Answer
$f(x)=\frac{2x-1}{5-3x}$, domain $(-\infty,\frac{5}{3})\cup(\frac{5}{3},\infty)$, range $(-\infty,-\frac{2}{3})\cup(-\frac{2}{3},\infty)$
$f^{-1}(x)=\frac{5x+1}{3x+2}$, domain $(-\infty,-\frac{2}{3})\cup(-\frac{2}{3},\infty)$, range $(-\infty,\frac{5}{3})\cup(\frac{5}{3},\infty)$
Work Step by Step
1. Given $f(x)=\frac{2x-1}{5-3x}$, we can identify it is one-to-one with domain $(-\infty,\frac{5}{3})\cup(\frac{5}{3},\infty)$ and range $(-\infty,-\frac{2}{3})\cup(-\frac{2}{3},\infty)$
2. Rewrite the function as $y=\frac{2x-1}{5-3x}$
3. Exchange $x,y$ to get $x=\frac{2y-1}{5-3y}$
4. Solve for $y$ to get $5x-3xy=2y-1\longrightarrow y=\frac{5x+1}{3x+2}$
5. Replace $y$ with $f^{-1}(x)$ to get $f^{-1}(x)=\frac{5x+1}{3x+2}$
6. For $f^{-1}(x)$, we can find its domain $(-\infty,-\frac{2}{3})\cup(-\frac{2}{3},\infty)$ and range $(-\infty,\frac{5}{3})\cup(\frac{5}{3},\infty)$