# EVNREV05 - 548 Chapter 5 Logarithmic Exponential and Other...

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Unformatted text preview: 548 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions Review Exercises for Chapter 5 2. f x ln x 3 3 2 1 x 1 −1 −2 −3 2 4 5 6 y 4. ln x2 x=3 1x 1 ln x2 1 ln x 1 Horizontal shift 3 units to the right Vertical asymptote: x 3 6. 3 ln x 2 ln x2 1 2 ln 5 3 ln x 6 ln x2 1 ln 52 ln x3 ln x2 xx x 1 6 ln 25 ln 25x3 x 1 2 6 8. ln x ln x ln x x xx x2 3x 3 3 3 1 x x 0 0 e0 0 3± 2 3 2 13 only since 3 2 13 < 0. 13 10. h x hx ln 1 x 1 2 1 1 ln x 1 x 2 ln x x2 x3 1 4x 3x2 ln x 2 2x 2 x 12. f x fx ln x x2 1 x 1 ax 1 ax 2 23 ln x 7x2 3x3 2 ln x2 3 6 6x 2 14. y dy dx 1 a b2 1 b b2 bx a ln a ab bx bx x a bx 2 2x 3 x2 2 b a bx ln a2 x b ln a a2 1 x2 a 16. y 18. u ln x 1 x 1 ax2 1 bx e x2 x x2 1 1, du dx 2x dx 2x 1 dx 2 x2 1 1 ln x2 2 1 C bx dy dx 1 a 1 ax2 b b a2 a bx b a a2 x a bx x2 a b ax a bx a bx bx ax2 a bx 1 dx x 1 2 ln x 20. u ln x, du ln x dx x 4 22. 1 ln x dx x e ln x 1 1 1 dx x 1 ln x 2 e 2 1 1 2 1 dx x 1 ln x 4 2 C 4 24. 0 tan 4 x dx ln cos 0 ln 4 1 2 x 0 1 ln 2 2 Review Exercises for Chapter 5 26. (a) fx y y 5 x 5 f 1 549 5x 5x x 7 7 (b) −10 6 f −1 6 7 7 x f −10 y x 5 ff 1 7 (c) f 1 fx x f f 1 5x 7 5 7 5 5x x 5 7 5 7 7 7 x x x 28. (a) fx y 3 3 x3 x3 x y 3 2 2 (b) −4 3 f f −1 5 y x f 1 2 2 x −3 x 2 (c) f 1 fx 1 f f 1 3 x3 x 2 2 3 3 x3 x 2 2 3 2 2 x x ff 30. (a) fx y y x f 1 x x2 x2 x y x 5, x ≥ 0 5 (b) f −1 −6 4 6 5 5 x f −6 5 (c) f 1 fx 1 f f 1 x2 x 5 5 x2 x 5 5 2 5 5 x for x ≥ 0. x ff 32. fx f4 fx f4 f 1 x xx 4 x 1 1 f4 e1 e1 1 1 1 1 x x 3 34. f fx 1 1 1 ln x ex ex e0 1 x x 0 3 2 1 3 3 1 xx 2 3 12 f f 4 36. (a) fx y ln y x y f 1 (b) 4 x ln y ln x ln x −4 f f −2 −1 5 (c) f 1 fx f 1 1 e1 1 x 1 x x e1 ln e1 x x ff 1 x f1 ln x 1 ln x eln x x 550 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions ex 1 ex ln 1 ex 1 ex 1 ex 1 ex x ln 1 ex 38. y 4e x2 y 5 4 40. g x ln ln ex gx 1 1 −5 − 4 −3 −2 −1 −2 −3 −4 −5 x 12345 42. h z hz e z2 2 44. y z2 2 3e 3e 3t 46. f 3t 2 1 sin 2 e 2 cos 2 esin 2 ze y 3t 9e 3 t2 t f 48. cos x2 2x sin x2 dy dx xey xey dy dx ey x2 ey 50. Let u e1 x dx x2 1 , du x e1 1 dx. x2 x 2x sin xey e dx 2x, 1 dx x2 e1 x C 52. Let u e2x e2x e2x e e 2x 2x du 2e2x e 2x dx. 54. Let u x2ex 3 x3 1 1, du 1 x3 e 3 3x2 dx. 1 1 2e2x 2 e2x 1 ln e2x 2 2e 2x dx e 2x e 2x dx 3x2 dx 1 x3 e 3 1 C C 56. e2x dx e2x 1 1 1 2e2x dx 2 e2x 1 1 ln e2x 2 1 C 58. (a), (c) 10,000 0 0 5 (b) V Vt V1 V4 8000e 0.6t ,0≤t≤5 4800e 0.6t 2634.3 dollars year 435.4 dollars year 2 2 60. Area 0 2e x dx 2e x 0 2e 2 2 2 2 e2 1.729 62. g x 62 y x2 64. y log4 x2 y 5 4 3 2 1 8 6 −5 − 4 −3 −2 −1 2 x 2 4 x 12345 −4 −2 −2 −3 −4 −5 Review Exercises for Chapter 5 66. f x fx 4xex 4xex ln 4 4xex x x 1 1 x 1 4xex 1 ln 4 68. y y 2 x4 4 1t x x 551 x 4 x ln 4 70. h x hx log5 log5 x log5 x 1 72. t2 dt 1 2 ln 2 1t C 11 ln 5 x 1 1 ln 5 x x 1 18,000 18,000 h 74. t 50 log10 18,000 18,000 h (c) 10t 18,000 t 50 50 log10 (a) Domain: 0 ≤ h < 18,000 (b) 100 80 60 40 20 h 4000 12,000 t 18,000 18,000 h 18,000 10 18,000 1 . 50 log10 18,000 h h 2 t 50 h h 10 t 50 As h → 18,000, t → (d) t 50 log10 18,000 dt 50 dh ln 10 18,000 d 2t dh2 50 ln 10 18,000 h Vertical asymptote: h 18,000 No critical numbers As t increases, the rate of change of the altitude is increasing. 1 2 1 2 t 1620 76. 2P 2 ln 2 r Pe10r e10r 10r ln 2 10 dy ds 1 0.012 dy y s Ce 50, y 28e0.6 e 1 1 2x 78. y y 600 5 5 600 1620 3.868 grams 6.93% 80. (a) 0.012y, s > 50 ds C1 0.012s (b) Speed(s) Miles per Gallon (y) 50 28 55 26.4 60 24.8 65 23.4 70 22.0 1 ln y 0.012 y When s y dy dx dy y 28 Ce 0.012 50 ⇒C 28e0.6 0.012s, s > 50 82. e e 2x 2x e 2x dx e 2x 1 2e 21 e C 2x 2x dx 1 ln 1 2 552 84. y Chapter 5 ey sin x dy dx e y Logarithmic, Exponential, and Other Transcendental Functions 0 ey sin x sin x dx cos x 1 cos x ln 1 cos x C C C1 C C1 ln cos x C dy y e ey y 86. dy dx 3x x 3x y y dx (homogeneous differential equation) x dy x dv 0 v dx. v dx x2 dv 2v dx 1 dx x ln x x x2 x3 y 0 0 x dv 1 3 1 ln 3 2 C2 3 C3 C 3x x3 2v dv C1 ln 3 2v 12 Let y 3x vx, dy vx dx 3x x x dv 2vx dx 3 2v 2v 2v 2y 12 ln C2 C3 3Cx 2 y x 2Cy 3Cx 2C 88. dv dt (a) kv 9.8 dv kv 9.8 9.8 9.8 9.8 v t kt ekt dt C1 C2 C2 (b) lim v t t→ 9.8 k 1 9.8 k 1 9.8t k kv0 1 kv k0 1 kv k2 0 9.8 1 kv k2 0 1 kv k2 0 9.8 ekt dt 9.8 ekt 9.8 ekt C⇒C 9.8 ekt 9.8 ekt s0 1 C s0 1 kv k2 0 1 kv k2 0 s0 9.8 9.8 C 1 ln kv k ln kv kv (c) s t C3ekt C3 ekt s0 C3 ⇒ C3 kv0 9.8 ekt kv0 9.8 st 1 9.8 k 1 9.8 k 1 9.8 k 9.8t k 1 kv k2 0 9.8t k 9.8t k At t 0, v0 v Note that k < 0 since the object is moving downward. Review Exercises for Chapter 5 90. h x 4 553 3 arcsin 2x y 92. (a) Let cot arccot 2 2 θ 5 1 tan arccot 2 −π 2 −π 4 −2 −4 tan 5 1 . 2 2 π 4 π 2 x (b) Let sec arcsec 5 5 5 2 cos arcsec cos 1 . 5 θ 1 94. y y arctan x2 1 2x x2 1 1 2 96. y x4 2x 2x2 2 y 1 arctan e2x 2 1 1 2e2x 2 1 e4x e2x 1 e4x 98. y y x2 x x 2 4 x 2 arcsec , 2 < x < 4 2 x2 5 dx. 1 5 1 3 2 4 1 x2 x 2 1 x 2 4 x 4 x2 x2 4 x x 2 4 4 x2 x 4 100. Let u 3 5x, du 1 dx 25x2 1 5x 2 5 dx 1 5x arctan 53 3 C 102. 16 4 4 x2 dx 1 x arctan 4 4 1 4 C 104. x dx x2 4 x2 1 1 2 dx 1 2 4 x2 12 2x dx 4 arcsin x 2 4 x2 C 106. Let u arcsin x, du x2 dx. 108. π 2 y arcsin x dx 1 x2 1 arcsin x 2 C π 4 y = arcsin x x 0.25 0.5 0.75 1 2 Since the area of region A is 1 0 1 sin y dy , 1 0.571. the shaded area is 0 arcsin x dx 2 110. y y x tanh x 2 1 1 2x tanh 1 112. Let u 2x 1 2x 4x2 tanh 1 x3, du 3x2 dx. 1 3 sech x3 2 4x2 2x x2 sech x3 2 dx 3x2 dx 1 tanh x3 3 C 554 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions Problem Solving for Chapter 5 2. (a) 2 1 x y (b) 4 3 2 1 y −1 −2 π 2 π 3π 2 2π π 2 π 3π 2 2π x 2 2 2 sin x dx 0 sin x dx ⇒ 0 sin x dx 0 0 sin x (d) 1 y 2 dx 22 4 (c) π y π 2 1 2 x −1 1 π 4 π 2 x 1 arccos x dx 1 2 2 y 1 , 1 tan x 1 . 42 2 is symmetric with respect to the 2 point 0 1 1 tan x 2 dx 1 22 4 4. y y 0.5x and y 2x 1.2x intersect y x. x. y = 0.5 x 6 y = 2x y = x does not intersect y y = 1.2 x Suppose y ax y x is tangent to y x1 x. x ax at x, y . −6 −2 6 x⇒a ax ln a 1 ⇒ x ln x1 e 1 ⇒ ln x 1⇒x e, a x. e1 e For 0 < a ≤ e1 6. (a) y sin y Area A fx x 1.445, the curve y ax intersects y arcsin x 4 4 sin y 6 dy cos y 6 2 2 3 2 3 2 2 0.1589 Area B 22 1 2 6 12 Area C 0.2618 2 2 3 2 2 8 1 12 2 2 A 2 12 3 0.1346 B (b) 12 arcsin x dx 4 2 8 —CONTINUED— Problem Solving for Chapter 5 6. —CONTINUED— ln 3 y 555 (c) Area A 0 ey dy ln 3 ln 3 y = ln x ey 0 3 3 ln x dx 1 3 ln 3 2 A 3 ln 3 2 ln 27 2 1.2958 A B ey = x Area B 1 x 1 2 3 (d) tan y Area A x 3 tan y dy 4 3 ln cos y 4 y y = arctan x 1 ln 2 3 2 ln 2 ln 2 3 1 ln 2 2 1 ln 2 2 4 1 π 3 π 4 A C x 1 3 Area C 1 arctan x dx 43 3 B 3 1 ln 2 2 12 8. y If y y y b y 0, eax bx x c Thus, a dy dt 1 y ln y aea ab a a c 1 1 a ex ex ea x 0.6818 10. Let u a aea b b b ea b Tangent line Area tan x, du 4 0 sec2 x dx 1 4 sin x 2 4 cos x 2 dx 0 1 0 sec2 x dx tan2 x 4 du eax u2 4 1 0 1 u arctan 2 2 1 1 arctan 2 2 a 1 1 4 dt t t et 1. (b) 1 y 12. (a) y1 1 1 y ,y 0 dy y y 1 y 1 y y y y y ln 1 ln C C C −6 −4 −2 x 2 4 6 C1et yC1et t dy dt d2y dt 2 t y1 y y y y y2 0 for y 1 2 C1et 1 2yy ⇒ y C1e C1et 3 1 1 C2e y0 1 4 1 1 1 C2 ⇒ C2 Hence, y 1 . 3e t d2y 1 1 d2y > 0 if 0 < y < and 2 < 0 if < y < 1. 2 dt 2 2 dt 1 , the Thus, the rate of growth is maximum at y 2 point of inflection. —CONTINUED— 556 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 12. —CONTINUED— (c) y y1 y ,y 0 1 1 1 C2 1 1 e 2 2 1 C2e t As before, y y0 Thus, y 1 2 ⇒ C2 2 t. 1 2 e t 2 The graph is different: y 6 4 2 −6 −4 −2 x 2 4 6 14. (a) u v 985.93 985.93 985.93 120,000 0.095 12 1 0.095 12 1 12t 0.095 12 12t 1000 u v 120,000 0.095 12 (b) The larger part goes for interest. The curves intersect when t (c) The slopes are negatives of each other. Analytically, u u 15 (d) t 985.93 v 15 du v ⇒ dt 14.06. dv dt 27.7 years. 0 0 35 12.7 years Again, the larger part goes for interest. ...
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## This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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