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Math Help - Small Challenge

  1. #16
    Senior Member Dinkydoe's Avatar
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    I see what you mean. They give f(x)=x^2\sin(1/x^3) for x in (0,1], f(0)=0.

    It's easy to check f'(x) is unbounded in this case. And then they conclude f'(x) is not integrable.....
    It wouldnt be too hard to make such function with f(1)=1 and f(0)=0.

    I'm confused right now. I think you may be right after all.
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  2. #17
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    Quote Originally Posted by Unbeatable0 View Post
    That's not true. A derivative need not in general be integrable over a closed interval.
    .
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  3. #18
    Senior Member Dinkydoe's Avatar
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    Anyway. To correct my solution in that case, I'll just 'steal' your idea. Again we assume f(x)\neq x

    since we only need to show the existence of a,b with f'(a)<1, f'(b)>1 we need to show that f'(x)\leq 1 and f'(x)\geq 1 for all x is impossible. Then we're done.

    Now, g(x)=f(x)-x is increasing/decreasing. And since g'(x)\neq 0 for all x we can't have g(1)=0.
    A contradiction.

    Now again the intermediate value theorem guarantuees f'(x),f'(y) \in (f'(a),f'(b)) such that  f'(x)f'(y)=1.
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