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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

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.

Ask a homework question - tutors are online