## Calculus (3rd Edition)

$$x\in [0,3) \cup (3, \infty).$$
We see that $\sqrt{x}$ is defined for all $x\geq 0$. Moreover, $$x^2-9=0 \Longrightarrow x=\pm 3.$$ Thus the domain of $f$ is $$x\in [0,3) \cup (3, \infty).$$