<s>[INST] Inverse function for $f(x)=\frac{x^2+1}{2x^2+1}$ 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.
 [/INST] 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 ?
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