In calculus, how can I tell when a function's discontinuities are removable or non removable?

I can tell when a function has discontinuities but how can I determine whether they are removable or non removable? Are there any ground rules for making this decision or do I have to look at a graph? Please help, my test is tomorrow!

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  • 1 decade ago
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    if you have factors on the top and bottom that cancel, then this is a removable discontinuity. You can find the y-value of the discontinuity by cancelling the common factors, and then plugging in the x-value of the hole to what's left

    In other words, a removable discontinuity looks like 0/0

    a non-removable discontinuity looks like 1 / 0

    ex: f(x) = (x - 2)(x + 3) /[(x - 2) (x - 4)]

    when x = 2, this becomes 0 / 0

    x = 2 is a removable discontinuity

    cancel the (x - 2) / (x - 2) to leave (x + 3) / (x - 4)

    when x = 2, this is 5 / -2 =-5/2

    so the removable discontinuity is the point (2 , -5/2) (that is, the "hole" in the graph)

    x = 4 is a non-removable discontinuity, and will be an asymptote of the graph

    ex: the graphs of y = x - 2 and y = (x^2 - 4) / (x + 2) are identical, except for the removable discontinuity at x = -2

    (this would be the point (-2 , -4) that's removed)

    that's because (x^2 - 4) / (x + 2) = (x - 2)(x + 2) / (x + 2)

    they look the same, but the (x + 2) / (x + 2) creates the hole that is removed from the function

  • 4 years ago

    Removable Discontinuity

  • Anonymous
    6 years ago

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    RE:

    In calculus, how can I tell when a function's discontinuities are removable or non removable?

    I can tell when a function has discontinuities but how can I determine whether they are removable or non removable? Are there any ground rules for making this decision or do I have to look at a graph? Please help, my test is tomorrow!

    Source(s): calculus function discontinuities removable removable: https://shortly.im/YVVpc
  • Janine
    Lv 4
    5 years ago

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    factor x^2 - x into x(x -1) so the function is now x / (x(x-1) = 1 / (x - 1) which will have a non-removable discontinuity at x =1 (you can't divide by 0)

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  • Anonymous
    1 decade ago

    Okay if you have a question, look at it. see if you can factor the denominator. once factored look at the numerator. if there is something you can cancel out, try to find what would make what you are planning on cancelling out equal to zero.

    once you cancel something that is your removable; the removable shows up on a graph as a hole.

    the one that is left after you cancel out the factor is called the non removable. which ussually results in becoming the "vertical asymptote"

    how you can tell is: after you cancel out the removable, think of a number that would make the bottom equal to zero. for example:

    1/x - 4

    the vertical asymptote would be x=4, because when plugged in. that is what makes the function undefined.

    Source(s): i'm taking calculus 1.
  • 1 decade ago

    i dont no what ur mean by removable

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