Question

What are the removable discontinuities of the following function?

f (x) = StartFraction x squared minus 36 Over x cubed minus 36 x EndFraction
x = –6, x = 0, and x = 6
x = –6 and x = 6
x = 0 and x = –6
x = 0 and x = 6

Answer 1

Factor: f(x) = (x^2 − 36)/(x^3 − 36x) = (x−6)(x+6)/[x(x−6)(x+6)] = 1/x for x ≠ −6,0,6.

The factors (x−6) and (x+6) cancel, so x = 6 and x = −6 are holes (removable). x = 0 remains a nonremovable vertical asymptote.

Answer: x = −6 and x = 6.