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Math Help - question with derivative

  1. #1
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    question with derivative

    f is a differntiable function at [0,1] which maintains 0<=f(x)<=1 to every x[0,1].
    i need to prove that there's a point x[0,1] so
    f(x)=3x/(sqrt(3x2+6))
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  2. #2
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    Re: question with derivative

    Hi there, if I'm not wrong you should do the following:

    first - find the derivative of f(x)

    second - use the Mean Value Theorem f(c) = (f(1) - f(0))/(1-0),

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  3. #3
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    Re: question with derivative

    But i dont have any written funcrion, so how would i find its derivative?
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  4. #4
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    Re: question with derivative

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  5. #5
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    Re: question with derivative

    The integral of your function g(x)=\frac{3x}{\sqrt{3x2+6}} from 0 to 1 is approximately 0.55, so it would seem that you could set 0 < f(0) < 0.45 and f'(x) slightly greater than g(x) to get a function which stays between 0 and 1, but the derivative is never equal to g(x).

    - Hollywood
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  6. #6
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    Re: question with derivative

    Hi,
    What you need is the intermediate value theorem for derivatives; you can look up Darboux's theorem on the web for a proof. Here's a proof of your problem:

    question with derivative-mhfcalc4.png
    Thanks from hollywood
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  7. #7
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    Re: question with derivative

    Yes, johng, you are correct. Thank you. I must have subconsciously changed the conditions from 0 \le f'(x) \le 1 to 0 \le f(x) \le 1.

    - Hollywood
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  8. #8
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    Re: question with derivative

    Hi again,
    I took my own advice and surfed the web for the intermediate value theorem for derivatives. The proof in wikapedia, Fermat's theorem (stationary points) - Wikipedia, the free encyclopedia, has an error or at least an incomplete argument. PlanetMath has the same proof. So here's a complete proof of the theorem.

    question with derivative-mhfcalc4a.png
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