Answer
$(-\infty,-2\sqrt 2]\cup[2\sqrt 2,\infty)$
Work Step by Step
We are given the function:
$f(x)=\sqrt{2x^2-16}$.
First determine the domain on which the function is defined:
$2x^2-16\geq 0$
$2(x^2-8)\geq 0$
$2(x-2\sqrt 2)(x+2\sqrt 2)\geq 0$
$D=(-\infty,-2\sqrt 2]\cup[2\sqrt 2,\infty)$
$f(x)$ is a composed function of the functions $2x^2-16$ and $\sqrt x$. Both functions are continuous, therefore their composition is also continuous on the domain.
The interval of continuity is:
$(-\infty,-2\sqrt 2]\cup[2\sqrt 2,\infty)$
