Find $f$ such that the contraction $\phi$ has a fixed-point $\rho= \sqrt{2}$

Question

I use the Newton method and the Banach fixed-point theorem and have:

Let $I \subset \mathbb{R}$ a closed interval and $\phi: I \rightarrow I$ Lipschitz continous and $f: I \rightarrow \mathbb{R}$ a function.

I know, that

\begin{equation} \phi(x_n) = x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \end{equation}

Now I want to find an $f$ such that the contraction $\phi$ has a fixed-point $\rho= \sqrt{2}$

Are there many $f$ or is there only one $f$ which I can use? I just do not know how to start.

Answer 1

Try with $$ f(x) = \frac 12 \left( x + \frac 2x \right) $$

This is not Lipschitz on $(0,\infty)$ but on any $[a,\infty)$. It remains to prove that $x$ is bounded from below to conclude with the theorem.