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Thread: Absolute Extrema Problem Stuck again

  1. #1
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    Absolute Extrema Problem Stuck again

    I though i had a grasp of these but this one is a little different from the other ones.

    Find the locations (ordered pairs) of all absolute extrema (min max) for the following function with the specified domain.

    $\displaystyle F(x)=\frac{x}{x^2+2}$ Domain [0,4]

    Heres what ive done so far:
    $\displaystyle f'(x)=\frac{(x^2+2)(1)-(x)(2x)}{(x^2+2)^2}$
    $\displaystyle F'(x)=\frac{(x^2+2)+(2x^2)}{(x^2+2)^2}$
    $\displaystyle F'(x)=\frac{3x^2+2}{(x^2+2)^2}$

    The bottom can never equal 0 so how do i test to find the absolute max and absolute min? Thanks for any help!
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  2. #2
    Senior Member Pinkk's Avatar
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    You suddenly switched signs in the numerator. This is what you should have:

    $\displaystyle f'(x)=\frac{2-x^{2}}{(x^{2}+2)^{2}}$

    $\displaystyle f'(x)=0$ when $\displaystyle 2-x^{2}=0$ and that happens when $\displaystyle x=\pm \sqrt{2}$. However, $\displaystyle x=-\sqrt{2}$ is not in the interval $\displaystyle [0,4]$, so the only critical point to be concerned with is $\displaystyle x=\sqrt{2}$. Evaluate $\displaystyle f(0),f(\sqrt{2}),f(4)$ and determine which are your max and min.
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  3. #3
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    Quote Originally Posted by Jim Marnell View Post
    I though i had a grasp of these but this one is a little different from the other ones.

    Find the locations (ordered pairs) of all absolute extrema (min max) for the following function with the specified domain.

    $\displaystyle F(x)=\frac{x}{x^2+2}$ Domain [0,4]

    Heres what ive done so far:
    $\displaystyle f'(x)=\frac{(x^2+2)(1)-(x)(2x)}{(x^2+2)^2}$
    $\displaystyle F'(x)=\frac{(x^2+2)+(2x^2)}{(x^2+2)^2}$
    What happened to the "-"?
    That should be
    $\displaystyle f'(x)= \frac{(x^2+ 2)- (2x^2)}{x^2+ 2)^2}$

    $\displaystyle F'(x)=\frac{3x^2+2}{(x^2+2)^2}$

    The bottom can never equal 0 so how do i test to find the absolute max and absolute min? Thanks for any help!
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  4. #4
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    f(0)=0
    f($\displaystyle \sqrt{2}$=0
    f(4)=$\displaystyle \frac{-7}{162}$

    Did i do this wrong, cause those numbers seem weird. If not would max be 0,0 and min be $\displaystyle (4,\frac{-7}{162})$
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  5. #5
    Senior Member Pinkk's Avatar
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    You have to plug those values into $\displaystyle f(x)$, not $\displaystyle f'(x)$

    $\displaystyle f(0)=0$
    $\displaystyle f(\sqrt{2})=\frac{\sqrt{2}}{4}$
    $\displaystyle f(4)=\frac{2}{9}$

    Your maximum is at $\displaystyle (\sqrt{2},\frac{\sqrt{2}}{4})$ and your minimum is at $\displaystyle (0,0)$.
    Last edited by Pinkk; Mar 23rd 2009 at 11:28 AM.
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