求y=x^(1/3) * (1-x)^（2/3）的极值

网友回答

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)>=03x^2-4x+1=(3x-1)(x-1)
x1 y'>0 单调递增1/3