Answer
(a) $ f^{-1}(x)=\frac{3x+1}{x-1} $
(b) see graph
(c) $f(x)$ domain $(-\infty,3)U(3,\infty)$ and range $(-\infty,1)U(1,\infty)$, $f^{-1}(x)$ domain $(-\infty,1)U(1,\infty)$ and range $(-\infty,3)U(3,\infty)$.
Work Step by Step
(a) This function $f(x)=\frac{x+1}{x-3}, x\ne3$ is one-to-one. Find the inverse as the following: $y=\frac{x+1}{x-3}\longrightarrow x=\frac{3y+1}{y-1}\longrightarrow f^{-1}(x)=\frac{3x+1}{x-1}, x\ne1$
(b) see graph
(c) $f(x)$ domain $(-\infty,3)U(3,\infty)$ and range $(-\infty,1)U(1,\infty)$, $f^{-1}(x)$ domain $(-\infty,1)U(1,\infty)$ and range $(-\infty,3)U(3,\infty)$.
