f'(x)=-x²+2x+m²-1=-(x-1)²+m²=-(x-1+m)(x-1-m)
極值點(diǎn)為x=1-m,1+m
討論m:
1)若m=0,則f'(x)=-(x-1)²>=0,函數(shù)在R上單調(diào)減,沒有極值；
2)若m>0,則單調(diào)增區(qū)間為：(1-m,1+m),
單調(diào)減區(qū)間為：(-∞,1-m)U(1+m,+∞)
極小值為f(1-m)=-(1-m)²(1+2m)/3
極大值為f(1+m)=-(1+m)²(1-2m)/3
3)若m