求函数y＝（x－1）三次根号下x^2的极值

网友回答

y'=[(x-1)x^2/3]'
=x^2/3+(x-1)x^-1/3
=x^2/3+x^2/3-x^-1/3
=2x^2/3-x^-1/3
=2x-1/x^-1/3
J=1/2======以下答案可供参考======
供参考答案1:
y＝(x-1)x^(2/3)
y＝x^(5/3)-x^(2/3)
y' ＝(5/3)x^(2/3)-(2/3)x^(-1/3)
令y'=0，可得 x/2=1，x=2 。
当 x=2 时，
有 y＝(2-1)×2^(2/3) ≈ 1.59