Answer
$f(x)=\dfrac{x^2-4}{x^2-9}$ is continuous for all real numbers except for $-3$ and $3$.
Work Step by Step
Given: $f(x)=\dfrac{x^2-4}{x^2-9}$
Re-write as: $\dfrac{x^2-4}{x^2-9}=\dfrac{x^2-4}{(x+3)(x-3)}$
When the denominator is $0$, then the fraction $\dfrac{x^2-4}{x^2-9}=\dfrac{x^2-4}{(x+3)(x-3)}$ is undefined. By the zero product rule, we have: $x+3 \ne0$ and $x-3 \ne0$. So, $x\ne3$ and $x\ne-3$,
Therefore, the fraction function is continuous everywhere, except for where it is undefined. So, $f(x)=\dfrac{x^2-4}{x^2-9}$ is continuous for all real numbers except for $-3$ and $3$.
