Answer
$(-\infty,0]\cup[6,\infty)$
Work Step by Step
We are given the function:
$$g(x)=\sqrt[4]{x^2-6x}.$$
The domain of the function is the set of value for $x$ so that the radical makes sense. Since the radical is of even order ($4$), the radicand must be positive. We solve the inequality:
$$\begin{align*}
x^2-6x&\geq 0\\
x(x-6)&\geq 0.
\end{align*}$$
The solution set is:
$$(-\infty,0]\cup[6,\infty).$$
