Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 349: 12

Answer

\sqrt[8] 500 ≈ 2.17455928

Work Step by Step

First we need to formulate a function to use in Newton's method x = \sqrt[8] 500 x^{8} - 500 = 0 Let f(x) = x^{8} - 500 = 0, The positive zero of this function is the solution to x = \sqrt[8] 500 f'(x) = 8x^{7} You could test some numbers or look at a graph to find a number to use as the initial approximation. Or just pick any number on interval [2,3] since it's a small interval. Since 2^{8} = 256 and 3^{8} = 6561 we can see that the number is closer to 2 than it is to 3. We'll start with the first guess of x_{n} = x_{1} = 2.2 and plug into x_{n+1} = x_{n} - (f(x_{n})/f'(x_{n}) until the numbers in the first 8 decimal places don't change anymore. Each x_{n+1} we get will get plugged back into the formula to get x_{n+2} x_{1} = 2.2 x_{2} = 2.2 - (f(x_{2.2})/f'(x_{2.2}) \approx 2.17556548 x_{3} = 2.17556548 - (f(x_{2.1755654874})/f'(x_{2.1755654874}) \approx 2.17456090 x_{4} = 2.17456090 - (f(x_{2.1745609034})/f'(x_{2.1745609034}) \approx 2.17455928 x_{5} = 2.17455928 - (f(x_{2.17455928})/f'(x_{2.17455928}) \approx 2.17455928 \sqrt[8] 500 = 2.17455928
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