Question

what are the removable discontinuities of the following function?
$f(x) = \frac{x^2 - 36}{x^3 - 36x}$
$x = -6$, $x = 0$, and $x = 6$
$x = -6$ and $x = 6$
$x = 0$ and $x = -6$
$x = 0$ and $x = 6$

Explanation:

Step1: Factor numerator and denominator

Numerator: $x^2 - 36 = (x-6)(x+6)$
Denominator: $x^3 - 36x = x(x^2 - 36) = x(x-6)(x+6)$
So $f(x)=\frac{(x-6)(x+6)}{x(x-6)(x+6)}$

Step2: Find domain restrictions

Set denominator equal to 0:
$x(x-6)(x+6)=0$
Solutions: $x=0$, $x=6$, $x=-6$

Step3: Identify removable discontinuities

Removable discontinuities occur where common factors cancel. The common factors are $(x-6)$ and $(x+6)$, so these correspond to $x=6$ and $x=-6$. $x=0$ is a non-removable (infinite) discontinuity as it does not cancel out.

Answer:

x = -6 and x = 6