Answer
$f^{-1}(x)=\displaystyle \frac{3-2x}{x-1}$
Domain of $f^{-1}$ = $\{x\in \mathbb{R}|x\neq 1\}$
Range of $f^{-1}=$ domain of f = $\{x\in \mathbb{R}|x\neq 2\}$
Work Step by Step
Domain of f = $\{x\in \mathbb{R}|x\neq 2\}.$
(1) Solve the equation for x, writing $y=f(x)$
$y=\displaystyle \frac{x+3}{x-2},\quad $... multiply with $(x-2)$
$xy-2y=x+3\quad $... add $(-x+2y)$
$xy-x=2y+3$
$x(y-1)=2y+3$
$x=\displaystyle \frac{2y+3}{y-1}$
(2) interchange $x\leftrightarrow y, \quad y=f^{-1}(x)$
$y=f^{-1}(x)=\displaystyle \frac{2x+3}{x-1}$
Domain of $f^{-1}$ = $\{x\in \mathbb{R}|x\neq 1\}$
Range of $f^{-1}=$ domain of f = $\{x\in \mathbb{R}|x\neq 2\}$
Check:
$f[f^{-1}(x)]=\displaystyle \frac{f^{-1}(x)+3}{f^{-1}(x)-2}=\frac{\frac{2x+3}{x-1}+3}{\frac{2x+3}{x-1}-2}$
$=\displaystyle \frac{\frac{2x+3+3(x-1)}{x-1} }{\frac{2x+3-2(x-1)}{x-1}}=\frac{5x}{5}=x$
$f^{-1}[f(x)]=\displaystyle \frac{2(f(x))+3}{f(x)-1}=\frac{2(\frac{x+3}{x-2})+3}{\frac{x+3}{x-2}-1}$
$=\displaystyle \frac{\frac{2(x+3)+3(x-2)}{x-2}}{\frac{x+3-(x-2)}{x-2}}=\frac{5x}{5}=x$
