Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.8 Newton's Method - 3.8 Exercises - Page 277: 29

Answer

$$ f(x)=x^{3}-3 x+6 $$ the tangent line used for approximating $x_{2}$ is horizontal. Attempting to find $x_{2}$ results in trying to divide by zero. Therefore, Newton’s method doesn’t work for finding the root of the given equation .

Work Step by Step

$$ f(x)=x^{3}-3 x+6 $$ We apply Newton’s method with $$ f(x)=x^{3}-3 x+6 $$ and $$ f^{\prime}(x)=3 x^{2}-3 $$ If $$x_{1}=1$$ then $$ f^{\prime}(x_{1})=3 (1)^{2}-3=0 $$ since $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ With $n= 1$ we have $$ x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)} $$ the tangent line used for approximating $x_{2}$ is horizontal. Attempting to find $x_{2}$ results in trying to divide by zero. Therefore, Newton’s method doesn’t work for finding the root of the given equation .
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