Answer
The domain of this function is $(-\infty,0]\cup[6,\infty)$
Work Step by Step
$g(x)=\sqrt[4]{x^{2}-6x}$
This function is not defined for negative numbers. Its domain can then be found by solving the following inequality:
$x^{2}-6x\ge0$
Find the intervals. Take out common factor $x$ from the left side:
$x(x-6)\ge0$
The factors are $x$ and $x-6$. Set them equal to $0$ and solve for $x$:
$x=0$
$x-6=0$
$x=6$
The factors are zero when $x=0,6$. These numbers divide the real line into the following intervals:
$(-\infty,0)$ $,$ $(0,6)$ $,$ $(6,\infty)$
Elaborate a diagram, using test points to determine the sign of each factor in each interval: (refer to the attached image below)
It can be seen from the diagram that the inequality is satisfied on the intervals $(-\infty,0)$ and $(6,\infty)$. Also, the inequality involves $\ge$, so the endpoints of these intervals also satisfy the inequality.
The domain of this function is $(-\infty,0]\cup[6,\infty)$
