0.2832031 0.0057745 9 0.281848 10 0.0009766 0.2822266 0.0017984 10 0.281762 11 0.0004883 0.2817383 −0.0001884 11 0.281793 12 0.0002442 0.2819824 0.0008049 12 0.281782 13 0.0001221 0.2818604 0.0003082 13 0.281786 14 0.0000611 0.2817994 0.0000600 14 0.281784 15 0.0000306 0.2817689 −0.0000641 15 0.281785 16 0.0000153 0.2817842 −0.0000018 16 0.281785 17 0.0000077 0.2817918 0.0000293 x ≈ 0.2818 18 0.0000039 0.2817880 0.0000138 52. y = x and y = tan x 19 0.0000020 0.2817861 0.0000061 20 0.0000010 0.2817852 0.0000022 21 0.0000005 0.2817847 0.0000004 22 0.0000003 0.2817845 −0.0000006 23 0.0000002 0.2817846 −0.0000000 x ≈ 0.281785 50. f(x) = 3x – cos 2x, f ′( x) = 3 + 2sin 2 x 11π Let x1 = 0.5 . Let x1 = . 8 n xn f(x) = x – tan x, f ′( x) = 1 – sec2 x . 1 0.5 2 0.2950652 n xn 3 0.2818563 11π 1 4 0.2817846 8 5 0.2817846 2 4.64661795 x ≈ 0.281785 3 4.60091050 4 4.54662258 5 4.50658016 6 4.49422443 7 4.49341259 8 4.49340946 x ≈ 4.4934 Instructor's Resource Manual Section 3.10 245 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• ∫ ( x − 3x + 3 x ) dx 53. 3 2 58. Let u = cos x ; then du = − sin x dx or −du = sin x dx . ( = ∫ x3 − 3x 2 + 3x1/ 2 dx ) cos 4 x sin x dx = ∫ ( cos x ) sin x dx ∫ 4 1 4 2 = x − x3 + 3 ⋅ x3/ 2 + C 4 3 = ∫ u 4 ⋅ −du 1 4 = x − x3 + 2 x3/ 2 + C = − ∫ u 4 du 4 1 = − u5 + C 2 x 4 − 3x 2 + 1 5 54. ∫ x2 dx 1 = − cos5 x + C ( = ∫ 2 x 2 − 3 + x −2 dx ) 5 2 3 59. u = tan(3 x 2 + 6 x ), du = (6 x + 6) sec2 (3x 2 + 6 x) = x − 3 x − x −1 + C ∫ ( x + 1) tan ( 3x + 6 x ) sec ( 3x + 6 x ) dx 3 2 2 2 2 2 x3 1 2 x4 − 9 x2 − 3 = − 3x − + C or +C 3 x 3x 1 2 1 6∫ = u du = u 3 + C 18 y 3 − 9 y sin y + 26 y −1 55. ∫ y dy 1 ( = tan 3 3 x 2 + 6 x + C 18 ) =∫ ( y 2 − 9sin y + 26 ) dy 60. u = t 4 + 9, du = 4t 3 dt 1 3 = y + 9 cos y + 26 y + C 1 du 3 t3 ∫ t4 + 9 dt = ∫ 4 u 1 56. Let u = y 2 − 4 ; then du = 2 ydy or du = ydy . 1 4∫ 2 = u −1/ 2 du 1 ∫ y y 2 − 4 dy = ∫ u ⋅ du 2 1 = ⋅ 2u1/ 2 + C 4 1 = ∫ u du 1/ 2 1 4 2 = t +9 +C 2 1 2 3/ 2 = ⋅ u +C 2 3 1 61. Let u = t 5 + 5 ; then du = 5t 4 dt or du = t 4 dt . 1 2 ( ) 3/ 2 = y −4 +C 5 3 ∫ t ( t + 5) 2/3 1 2/3 4 5 dt = ∫ u du 1 5 57. Let u = 2 z 2 − 3 ; then du = 4 zdz or du = zdz . 1 = ∫ u du 2/3 4 5 ∫ z ( 2 z − 3) 1/ 3 1 2 dz = ∫ u1/ 3 ⋅ du 1 3 5/3 4 = ⋅ u +C 1 5 5 = ∫ u1/ 3 du 3 5 ( ) 5/ 3 4 = t +5 +C 1 3 4/3 25 = ⋅ u +C 4 4 3 ( ) 4/3 = 2z2 − 3 +C 16 246 Section 3.10 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 62. Let u = x 2 + 4 ; then du = 2 x dx or 1 du = xdx . 66. Let u = y 3 − 3 y ; then x 1 du 2 ( ) du = 3 y 2 − 3 dy = 3 y 2 − 1 dy . ( ) ∫ x +4 dx = ∫ 2 y2 −1 1 du ∫ 3 ∫ u2 2 u dy = (y ) 2 1 3 − 3y 2∫ = u −1/ 2 du 1 3∫ 1 = u −2 du = ⋅ 2u1/ 2 + C 2 1 = ⋅ − u −1 + C = x2 + 4 + C 3 1 1 1 =− ⋅ 3 +C 63. Let u = x3 + 9 ; then du = 3 x 2 dx or du = x 2 dx . 3 y − 3y 3 1 x2 1 du =− +C ∫ dx = ∫ 3y − 9 y 3 x +9 3 3 u 1 67. u = 2 y 3 + 3 y 2 + 6 y, du = (6 y 2 + 6 y + 6) dy 3∫ = u −1/ 2 du 1 5 u −1/ 5 du = 6∫ 1 (2 y 3 + 3 y 2 + 6 y )4 / 5 + C = ⋅ 2u1/ 2 + C 24 3 = 2 3 3 x +9 +C 68. ∫ dy = ∫ sin x dx y = − cos x + C 64. Let u = y + 1 ; then du = dy . y = –cos x + 3 1 1 ∫ ( y + 1)2 dy = ∫ u2 du 69. ∫ dy = ∫ 1 dx x +1 = ∫ u −2 du y = 2 x +1 + C y = 2 x + 1 + 14 = −u −1 + C 1 =− y +1 +C 70. ∫ sin y dy = ∫ dx − cos y = x + C 65. Let u = 2 y − 1 ; then du = 2dy . x = –1 – cos y 2 du ∫ ( 2 y − 1)3 dy = ∫ u3 71. ∫ dy = ∫ 2t − 1 dt 1 y = (2t − 1)3 2 + C = ∫ u −3 du 3 1 1 = − u −2 + C y = (2t − 1)3 2 − 1 2 3 1 =− +C −4 2 ( 2 y − 1) 2 72. ∫y dy = ∫ t 2 dt 1 t3 − = +C 3 y3 3 1 t3 2 − = − 3 y3 3 3 1 y=3 2 − t3 Instructor's Resource Manual Section 3.10 247 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• 7. Aregion = 0.5 (1 + 1.5 + 2 + 2.5 ) = 3.5 ∫ 2 y dy = ∫ (6 x − x 3 73. )dx 1 4 y 2 = 3x 2 − x +C 8. Aregion = 0.5 (1.5 + 2 + 2.5 + 3) = 4.5 4 1 y 2 = 3x 2 − x 4 + 9 1 1 4 9. Aregion = Arect + Atri = 1x + x ⋅ x = x2 + x 2 2 1 y = 3x 2 − x 4 + 9 4 1 1 1 10. Aregion = bh = x ⋅ xt = x 2t 2 2 2 74. ∫ cos y dy = ∫ x dx 11. y = 5 − x; Aregion = Arect + Atri x2 sin y = +C 1 2 = 2 ( 2) + ( 2 )( 2 ) = 6 ⎛ x2 ⎞ 2 y = sin −1 ⎜ ⎟ ⎜ 2 ⎟ ⎝ ⎠ 12. Aregion = Arect + Atri 1 75. s ( t ) = −16t 2 + 48t + 448; s = 0 at t = 7; = 1(1) + (1)( 7 ) = 4.5 2 v ( t ) = s ' ( t ) = −32t + 48 when t = 7, v = –32(7) + 48 = –176 ft/s Review and Preview Problems 1 1 3 2 1. Aregion = bh = aa sin 60o = a 2 2 4 ⎛1 ⎞ ⎛ 1 ⎞⎛ 3 ⎞ 2. Aregion = 6 ⎜ base × height ⎟ = 6 ⎜ a ⎟ ⎜ a⎟ ⎝2 ⎠ ⎝ 2 ⎠⎜ 2 ⎟ ⎝ ⎠ 3 3 2 = a 2 2 ⎛1 ⎞ a 3. Aregion = 10 ⎜ base × height ⎟ = 5 cot 36 ⎝ 2 ⎠ 4 5 = a 2 cot 36 4 1 ⎛ 8.5 ⎞ 4. Aregion = Arect + Atri = 17 ( 8.5 ) + 17 ⎜ 2 ⎝ tan 45o ⎟ ⎠ = 216.75 1 5. Aregion = Arect + Asemic. = 3.6 ⋅ 5.8 + π (1.8 ) 2 2 ≈ 25.97 ⎛1 ⎞ 6. Aregion = A#5 + 2 Atri = 25.97 + 2 ⎜ ⋅1.2 ⎟ 5.8 ⎝ 2 ⎠ = 32.93 248 Review and Preview Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
• −1/ 2 33. a. b. ∫ 10V dV = ∫ C1dt ; 20 V = C1t + C2 ; V(0) = 1600: C2 = 20 ⋅ 40 = 800; 800 V(40) = 0: C1 = − = −20 40 1 2 V (t ) = (−20t + 800)2 = ( 40 − t ) 400 2 b. Since the trip that involves 1 min more travel c. V (10) = ( 40 − 10 ) = 900 cm3 time at speed vm is 0.6 mi longer, dP vm = 0.6 mi/min 36. a. = C1 3 P , P(0) = 1000, P(10) = 1700 = 36 mi/h. dt where t is the number of years since 1980. c. From part b, vm = 0.6 mi/min. Note that the b. −1/ 3 3 average speed during acceleration and ∫P dP = ∫ C1dt ; P 2 / 3 = C1t + C2 v 2 deceleration is m = 0.3 mi/min. Let t be the 3 2 P(0) = 1000: C2 = ⋅10002 / 3 = 150 time spent between stop C and stop D at the 2 constant speed vm , so 3 ⋅17002 / 3 − 150 2 P(10) = 1700: C1 = 0.6t + 0.3(4 – t)= 2 miles. Therefore, 10 2 ≈ 6.3660 t = 2 min and the time spent accelerating 3 P = (4.2440t + 100)3 / 2 4−22 3 2 is = min. 2 3 c. 4000 = (4.2440t + 100)3 / 2 0.6 − 0 a= = 0.9 mi/min 2 . 40002 / 3 − 100 2 t= ≈ 35.812 3 4.2440 t ≈ 36 years, so the population will reach dh 4000 by 2016. 34. For the balloon, = 4 , so h(t ) = 4t + C1 . Set dt t = 0 at the time when Victoria threw the ball, and 37. Initially, v = –32t and s = −16t 2 + 16 . s = 0 when height 0 at the ground, then h(t) = 4t + 64. The t = 1. Later, the ball falls 9 ft in a time given by height of the ball is given by s (t ) = −16t 2 + v0t , 3 0 = −16t 2 + 9 , or s, and on impact has a since s0 = 0 . The maximum height of the ball is 4 ⎛3⎞ v velocity of −32 ⎜ ⎟ = −24 ft/s. By symmetry, when t = 0 , since then s ′(t ) = 0 . At this time ⎝4⎠ 32 24 ft/s must be the velocity right after the first 2 ⎛v ⎞ ⎛v ⎞ ⎛v ⎞ bounce. So h(t) = s(t) or 4 ⎜ 0 ⎟ + 64 = −16 ⎜ 0 ⎟ + v0 ⎜ 0 ⎟ . ⎝ 32 ⎠ ⎝ 32 ⎠ ⎝ 32 ⎠ ⎧−32t for 0 ≤ t < 1 Solve this for v0 to get v0 ≈ 68.125 feet per a. v(t ) = ⎨ second. ⎩ −32(t − 1) + 24 for 1 < t ≤ 2.5 dV b. 9 = −16t 2 + 16 ⇒ t ≈ 0.66 sec; s also equals 9 35. a. = C1 h where h is the depth of the dt at the apex of the first rebound at t = 1.75 sec. V water. Here, V = πr 2 h = 100h , so h = . 100 dV V Hence = C1 , V(0) = 1600, dt 10 V(40) = 0. 234 Section 3.9 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.