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A question for you all

  1. Mar 22, 2005 #1
    Here is a problem that I am having trouble with.

    Let f be a function such that:

    Limit as h -->0 [f(2 + h) - f(2)] / h = 5

    (essentially that: f ' (2) = 5 ).

    It then asks: which of the following must be true?

    1) f is continuous at x = 2.

    2) f is differentiable at x = 2.

    3) The derivative of f is continuous at x = 2.

    THE ANSWERS: The first two are true, but the last one is false.

    I understand the first two statements and how they must be true, but I also think the third statement must also be true. I can't think of an example of a function that disproves number 3. Can you?

    I've tried drawing some possible functions for f ' that has a discontinuity (asymptote, jump, point) at x = 2 but whose antiderivative f is still continuous and differentiable at x = 2, but I can't come up with an example that works!

    Thanks for any help
  2. jcsd
  3. Mar 22, 2005 #2


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    Unless I'm missing something, this doesn't seem like a simple problem. Consider a function that has derivatives only at 0, and points of the form 1/n, where n is a non-zero integer. Suppose its derivative at 0 is 0, and is 1 at all those other points. Then such a function has a derivative at 0, but it is discontinuous there, since any open set around 0 will contain some of the points of the form 1/n, but the derivatives there will always be 1, so the limit does not exist. Now, can a continuous function have such a weird "derivative"? Well, the Weierstrass Function is continuous but only has derivatives on a set of measure 0. The set consisting of 0 and 1/n for all non-zero integers n is a set of measure 0 (this is not a definition of measure 0, just an example of one). So there is a continuous function that has a derivative that is weird in some related sense, but that doesn't mean that it is "weird enough", and the Weierstrass function may still have a continuous derivative. Also, the discontinous derivative I suggested is one way it may be discontinuous, there may be other ways (like a jump discontinuity, etc.).
  4. Mar 22, 2005 #3
    I googled this up to the surface:

    An example of a function which is differentiable at 0, but whose derivative is not continuous at 0

    [itex]f(x) = (x^2) * sin(1/x), with f(0) = 0[/itex]

    I haven't worked out the details. Perhaps someone else would like to try.
    Last edited: Mar 22, 2005
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