Question

discuss the continuity of the function. (enter your answers as a comma - separated list.)
$f(x)=\frac{1}{x^{2}-16}$
$f$ is discontinuous at $x = - 4,4$

Explanation:

Step1: Find denominator zeros

Set \( x^2 - 16 = 0 \)

Step2: Solve for x

\( x^2 = 16 \implies x = \pm 4 \)

Answer:

-4, 4