设函数=-1/3x3+2ax2-3a2x+1,0
问题描述：
设函数=-1/3x3+2ax2-3a2x+1,0
答
(1):y`=-x²+4ax-3a²,令y`=0→x1=a,x2=3a→当a0,f(x)↑所以：x=a,f(a)是最大值,得：f(a)=1-4a³/3(2):-a≤f`(x)≤a→f`(x)≤|a|→f`(x)≤a→-x²+4ax-3a²-a≤0→△≤0→16a²≤4(3a²...