If _{$f\left(x\right)=x-x^2+x^3-x^4+\mathrm{.......}\infty$}  when   _{$\left|x\right|<1$}  then _{$f^{-1}\left(\frac{1}{2}\right)=\_\_\_\_$}

$f\left(x\right)= \frac{x}{1-\left(-x\right)} \left(sum of infinite terms in G.P =\frac{a}{1-r}\right)$ 

$Let f\left(x\right)=\frac{x}{1+x}=y\Rightarrow x=y+xy \Rightarrow x\left(1-y\right)=y \Rightarrow x=\frac{y}{1-y}$

   $therefore f^{-1}\left(x\right)=\frac{x}{1-x}$

   $f^{-1}\left(\frac{1}{2}\right)=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1-{\displaystyle \frac{1}{2}}}=1$