Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.6 - Newton's Method - Exercises 4.6 - Page 230: 5

Answer

$x_2\approx1.19$

Work Step by Step

Step 1. Given $f(x)=x^4-2$, we have $f'(x)=4x^3$. Step 2. Use Newton’s method $x_{n+1}=x_n-\frac{f(x)}{f'(x)}$ to estimate the solution up to $x_2$. Step 3. Start with $x_0=1$; we can get the result $x_2\approx1.19$ as shown in the table.
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