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Math Help - Extreme values

  1. #1
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    Extreme values

    Just double checking to make sure I got the correct answer because it is not the answer the book gives.

    I need to find the maximum and minimum value

    \frac{4x}{x^2+1}, -2<x<4

    so I got the derivative and solved for f'(x) = 0

    the derivative being \frac{4x^2-8x+4}{(x^2+1)^2} \Rightarrow x = 1

    solving f(x) for x = -2,1,4, I got the answers -\frac{8}{5}, 2, \frac{16}{17}

    and so the max is 2 and min is -8/5 but the book has the max at 2 and min at -2

    Is my work correct?

    thank you!
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  2. #2
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    Re: Extreme values

    No, you are attempting to use the closed interval method on an open interval. Instead, the maximum is 2, but there is no minimum value.

    On the other hand, if you change the interval to -2 \le x \le 4, then your answer is correct and the book is wrong.
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  3. #3
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    Re: Extreme values

    Sorry I did mean -2  \le x \le 4
    Last edited by Jonroberts74; October 17th 2013 at 03:22 PM.
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  4. #4
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    Re: Extreme values

    Oops. I messed up. You miscalculated the derivative.

    \dfrac{d}{dx}\dfrac{4x}{x^2+1} = \dfrac{4(x^2+1) - (4x)(2x)}{(x^2+1)^2} = \dfrac{4-4x^2}{(x^2+1)^2}

    So, x = \pm 1.

    Now, you have -\dfrac{8}{5},-2,2,\dfrac{16}{17}

    So, the book was correct after all.
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  5. #5
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    Re: Extreme values

    Ah okay
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