Answer
$(-\infty,-2)\cup(7,\infty)$
Work Step by Step
$(\frac{a}{b})^n=\frac{a^n}{b^n}$,
$\left(\frac{1}{x^2-5x-14}\right)^{1/2}=\frac{1}{\sqrt{x^2-5x-14}},$
solving for trinomial,
$x^2-5x-14=x^2+2x-7x-14=x(x+2)-7(x+2)=(x-7)(x+2)=0,$
thus, $x=7$ or $x=-2$
Since Domain of square root is greater than or equal to zero and the denominator cannot be zero, we check our interval as follows.
Therefore,
$\begin{array}{lll}
Intervals & (x-7)(x+2)& (x-7)(x+2)>0?\\
(-\infty, -2) & (-)(-)=(+)& True\\
(-2, 7) & (-)(+)=(-)& False\\
(7,\infty)&(+)(+)=(+)& True
\end{array}$
Thus, The solution set: $(-\infty,-2)\cup(7,\infty)$
