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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

3 answers

x^2+x-2=0
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.
x^2+x-2=0
(1)^2 + 1 - 2=0
1+1-2=0
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