## Precalculus: Mathematics for Calculus, 7th Edition

$[-2, 3) \cup (3, +\infty)$
(i) The denominator cannot be zero therefore $x$ cannot be equal to 3, (ii) 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 in the given function, the value of $2+x$ must be greater than or equal to 0. Thus, $2+x \ge 0 \\x \ge -2$ Note that from part (i), $x \ne 3$. Therefore, the domain of the given function is all real numbers greater than or equal to -2 except 3. In interval notation, the domain is: $[-2, 3) \cup (3, +\infty)$