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Math Help - Extrema on an interval find critical numbers

  1. #1
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    Extrema on an interval find critical numbers

    I am having some trouble with this problem:

    (x-1)/(x+3)

    After I have evaluated the problem, no critical numbers exist. But according to the multiple choice answer it states there a critical number.

    This is my understanding of critical numbers:

    if f'(c)= 0 or f'(c)= undefined and c lies in the domain of the function of f(x)

    a critical number exist.

    According to everything I done, f'(x) = -3 this number is undefined and also undefined at f(x) so the number is not a critical number.

    Could someone verify this for me before I present this to the teacher?

    Thanks!
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  2. #2
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    Alright so f(x)=(x-1)/(x+3), so to find f '(x) use the quotient rule.

    So f '(x)=[(x+3)(1)-(x-1)(1)]/(x+3)^2 = 4/(x+3)^2

    so from there you should be able to identify critical points
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  3. #3
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     f(x) = \frac{x-1}{x+3}

     f '(x) = \frac{(x+3)(1)-(1)(x-1)}{(x+3)^2} =  \frac{4}{(x+3)^2}

    Now you should be able to continue. I am not sure how you got your derivative to be -3. Because this is a fraction you should use the quotient rule.
    Last edited by Soriano; May 9th 2009 at 04:25 PM. Reason: Latex not working
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