Answer
$$f^{-1}(x)=\frac{2x+3}{x-1}$$
- Domain: $(-\infty,1)\cup(1,\infty)$
- Range: $(-\infty,2)\cup(2,\infty)$
Work Step by Step
$$y=f(x)=\frac{x+3}{x-2}\hspace{1cm}$$
To find its inverse:
1) Solve for $x$ in terms of $y$:
$$y=\frac{x+3}{x-2}$$ $$y = \frac{(x-2)+5}{x-2}$$ $$y =1+\frac{5}{x-2}$$ $$\frac{5}{x-2}=y-1$$ $$x-2=\frac{5}{y-1}$$ $$x=\frac{5}{y-1}+2=\frac{5+2y-2}{y-1}=\frac{2y+3}{y-1}$$
2) Interchange $x$ and $y$:
$$y=\frac{2x+3}{x-1}$$
Therefore, $$f^{-1}(x)=\frac{2x+3}{x-1}$$
- Domain: $x$ is defined in $R$ except where $x-1=0$ or $x=1$.
So the domain of $f^{-1}$ is $(-\infty,1)\cup(1,\infty)$.
- Range:
$$f^{-1}(x)=\frac{2x+3}{x-1}=\frac{2x-2+5}{x-1}=2+\frac{5}{x-1}$$
We know that the range of $(x-1)$ is $(-\infty,0)\cup(0,\infty)$. So the range of $\frac{5}{x-1}$ would also be $(-\infty,0)\cup(0,\infty)$ ($\frac{5}{x-1}$ cannot be $0$).
Therefore, the range of $f^{-1}(x)=2+\frac{5}{x-1}$ is $(-\infty,2)\cup(2,\infty)$
*Check:
$$f(f^{-1}(x))=\frac{\frac{2x+3}{x-1}+3}{\frac{2x+3}{x-1}-2}=\frac{\frac{2x+3+3x-3}{x-1}}{\frac{2x+3-2x+2}{x-1}}=\frac{5x}{5}=x$$
$$f^{-1}(f(x))=\frac{2\times\frac{x+3}{x-2}+3}{\frac{x+3}{x-2}-1}=\frac{\frac{2x+6+3x-6}{x-2}}{\frac{x+3-x+2}{x-2}}=\frac{5x}{5}=x$$
Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$
