Answer
(a) $f^{-1}(x)=log_2(\frac{x}{1-x})$
(b) $(0,1)$
Work Step by Step
(a) To find the inverse, let $y=\frac{2^x}{1+2^x}$,
so $y+y2^x=2^x$ and $(1-y)2^x=y$ which leads to $2^x=\frac{y}{1-y}$
and $x=log_2(\frac{y}{1-y})$ switch $x,y$, we have $f^{-1}(x)=log_2(\frac{x}{1-x})$
(b) The domain requirement for $f^{-1}(x)$ is that
$x\ne1$ and $\frac{x}{1-x}\gt0$. The cut points of the inequality are $0,1$
make a table as shown to solve the inequality and we can obtain the domain as $(0,1)$
