## Calculus 8th Edition

Published by Cengage

# 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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