## anonymous 4 years ago what is the domain of the function the square root of 64-x^2

1. anonymous

Quantities under root sign can't be less than 0. Set >= 0: $64-x^2 \ge 0$ $64 ≥ x^2$ $64 ≥ x^2$ $x \le 8$

2. anonymous

sorry, there is a bit more. Because you end up with a + and - when you solve a squareroot you should instead get for the domain: . . $64 \ge x^2$ $x \le \left| 8 \right|$ $-8 \le x \le 8$