Question

求函数y＝（x－1）三次根号下x^2的极值

Answer 1

y=2x-√(x-1)
x-1>=0
设√(x-1)=t
x=1+t
^2
t>=0
y=2t^2+2-t
根据2次函数性质当t=1/4取得最小值
y>=2*1/4*1/4+2-1/4=15/8

Answer 2

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