SalasSV_07_02_ex_ans

# SalasSV_07_02_ex_ans - A-54 ANSWERS TO ODD-NUMBERED...

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Unformatted text preview: A-54 ANSWERS TO ODD-NUMBERED EXERCISES ( f −1 ) ( π ) = 1 1 6 39. f (x) = 2 − sin x > 0 on (−∞, ∞); 43. f (x) = 3x2 + 49. (a) f (x) = √ 41. f (x) = sec2 x > 0 on (−π/2, π/2); ( f −1 ) ( 3) = 45. ( f −1 ) (x) = 1 x 1 47. ( f −1 ) (x) = √ 1 − x2 51. (a) f (x) = (b) g (x) = 0 √ 1 + x2 > 0 1 4 3 > 0 on (0, ∞); ( f −1 ) (2) = x4 iff ad − bc =0 (cx + d )2 ad − bc = 0 (b) f −1 (x) = dx − b a − cx 1 (b) ( f −1 ) (0) = √ 5 1 g (x ) 53. (a) g (t ) ≥ 0 or g (t ) ≤ 0 on [a, b], with g (t ) = 0 only at “isolated" points. 55. (a) g (x) = 1 f ( g (x))g (x) f ( g (x)) =− ; g (x ) = − f ( g (x)) [ f ( g (x))]2 [ f ( g (x))]3 (c) ( f −1 ) (x) = (b) If f is increasing, then the graphs of f and g have opposite concavity; if f is decreasing, then the graphs of f and g have the same concavity. 1 57. ( f −1 ) (x) = √ . 1 − x2 65. y f 59. f −1 (x) = −1 + (2 − x)1/3 67. 61. f −1 (x) = x2 − 8x + 25 , x≥4 9 63. f −1 (x) = 1−x 1+x y 4 2 f –1 0 2 x –4 –4 π 4 π 4 x –4 SECTION 7.2 1. ln 2 + ln 10 ∼ 2. 99 = 3. 2 ln 4 − ln 10 ∼ 0. 48 = 5. −ln 10 ∼ −2. 30 = 7. ln 8 + ln 9 − ln 10 ∼ 1. 98 = 9. 1 2 area = In n m mn x 11. y y= 1 x ln 2 ∼ 0. 35 = 13. 0.406 15. (a) 1.65 (b) 1.57 (c) 1.71 17. x = e2 19. x = 1, e2 21. x = 1 23. lim x→1 d (ln x) ln x = x−1 dx x =1 =1 25. (a) P = {1, 2, . . . , n} is a regular partition of [1, n]; Lf (P ) = n 1 1 1 1 1 1 + + ··· + < dt = ln n < 1 + + · · · + = Uf (P ). 2 3 n 2 n−1 1t n 1 1 1 (b) Sum of shaded areas = Vf (P ) − dt = 1 + + · · · + − ln n. 2 n−1 1t 1 (c) Connect the points (1,1), (2, 1 ),. . . , n, by straight-line segments. The sum of the areas of the triangles that are formed is 2 n 1 ·1 2 so 1− 1 2 + 1 1 − 2 3 + ··· + 1 1 − n−1 n = 1 1 1− , 2 n 1 1 1− 2 n The sum of the areas of the indicated rectangles is 1 1− 1 2 + 1 1 − 2 3 + ··· + < γ. 1 1 − n−1 n =1− 1 , n ANSWERS TO ODD-NUMBERED EXERCISES so 1 . n A-55 γ <1− Letting n → ∞, we have 1 2 < γ < 1. (b) r ∼ 2. 2191 = 29. 1 27. (a) ln 3 − sin 3 ∼ 0. 96 > 0; ln 2 − sin 2 ∼ −0. 22 < 0 = = 31. (b) eπ/2 ∼ 4. 81048; e3π/2 ∼ 111. 318 = = SECTION 7.3 1. domain (0, ∞), f (x) = 1 x x4 3. domain (−1, ∞), f (x) = 4x 3 −1 (c) eπ/2+2nπ ; e3π/2+2nπ 3x 2 x3 + 1 5. domain (−∞, ∞), f (x) = x 1 + x2 7. domain all x = ±1, f (x) = 9. domain (− 1 , ∞), f (x) = 2(2x + 1) 1 + 2 ln (2x + 1) 2 13. domain (0, ∞); f (x) = 23. 1 +C 2(3 − x2 ) 1 cos (ln x) x 15. ln |x+1|+C 17. − 1 ln |3 − x2 | + C 2 27. ln | ln x| + C 11. domain (0, 1) ∪ (1, ∞), f (x) = − 19. 1 3 1 x(ln x)2 ln | sec 3x| + C 21. 1 2 ln | sec x2 + tan x2 | + C 33. 45. 1 2 2 3 25. − ln |2 + cos x| + C 29. −1 +C ln x 39. 1 31. − ln | sin x + cos x| + C 41. 1 2 √ ln |1 + x x| + C 35. x + 2 ln | sec x + tan x| + tan x + C 37. 1 ln 8 5 43. ln 4 3 ln 2 4x 5 3 + + +1 x−1 x 61. 2π ln (2 + √ 3) 51. g (x) = x4 (x − 1) 1 1 2x 4 + − −2 (x + 2)(x2 + 1) x x−1 x+2 x +1 65. (−1)n−1 (n − 1)! xn 47. The integrand is not deﬁned at x = 2. 53. 67. 1 π 3 49. g (x) = (x2 +1)2 (x−1)5 x3 57. 15 − 8 x2 − 1 2 ln 3 55. 1 π 4 − 1 2 ln 2 ln 4 59. π ln 9 63. ln 5 ft csc x dx = csc x( csc x − cot x) dx csc x − cot x Let u = csc x − cot x, du = csc x( csc x − cot x)dx. Then csc x dx = 1 du = ln |u| + C = ln | csc x − cot x| + C . u (v) y x=4 69. (i) domain (−∞, 4) (ii) decreases throughout (iii) no extreme values (iv) concave down throughout; no pts of inﬂection (3, 0) x 71. (i) domain (0, ∞) (ii) decreases on (0, e−1/2 ), increases on [e−1/2 , ∞] (iii) f (e−1/2 ) = − 21e local and absolute min. (iv) concave down on (0, e−3/2 ); concave up on (e−3/2 , ∞) pt of inﬂection at (e−3/2 , − 3 e−3 ) 2 (v) y (e –1/2, –1/2e) x ...
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## This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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