Answer
(a) $f^{-1}(x)=\frac{1}{x}-2$
(b) see graph
(c) $f(x)$ domain $(-\infty,-2)U(-2,\infty)$ and range $(-\infty,0)U(0,\infty)$, $f^{-1}(x)$ domain $(-\infty,0)U(0,\infty)$ and range $(-\infty,-2)U(-2,\infty)$.
Work Step by Step
(a) This function $f(x)=\frac{1}{x+2}, x\ne-2$ is one-to-one. Find the inverse as the following: $y=\frac{1}{x+2}\longrightarrow x=\frac{1}{y}-2\longrightarrow f^{-1}(x)=\frac{1}{x}-2$
(b) see graph
(c) $f(x)$ domain $(-\infty,-2)U(-2,\infty)$ and range $(-\infty,0)U(0,\infty)$, $f^{-1}(x)$ domain $(-\infty,0)U(0,\infty)$ and range $(-\infty,-2)U(-2,\infty)$.