absolute minimum math

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x^2/x-3 on the interval from 3 until infinity. How would you do absolute minimum on this?

  • absolute minimum math -

    y = x^2/(x-3)

    dy/dx = ( (x-3)(2x) - x^2) /(x-3)^2
    = (2x^2 - 6x - x^2)/(x+3)^2
    = (x^2 - 6x)/(x-3)^2

    for a max/min, dy/dx = 0
    x^2 - 6x = 0
    x(x-6) = 0
    x = 0 or x = 6
    so in the interval given, we need x = 6


    if x = 6
    y = 36/3 = 12
    The minimum is 12

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