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Math Help - Help with Calculus (undergraduate mathematics)

  1. #1
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    Help with Calculus (undergraduate mathematics)

    Hi, I'm a mathematics student and I need help with a simple problem:

    Given a C1 function f:[a,b]-->R, with a right-hand derivative f'(a) on a. Proof if the following statement is true or false.
    If f'(a)=0, then f has a local maximum or a local minimum on a.

    Thanks very much!!
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  2. #2
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    Re: Help with Calculus (undergraduate mathematics)

    Quote Originally Posted by yannt View Post
    Hi, I'm a mathematics student and I need help with a simple problem:

    Given a C1 function f:[a,b]-->R, with a right-hand derivative f'(a) on a. Proof if the following statement is true or false.
    If f'(a)=0, then f has a local maximum or a local minimum on a.
    Think about f(x)=x^3 on [0,1].
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  3. #3
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    Re: Help with Calculus (undergraduate mathematics)

    That is exactly what I did in the first place, but the teacher wrote the following:

    You have to place the function on an interval of the type [0,b], in your example, there is a local minimum on 0 in the interval [0,b]
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  4. #4
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    Re: Help with Calculus (undergraduate mathematics)

    Quote Originally Posted by yannt View Post
    That is exactly what I did in the first place, but the teacher wrote the following: You have to place the function on an interval of the type [0,b], in your example, there is a local minimum on 0 in the interval [0,b]
    The function f(x) = \left\{ {\begin{array}{rl} {{x^2}\sin \left( {{x^{ - 2}}} \right),}&{0 < x \leqslant b} \\   {0,}&{x = 0}\end{array}} \right. has that property.
    f(0)=0~\&~f'(0)=0 but f(0) is neither a min nor a max in any neighborhood of 0.
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  5. #5
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    Re: Help with Calculus (undergraduate mathematics)

    Ok, I can see it now. Thanks!!
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  6. #6
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    Re: Help with Calculus (undergraduate mathematics)

    need some help in this please!!
    is given the function {y=sin2x/x for x different from 0} or {y=2a for x=o}. find "a" if this function is continuous everywhere.
    please help
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