Answer
$(-\infty,-4)\cup(4,\infty)$
Work Step by Step
The domain is the set of all possible values of x
for which the value f(x) exists (is defined).
For f(x) to be defined, the following conditions must be met:
1. The radicand $\displaystyle \frac{2}{x^{2}-16}$ must not be negative (because of the square root)
Since the numerator is positive, the denominator must not be negative:
So, $\qquad x^{2}-16 \geq 0.$
2. The denominator must not be zero $x^{2}-16\neq 0$
Combined, these two conditions produce
$x^{2}-16 > 0$
$x^{2} > 16$
This is true for $x < -4$ and for $x > 4.$
In interval notation, the domain is
$(-\infty,-4)\cup(4,\infty)$
