Answer
$[3,+\infty)$
Work Step by Step
The domain is the set of all possible values of x
for which a value f(x) exists.
$(x-3)^{1/2}=\sqrt{x-3}$
produces a real number, that is, f(x) is defined
only when the radicand $x-3 $ is non-negative.
So
$x-3 \geq 0$
$x \geq 3,$
In interval notation, the domain is
$[3,+\infty)$
