求函数y=(1/3)x^3 - x^2 + 2 的单调区间和极值

Question

求函数y=(1/3)x^3 - x^2 + 2 的单调区间和极值，这到题的解答过程

Answer 1

y=(1/3)x³ - x² + 2
y ′ = x² -2x = x(x-2)
x＜0或x＞2时单调增
0＜x＜2时单调减
极大值f(0)=0-0+2=2
极小值f(2)=1/3*8-4+2 = 2/3

Answer 2

y=(1/3)x³ - x² + 2
求导得y ′ = x² -2x = x(x-2)
单调增区间{x|x＜0或x＞2}
单调减区间{x|0＜x＜2}
极大值f(0)=0-0+2=2
极小值f(2)=1/3*8-4+2 = 2/3

Answer 3

y=x^(1/3)*(1-x)^(2/3)
=[x(1-x)^2]^(1/3)
=[x^3-2x^2+x]^(1/3)
y'=(1/3)(3x^2-4x+1)*(x^3-2x^2+x)^(-2/3)
=(1/3)(3x^2-4x+1)*[(x^3-2x^2+x)^2]^(-1/3)
[(x^3-2x^2+x)^2]^(-1/3)>=0
3x^2-4x+1=(3x-1)(x-1)
x1 y'>0 单调递增
1/3