Answer
$ 0\leq x \leq 6$
Work Step by Step
For $\sqrt{6x - x^{2}}$ to be defined as a real number:
$6x - x^{2}\geq0$ <=> $x(6-x)\geq0$.
The expression on the left of the inequality changes signs when $x = 0$ and $x = 6$. Thus we must check the intervals in the following table.
From the table, we see that
$6x - x^{2}$ is defined when $ 0\leq x \leq 6$.
