(1)运用一阶导数
f‘(x)=3x^2-6x-9
=3(x^2-2x-3)
=3(x-3)(x+1)
令f’(x)=0则x=3,x=-1
由以上可知:-1<x<3时,f'(x)<0,故f(x)在(-1,3)单调递减;当x<=-1,x>=3时,f'(x)>=0,故f(x)在(-无穷,-1],[3,+无穷)单调递增
(2)由于:(1)f'(-1)=0,f'(3)=0
(2)x<-1时,f'(x)>0,-1<x<3时,f'(x)<0;-1<x<3时,f'(x)<0,x>3时,f'(x)>0
故x=-1为f(x)的极大值点,x=3为f(x)的极小值点且f(-1)=5,f(3)=-27
温馨提示:答案为网友推荐,仅供参考