Question

设函数f（x）=-1\3x3+x2+（m2-1）x x属于R 求函数单调区间与极值

Answer 1

f'(x)=-x²+2x+m²-1=-(x-1)²+m²=-(x-1+m)(x-1-m)

极值点为x=1-m,1+m

讨论m:

1)若m=0,则f'(x)=-(x-1)²>=0,函数在R上单调减,没有极值；

2)若m>0,则单调增区间为：(1-m,1+m),

单调减区间为：(-∞,1-m)U(1+m,+∞)

极小值为f(1-m)=-(1-m)²(1+2m)/3

极大值为f(1+m)=-(1+m)²(1-2m)/3

3)若m