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: 7

Answer

$x_{3}=1.5215$

Work Step by Step

$f(x)=\frac{2}{x}-x^{2}+1$ $f′(x)=\frac{-2}{x^{2}}-2x$ $x_{1}=2$ $x_{n+1}=x_{n}-\frac{\frac{2}{x_{n}}-x^{2}_{n}+1}{\frac{-2}{x^{2}_{n}}-2x_{n}}$ $x_{2}=2-\frac{\frac{2}{(2)}-(2)^{2}+1}{\frac{-2}{(2)^{2}}-2(2)}$ $x_{2}=2-\frac{-2}{-4.5}$ $x_{2}=\frac{14}{9}$ $x_{3}=\frac{14}{9}-\frac{\frac{2}{(\frac{14}{9})}-(\frac{14}{9})^{2}+1}{\frac{-2}{(\frac{14}{9})^{2}}-2(\frac{14}{9})}$ $x_{3}=\frac{14}{9}-0.0340$ $x_{3}=1.5215$
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