Answer
(a) $ f^{-1}(x)=\frac{2x-4}{x+3}$.
(b) $f$: $(-\infty,2)U(2,\infty)$ and $(-\infty,-3)U(-3,\infty)$.
$f^{-1}$: $(-\infty,-3)U(-3,\infty)$ and $(-\infty,2)U(2,\infty)$.
Work Step by Step
(a) $f(x)=\frac{-3x-4}{x-2} \longrightarrow y=\frac{-3x-4}{x-2} \longrightarrow x=\frac{-3y-4}{y-2} \longrightarrow y=\frac{2x-4}{x+3} \longrightarrow f^{-1}(x)=\frac{2x-4}{x+3}$. Check $f(f^{-1})=\frac{-3(\frac{2x-4}{x+3})-4}{\frac{2x-4}{x+3}-2}=x$ and $f^{-1}(f)=\frac{2(\frac{-3x-4}{x-2})-4}{\frac{-3x-4}{x-2}+3}=x$
(b) Domain and range of $f$: $(-\infty,2)U(2,\infty)$ and $(-\infty,-3)U(-3,\infty)$. Domain and range of $f^{-1}$: $(-\infty,-3)U(-3,\infty)$ and $(-\infty,2)U(2,\infty)$.
