Answer
(a) $f^{-1}(x)=\frac{x}{x+2}$.
(b) $f$: $(-\infty,1)U(1,\infty)$ and $(-\infty,-2)U(-2,\infty)$. $f^{-1}$: $(-\infty,-2)U(-2,\infty)$ and $(-\infty,1)U(1,\infty)$.
Work Step by Step
(a) $f(x)=-\frac{2x}{x-1} \longrightarrow y=-\frac{2x}{x-1} \longrightarrow x=-\frac{2y}{y-1} \longrightarrow y=\frac{x}{x+2}\longrightarrow f^{-1}(x)=\frac{x}{x+2}$. Check $f(f^{-1})=-\frac{2(\frac{x}{x+2})}{(\frac{x}{x+2})-1} =x$ and $f^{-1}(f)=\frac{-\frac{2x}{x-1}}{-\frac{2x}{x-1}+2}=x$
(b) Domain and range of $f$: $(-\infty,1)U(1,\infty)$ and $(-\infty,-2)U(-2,\infty)$. Domain and range of $f^{-1}$: $(-\infty,-2)U(-2,\infty)$ and $(-\infty,1)U(1,\infty)$.
