## Precalculus: Mathematics for Calculus, 7th Edition

Domain: $(-\infty, -2] \cup [0, +\infty)$
RECALL: 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: $x^2 \ge 4$ Note that the value of $x^2$ is greater than or equal to $4$ 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: $(-\infty, -2] \cup [0, +\infty)$