## Calculus 8th Edition

The inverse function is given by $$f^{-1}(x)=\left(\frac{1-x}{1+x}\right)^2.$$
We have to express $x$ in terms of $y$. Then $x=f^{-1}(y)$. $$y=\frac{1-\sqrt{x}}{1+\sqrt{x}}\Rightarrow(1+\sqrt{x})y=1-\sqrt{x}\Rightarrow y+y\sqrt{x}=1-\sqrt{x}.$$ Rearranging terms $$y\sqrt{x}+\sqrt{x}=1-y\Rightarrow \sqrt{x}(y+1)=1-y\Rightarrow \sqrt{x}=\frac{1-y}{y+1},$$ which finally gives $$x=\left(\frac{1-y}{1+y}\right)^2$$ so $$f^{-1}(y)=\left(\frac{1-y}{1+y}\right)^2.$$ Renaming $y$ back to $x$ we have $$f^{-1}(x)=\left(\frac{1-x}{1+x}\right)^2.$$