What is the discontinuity of the function #f(x)=(3x^2+x-4)/(x-1)#?

1 Answer
George C.
Aug 6, 2015

There is a removable discontinuity at #x=1#, where the numerator and denominator are both #0#, since:

#(3x^2+x-4)/(x-1) = ((x-1)(3x+4))/(x-1) = 3x+4#

with exclusion #x != 1#

Explanation:

Since we have this factorisation, #lim_(x->1) f(x)# exists and is equal to #3(1)+4 = 7#.

So the discontinuity is removable by redefining #f(1) = 7#

Related questions
See all questions in Classifying Topics of Discontinuity (removable vs. non-removable)
Impact of this question
5256 views around the world
You can reuse this answer
Creative Commons License