Answer
$x_{1}=x_0$
$x_n=x_0$
Work Step by Step
Step 1. Given a good guess $f(x_0)=0$, we use Newton’s method $x_{n+1}=x_n-\frac{f(x)}{f'(x)}$, to calculate the next few terms.
Step 2. As $f'(x_0)\ne0$, we have $x_{1}=x_0-\frac{f(x_0)}{f'(x_0)}=x_0$
Step 3. Similarly, we have $x_{2}=x_1-\frac{f(x_1)}{f'(x_1)}=x_0-\frac{f(x_0)}{f'(x_0)}=x_0$
Step 4. We can generalize the process to get $x_n=x_0$
