Practicing Success
If $f: R→(-1, 1)$ is defined by $f (x)=\frac{-x|x|}{1+x^2}$, then $f^{-1}(x)$ equals |
$\sqrt{\frac{|x|}{1-|x|}}$ $-sgn(x)\sqrt{\frac{|x|}{1-|x|}}$ $-\sqrt{\frac{x}{1-x}}$ none of these |
$-sgn(x)\sqrt{\frac{|x|}{1-|x|}}$ |
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, $f^{-1}(x)<0$ $⇒f^{-1}(x)=\left\{\begin{matrix}\sqrt{\frac{-x}{1+x}},&if,\,x≤0\\-\sqrt{\frac{x}{1-x}},&if,\,x>0\end{matrix}\right.$ $⇒f^{-1}(x)=\left\{\begin{matrix}\sqrt{\frac{|x|}{1+|x|}},&if,\,x≤0\\-\sqrt{\frac{|x|}{1-|x|}},&if,\,x>0\end{matrix}\right.$ $⇒f^{-1}(x)=-sgn(x)\sqrt{\frac{|x|}{1-|x|}}$ |