Answer
(a). $f^{-1}=y=\frac{\log x - \log (1-x)}{\log 2}$
(b). $D_\left({\frac{\log x - \log (1-x)}{\log 2}}\right)=\{x|x \in 0\lt x \lt 1\}=(0,1)$
Work Step by Step
$f(x)=\frac{2^x}{1+2^x},$
(a).
$y=\frac{2^x}{1+2^x},$
$x=\frac{2^y}{1+2^y},$
$x(1+2^y)=2^y,$
$x+x2^y=2^y,$
$x=2^y-x2^y,$
$x=2^y(1-x),$
$2^y=\frac{x}{1-x},$
$\log 2^y=\log (\frac{x}{1-x}),$
$y\log 2=\log x - \log (1-x),$
$f^{-1}=y=\frac{\log x - \log (1-x)}{\log 2}$
(b).Domain of $f^{-1}=y=\frac{\log x - \log (1-x)}{\log 2}$ is$,$
$D_{\log x}=\{x|x\gt 0\}$ and $D_{\log (1-x)}=\{x|(1-x) \gt0, x \lt 1\}$. Therefore, $D_\left({\frac{\log x - \log (1-x)}{\log 2}}\right)=\{x|x \in 0\lt x \lt 1\}=(0,1)$
