Answer
(a) $ f^{-1}(x)=2-\frac{4}{x}$.
(b) $f$: $(-\infty,2)U(2,\infty)$ and $(-\infty,0)U(0,\infty)$. $f^{-1}$: $(-\infty,0)U(0,\infty)$ and $(-\infty,2)U(2,\infty)$.
Work Step by Step
(a) $f(x)=\frac{4}{2-x} \longrightarrow y=\frac{4}{2-x}\longrightarrow x=\frac{4}{2-y}\longrightarrow y=2-\frac{4}{x}\longrightarrow f^{-1}(x)=2-\frac{4}{x}$. Check $f(f^{-1})=\frac{4}{2-2+\frac{4}{x}}=x$ and $f^{-1}(f)=2-\frac{4}{\frac{4}{2-x}}=x$
(b) Domain and range of $f$: $(-\infty,2)U(2,\infty)$ and $(-\infty,0)U(0,\infty)$. Domain and range of $f^{-1}$: $(-\infty,0)U(0,\infty)$ and $(-\infty,2)U(2,\infty)$.
