# 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 # n1 n x3 x1 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± ± ² %# 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 Ê œ ± Êœ ± œ Ê œ w# !" n tan x 2x sec x ab nn n ± # x 1.155327774 x x 1.165561185 Ê œœ # 16 17 16. f(x) x 2x x 2x 2 f (x) 4x 6x 2x 2 x x ; œ± ±±²Ê œ ± ±±Ê œ± %\$ # w \$# n x 2 xx2 x2 4x 6x 2x 2 # n n n ±±±² ±±± if x 0.5, then x 0.630115396; if x 2.5, then x 2.57327196 !%

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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 # 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 near x 0.4. œ (b) f(x) sin 3x 0.99 x f (x) 3 cos 3x 2x ² Ê œ ² #w x x and the solutions Êœ ± n1 n sin (3x ) 0.99 x 3 cos (3x ) 2x n n nn ±² ² # 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) Ê w 3 sin 3x 1 x ± Ê x; a t n cos 3x x 3 sin 3x 1 ab 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±² Ê œ±Ê œ ± œ ± œ ± %# w \$ !' n 2x 4x 1 8x 8x \$ n n ± x 0.5, then x 0.5411961; the roots are approximately 0.5411961 and 1.30656296 because f(x) is !\$ an even function. 20. f(x) tan x f (x) sec x x x ; x 3 x 3.13971 x 3.14159 and we œÊœ Ê œ ± œ Ê œ Ê œ w# !" # n tan x sec x n n # approximate to be 3.14159. 1 21. From the graph we let x 0.5 and f(x) cos x 2x ! œœ ± x x x .45063 ± Ê œ n cos x 2x sin x 2 n ± " x .45018 at x 0.45 we have cos x 2x. Ê ¸ œ # 22. From the graph we let x 0.7 and f(x) cos x x ! œ ² x x x .73944 ± Ê œ ± n x cos x 1s i n x n ² ± " x .73908 at x 0.74 we have cos x x. Ê Ê ¸± #
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±² œ ²³ œ´Ê \$ x 2x 4 0 has a solution between 1 and 2. Consequently, f (x) 3x 2 and x x . \$ w# ±²œ œ ± œ² n1 n x2 x4 3x 2 \$ # n n n ±² ± Then x 1 x 1.2 x 1.17975 x 1.179509 x 1.1795090 the root is approximately !" # \$ % œÊ 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 %\$ # # ²² ± ² œ œ ± ² f (x) 32x 42x 18x 11 x x . w\$ # œ²² ± Êœ ² n 8x 14x 9x 11x 1 3 x 42x 18x 11 # \$# nn n n n # ²²± 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 ² Ê œ ² Ê œ² %# w \$ i 1i i fx xx x ab i i i i i w \$ # ² # procedure in problem 13 in this section.

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

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