Answer
$(-\infty, -2] \cup [2, +\infty)$
Work Step by Step
The elementary radical function $y=\sqrt{x}$ is defined only when the radicand, which is $x$, is greater than or equal to 0.
This means that the given function $g(x)=\sqrt{x^2-4}$ is defined only when $x^2-4\ge 0$.
Solve the inequality to have:
$(x-2)(x+2) \ge 0$
Note that the value of $g$ is greater than or equal to $0$ when:
$x \ge 2$ or when $x \le -2$
The given function is defined when $x \ge 2$ or when $x \le -2$.
Therefore the domain of the given function is:
$(-\infty, -2] \cup [2, +\infty)$