Answer
(a) $ f^{-1}(x)=\sqrt {x-9}$.
(b) $f$: $[0,\infty)$ and $[9,\infty)$. $f^{-1}$: $[9,\infty)$ and $[0,\infty)$.
(c) See graph.
Work Step by Step
(a) $f(x)=x^2+9, (x\ge0) \longrightarrow y=x^2+9\longrightarrow x=y^2+9\longrightarrow y=\sqrt {x-9}\longrightarrow f^{-1}(x)=\sqrt {x-9}$. Check $f(f^{-1})=x$ and $f^{-1}(f)=x$
(b) Domain and range of $f$: $[0,\infty)$ and $[9,\infty)$. Domain and range of $f^{-1}$: $[9,\infty)$ and $[0,\infty)$.
(c) See graph.
