Posted by **Eric** on Thursday, September 29, 2011 at 8:23pm.

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!!!

f(x)=4x/x^2+x-2

- Calculus -
**Brianna**, Thursday, September 29, 2011 at 8:32pm
x^2+x-2=0

x=1 is your discontinuity

- Calculus -
**Eric**, Thursday, September 29, 2011 at 8:44pm
Why would it be 1?? I'm so confused...

- Calculus -
**Brianna**, Thursday, September 29, 2011 at 8:57pm
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.

x^2+x-2=0

(1)^2 + 1 - 2=0

1+1-2=0

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