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Thread: Constructing functions on closed intervals whose derivatives are unbounded

  1. #1
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    Constructing functions on closed intervals whose derivatives are unbounded

    Hello folks,

    I'm looking for a way to construct a function F on a closed interval [a, b] whose derivative is unbounded. Why properties does F need to possess?

    Is there a discontinuity requirement?
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  2. #2
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    Re: Constructing functions on closed intervals whose derivatives are unbounded

    It is already discontinuous at $a$ and $b$. So, make it unbounded at the endpoints. For example: $f(x) = \begin{cases}(x-a)\sin\left( \dfrac{1}{x-a} \right) & x > a \\ 0 & x = a\end{cases}$. Note that this function is continuous, but its derivative at a is not defined. In fact, its derivative approaches all real numbers as x approaches a. But, this function is differentiable on $(a,b)$.
    Last edited by SlipEternal; Feb 1st 2018 at 12:00 PM.
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    Re: Constructing functions on closed intervals whose derivatives are unbounded

    Hey jumbo1985.

    You will have a discontinuity if you allow unbounded derivatives.

    You can think about integrating this to show that continuity is violated [Fundamental theorem of calculus doesn't work over interval].
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    Re: Constructing functions on closed intervals whose derivatives are unbounded

    Quote Originally Posted by chiro View Post
    Hey jumbo1985.

    You will have a discontinuity if you allow unbounded derivatives.

    You can think about integrating this to show that continuity is violated [Fundamental theorem of calculus doesn't work over interval].
    This is false. The function I gave above is continuous (no discontinuities) but it's derivative is unbounded at x=a.

    You can extend the function to the entire real line and it will still be continuous at x=a and its derivative still unbounded and discontinuous at x=b. Continuity does not imply differentiability.
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