Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.7 Newton's Method - Exercises - Page 219: 4


\begin{aligned} x_{1} &= -0.75 \\ x_{2} & \approx-0.6860465 \\ x_{3} & \approx-0.6823396 \end{aligned}

Work Step by Step

Given $$ f(x)=x^{3}+x+1, \quad x_{0}=-1 $$ Since $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ Then \begin{align*} x_{n+1}&= x_n -\frac{x_n^{3}+x_n+1}{3x_n^2+1}\\ &=\frac{2 x_n^{3}-1}{3 x^{2}_ n+1} \end{align*} Hence \begin{aligned} x_{1}=& \frac{2 x_{0}^{3}-1}{3 x_{0}^{2}+1}=\frac{2(-1)^{3}-1}{3(-1)^{2}+1} =-0.75 \\ x_{2}=& \frac{2 x_{1}^{3}-1}{3 x_{1}^{2}+1}=\frac{2(-0.75)^{3}-1}{3(-0.75)^{2}+1} \approx-0.6860465 \\ x_{3}=&\frac{2 x_{2}^{3}-1}{3 x_{2}^{2}+1} =\frac{2(-0.6860465)^{3}-1}{3(-0.6860465)^{2}+1} \approx-0.6823396 \end{aligned}
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