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Limit Questioin

  1. Jan 13, 2012 #1
    The paragraph says, " Even if the function f is an everywhere differentiable function, it is still possible for f ' to be discountinuous. However, the graph of f ' can never exhibit a discountinuity of ...." picture is in paint document...

    What type of discountinuity is that? a hole discountinuity?

    second Question: Is my understanding correct? This is my explanation why f is not differentiable at some x value.

    f ' (x) =
    x , x>0
    x^2 x<0
    7 , x = 0.

    Therefore, the graph of the differnetiable function should have a similar discontinuity as the one shown in the paint document. (Determining the limit)
    lim f ' (x) as x → 0- = lim f ' (x) as x → 0+ = 0 however because lim Δx → 0 [ f(0 + Δx) - f(0) ]/Δx is equal to f ' (0) = 7 ≠ 0, therefore the function f non differentiable at x = 0 because inorder for f to be diff at 0, f'(0) must equal 0

    Is my understanding why f is not differentiable at x = 0 correct?

    Next, If I wanted to make f differentiable at x could I do this by defining one of two functions....

    f ' (x) =
    x , x>0
    x^2 x<0
    0 , x = 0.

    f ' (x) =
    x , x>0
    x^2 x<0

    Would this work to make f diff at x = 0?

    Attached Files:

    Last edited: Jan 13, 2012
  2. jcsd
  3. Jan 13, 2012 #2
    The answer to your last question is yes for the first function. The second function is not defined in 0.
    Last edited: Jan 13, 2012
  4. Jan 13, 2012 #3


    User Avatar
    Science Advisor

    While the derivative of a function is not necessarily continuous, it must have the "intermediate value propert"- if f'(x0)= a and f'(x1)= then, between x0 and x1, f' must take on all values between a and b. From that, it follows that a derivative cannot have a "jump" discontinuity (the left and right limits exist but are different).
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