91Ó°ÊÓ

We have been given a function $f\left(x\right)=\frac{x^2-4}{x^2-9}$.

We have to find the numbers at which f is continuous and at which numbers f is discontinuous.

A function f is discontinuous at the number where:

• f is undefined
• the limit does not exist, i.e., its one-sided limits are not equal
• the limit exists, i.e., its one-sided limits equal, but is not equal to the function value

Analyzing the given function,

$f\left(x\right)=\frac{x^2-4}{x^2-9}$

We know that it is a rational function in the form of $R\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}$whose domain is

$\left\{x\left|Q\right(x)≠0\right\}$

Solve for the $Q\left(x\right)$,

$f\left(x\right)=\frac{x^2−4}{x^2−9}=\frac{x^2−4}{(x+3)(x−3)}$

Thus,

role="math" $Q\left(x\right)=(x+3)(x-3)$

which implies that,

$x≠3;x≠-3$

According to the properties of continuities, if a function is rational, then the function's graph must be continuous at every number in the domain and has a hole or vertical asymptote where the function is undefined.

This implies that at x = -3 and x = 3 f is undefined.