Question

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

Explanation:

Step1: Factor numerator and denominator

Factor the numerator \(x^2 - 36\) as a difference of squares: \(x^2 - 36=(x - 6)(x + 6)\).
Factor the denominator \(x^3 - 36x\) by first factoring out \(x\): \(x^3 - 36x=x(x^2 - 36)\), then factor \(x^2 - 36\) as \((x - 6)(x + 6)\), so the denominator becomes \(x(x - 6)(x + 6)\).

The function is now \(f(x)=\frac{(x - 6)(x + 6)}{x(x - 6)(x + 6)}\).

Step2: Identify removable discontinuities

Removable discontinuities (holes) occur where a factor cancels out from both numerator and denominator.
Cancel the common factors \((x - 6)\) and \((x + 6)\) (note: we must exclude \(x = 6\) and \(x=-6\) from the domain initially, but after canceling, the hole is where the canceled factors are zero, i.e., \(x = 6\) and \(x=-6\)? Wait, no—wait, let's re - examine. Wait, the denominator is \(x(x - 6)(x + 6)\), numerator is \((x - 6)(x + 6)\). So the common factors are \((x - 6)\) and \((x + 6)\)? Wait, no, the numerator is \((x - 6)(x + 6)\) and the denominator is \(x(x - 6)(x + 6)\). So when we cancel \((x - 6)\) and \((x + 6)\), we have to consider that \(x
eq6\), \(x
eq - 6\), and \(x
eq0\) (since \(x = 0\) makes the original denominator zero but does not cancel). Wait, no—removable discontinuities are where a factor is in both numerator and denominator. So the factors \((x - 6)\) and \((x + 6)\) are in both numerator and denominator. So the values of \(x\) that make these canceled factors zero are \(x = 6\) and \(x=-6\)? Wait, no, wait, let's check the options. Wait, maybe I made a mistake. Wait, the denominator is \(x(x^2 - 36)=x(x - 6)(x + 6)\), numerator is \((x - 6)(x + 6)\). So when we simplify, we get \(f(x)=\frac{1}{x}\) for \(x
eq6\), \(x
eq - 6\), \(x
eq0\). Wait, no—wait, the canceled factors are \((x - 6)\) and \((x + 6)\), so the holes are at \(x = 6\) and \(x=-6\)? But wait, the options have \(x=-6\) and \(x = 6\) as an option (the bottom left: \(x=-6\) and \(x = 6\)). Wait, let's re - evaluate.

Wait, removable discontinuities are points where the function is undefined (due to a zero in the denominator) but the limit exists. For a rational function, a hole occurs when a factor is present in both numerator and denominator. So in \(f(x)=\frac{(x - 6)(x + 6)}{x(x - 6)(x + 6)}\), the factors \((x - 6)\) and \((x + 6)\) are in both numerator and denominator. So when \(x = 6\) or \(x=-6\), the factor \((x - 6)\) or \((x + 6)\) is zero, but since it cancels, the limit as \(x\) approaches 6 or - 6 exists (it's \(\frac{1}{x}\) evaluated at \(x = 6\) or \(x=-6\), i.e., \(\frac{1}{6}\) or \(-\frac{1}{6}\)). The factor \(x\) is only in the denominator, so \(x = 0\) is a non - removable discontinuity (vertical asymptote). So the removable discontinuities are at \(x=-6\) and \(x = 6\).

Answer:

\(x=-6\) and \(x = 6\) (the option: \(x=-6\) and \(x = 6\))