Answer
$(-\infty, -1)\cup(5, \infty)=\{x|x\leq-1, x\geq5\}$
Work Step by Step
The function has a square root. All values within a square root must be greater than or equal to $0$. Therefore, $x^2-4x-5\geq0$. To determine the domain, let us determine the values when $x^2-4x-5=0$.
$(x-5)(x+1)=0$.
$x=5$ and $x=-1$
Let us take a value within this interval and check its sign. When $x=0$, $x^2-4x-5<0$. Therefore, we need the interval outside.
$(-\infty, -1]\cup[5, \infty)$
