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Math Help - function differentiable, but derivative not bounded

  1. #1
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    function differentiable, but derivative not bounded

    Question: Give an example of a function f that is differentiable on [0,1] but its derivative is not bounded on [0,1].



    Ok, I know that the derivative f' cannot be continuous, because then it would be bounded on [0,1]. I also know that it cannot be increasing with jump discontinuities, because the derivative has to have the intermediate value property. I also do not think (but am not 100 percent sure) that it can have any infinite discontinuities, because I think that would make f unbounded, which it cannot be on [0,1] if its differentiable on [0,1].

    Any ideas?
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  2. #2
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    No, that's impossible. Saying that the function is differentiable on [0, 1] means that it has a finite derivative at each point on [0, 1]. Since [0, 1] is a closed and bounded set, that derivative must be bounded.
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  3. #3
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    Halls, what about  f(x) = x^2 sin(1/x^2) when x isn't 0, and f(0) = 0.

    f is differentiable at all non-zero points, and note that  f(0 +h) - f(0) = h^2 sin(1/h^2) . Thus  \frac{ f(0+h) - f(0) } {h} = h sin(1/h^2) \rightarrow 0 . Thus f is also differentiable at 0 and f(0) = 0.

    Note that on (0, 1],  f'(x) = 2x sin(1/x^2) -  \frac{2cos(1/x^2)}{x} , which goes to -inf as x goes to 0 through points of (0, 1], thus f' is not bounded on [0,1]
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