设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0
问题描述：
设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0
答
(f(x))'=(-1/3x^3+2ax^2-3a^2x+b)'
=-x^2+4ax-3a^2
令(f(x))'=0
得x=3a,或x=a
所以当x=3a,x=a时,f(x)可取得极值
又题意可知定义域是R
a>3a
当x