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Critical numbers

  1. Nov 11, 2005 #1
    Im supposed to find the critical numbers for the function:
    F(x)= x^3-12x^2
    I think I did it right I just needed some reassurance, I got
    F'(x)= 3x^2-24
    =3x=0 and x-8=0

    so the critcal numbers are x=0 and x=8 I think:shy: I hope someone can tell me if I did this right

  2. jcsd
  3. Nov 11, 2005 #2


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    yes, correct [save typo F'(x)= 3x^2-24x = 3x(x-8)]
  4. Nov 11, 2005 #3
    Thank you I might need more help later
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