# How do you find the domain of f(x) = - sqrt(2 / (x^2 - 16))?

May 18, 2015

You have to check:

• the denominator must be different from zero
• the expression under the square root must be positive

So we must have ${x}^{2} - 16 \ne 0 \implies x \ne \pm 4$

And (the numerator doesn't interfere):

${x}^{2} - 16 \ge 0 \implies x \ge 4 \mathmr{and} x \le - 4$

So, intersecting, we have:
$D : = \left\{x \in \mathbb{R} : x > 4 \mathmr{and} x < - 4\right\}$