posted by Eric on .
Could anyone please help me figure out the discontinuity that is removable of the following function?? Any help would be greatly appreciated! Thanks in advance!!!
x=1 is your discontinuity
Why would it be 1?? I'm so confused...
because having 0 in the denominator would make your function be undefined and therefore be a discontinuity at that point. If you set the denominator equal to 0 and solve your answer would be 1.
(1)^2 + 1 - 2=0