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