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SalasSV_07_07_ex_ans

# SalasSV_07_07_ex_ans - A-58 39 1 e ANSWERS TO ODD-NUMBERED...

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Unformatted text preview: A-58 39. 1 e ANSWERS TO ODD-NUMBERED EXERCISES 41. f (x) = px ln f (x) = x ln p f (x ) = ln p f (x ) f (x) = px ln p 43. (x + 1)x x + ln (x + 1) x+1 45. (ln x)ln x 1 + ln (ln x) x 47. xsin x cos x ln x + sin x x 49. ( sin x)cos x cos2 x − sin x ln ( sin x) sin x 51. x2 x 2x + 2x (ln x)(ln 2) x y In 53. y x 55. 2x 57. log 3 x x y 2x x log 2 x x 59. 1 4 ln 2 61. 2 63. 45 ln 10 65. 1 1 + 3 ln 2 67. approx. 16. 999999; 5(ln17)/(ln 5) = (eln 5 )(ln17)/(ln 5) = eln 17 = 17 69. (b) the x-coordinates of the points of intersection are: x ∼ −1. 198, x = 3 and x ∼ 3. 408. = = (c) for the interval [ − 1. 198, 3], A ∼ 5. 5376; for the interval [3, 3. 408], A ∼ 0. 1373 = = SECTION 7.6 1. (a) \$411.06 (b) \$612.77 (c) \$859.14 3. about 5 1 % : (ln 3)/20 ∼ 0. 0549 = 2 5. 16, 000 9. (a) e0.35 (b) k = liters 17. 5 4 5/2 5 7. (a) P (t ) = 10, 000 et ln 2 = 10, 000(2)t (b) P (26) = 10, 000(2)26 , P (52) = 10, 000(2)52 13. in the year 2112 15. 200 4 t /5 5 ln 2 15 ∼ 2. 86 gms = 11. P (20) ∼ 317. 1 million; P (11) ∼ 284. 4 million = = 19. 100[1 − 1 1/n 2 % 21. 80.7%, 3240 yrs 23. (a) x1 (t ) = 106 t , x2 (t ) = et − 1 d d [x1 (t ) − x2 (t )] = [106 t − (et − 1)] = 106 − et . (b) dt dt This derivative is zero at t = 6 ln 10 ∼ 13. 8. After that the derivative is negative = ∼ (c) x2 (15) < e15 = (e3 )5 = 205 = 25 (105 ) = 3. 2(106 ) < 15(106 ) = x1 (15) x2 (18) = e18 − 1 = (e3 )6 − 1 ∼ 206 − 1 = 64(106 ) − 1 > 18(106 ) = x1 (18) = ∼ 64(106 ) − 1 − 18(106 ) ∼ 46(106 ) x2 (18) − x1 (18) = = (d) If by time t1 EXP has passed LIN, then t1 > 6 ln 10. For all t ≥ t1 the speed of EXP is greater than the speed of LIN: for t ≥ t1 > 6 ln 10, v2 (t ) = et > 106 = v1 (t ). 25. (a) 15( 2 )1/2 ∼ 12. 25 lb/in.2 = 3 31. 176/ln 2 ∼ 254 ft = (b) 15( 2 )3/2 ∼ 8. 16 lb/in.2 = 3 35. f (t ) = Cet 2 /2 27. 6. 4% 29. (a) \$18,589.35 (b) \$20,339.99 (c) \$22,933.27 33. 11,400 years 37. f (t ) = Cesin t SECTION 7.7 1. (a) 0 (b) − π 3 13. 3. (a) √ 2 2π 3 √ (b) 3π 4 5. (a) 1 2 (b) π 4 7. (a) does not exist 2 sin−1 x 17. √ 1 − x2 (b) does not exist 9. (a) 3 2 (b) − 7 25 11. 1 x 2 + 2x + 2 1 x[1 + (ln x)2 ] x 4x4 − 1 r √ |r | 1 − r 2 2x 15. √ + sin−1 2x 1 − 4x2 27. 2x sec−1 1 x −√ x 2 19. x − (1 + x2 ) tan−1 x x2 (1 + x2 ) 1 21. (1 + 4x 2 ) 1 √ tan−1 2x c−x c+x 23. 25. − 1 − x2 29. cos [ sec−1 (ln x)] · x|ln x| (ln x)2 − 1 31. 33. Set au = x + b, a du = dx. dx a2 − (x + b)2 = a du = √ a2 − a 2 u 2 √ du 1 − u2 = sin−1 u + C = sin−1 x+b a +C ANSWERS TO ODD-NUMBERED EXERCISES 35. Set au = x + b, a du = dx. dx ( x + b) ( x + b) 2 − a 2 37. domain ( − ∞, ∞), range (0, π ) 51. tan−1 2 − 1 π ∼ 0. 322 = 4 61. sin−1 (ln x) + C 71. 53. 1 −x 2 1 2 A-59 = 1 a du = √ a au a2 u2 − a2 43. 1 π 20 1 du |x + b| = sec−1 + C. √ a a u u2 − 1 47. 1 π sec−1 4 − 3 9 59. 49. 1 π 6 39. 1 π 4 41. 1 π 4 1 2 45. 57. 1 π 24 1 3 sin−1 x2 + C 55. tan−1 x2 + C 65. π 3 tan−1 ( 1 tan x) + C 3 4 3 1 ( sin−1 2 x)2 + C 63. √ 1 is not deﬁned for x ≥ 1. 67. 2π − √ 69. 4π ( 2 − 1) √ s2 + sk feet from the point where the line of the sign intersects the road. 1 π a2 ; 2 73. (b) area of semicircle of radius a 75. (a) There exist constants C1 , C2 such that f ( x ) + g ( x ) = C1 (b) lim f (x) = x→0+ π 2 for x < 0; f (x) + g (x) = C2 for x > 0. (e) C1 = π 2 ; x→0− lim f (x) = − π 2 (d) This is clear from the graphs in (a). ; C2 = − π 2 77. estimate ∼ 0. 523, sin 0. 523 ∼ 0. 499 explanation: the integral = sin−1 0. 5; therefore sin (integral)= 0. 5 = = 79. (a) 16 87 81. (a) (0.78615, 0.66624) (b) A ∼ 0. 37743 = (b) 0 (c) − 120 169 SECTION 7.8 1. 2x cosh x2 3. a sinh ax √ 2 cosh ax 5. 1 1 − cosh x 7. ab( cosh bx − sinh ax) 9. a cosh ax sinh ax 11. 2 e2x cosh (e2x ) 13. −e−x cosh 2x + 2 e−x sinh 2x 19. cosh2 t − sinh2 t = e t + e −t 2 2 15. tanh x − e t − e −t 2 2 17. ( sinh x)x [ln ( sinh x) + x coth x] = e2t − 2 + e−2t e 2 t − 2 + e −2 t − =1 4 4 + e t − e −t 2 e s − e −s 2 21. cosh t cosh s + sinh t sinh s = e t + e −t 2 e s + e −s 2 = 1 ( e t −s + e s −t + e t −s + e −t −s + e t +s − e s −t − e t −s + e −t −s ) 4 = 1 (et −s + e−(t +s) ) = cosh (t + s) 2 23. cosh2 t + sinh2 t = 25. sinh ( − t ) = e t + e −t 2 2 + e t + e −t 2 2 = 1 (e2t + 2 + e−2t + e2t − 2 − e−2t ) = 4 27. absolute max −3 e2t + e−2t = cosh 2t 2 e −t − e −( − t ) e −t − e t = = − sinh t 2 2 e x + e −x e x − e −x + 2 2 33. n 29. [ cosh x + sinh x]n = 31. A = 2, B = 1 , C = 3 3 41. 1 ( sinh x cosh x 2 1 4 = [ex ]n = enx = enx + e−nx enx − e−nx + = cosh nx + sinh nx 2 2 39. − 1 +C a cosh ax + x) + C 1 1 1 sinh ax + C 35. sinh3 ax + C 37. ln ( cosh ax) + C a 3a a √ 43. 2 cosh x + C 45. sinh 1 ∼ 1. 175 47. 81 49. π = 20 53. (a) (0. 69315, 1. 25) (b) A ∼ 0. 38629 = 51. π [ln 5 + sinh (4 ln 5)] ∼ 250. 492 = SECTION 7.9 1. 2 tanh x sech2 x 3. sech x csch x 5. 2e2x cosh( tan−1 e2x ) 1 − e 4x 7. √ −x csch2 ( x2 + 1) √ x2 + 1 9. − sech x( tanh x + 2 sinh x) (1 + cosh x)2 ...
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