Answer
(a) $ f^{-1}(x)=x^2-6, x\ge0$
(b) see graph
(c) $f(x)$ domain $[-6,\infty)$ and range $[0,\infty)$, $f^{-1}(x)$ domain $[0,\infty)$ and range $[-6,\infty)$.
Work Step by Step
(a) This function $f(x)=\sqrt {x+6}, x\ge-6$ is one-to-one. Find the inverse as the following: $y=\sqrt {x+6}\longrightarrow x=y^2-6\longrightarrow f^{-1}(x)=x^2-6, x\ge0$
(b) see graph
(c) $f(x)$ domain $[-6,\infty)$ and range $[0,\infty)$, $f^{-1}(x)$ domain $[0,\infty)$ and range $[-6,\infty)$.