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Math Help - Finding Relative max...

  1. #1
    Junior Member
    Joined
    Dec 2009
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    Finding Relative max...

    How do I find the relative max on the interval 0<(or equal) X <(or equal) 3
    when given the first derivative as:

    f ' (x) = (e^(-x/4))(sinx^2)

    where x is greater than or equal to zero
    & f(5)=0



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  2. #2
    Senior Member
    Joined
    Dec 2008
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    Because f is differentiable everywhere and the domain (0,3) contains no boundary points, relative maxima will only occur where

    f'(x)=e^{-\frac{x}{4}}\sin x^2=0.

    By a theorem in algebra, this occurs only where

    e^{-\frac{x}{4}}=0\quad\mbox{or}\quad \sin x^2=0.

    Because e^{-\frac{x}{4}} is always positive, we are left with the cases in which

    \sin x^2=0\quad\quad\quad 0 < x < 3.
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