Approximate root of $f(x)=x^2-2$ by the iteration $x_{n+1}=x_x+A{x_n^2-2\over x_n}+B{x_n^2-2\over x_n^3},A,B\in \Bbb{R}$. Find $A,B$ for which the order of convergence is maximal.
According to our course, defining $g(x)=x+A{x^2-2\over x}+B{x^2-2\over x^3}$, a fixed point of $g$ guarantees a root of $f$ (generally speaking), and there is a theorem that states that for a fixed point $r$, if $|g'(r)|<1$, there is a neighborhood of $r$ in which $x_{n+1}=g(x_n)$ converges to $r$. There is another theorem saying that if $|g'(r)|<1$, the order of convergence is linear, and if $g'(r)=0$, the order of convergence is at least quadratic.
(Definition of order of convergence, in case needed, is: if $\lim\limits_{n\to \infty}{|x_{n+1}-r|\over |x_n-r|^{\alpha}}=\lambda\ne 0$, it is said that $x_n$ converges to $r$ at rate $\alpha$ with asymptotic error $\lambda$.).
What I have done so far is: for points $r_1,r_2=\pm \sqrt 2$, require $|g'(r_i)|=0$ (so that $|g'(r_i)|<1$ and the iteration converges in some neighborhood, and $|g'(r_i)|=0$ so that the order of convergence is at least $2$). So $g'(x)=1+A{2x^2-(x^2-2)\over x^2}+B{2x^4-3x^2(x^2-2)\over x^6}$, $g'(r_1)=g'(r_2)=1+2A+B$, and $2A+B+1=0$, and $B=-1-2A$. This is as far as I could go actually knowing what I am doing. I have no idea how to proceed. What should I be doing next or generally?