y=x^3/(x-1)^2 求增减区间极值

1个回答

  • y = x^3/(x-1)^2

    分母不为零,定义域x≠1

    y' = { 3x^2*(x-1)^2 - x^3*2(x-1) } / (x-1)^4 = x^2(x-1)(x-3)/(x-1)^4 = x^2(x-3)/(x-1)^3

    x<1,或x>3时,y'>0,单调增;1<x<3时,y'<0,单调减

    单调增区间(-∞,1);(3,+∞)

    单调减区间:(1,3)

    x≠1,所以极大值不存在;

    x=3时,极小值=3^3/(3-1)^2=27/4

    y' = (x^3-3x^2)/(x-1)^3

    y'' = {(x-1)^3*(3x^2-6x) - (x^3-3x^2)*3(x-1)^2 } /(x-1)^6 = 6x/(x-1)^4

    凸区间 (-∞,0)

    凹区间(0,1),(1,+∞)

    拐点 x=0