Question

已知函数f(X)=-(1/3)x^3+2ax^2-3(a^2)x(a不等于0）当a>0时求函数y=f(x)的单调区间和极

Answer 1

f(x)=-(1/3)x^3+2ax^2-3(a^2)x

f'(x)=-x^2+4ax-3a^2=-(x-3a)(x-a)=0

x=3a或x=a

∵a>0

∴3a>a

单调增区间(-∞,a],[3a,+∞)

单调减区间[a,3a]

极大值f(a)=-4a^3/3

极小值f(3a)=0