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