Question

How do you find the removable discontinuities of `f(x ) = \frac { x ^ { 2} - 36} { x ^ { 3} - 36x }`?

Answer 1

`x = +-6`

Explanation:

A removable discontinuity (also called a hole) in a function is where there is no `f(x)` value at a certain `x`-point, but it is not an asymptote.

These "holes" originate when two terms (i.e. factors) are the same in the numerator and denominator. Essentially, at that `x`-value, the denominator equals zero (which means the function is undefined), but at the same time that very factor can be eliminated because it also exists in the numerator.

To find any removable discontinuities in this function, let's factor the numerator and denominator:

`f(x) = ((x-6)(x+6))/(x(x-6)(x+6))`

Notice how the two factors `(x-6)` and `(x+6)` cancel out, but are also points where the function is undefined.

Therefore, the removable discontinuities of this function are located at `x = -6` and `x = 6`.