How do i find inverse of this function?

Question

I need to find $f^{-1} $ of $f(x)= -x|x|/(1+x^2)$

I divided function into parts x>0 and x<0.

For $ x>0 $, $f(x)= -x^2/(1+x^2)$

Replace x with y and after solving the quadratic i get $y=-\sqrt{-x/(x+1)}$ , which makes it undefined.

For $ x<0 $, $f(x)= x^2/(1+x^2)$

Replace x with y and after solving the quadratic i get $y=\sqrt{x/(-x+1)}$.

The correct answer given in my book is $y=sig(-x)\sqrt{|x|/(-|x|+1)}$.

Can anybody point out my error ?

Answer 1

I think you're getting confused, because $x$ and $y$ have opposite signs, and then you switched them.

For $x>0$, $y<0$, so the solution would be $x=\sqrt{-y/(y+1})$, which makes sense with the signs,

and then you could switch $x$ and $y$ to get $y=\sqrt{-x/(x+1)}$,

keeping in mind that, after the switch, $y>0$ and $x<0$;

that agrees with the answer in your book for $x<0$.

Answer 2

$\newcommand{\sign}{\text{sign}}$ Since $\;y=\cfrac{-x|x|}{1+x^2}\;,\;$ it results that $\;\color{blue}{\sign(x)=\sign(-y)}\;.$

Moreover,

$|y|=\cfrac{x^2}{1+x^2}\;,$

$\left(1-|y|\right)x^2=|y|\;,$

$x=\color{blue}{\sign(x)}\sqrt{\cfrac{|y|}{1-|y|}}\;,$

$x=\color{blue}{\sign(-y)}\sqrt{\cfrac{|y|}{1-|y|}}\;.$