Answer
$f(x)=\sqrt {x^2-9}, x\geq3$, domain $[3,\infty)$, range $[0,\infty)$
$f^{-1}(x)=\sqrt {x^2+9}$, domain $[0,\infty)$, range $[3,\infty)$
Work Step by Step
1. Given $f(x)=\sqrt {x^2-9}, x\geq3$, we can identify it is one-to-one with domain $[3,\infty)$ and range $[0,\infty)$
2. Rewrite the function as $y=\sqrt {x^2-9}$
3. Exchange $x,y$ to get $x=\sqrt {y^2-9}$
4. Solve for $y$ to get $y^2=x^2+9\longrightarrow y=\sqrt {x^2+9}$ (choose positive based on the result of step-1)
5. Replace $y$ with $f^{-1}(x)$ to get $f^{-1}(x)=\sqrt {x^2+9}$
6. For $f^{-1}(x)$, we can find its domain $[0,\infty)$ and range $[3,\infty)$
