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Math Help - 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.

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

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

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

    f'(x)=0 when 2-x^{2}=0 and that happens when x=\pm \sqrt{2}. However, x=-\sqrt{2} is not in the interval [0,4], so the only critical point to be concerned with is x=\sqrt{2}. Evaluate 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.

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

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

    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( \sqrt{2}=0
    f(4)= \frac{-7}{162}

    Did i do this wrong, cause those numbers seem weird. If not would max be 0,0 and min be (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 f(x), not f'(x)

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

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