Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.8 Exercises - Page 229: 1

Answer

$x = 2.236067997$

Work Step by Step

$f(x) = x^2 - 5$ and $x_1 = 2.2$ 1.) Use the formula $x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)}$ to find $x_2$ $x_2 = 2.2 - \dfrac{f(2.2)}{f'(2.2)} = 2.2 - \dfrac{-0.16}{4.4}$ = 2.2363636 2.) Use the same formula to calculate $x_3$ $x_3 = x_2 - \dfrac{f(2.2363636)}{f'(2.2363636)} = 2.2363636 - \dfrac{0.001322314}{4.72727273}$ = 2.236067997 We have now completed two iterations of Newton's Method, and have found the approximation to be x = 2.236067997
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