Find the range of initial values for which Newton–Raphson's method for $f(x) = a - \frac{1}{x}$ converges I'm tryig to use Newton–Raphson's method to compute the inverse of $a$. Given $f(x) = a - \frac{1}{x}, a \neq 0$ we are looking for $x^*$ such that $f(x^*) = 0$. Newton–Raphson's  method gives the sequence $x_{n+1} = -ax_n^2 + 2x_n$. For example let $a = 2$, so $x_{n+1} = -2x_n^2 + 2x_n$. So want to find for which initial values $x_0$ the method converges. Playing around with an algorithm I implemented I found that $x_0$ must be in $(0, 1)$ for it to converge.
But how can I prove that? And in the more general case of $a$. Any help would be appreciated.
 A: Have you heard of Kantorovich's theorem for Newton's method? 

Theorem of  Kantorovich. Let $I\subseteq \mathbb{R}$ a interval 
    and $F:{I}\to \mathbb{R}$ a continuous function, continuously
    differentiable on $\mathrm{int}(I)$. Take $x_0\in \mathrm{int}(I)$,
    $L,\, b>0$ and suppose that
  
  
*
  
*$F '(x_0)$ is non-singular,
  
*$ \|  F'(y)- F'(x)
    \| \leq L\|x-y\|
    \;\;$  for any $x,y\in I$,
  
*$ \|F'(x_0)^{-1}\cdot F(x_0)\|\leq b$,
  
*$2\cdot b\cdot L\leq 1$.
Define
    $
    t_*:=\frac{1-\sqrt{1-2bL}}{L}. 
  $
    If 
    $
  [x_0-t_*,x_0+t_*]\subset I,
  $
    then  the sequences $\{x_k\}$ generated by Newton's Method for
    solving $F(x)=0$ with starting point $x_0$,
    $$
    x_{k+1} ={x_k}-F'(x_k) ^{-1}F(x_k), \qquad k=0,1,\cdots, 
  $$
    is well defined, is contained in $(x_0-t_*,x_0+t_*)$, converges to a
    point $x_*\in [x_0-t_*,x_0+t_*]$ which is the unique zero of $F$ in
    $[x_0-t_*,x_0+t_*]$. 

Here a expository Article: Kantorovich's s theorem on  Newton's method.. See too other expository Article here.. See Ortega's paper here for a simple proof. 
See personal home page of  Prof. Ferreira for articles of Newton-Kantorovich method and variants in Banach spaces.
A: Hint.
A coarse appreciation. 
The Newton-Raphson recurrence formula is $x_{k+1} = \phi(x_k)$ then
$$
x_{k+1}-x_k = \phi(x_k)-\phi(k_{k-1})=\phi'(\zeta)(x_k-x_{k-1}),\ \ \zeta\in B(x_k,x_{k-1})
$$
For convergence it is sufficient then that $|\phi'(\zeta)| < 1$ for $\zeta\in B(x_k,x_{k-1})$
Here $B(x_k,x_{k-1})$ is an open neighbourn containing $(x_k,x_{k-1})$.
In our case $\phi(x) = -ax^2+2x$ and $\phi'(x) = -2ax+2$
