Answer
$[0,3]$
Work Step by Step
Let $u(x)=\sqrt{x}.$ Its domain is $[0,+\infty)$ , as the radicand must not be negative.
Let $v(x)=\sqrt{3-x}.$
Its domain is $(-\infty,3]$, as the radicand must not be negative.$\left(\begin{array}{ll}
3-x & \geq 0\\
3 & \geq x
\end{array}\right)$
The intersection of the domains is $[0,3].$
This interval is the domain of $(u+v)(x)=f(x)$
