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