Pre-Calc Homework Solutions 171

Pre-Calc Homework Solutions 171 - Section 4.5 13. Center (x...

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Unformatted text preview: Section 4.5 13. Center (x 171 1 17. Let f (x) Then f (x) xn 1) 1 x 4 3 2 3 2 1 4 4 . 25 4 x 25 1 x2 2x 2x 2 1 sin x. cos x and xn2 2xn 1)(1) (x)(1) 1 f (x) (x 1)2 (x 1)2 1 1 We have f (1) and f (1) 2 4 1 1 L(x) f (1) f (1)(x 1) (x 2 4 xn f (xn) f (xn) xn 2xn 2 1 sin xn cos xn The graph of y solutions f (x) shows that f (x) 0 has two Alternate solution: Using center L(x) 14. Center f (x) We have f L(x) f 2 3 3 , we have f 2 2 3 3 3 f x 2 2 5 3 and f 5 4 x 25 f 3 2 9 25 2 [ 4, 4] by [ 3, 3] sin x 2 0 and f f 2 2 2 1. 0 1x 3x 2 1 1 2 x x 2 x1 x2 x3 x4 x5 0.3 0.3825699 0.3862295 0.3862369 0.3862369 x1 x2 x3 x4 x5 2 1.9624598 1.9615695 1.9615690 1.9615690 15. Let f (x) xn 1 x3 xn Solutions: x 1 and 18. Let f (x) xn 1 0.386237, 1.961569 2. Then f (x) f (xn) f (xn) x f (xn) f (xn) 1. Then f (x) xn xn3 3xn2 xn . x4 xn 4x 3 and 2 Note that f is cubic and f is always positive, so there is exactly one solution. We choose x1 0. x1 0 x2 1 x3 0.75 x4 0.6860465 x5 0.6823396 x6 0.6823278 x7 0.6823278 Solution: x 16. Let f (x) xn 1 xn xn4 4xn3 . 0.682328 x4 x f (xn) f (xn) 3. Then f (x) xn xn4 4xn 3 4x 3 3 1 1 and Note that f (x) 0 clearly has two solutions, namely 4 x 2. We use Newton's method to find the decimal equivalents. x1 1.5 x2 1.2731481 x3 1.1971498 x4 1.1892858 x5 1.1892071 x6 1.1892071 Solutions: x 1.189207 3x 2 3, dy (3x 2 3) dx. dy dx xn xn The graph of y solutions. f (x) shows that f (x) 0 has two 19. (a) Since (b) At the given values, dy (3 22 3)(0.05) 20. (a) Since dy 9(0.05) 0.45. 2 (1 2x 2 , x2)2 (1 x 2)(2) (2x)(2x) dy (1 x 2)2 dx 2 2 2x dx. (1 x2)2 [ 3, 3] by [ 4, 4] (b) At the given values, x1 x2 x3 x4 x5 1.2 1.6541962 1.1640373 1.1640351 1.1640351 dy 2 [1 2( 2)2 (0.1) ( 2)2]2 2 52 8 x1 x2 x3 x4 x5 1.5 1.455 1.4526332 1.4526269 1.4526269 (0.1) 0.024. 21. (a) Since dy dy dx Solutions: x 1.452627, 1.164035 (x 2) 1 x (ln x)(2x) 2x ln x x, (2x ln x x) dx. (b) At the given values, dy [2(1) ln (1) 1](0.01) 1(0.01) 0.01 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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