Answer
(a) $ f^{-1}(x)=\sqrt {x^2+16}, x\le0$
(b) see graph
(c) $f(x)$ domain $[4,\infty)$ and range $(-\infty,0]$, $f^{-1}(x)$ domain $(-\infty,0]$ and range $[4,\infty)$.
Work Step by Step
(a) This function $f(x)=-\sqrt {x^2-16}, x\ge4$ is one-to-one. Find the inverse as the following: $y=-\sqrt {x^2-16}\longrightarrow x=\sqrt {y^2+16}\longrightarrow f^{-1}(x)=\sqrt {x^2+16}, x\le0$
(b) see graph
(c) $f(x)$ domain $[4,\infty)$ and range $(-\infty,0]$, $f^{-1}(x)$ domain $(-\infty,0]$ and range $[4,\infty)$.
