Answer
$1.\ \ \mathrm{Domain:} \ \ (-\infty,0] \cup [3,\infty)$
$2.\ \ \mathrm{Range:} \ \ [0,\infty)$
Work Step by Step
$1.$ We can take the square root only for positive numbers.
$x^2-3x>=0$, $\ \ $ solve for $x$.
$x(x-3)>=0$
The product of two numbers will be positive provided either both are negative or both are positive.
$\mathrm{I}.$ When both are negative:
$x<=0 \ \ and \ \ x-3<=-0$
All numbers that are $\ <=0\ $ satisfy both inequalities.
So, $\ \ x\in(-\infty,0]$
$\mathrm{II}.$ When both are positive:
$x>=0 \ \ and \ \ x-3>=0$
All numbers that are $\ >=3\ $ satisfy both inequalities.
So, $\ \ x\in[3,\infty)$
Therefore, domain of the function is, $\ \ (-\infty,0] \cup [3,\infty)$
$2.$ Range of function is $\ \ [0,\infty]\ \ $ because the square root from any positive number is positive.
