F x 12 cos x critical points x 3 53 f x 00

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Unformatted text preview: 1)2 . If x = f (y ) = 3/(y + 1) then y = f −1 (x) = (3/x) − 1, so and (b) 17. 1 f (f −1 (x)) =− (f −1 (x) + 1)2 (3/x)2 3 =− = − 2. 3 3 x d −1 −3 f (x) = 2 ; dx x d −1 2 f (x) = ; and f (x) = ex/2 , f (x) = 1 ex/2 . If x = f (y ) = ey/2 then y = f −1 (x) = 2 ln x, so 2 dx x 2 1 −1 = 2e−f (x)/2 = 2e− ln x = 2x−1 = f (f −1 (x)) x y (a) (b) 2 2 -2 4 x The curve y = e−x/2 sin 2x has x-intercepts at x = 0, π/2. It intersects the curve y = e−x/2 at x = π/4, and it intersects the curve y = −e−x/2 at x = −π/4, 3π/4. 137 18. Chapter 4 (a) (b) y y π/2 π/2 x x 1 (c) 1 (d) y y π/2 x x 1 5 −π/2 19. (a) (b) x/(1 + x) − ln(1 + x) ln(1 + x) ln(1 + x) y 1 , = − = , x y x2 x(1 + x) x2 (1 + x)(1/x) 1 dy ln(1 + x) = (1 + x)(1/x)−1 − dx x x2 dy y 1 1 x x x = ex + ln x , = xe ex + ln x = ex xe −1 + xe ln x ln y = ex ln x, y x dx x ln y = (c) y = x3 + 1 so y = 3x2 . (d) y= (e) abe−x (1 + be−x )2 dy 2 −1/3 dy 2 xy + y 2/3 + yx−1/3 + x2/3 = 2x. Multiply by 3x1/3 y 1/3 : 3 dx 3 dx dy dy + 3x1/3 y + 2y 4/3 + 3xy 1/3 = 6x4/3 y 1/3 . Regroup: 2x4/3 dx dx 6x4/3 y 1/3 − 3x1/3 y − 2y 4/3 dy dy 2x4/3 + 3xy 1/3 = 6x4/3 y 1/3 − 3x1/3 y − 2y 4/3 , = . dx dx 2x4/3 + 3xy 1/3 1 1 ln x + ln(x + 1) − ln sin x + ln cos x, so 2 3 1 cos x sin x 5x + 3 1 y= + − − = − cot x − tan x. 2x 3(x + 1) sin x cos x 6x(x + 1) (f ) 21. (a) Find x when y = 5 · 12 = 60 in. Since y = log x, x = 10y = 1060 in. This is approximately 2.68 × 1042 light-years, so even in astronomical terms it is a fabulously long distance. (b) 20. y= Find x when y = 100(5280)(12) in. Since y = 10x , x = log y = 6.80 in or 0.57 ft, approximately. (a) The function ln x − x0.2 is negative at x = 1 and positive at x = 4, so it must be zero in between (IVT). (b) x = 3.654 Supplementary Exercises 4 22. 138 ln x 1 = . The steps are reversible. x k (b) By zooming it is seen that the maximum value of y is approximately 0.368 (actually, 1/e), so there are two distinct solutions of xk = ex whenever k > 1/0.368 ≈ 2.717. (a) If xk = ex then k ln x = x, or y x 2 -2 (c) 23. 24. x = 1.155 1 1 = 1 so x = . The x ln b ln b curves intersect when (x, x) lies on the graph of y = logb x, so ln x from which x = logb x. From Formula (9), Section 4.2, logb x = ln b 1/e ln x = 1, x = e, ln b = 1/e, b = e ≈ 1.4447. y Set y = logb x and solve y = 1: y = (a) (b) √ Find the point of intersection: f (x) = x + k = ln x. The 1 1√ slopes are equal, so m1 = = m2 = √ , x = 2, x = 4. x 2x √ Then ln 4 = 4 + k , k = ln 4 − 2. √ k 1 Since the slopes are equal m1 = √ = m2 = , so k x = 2. x 2x √ At the point of intersection k x = ln x, 2 = ln x, x = e2 , k = 2/e. 2 2 x y 2 x 2 y 2 0 x 5 CHAPTER 5 Analysis of Functions and Their Graphs EXERCISE SET 5.1 1. (a) f > 0 and f > 0 (b) y f > 0 and f < 0 y x x (c) f < 0 and f > 0 (d) f < 0 and f < 0 y y x x 2. (a) (b) y y x (c) x (d) y y x 3. A: dy/dx < 0, d2 y/dx2 > 0 B : dy/dx > 0, d2 y/dx2 < 0 C : dy/dx < 0, d2 y/dx2 < 0 x 4. A: dy/dx < 0, d2 y/dx2 < 0 B : dy/dx < 0, d2 y/dx2 > 0 C : dy/dx > 0, d2 y/dx2 < 0 139 Exercise Set 5.1 140 5. An inflection point occurs when f changes sign: at x = −1, 0, 1 and 2. 6. (a) f (0) < f (1) since f > 0 on (0, 1). (c) f (0) > 0 by inspection. (e) f (0) < 0 since f is decreasing there. 7. (a) [4, 6] (d) (2, 3) and (5, 7) 8. f f (1, 2) − + (2, 3) − − (3, 4) − + (b) f (1) > f (2) since f < 0 on (1, 2). (d) f (1) = 0 by inspection. (f ) f (2) = 0 since f has a minimum there. (b) [1, 4] and [6, 7] (e) x = 2, 3, 5 (4, 5) + + (5, 6) + − (c) (1, 2) and (3, 5) (6, 7) − − f (x) = 2x − 5 f (x) = 2 (a) [5/2, +∞) (c) (−∞, +∞) (e) none (b) (−∞, 5/2] (d) none f (x) = −2(x + 3/2) f (x) = −2 (a) (−∞, −3/2] (c) none (e) none (b) (d) 11. f (x) = 3(x + 2)2 f (x) = 6(x + 2) (a) (−∞, +∞) (c) (−2, +∞) (e) −2 (b) none (d) (−∞, −2) 12. f (x) = 3(4 − x2 ) f (x) = −6x (a) [−2, 2] (c) (−∞, 0) (e) 0 (b) (−∞, −2], [2, +∞) (d) (0, +∞) 9. 10. [−3/2, +∞) (−∞, +∞) 13. f (x) = 12x2 (x − 1) f (x) = 36x(x − 2/3) (a) [1, +∞) (c) (−∞, 0), (2/3, +∞) (e) 0, 2/3 (b) (d) 14. f (x) = 4x(x2 − 4) f (x) = 12(x2 − 4/3) (a) [−2, 0], [2, +∞) √ √ (c) (−∞, −2/ 3), (2/ 3, +∞) √ √ (e) −2/ 3, 2/ 3 (b) (−∞, −2], [0, 2] √ √ (d) (−2/ 3, 2/ 3) 15. f (x) = 4x 2 + 2)2 (x (a) [0, +∞) (d) (−∞, − 16. 17. f (x) = −4 (−∞, 1] (0, 2/3) 3x2 − 2 (x2 + 2)3 2/3), (+ 2/3, +∞) (b) (−∞, 0] (e) − 2/3, (c) (− 2/3, + 2/3) 2/3 2 − x2 2x(x2 − 6) f (x) = 2 + 2)2 (x (x2 + 2)3 √√ √ √ (a) [− 2, 2] (b) (−∞, − 2], [ 2, +∞) √ √ √ √ (d) (−∞, − 6), (0, 6) (e) − 6, 0, 6 f (x) = f (x) = 1 (x + 2)−2/3 3 f (x) = − 2 (x 9 −5/3 + 2) (a) (−∞, +∞) (c) (−∞, −2) (e) −2 √ √ (c) (− 6, 0), ( 6, +∞) (b) none (d) (−2, +∞) 141 Chapter 5 f (x) = 2 x−1/3 3 f (x) = − 2 x−4/3 9 18. (a...
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