Answer
The domain of this function is $(-\infty,-2]\cup[4,\infty)$
Work Step by Step
$g(x)=\sqrt{x^{2}-2x-8}$
This function is not defined when the expression inside the square root is a negative number. Its domain can then be found by solving the following inequality:
$x^{2}-2x-8\ge0$
Find the intervals. Factor the left side of the inequality:
$(x+2)(x-4)\ge0$
The factors are $x+2$ and $x-4$. Set them equal to $0$ and solve for $x$:
$x+2=0$
$x=-2$
$x-4=0$
$x=4$
The factors are zero when $x=-2,4$. These numbers divide the real line into the following intervals:
$(-\infty,-2)$ $,$ $(-2,4)$ $,$ $(4,\infty)$
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 $(-\infty,-2)$ and $(4,\infty)$. Also, the inequality involves $\ge$, so the endpoints of these intervals satisfy the inequality too.
The domain of this function is $(-\infty,-2]\cup[4,\infty)$
