Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 26

Prove that $\displaystyle g(x) = \frac{\sqrt{x^2 - 9}}{x^2 - 2}$ is continuous on its domain. State the domain

We can rewrite $\displaystyle g(x) = \frac{\sqrt{x^2 - 9}}{x^2 - 2}$ as

$\qquad \displaystyle g(x) = \frac{F(H(x))}{I(x)}$

Where

$\qquad F(x) = \sqrt{x}, \qquad H(x) = x^2 - 9 \qquad$ and $\qquad I(x) = x^2 - 2$

$\qquad H(x)$ and $I(x)$ are one example of function that is continuous on every values of $x$ according to the definition. However, $F(x)$ is a root function and $H(x)$ is inside $F(x)$ so $x^2 - 9 \geq 0$ .

Therefore,

$\qquad$ The domain of $g(x)$ is $(-\infty, -3] \bigcup [3, \infty)$