Answer
$|x_n|=2^n$
$n\to\infty$, $|x_n|\to\infty$
See graph.
Work Step by Step
Step 1. Given $f(x)=x^{1/3}$, we have $f'(x)=\frac{1}{3}x^{-2/3}$
Step 2. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$, start with $x_0=1$; we can get the result up to $x_4$ as shown in the table. It seems that $|x_n|=2^n$.
Step 3. Algebraically, $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^{1/3}}{x_n^{-2/3}/3}=x_n-3x_n=-2x_n$ or $|x_{n+1}|=2|x_n|$, which represents a geometric series with $|x_n|=2^n$
Step 4. When $n\to\infty$, we have $|x_n|\to\infty$.
Step 5. See the graph for the function with tangent lines and the first few points $x_0, x_1, x_2$.
