Answer
The domain of $~~g(x) = \frac{\sqrt{x^2-9}}{x^2-2}~~$ is $~~(-\infty, -3]\cup [3, \infty)~~$ and $~~g(x)~~$ is continuous on its domain.
Work Step by Step
Let $f(x) = \sqrt{x^2-9}$
We can find the domain of $f(x)$:
$x^2-9 \geq 0$
$x^2 \geq 9$
$x \leq -3~~$ or $x \geq 3$
The domain is $(-\infty, -3]\cup [3, \infty)$
Since $x^2-9$ is a polynomial, it is continuous for all real numbers.
A root function is also continuous on its domain.
Thus $f(x)$ is continuous on its domain $(-\infty, -3]\cup [3, \infty)$
Let $h(x) = x^2-2$
Since $h(x)$ is a polynomial, it is continuous for all real numbers.
$\frac{f(x)}{h(x)}$ is continuous on its domain, except when $h(x) = 0$
$h(x) = 0$ when $x = \pm \sqrt{2}$
Note that $\pm \sqrt{2}$ are not included in the domain $(-\infty, -3]\cup [3, \infty)$
Therefore, the domain of $~~g(x) = \frac{\sqrt{x^2-9}}{x^2-2}~~$ is $~~(-\infty, -3]\cup [3, \infty)~~$ and $~~g(x)~~$ is continuous on its domain.
