Answer
$x_{100}\approx1.07951729$
Work Step by Step
Step 1. Given $f(x)=(x-1)^{40}$, we have $f'(x)=40(x-1)^{39}$.
Step 2. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a starting point $x_0=2$, we can obtain a value of $x_{100}\approx1.07951729$ which is still about $8\%$ larger than $1$, as shown in the table.
