# How do you find the domain of square root of 16-x^2?

Jun 15, 2015

$\sqrt{16 - {x}^{2}}$ may never have a negative "under the root".
$16 - {x}^{2} \ge 0 \to {x}^{2} \le 16 \to$
$| x | \le 4 \to - 4 \le x \le 4$