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(1)a=-8时,f(x)=[1/3]x3+x2-8x+1,x∈R,∴f′(x)=x2+2x-8,
令f′(x)=0,得x1=-4,x2=2,
当x<-4时,f′(x)>0;当-4<x<2时,f′(x)<0;当x>2时,f′(x)>0,
∴f(x)的单调递减区间为[-4,2],单调递增区间为(-∞,-4)和(2,+∞);
(2)(方法一)∀a∈(-24,-10),f(−4)=−
13
3−4a>−
13
3+40>0,f(4)=
115
3+4a<
115
3−40<0,
因为y=f(x)在区间[-4,4]上是连续不断的曲线,且f(-4)•f(4)<0,
所以f(x)在区间[-4,4]上有零点;
解f′(x)=x2+2x+a=0(a∈(-24,-10))得x1=−1−
1−a<−4(舍去),x2=−1+
1−a∈(−4,4),
当−4<x<−1+
1−a时,f′(x)<0;
当−1+
1−a<x<4时,f′(x)>0;
因为f(4)<0,所以∀x∈[−1+
1−a,4],f(x)<0,f(x)在区间[−1+
1−a,4]上无零点;
f(−4)•f(−1+
点评:
本题考点: 利用导数研究函数的单调性;函数的零点.
考点点评: 本题考查了利用导数研究函数的单调区间以及根据函数的增减性研究函数的零点问题,是易错题.
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