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Find the critical values

  1. Oct 20, 2009 #1
    F(x) = x^5/7(x-5)^2

    so you have to find the derivative and you end up with

    f'(x) = x^5/7(2(x-5)) + 5/7x^-2/7(x-5)^2

    simplified it to f'(x) = (19x^2-120x+125)/7x^2/7

    so you use quadratic to find the zeroes of the numerator

    which is 120 +/- sqrt(120^2-4(19)(125))/2(19)

    simplifies out to (120 +/- 70)/38

    so the zeros are 95/19 and 35/19

    ive worked out the problem a couple times and came up with the same wrong answer.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 20, 2009 #2


    Staff: Mentor

    Your function is written correctly, but would be better written as (1/7)x^5/(x - 5)^2. Also, you started with F as the name of the function, so should keep the same letter to denote the derivative (i.e., not f').
    Mistake in the line above. It should be
    F(x) = x^5/7(2(x-5)) + 5/7x^4/7(x-5)^2
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