To test x0 rst rewrite f x as 27 cos xtan3 x 6427 27

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Unformatted text preview: ) can change sign only if the factor L − 2P changes sign, L 1 L = ln A. , 1 = Ae−kLt , t = which it does when P = L/2. From (5) we have 2 1 + Ae−kLt Lk SUPPLEMENTARY EXERCISES FOR CHAPTER 5 4. (a) x3 x2 − on [−2, 2]; x = 0 is a relative maximum and x = 1 is a relative 3 2 1 minimum, but y = 0 is not the largest value of y on the interval, nor is y = − the smallest. 6 False; an example is y = Supplementary Exercises 168 (b) (c) 6. true False; for example y = x3 on (−1, 1) which has a critical point but no relative extrema (a) (b) y 4 (c) y 4 4 x 2 7. (a) y x x 2 2 7(x − 7)(x − 1) ; critical points at x = 0, 1, 7; 3x2/3 neither at x = 0, relative max at x = 1, relative min at x = 7 (first derivative test) f (x) = f (x) = 2 cos x(1 + 2 sin x); critical points at x = π/2, 3π/2, 7π/6, 11π/6; relative max at x = π/2, 3π/2, relative min at x = 7π/6, 11π/6 √ 3 x−1 ; critical points at x = 5; relative max at x = 5 (c) f (x) = 3 − 2 (b) x−9 27 − x , f (x) = ; critical point at x = 9; f (9) > 0, relative min at x = 9 18x3/2 36x5/2 (a) f (x) = (b) 8. x3 − 4 x3 + 8 , f (x) = 2 ; x2 x3 1/3 1/3 critical point at x = 4 , f (4 ) > 0, relative min at x = 41/3 (c) 9. f (x) = 2 f (x) = sin x(2 cos x + 1), f (x) = 2 cos2 x − 2 sin2 x + cos x; critical points at x = 2π/3, π, 4π/3; f (2π/3) < 0, relative max at x = 2π/3; f (π ) > 0, relative min at x = π ; f (4π/3) < 0, relative max at x = 4π/3 lim f (x) = +∞, lim f (x) = +∞ x→−∞ y x→+∞ f (x) = x(4x2 − 9x + 6), f (x) = 6(2x − 1)(x − 1) relative min at x = 0, points of inflection when x = 1/2, 1, no asymptotes 4 3 2 1 (1,2) (0,1) ( 1 , 23 ) 2 16 1 10. lim f (x) = −∞, lim f (x) = +∞ x→−∞ x→+∞ f (x) = x3 (x − 2)2 , f (x) = x2 (5x − 6)(x − 2), f (x) = 4x(5x2 − 12x + 6) √ 8 ± 2 31 critical points at x = 0, √5 8 − 2 31 = −0.63 relative max at x = 5 √ 8 + 2 31 = 3.83 relative min at x = 5 √ 6 ± 66 = 0, −0.42, 2.82 points of inflection at x = 0, 5 no asymptotes (-0.42,0.16) 2 x y (0,0) (-0.63,0.27) x -1 1 2 3 4 -100 -200 (2.82,-165.00) (3.83,-261.31) 169 11. Chapter 5 y lim f (x) doesn’t exist x→±∞ f (x) = 2x sec2 (x2 + 1), 2 4 2 2 2 f (x) = 2 sec (x + 1) 1 + 4x tan(x + 1) 2 critical point at x = 0; relative min at x = 0 2 x -2 -1 1 2 point of inflection when 1 + 4x tan(x + 1) = 0 12. -2 π (n + 1 ) − 1, n = 0, 1, 2, . . . 2 vertical asymptotes at x = ± -4 lim f (x) = −∞, lim f (x) = +∞ x→−∞ 2 y x→+∞ f (x) = 1 + sin x, f (x) = cos x 4 critical points at x = 2nπ + π/2, n = 0, ±1, ±2, . . ., 2 −π no extrema because f ≥ 0 and by Exercise 51 of Section 5.1, x π -2 f is increasing on (−∞, +∞) -4 inflections points at x = nπ + π/2, n = 0, ±1, ±2, . . . -6 no asymptotes 13. x(x + 5) 2x3 + 15x2 − 25 , f (x) = −2 2 (x2 + 2x + 5)2 (x + 2x + 5)3 critical points at x = −5, 0; y f (x) = 2 1 relative max at x = −5, 0.8 relative min at x = 0 0.6 points of inflection at x = −7.26, −1.44, 1.20 0.4 0.2 horizontal asymptote y = 1 as x → ±∞ 14. x -20 -10 3x2 − 25 3x2 − 50 , f (x) = −6 x4 x5 √ critical points at x = ±5 3/3; √ relative max at x = −5 3/3, √ relative min at x = +5 3/3 10 f (x) = 3 inflection points at x = ±5 20 y 5 x -4 2/3 6 -5 horizontal asymptote of y = 0 as x → ±∞, vertical asymptote x = 0 15. lim f (x) = +∞, lim f (x) = −∞ x→−∞ y x→+∞ x x≤0 if −2x x>0 critical point at x = 0, no extrema 2 f (x) = inflection point at x = 0 (f changes concavity) no asymptotes 1 -2 1 -2 x Supplementary Exercises 16. 170 5 − 3x , 3(1 + x)1/3 (3 − x)2/3 −32 f (x) = 9(1 + x)4/3 (3 − x)5/3 critical point at x = 5/3; y f (x) = 4 2 relative max at x = 5/3 -4 -2 x 2 -1 cusp at x = −1; -3 point of inflection at x = 3 oblique asymptote y = −x as x → ±∞ 17. lim f (x) = +∞ y x→+∞ f (x) = 1 + ln x, f (x) = 1/x lim f (x) = 0, lim+ f (x) = −∞ x→0+ 1 x→0 critical point at x = 1/e; relative min at x = 1/e 1 2 x no points of inflection, no asymptotes 18. lim f (x) = +∞ y x→+∞ f (x) = x(2 ln x + 1), f (x) = 2 ln x + 3 0.2 0.4 x 0.6 lim f (x) = 0, lim+ f (x) = 0 x→0+ x→0 critical point at x = e−1/2 ; -0.1 −1/2 relative min at x = e point of inflection at x = e−3/2 19. 1 − 2 ln x 6 ln x − 5 , f (x) = x3 x4 critical point at x = e1/2 , y f (x) = relative max at x = e1/2 x 1 2 -1 5/6 point of inflection at x = e -2 horizontal asymptote y = 0 as x → +∞ 20. lim f (x) = +∞ y x→±∞ 2x 1 − x2 , f (x) = 2 2 +1 (x + 1)2 critical point at x = 0; f (x) = 2 x2 1 relative min at x = 0 points of inflection at x = ±1 no asymptotes x -2 2 171 21. Chapter 5 lim f (x) = +∞ y x→+∞ x−1 x2 − 2x + 2 , f (x) = ex 2 x x3 critical point at x = 1; f (x) = ex 6 relative min at x = 1 2 no points of inflection -2 1 -2 x 3 vertical asymptote x = 0, horizontal asymptote y = 0 for x → −∞ 22. f (x) = (1 − x)e−x , f (x) = (x − 2)e−x y critical point at x = 1; relative max at x = 1 0.2 point o...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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