对f（x）求导,并提取公因子,写成连乘的形式,可得
f‘（x）=(2/3)*x^(-1/3)*(x-2)^2+2*x^(2/3)*(x-2)
=2(x-2)*x^(1/3)*[(1/3)(x-2)+x]
倒数为0的点即为驻点
x-2=0,即x=2
x^(1/3)=0,即x=0
(1/3)(x-2)+x=0,即x=1/2
对f（x）求导,并提取公因子,写成连乘的形式,可得
f‘（x）=(2/3)*x^(-1/3)*(x-2)^2+2*x^(2/3)*(x-2)
=2(x-2)*x^(1/3)*[(1/3)(x-2)+x]
倒数为0的点即为驻点
x-2=0,即x=2
x^(1/3)=0,即x=0
(1/3)(x-2)+x=0,即x=1/2