Question

y=（2／3）x-3次根号x 求极值

Answer 1

∵y=（2／3）x-3次根号x
=2x/3-x^(1/3)
∴y'=2/3-(1/3)*x^(-2/3)
令y'=0,则2/3-(1/3)*x^(-2/3)=0
x^(-2/3)=2
x^(2/3)=1/2
x^2=1/8
x=±√2/4
∴当x