求y=x^(1/3) * (1-x)^(2/3)的极值

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 23:27:00
求y=x^(1/3) * (1-x)^(2/3)的极值
x){Ҷ"NPXSAKAPB3#}{:jy6i"}ِf2\DP&X6"d Q353Ү@Ts4R&چ@J"LPd.drv`i)@u 55@XPR@i_6LzhXF 1`9

求y=x^(1/3) * (1-x)^(2/3)的极值
求y=x^(1/3) * (1-x)^(2/3)的极值

求y=x^(1/3) * (1-x)^(2/3)的极值
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