## Precalculus (6th Edition) Blitzer

$[3, 6) \cup (6, +\infty)$
(i) The radicand (expression inside a radical sign) of a square root cannot be negative as its root is an imaginary number. This means that the radicand in the numerator, which is $x-3$, must be greater than or equal to 0. Thus, $x$ can be any real number greater than or equal to $3$ (ii) The denominator cannot be zero as division of zero leads to an undefined expression. This means that the value of $x$ cannot be $6$. From (i) and (ii) above, the restrictions to the value of $x$ are: (i) $x \ge 3$; and (ii) $x \ne 6$ Both of the restrictions above must be satisfied. Thus, the value of $x$ must be greater than or equal to $3$ but not including $6$. Therefore, the domain of the given function is: $[3, 6) \cup (6, +\infty)$