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Calculus

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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

  • Calculus - ,

    x^2+x-2=0
    x=1 is your discontinuity

  • Calculus - ,

    Why would it be 1?? I'm so confused...

  • Calculus - ,

    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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