Answer
Range: $[0,\infty)$
Domain: $(-\infty,0] \cup [3, \infty)$
Work Step by Step
The domain of the function should match where the function is defined. The argument of a square root must be positive in ${\rm I\!R}$, thus we solve$$x^2-3x\geq0$$
Factoring yields
$$x(x-3) \geq 0$$
Thus we want both factors to be nonnegative or both to be less than or equal to zero.
This occurs when we are on either extreme,
$x-3\geq 0 $ or $ x \leq 0 $.
That gives us the domain of $(-\infty,0] \cup [3, \infty)$.
The range is all nonnegative real numbers, as the square root can only return nonnegative values.
