newtons method self-map

Question

Let $f\in C^2(\mathbb R,\mathbb R),\alpha \in \mathbb R$ with $f(\alpha)=0$ and $f'(\alpha)\neq 0$. Show, that there is a compact neighborhood $D$ of $\alpha$, so that the function $x \mapsto x-\frac{f(x)}{f'(x)}$ is a self-map, which is a contraction. I have no idea for this exercise, can you please help me?

Answer 1

$$g(x) =x-\frac{f(x)}{f' (x)} $$ $$g'(x) = 1-\frac{(f' (x))^2 -f(x) f'' (x) }{(f' (x) )^2} =\frac{f'' (x) f (x) }{(f' (x) )^2}$$ Since $f(\alpha ) =0 $ we get $g(\alpha ) =\alpha $ and since $g' (\alpha ) = 0 $ and the continuity of the functions $f, f' ,f''$the there exist a closed interval $D=[\alpha -m , \alpha +m ]$ such that $g' (x) <\frac{1}{2} $ and $f'(x)\neq 0$ for $x\in D$ So we have by Lagrange mean value theorem we get $$|g(x) - g(y)|\leq \frac{1}{2}|x-y|$$ for all $x,y\in D.$ So $g:D\to D$ and $g$ is contraction.