ODDREV05 - 272 Chapter 5 Logarithmic, Exponential, and...

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Unformatted text preview: 272 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 85. As k increases, the time required for the object to reach the ground increases. ex 2 x 87. y y cosh x ex 2 e e x 89. y cosh y cosh x 1 1 sinh y 1 x sinh x sinh y y y 1 cosh2 y 1 1 x2 1 91. y y sech x 2 ex 2 ex e x e 2 x ex e x e x 2 e x ex ex e e x x sech x tanh x Review Exercises for Chapter 5 1. f x ln x 3 5 4 3 2 1 x 1 2 3 4 5 y Vertical shift 3 units upward Vertical asymptote: x 0 x=0 3. ln 5 4x2 4x2 1 1 1 2x 1 2x ln 5 4x2 1 1 1 ln 2x 5 1 ln 2x 1 ln 4x2 1 5. ln 3 1 ln 4 3 x2 ln x ln 3 ln 3 4 x2 ln x ln 33 4 x x2 7. ln x x x 1 1 1 x 2 e2 e4 e4 1 53.598 9. g x gx ln x 1 2x 1 ln x 2 11. f x fx x ln x x ln x 2 1 2 ln x 1 a bx ln a x 1 b a a bx 1 x 12 13. 1 x ln x 1 ln x 2 ln x 2 ln x y dy dx 1 ln a b2 b 1 b2 a bx bx a a a ab bx bx x 2 a bx 2 15. y dy dx 1 ln a a 1 xa bx bx ln x 17. u 7x 7x 1 2 2, du dx 7dx 1 1 7 dx 7 7x 2 1 ln 7x 7 2 C R eview Exercises for Chapter 5 sin x dx 1 cos x sin x dx 1 cos x ln 1 3 4 273 19. 21. 1 x x 1 4 dx 1 1 1 dx x 4 x ln x 1 3 ln 4 cos x C 3 23. 0 sec d ln sec 1 2x 1 2x tan 0 ln 2 3 25. (a) fx y 2y 2x f 1 3 3 (b) f − 11 −1 7 10 3 3 x x y 2x 6 (c) f 1 −7 f fx 1 f 11 2x 3 6 1 2 2 1x 2 2x 6 3 3 6 x x ff 27. (a) fx y y2 x2 f 1 x 4 f 2x x x x y x2 1 1 (b) f −1 f −3 6 1 1 x −2 1, x ≥ 0 (c) f 1 fx 1 f 1 x 1 1 x2 x2 1 1 2 1 x ff x f x2 x2 1 x for x ≥ 0. 29. (a) fx y y3 x3 f 1 3 3 x x 1 1 (b) 4 f −1 f 1 1 x x y x3 1 −4 5 −2 (c) f 1 fx 1 f f fx f 6 fx 13 x 1 1 3 3 x 1 1 3 1 x x ff 31. f f f 1 1 x x3 x3 1 fx 1 x3 x 1 x 3 1 3 1 35 3 2 2 13 33. tan x 3 3 sec2 x 4 3 1 f 6 3 4 x x 1 2 1 23 2 23 3 1 3 23 f f 1 6 3 3 0.160 274 35. (a) Chapter 5 fx y ey e2y e f 1 2x Logarithmic, Exponential, and Other Transcendental Functions x (b) −3 2 ln f −1 f 3 ln x x x y e2x (c) f 1 −2 fx 1 f f ln e 2x x2 1 ln x ln x2 e2 ln e2x x e ln x ex x x x ff 39. f x fx x e2x ln 37. y e x2 y 6 4 2 x 2 −2 4 −2 41. g t gx t2et t2et 2tet tet t 2 43. y y e2x 1 2x e 2 e e 2x 2x 12 2e2x 2e 2x e2x e 2x e2x e 2x 45. g x gx x2 ex ex 2x e2x x2ex x2 ex x 47. y 1 x ln x dy dx y ln x 2y y2 dy dx dy dx dy dx 0 0 y x y x 2y ln x 2y ln x 49. Let u xe 3x2 3x2, du dx 1 e 6 6x dx. 3x2 51. 1 e 6 3x2 e4x e2x ex 1 dx e3x 1 3x e 3 e4x ex ex 3e2x 3ex e e x x dx C C 6x dx C 3 53. xe1 x2 dx 11 e 2 11 e 2 x2 x2 2x dx C 55. Let u ex ex 1 ex dx 1, du ln ex ex dx. 1 C 57. y y ex a cos 3x e x b sin 3x 3b cos 3x a 3a 8a 3a b ex a cos 3x 3b cos 3x 3b sin 3x 6b cos 3x 10b sin 3x 8a 6b 2a 3b 10a cos 3x 0 ex 3a b sin 3x a 3b cos 3x b sin 3x 3a sin 3x 3a 3a 6a 6a ex y ex 3 ex y 2y 10y e x b sin 3x b cos 3x 8b sin 3x 8b 2 R eview Exercises for Chapter 5 4 275 59. Area 0 xe x2 dx 1 e 2 4 x2 0 1 e 2 16 1 0.500 61. y 33 2 y 63. y y log2 x 4 3 2 1 1 6 5 4 3 2 x −1 1 2 34 −2 −3 −4 x 1 2 34 5 67 −4 −3 −2 −1 −2 65. f x fx 3x 3x 1 1 67. ln 3 y ln y y y y x2x 2x 2x x y 1 1 ln x 1 2 ln x 1 x 1 2 2x 2 ln x x2x 1 2x x 1 2 ln x 69. g x gx log3 1 1 21 xa ax a 1 x 1 log3 1 2 2x x 71. x 1 5x dx 11 5x 2 ln 5 1 2 C 1 x ln 3 1 1 ln 3 (b) y y ax ln a ax (c) y ln y 1 y y y y xx x ln x x y1 xx Ph P 18,000 k Ph P 35,000 dy dx dy y x2 2 x2 x x 3 dx x 3 ln x C 1 1 x 1 ln x ln x ln x 30ekh 30e18,000k ln 1 2 18,000 30e 30e 3 15 ln 2 18,000 (d) y y aa 0 73. (a) y y 75. 10,000 P Pe 0.07 10,000 e1.05 15 77. $3499.38 h ln 2 18,000 35,000 ln 2 18,000 7.79 inches 79. P 2C 2 ln 2 t Ce0.015t Ce0.015t e0.015t 0.015t ln 2 0.015 46.21 years 81. 276 83. y Chapter 5 2xy dy dx 1 dy y ln y e x2 C1 Logarithmic, Exponential, and Other Transcendental Functions 0 2xy 2x dx x2 y Cex y2 2 C1 y dy dx x2 2xy x2 Let y x2 x2 85. (homogeneous differential equation) 2xy dy x dv dx 0 y2 dx vx, dy v2x2 v dx. v dx 2x3v dv x2v2 dx 1 v2 dx dx x ln x x 1 1 x2 0 0 2x3v dv 2x dv 2v 1 ln 1 C v2 Cx y2 1 v2 dv v2 C1 C yx 2 2x vx x dv 2x2v2 dx x2 v2x2 ln 1 Cx2 x2 x x2 y2 v2 y2 ln C or C1 87. y y y x2 y C1x C1 6C2 x 3xy C2 x3 3C2 x2 3y x2 6C2 x 6C2 x3 3x C1 3C1x 3C2 x2 9C2 x3 4C2 3C1x C1x C2 x3 0 3C2 x3 x x 2, y 2, y 0: 0 4: 4 4 2C1 C1 8C2 ⇒ C1 12C2 4C2 12C2 8C2 ⇒ C2 1 ,C 21 2 y 2x 13 x 2 3 4 y 89. f x 2 arctan x −6 −4 −2 −2 −4 x 2 R eview Exercises for Chapter 5 1 2 2 1 277 91. (a) Let arcsin 1 2 sin arcsin 1 2 cos sin 1 sin arcsin 2 (b) Let θ 1 . 2 1 2 3 sin cos arcsin 1 2 3 . 2 x 93. y y tan arcsin x 1 x2 12 1 x2 x2 12 95. y 1 x2 32 x arcsec x x x x2 1 arcsec x x2 1 1 x2 2x y 97. y y x arcsin x 2 21 2 x2 arcsin x 2 21 1 x2 x2 2x arcsin x 1 x2 arcsin x 2 2x arcsin x 1 x2 e2x, du 1 e2x e 2x arcsin x 99. Let u 2e2x dx. 1 e2x dx e4x 1 21 1 e2x 2 dx 2e2x dx 1 arctan e2x 2 103. Let u C 101. Let u x 1 x2, du x4 dx 2x dx. 1 2 x , du 2 1 2 1 1 x2 2 4 arctan k dt m k t m C 0. Thus, x2 x 2 2 16 x x2 dx x2, du 2x dx. 1 ln 16 2 x2 C 2x dx 1 arcsin x2 2 C 16 1 1 2x dx 2 16 x2 105. Let u arctan dx. 2 4 x 2 arctan x 2 dx 4 x2 107. dy A2 y2 y arcsin A Since y sin 0 when t k t m y dx 1 x arctan 4 2 2 C 109. y y 2x 2 cosh x 1 sinh 2x x 2 sinh x 2x 0, you have C y A A sin k t m 111. Let u x x4 x2, du 1 dx 2x dx. 1 2 1 x2 2 1 2x dx 1 ln x2 2 x4 1 C 278 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions Problem Solving for Chapter 5 1. tan tan 3 x 6 2 1 10 1 x 2: 6 3 Minimize fx fx 1 1 2 arctan 3 x2 3 x arctan 1 6 10 10 x 6 x 3 x2 9 36 36 118 x 10 2x 2 2 θ1 0 x θ a θ2 10 1 9 x2 1 36 10 x 2 0 6 x2 9 18 36 10 100 20x x2 x 2 x2 20x 2x2 0 20 ± 202 2 4 118 10 ± 218 a 10 1 218 2 4.7648 1.7263 1.0304 1.2793 10 fa 1.4153 or 98.9 Endpoints: a a 0: 10: Maximum is 1.7263 at a 3. f x sin ln x or 0, 218 4.7648. (a) Domain: x > 0 (b) f x 1 (f) 2 . 3 2 2k . 0 2 sin ln x ⇒ ln x e 2 5 Two values are x (c) f x 1 ,e 2 2 sin ln x ⇒ ln x e 2, −2 2k . x→0 Two values are x e3 2. lim f x seems to be e 1. e 1. 2, 1 . (This in incorrect.) 2 e 3 2, 1, 1 , (d) Since the range of the sine function is 1, 1 . parts (b) and (c) show that the range of f is (e) f x fx 1 cos ln x x 0 ⇒ cos ln x 0 ⇒ ln x x fe f1 f 10 2 (g) For the points x we have f x For the points x e 7 2, ... 2 ,e 5 2 ,e 9 2 ,... 2 e 2 k ⇒ we have f x on 1, 10 1 0 0.7440 Maximum is 1 at x e 2 That is, as x → 0 , there is an infinite number of 1, and an infinite number where points where f x fx 1. Thus lim sin ln x does not exist. x→0 4.8105 You can verifiy this by graphing f x on small intervals close to the origin. P roblem Solving for Chapter 5 5. (a) Area sector Area circle t ⇒ Area sector 2 1 base height 2 1 cosh t 2 1 cosh2 t 2 1 cosh2 t 2 1 cosh2 t 2 At 1 t 2 C. But, A 0 1 t or t 2 sinh t 1 279 t 2 x2 t 2 1 dx 1 dx 1 sinh t cosh t 1 cosh t (b) Area AOP At At x2 cosh2 t sinh2 t 1 2 0⇒C sinh2 t sinh2 t sinh2 t C 2A t . 0 Thus, A t 7. y y y b y If x ln x 1 x 1 x a 1 x a 0, c b a b 1 Tangent line 1. Thus, b c b b 1 1. 9. Let u dx Area 1 2u 4 1 x, x u 1, x u2 2u 1, 2 du. 1 dx xx 3 2 2u u u u2 du 3 2 u2 2u 1 1 du 32 2 1 du u 32 2 u 2 ln u 2 2 ln 3 0.8109 dy dt y 1.01 2 ln 2 2 ln 3 2 11. (a) y1.01 dt t C1 0.01t 1 0.01t 1 0.01t 100 (b) y 1 dy y y y kt k dt C1 kt 1 C 1 C1 1 y0 . kt 1 dy y 0.01 C 0.01 1 y0.01 y0.01 y y0 1: 1 C y0 y0 ⇒ C1 1 1 1 ⇒C y0 . 1 y0 C C Hence, y kt 1 ⇒C C100 1 t→T 1 For t → 1 , y→ y0 k Hence, y For T 1 0.01t 100 . 100, lim y . 280 Chapter 5 dy dt 1 y ln y 20 Logarithmic, Exponential, and Other Transcendental Functions 13. Since ky dy 20 , k dt kt Cekt C 20. 52. 52ek 22.35 . dS dt d 2S dt2 4 S 100 9 4 9 S dS dt 2S 50 or 20, ek 28 52 7 13 , and k ln 7 13 . 20 y When t When t Thus, y When t dS dt S dS dt 0, y 72. Therefore, C 1, y 48. Therefore, 48 52e ln 7 13 t 20. 5, y 52e5 ln 7 13 20 15. (a) k1S L L Ce L1 LC ke 1 Ce k L1 k L1 k1S L S is a solution because kt 2 (b) ln ln ln S 100 dS dt dS dt 0. S dS dt 1 kt Ce kt kt 2 C ke kt 4 100 9 0 when S C Le kt 1 Ce kt L 1 k . L 9. And, (d) 140 120 100 80 60 40 20 S L Ce L Ce Choosing S 50 50, we have: 1 1 t 100 9eln 4 9eln 4 9t 9t kt kt L Ce kt 2 ln 1 9 ln 4 9 t S , where k1 L S 100. Also, S 10 when t 0 ⇒ C 20 when t 1 ⇒ k ln 4 9 . 1 1 100 9eln 4 9t 2.7 months Particular Solution. S 100 9e 0.8109t (c) 125 1 2 3 4 t (e) Sales will decrease toward the line S 0 0 10 L. ...
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