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Critical point

  1. Nov 20, 2003 #1
    V(x) = x(10-2x)(16-2x)

    When i V'(x) = 0... x is always a non real number... I don't know why, can someone help me thanks.
     
  2. jcsd
  3. Nov 21, 2003 #2

    ShawnD

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    V(x) = (10x-2x^2)(16-2x)
    V(x) = 160x - 20x^2 - 32x^2 + 4x^3
    V(x) = 160x - 52x^2 + 4x^3
    V'(x) = 160 - 104x + 12x^2
    0 = 160 - 104x + 12x^2 I don't feel like doing quadratic equation so I just graphed it
    x = 2, x = 6.6666666 (which is 20/3)

    How did you go about getting your answers?
     
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