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数学
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求y=x^(1/3) * (1-x)^(2/3)的极值
人气:147 ℃ 时间:2020-09-10 19:37:58
解答
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)>=0
3x^2-4x+1=(3x-1)(x-1)
x1 y'>0 单调递增
1/3x=0无极值吧,不过还是谢谢了,我会做了
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