Question

The function $f$ is defined for $x\le 0$ by $f\left(x\right)=\frac{x^2-1}{x^2+1}$. Find its inverse

Answer 1

Given $f\left(x\right)=\frac{x^2-1}{x^2+1}$

Let $f\left(x\right)=y⟹x=f^{-1}\left(y\right)$

$y=\frac{x^2-1}{x^2+1}{x}^2-1=y(x^2+1)x^2-1=x^{2}y+y{x}^{2}y-x^2+y+1=0x^2(y-1)+y+1=0x^2(y-1)=-y-1x^2=\frac{1+y}{1-y}x=\sqrt{\frac{1+y}{1-y}}$

So $f^{-1}\left(x\right)=\sqrt{\frac{1+x}{1-x}}$