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ISM_T11_C04_H - Section 4.7 Newton's Method 13 For x c!$...

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Section 4.7 Newton's Method 265 13. For x , the procedure converges to the root ! œ !Þ$ !Þ$##")&$&ÞÞÞÞ (a) (b) (c) (d) Values for x will vary. One possible choice is x 1. ! œ !Þ (e) Values for x will vary. 14. (a) f(x) x 3x 1 f (x) 3x 3 x x the two negative zeros are 1.53209 œ Ê œ Ê œ Ê $ w # n 1 n x 3x 1 3x 3 n n n and 0.34730 (b) The estimated solutions of x 3x 1 0 are $ œ 1.53209, 0.34730, 1.87939. (d) The estimated x-values where g(x) 0.25x 1.5x x 5 has horizontal tangents œ % # are the roots of g (x) x 3x 1, and these are w $ œ 1.53209, 0.34730, 1.87939. 15. f(x) tan x 2x f (x) sec x 2 x x ; x 1 x 12920445 œ Ê œ Ê œ œ Ê œ w # ! " n 1 n tan x 2x sec x a b a b n n n x 1.155327774 x x 1.165561185 Ê œ Ê œ œ # 16 17 16. f(x) x 2x x 2x 2 f (x) 4x 6x 2x 2 x x ; œ Ê œ Ê œ % $ # w $ # n 1 n x 2x x 2x 2 4x 6x 2x 2 n n n n n n n if x 0.5, then x 0.630115396; if x 2.5, then x 2.57327196 ! % ! % œ œ œ œ
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266 Chapter 4 Applications of Derivatives 17. (a) The graph of f(x) sin 3x 0.99 x in the window œ # 2 x 2, 2 y 3 suggests three roots. Ÿ Ÿ Ÿ Ÿ However, when you zoom in on the x-axis near x 1.2, œ you can see that the graph lies above the axis there. There are only two roots, one near x 1, the other œ near x 0.4. œ (b) f(x) sin 3x 0.99 x f (x) 3 cos 3x 2x œ Ê œ # w x x and the solutions Ê œ n 1 n sin (3x ) 0.99 x 3 cos (3x ) 2x n n n n are approximately 0.35003501505249 and 1.0261731615301 18. (a) Yes, three times as indicted by the graphs (b) f(x) cos 3x x f (x) œ Ê w 3 sin 3x 1 x œ Ê n 1 x ; at œ n cos 3x x 3 sin 3x 1 a b a b n n n approximately 0.979367, 0.887726, and 0.39004 we have cos 3x x œ 19. f(x) 2x 4x 1 f (x) 8x 8x x x ; if x 2, then x 1.30656296; if œ Ê œ Ê œ œ œ % # w $ ! ' n 1 n 2x 4x 1 8x 8x n n n n x 0.5, then x 0.5411961; the roots are approximately 0.5411961 and 1.30656296 because f(x) is ! $ œ œ an even function. 20. f(x) tan x f (x) sec x x x ; x 3 x 3.13971 x 3.14159 and we œ Ê œ Ê œ œ Ê œ Ê œ w # ! " # n 1 n tan x sec x a b a b n n approximate to be 3.14159. 1 21. From the graph we let x 0.5 and f(x) cos x 2x ! œ œ x x x .45063 Ê œ Ê œ n 1 n cos x 2x sin x 2 a b a b n n n " x .45018 at x 0.45 we have cos x 2x. Ê œ Ê ¸ œ # 22. From the graph we let x 0.7 and f(x) cos x x ! œ œ x x x .73944 Ê œ Ê œ n 1 n x cos x 1 sin x n n n a b a b " x .73908 at x 0.74 we have cos x x. Ê œ Ê ¸ œ #
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Section 4.7 Newton's Method 267 23. If f(x) x 2x 4, then f(1) 1 0 and f(2) 8 0 by the Intermediate Value Theorem the equation œ œ œ Ê $ x 2x 4 0 has a solution between 1 and 2. Consequently, f (x) 3x 2 and x x . $ w # œ œ œ n 1 n x 2x 4 3x 2 n n n Then x 1 x 1.2 x 1.17975 x 1.179509 x 1.1795090 the root is approximately ! " # $ % œ Ê œ Ê œ Ê œ Ê œ Ê 1.17951. 24. We wish to solve 8x 14x 9x 11x 1 0. Let f(x) 8x 14x 9x 11x 1, then % $ # % $ # œ œ f (x) 32x 42x 18x 11 x x . w $ # œ Ê œ n 1 n 8x 14x 9x 11x 1 3 x 42x 18x 11 n n n n n n n # x approximation of corresponding root 1.0 0.976823589 0.1 0.100363332 0.6 0.642746671 2.0 1.983713587 ! 25. f(x) 4x 4x f (x) 16x 8x x x x . Iterations are performed using the œ Ê œ Ê œ œ % # w $ i 1 i i f x f x x x x a b a b i i i i i % # procedure in problem 13 in this section. (a) For x or x , x as i gets large. ! ! œ # œ !Þ) Ä " i (b) For x or x , x as i gets large.
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