Inverse function for $f(x)=\frac{x^2+1}{2x^2+1}$

Question

Given that $f(x)=\frac{x^2+1}{2x^2+1}$ and $x\geq2$, find $f^{-1}(x)$. Prove that $f^{-1}f(x)=x$

My attempt,

Let $f^{-1}(x)=a$

$x=f(a)$

$x=\frac{a^2+1}{2a^2+1}$

$(2x-1)a^2+x-1=0$

$a=\pm\frac{\sqrt{-(2x-1)(x-1})}{2x-1}$

Since $x\geq2$ for $f(x)$

So, $f^{-1}(x)=\frac{\sqrt{-(2x-1)(x-1})}{2x-1}$

Am I correct for my $f^{-1}(x)?$ If I do, how do I proceed for proving part? It's tedious.

Answer 1

HInt:$$f^{-1}(x)=\sqrt{\dfrac{-(x-1)}{2x-1}}\\\to\\f^{-1}f(x)=f^{-1}(\dfrac{x^2+1}{2x^2+1})=\\\sqrt{\dfrac{-(\dfrac{x^2+1}{2x^2+1}-1)}{2\dfrac{x^2+1}{2x^2+1}-1}}$$ can yoou go further ?