Answer
$(-\infty,-1] \cup [5,\infty)$
Work Step by Step
The domain is the set of all possible values of x
for which the value f(x) exists (is defined).
For f(x) to be defined, the radicand must not be negative,
$x^{2}-4x-5 \geq 0$
Factoring the LHS, we find two factors of -5 whose sum is -4.
(These are $-5$ and $+1)$
$(x+1)(x-5) \geq$ 0
The zeros of the LHS, $x=-1$ and $x=5$
divide the real number line into intervals.
Interval: $(-\infty,-1]$, test point x=$-2$,
$(-2+1)(-2-5)=-1(-7)=+7$ (positive)
Interval: $(-1, 5)$, test point x=$0$,
$(0+1)(0-5)=1(-5)=-5$ (negative)
Interval $[5,\infty)$, test point x=$10$,
$(10+1)(10-5)=11(5)=55$ (positive)
So, the domain is
$(-\infty,-1] \cup [5,\infty)$
