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Math Help - Finding Critical Numbers

  1. #1
    Junior Member
    Joined
    Oct 2009
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    72

    Finding Critical Numbers

    Find the critical numbers of the function and describe the behavior of f at these numbers.

    f(x) = x^6(x - 2)^5

    I took the derivative using product rule

    f'(x) = 6x^5 (x-2)^5 + x^6 5(x-2)^4

    Then i simplified and set f'(x) = 0

    f'(x) = x^5(x-2)^4 [6(x-2) + 5x] = 0
    f'(x) = x^5(x-2)^4 [6x-12 + 5x] = 0
    f'(x) = x^5(x-2)^4 (x-12) = 0

    and the three points i got were

    x=0 x=2 and x=12

    i know 0 is right, but 2 and 12 is wrong.


    what am i doing wrong?
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  2. #2
    Newbie
    Joined
    Jul 2009
    Posts
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    When finding f'(x)=0,
    6x - 12 + 5x is actually 11x - 12. This root gives you x = 12/11.

    The critical points are x = {0, 12/11, 2}.
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