1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mark44

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook