Question

Explain why Newton’s method will fail if you choose a value of $x_0$ for which $f\text{'}\left(x_0\right)=0$.

Short answer

Because when $f\text{'}\left(x_0\right)=0$, then the denominator becomes zero. So the value for next iteration will be undefined. So that while we choosing such type of value of$x_0$, then Newton's formula is fail.

Step-by-step solution

Step 1. Given Information

We have to prove and explain that when we chose a value of $x_0$, such that $f\text{'}\left(x_0\right)=0$, then Newton's formula is fail in this case.

Step 2. Required proof 

Newton's formula to approximate the root of a function is as following :-

$x_{k+1}=x_k-\frac{f\left(x_k\right)}{f\text{'}\left(x_k\right)}$

Then the formula for first iteration is as following :-

$x_1=x_0-\frac{f\left(x_0\right)}{f\text{'}\left(x_0\right)}$

Here if we put $f\text{'}\left(x_0\right)=0$, then we have :-

$$\begin{gathered} x_1=x_0-\frac{f\left(x_0\right)}{0} \\ \Rightarrow x_1=x_0-\infty \\ \Rightarrow x_1=\infty \end{gathered}$$

This is not possible. That is he newton's formula fails here.

So we can conclude that :-

when $f\text{'}\left(x_0\right)=0$, then the denominator becomes zero. So the value for next iteration will be undefined. So that while we choosing such type of value of $x_0$, then Newton's formula is fail.