Fixed point iteration contractive interval Consider the function $F(x) = x^2-2x+2$.

Find an interval in which the function is contractive and find the fixed point in this interval.

What is the convergence rate of the fixed point iteration: $x_{n+1} = F(x_n)$ in that interval?

I'm lost on fixed point iteration. I've watched a few youtube videos, but I'm still not understanding contractive intervals and methods for finding fixed points in functions.
You need an interval $[a,b]$ so that there's a $0\leq c&lt;1$ such that $|f(x)-f(y)|\leq c|x-y|$ for all $x,y\in[a,b]$. In our case, we have
$$
|f(x)-f(y)|=|x^2-y^2-(2x-2y)|\leq c|x-y|.
$$
Notice that, in fact, $f(1)=1$, so you're going to want to choose an interval somewhere around $1$.
You need to consider the recurrence relation

$$ x_{n+1}= x_{n}^2-x_n+2 ,$$

which suggests considering the function

$$f(x) = x^2-x+2  \implies  f'(x) = 2x-1. $$

Now, to find the interval, use the condition $|f'(x)|&lt;1$. See a related problem.