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