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