#### Answer

$[-2, 3) \cup (3, +\infty)$

#### Work Step by Step

(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:
$[-2, 3) \cup (3, +\infty)$