f is discontinuous at x = ±2
since the numerator (x) is not zero there, f(x) has vertical asymptotes at x=±2.
These are not removable.
Removable discontinuities occur where f(x) = 0/0 and the common factor can be removed, such as
f(x) = (x-2)/(x^2-4)
= (x-2)/(x-2)(x+2)
At x=2, f(x) = 0/0
but everywhere else, f(x) = 1/(x+2) because you can remove the common factor of (x-2). The "hole" is removable by defining
f(2) = 1/4 which is 1/(x+2)
Find the x-values (if any) at which f is not continuous. Which of the discontinuties are removable?
f(x)=(x)/(x^2-4)
I'm not sure how to do this problem but I know that it is not continnous at x=2?
2 answers
Got it, thanks for the help!
1 answer