## Precalculus (6th Edition)

$(-\infty, 2]$
The radicand (expression inside the radical sign) of a square root is not allowed to be negative since negative real numbers do not have real number square roots. Thus, the domain of the given function can be found by setting $6-3x$ as greater than or equal to zero: $6-3x \ge 0 \\-3x \ge 0-6 \\-3x \ge -6 \\\frac{-3x}{-3} \le \frac{-6}{-3} \text{(inequality becomes$\le$because a negative number was divided to both sides)}$ $x \le 2$ Therefore, the domain of the function is $(-\infty, 2]$.