求函数y=(x-1)*x^(2/3)的极值

题目

求函数y=(x-1)*x^(2/3)的极值

答案

其导数为y'=x^(2/3)+(2/3)·x^(-1/3)(x-1)=(5/3)·x^(2/3) - (2/3)·x^(-1/3)
令y'=0,则(5/3)·x^(2/3) - (2/3)·x^(-1/3)=0
5·x^(2/3) - 2·x^(-1/3)=0
两边乘x^(1/3)得
5x - 2 =0
x= 2/5
y''=(10/9)·x^(-1/3) + (2/9)·x^(-4/3)
则y''(2/5)恒>0.
说明y(2/5)是极小值,为 (-3/5)·(2/5)^(2/3)= -3·2^(2/3) /5^(5/3)
当x=0时,y''=(10/9)·x^(-1/3) + (2/9)·x^(-4/3)=0,是拐点,不是极值点