If f: R→(-1, 1) is defined by f (x)=(-x|x|)/(1+x^2), then f^-1(x) equ…

Clearly, $f: R→(-1, 1)$ given by $f (x)=\frac{-x|x|}{1+x^2}$ is a bijection.

Now, 

$fof^{-1}(x)=x$

$⇒ f(f^{-1}(x)) = x$

$⇒-\frac{f^{-1}(x)|f^{-1}(x)|}{1+\{f^{-1}(x)\}^2}=x$

$⇒\frac{-\{f^{-1}(x)\}^2}{1+\{f^{-1}(x)\}^2}=x$, if, $f^{-1}(x)≥0$

and,

$\frac{-\{f^{-1}(x)\}^2}{1+\{f^{-1}(x)\}^2}=x$, if, $f^{-1}(x)<0$

$⇒f^{-1}(x)=\left\{\begin{matrix}\sqrt{\frac{-x}{1+x}},&if,\,x≤0\\-\sqrt{\frac{x}{1-x}},&if,\,x>0\end{matrix}\right.$

$⇒f^{-1}(x)=\left\{\begin{matrix}\sqrt{\frac{|x|}{1+|x|}},&if,\,x≤0\\-\sqrt{\frac{|x|}{1-|x|}},&if,\,x>0\end{matrix}\right.$

$⇒f^{-1}(x)=-sgn(x)\sqrt{\frac{|x|}{1-|x|}}$