Question

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

Explanation:

Step1: Factor 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: Simplify the function

Cancel out \((x + 6)(x - 6)\) (for \(x
eq - 6,x
eq6\)), we get \(f(x)=\frac{1}{x}\), \(x
eq - 6,x
eq6,x
eq0\)

Step3: Identify removable discontinuities

Removable discontinuities occur when a factor can be canceled. The factors \((x + 6)\) and \((x - 6)\) are canceled. But when \(x=-6\) or \(x = 6\), the original function is undefined. However, the limit exists at \(x=-6\) and \(x = 6\) (after canceling the common factors). While at \(x = 0\), it is a non - removable discontinuity (vertical asymptote as the simplified function \(\frac{1}{x}\) has a vertical asymptote at \(x = 0\))

Answer:

\(x=-6\) and \(x = 6\)