Answer
$(-\infty, 6)$
Work Step by Step
(i) The denominator cannot be equal to zero thus,
$\sqrt{6-x}\ne 0
\\6-x \ne 0^2
\\6-x \ne 0
\\6 \ne x$
(ii) The radicand (number inside the square root symbol) cannot be negative. Thus,
$6-x \ge 0
\\-x \ge -6
\\-x(-1) \le -6(-1)
\\x \le 6$
This means that the value of $x$ has to be less than 6.
Therefore, the domain of the function is:
$(-\infty, 6)$