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March 28, 2017

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Im finding local Min and max of x^2/x-2

I know I take the derivative of the first one but how do i derive this one. And the second one especially don't know how to do.

  • Calculus - ,

    Use the quotient rule to differentiate (not derive):

    if y = f/g
    y' = (f'g - fg')/g^2

    so, we have

    y = x^2/(x-2)
    y' = [(2x)(x-2) - x^2(1)]/(x-2)^2
    = x(x-4)/(x-2)^2

    Or, you can divide first
    y = x+2 + 4/(x-2)
    y' = 1 - 4/(x-2)^2

    anyway, min/max is where y'=0, so x=0 or 4. (This is easiest to see looking at the first form of the solution.)

    which is min, which is max?
    y'' = 8/(x-2)^3
    y''(0) = -1, so y is concave down (max)
    y''(4) = +1, so y is concave up (min)

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