Question

what are the removable discontinuities of the following function? f(x) = (x^2 - 36)/(x^3 - 36x) x = 0 and x = 6 x = 0 and x = -6 x = -6 and x = 6 x = -6, x = 0, and x = 6

Explanation:

Step1: Factor the function.

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

Step2: Identify removable discontinuities.

Removable discontinuities occur where factors cancel. Here, $x
eq - 6,0,6$ initially. After canceling $(x + 6)$ and $(x - 6)$ (for $x
eq\pm6$), removable at $x=-6,6$. And $x = 0$ is a non - removable (vertical asymptote) as the factor $x$ in the denominator remains.

Answer:

$x=-6,x = 6$