Answer
The domain of definition is:
$(-\infty, -2) ∪ (5, +\infty)$
Work Step by Step
$(\frac{1}{x^2-3x-10})^{\frac{1}{2}}= \sqrt{(\frac{1}{x^2-3x-10})}$
We have 2 restrictions: the expression under square root must be non-negative and denominator must not be 0.
$\left\{ \begin{array}{ll}
(\frac{1}{x^2-3x-10})\geq0 \\
x^2-3x-10\ne0
\end{array} \right. $
Both of the conditions will be satisfied if:
$x^2-3x-10>0$
It is defined when we have $(-\infty, -2) ∪ (5, +\infty)$
