Understanding fixed point iteration

Question

Claim that if at some interval $I$ show off

$|\frac{f(x)f''(x)}{f'(x)^{2}}|\leq c < 1$

So then $x_0$ chosen from Interval $I$ results in an iteration of the fixed point of $g$.

The base knowledge is that we know with Newton's method we can find the x-intercept for a function $f$ by iterating the function

$g(x) = x - \frac{f(x)}{f'(x)}$

Then we are asked why is this at the same time the x-intercept for $f$ ?

I get confused by the $I$ and $c$ and the word fixed point. I know how iteration works but not how to use it on this task. Thank you in advance.

Answer 1

1. Are you aware of the fact that:

$$g'(x)=\frac{f(x)f''(x)}{f'(x)^{2}}$$

2. that the fixed point equation:

$$x=g(x)$$

has a solution under the condition that $|g'(x)|\le c<1$ (see here).