解:f'(x)=3x2+2ax+1=3(x+a/3)2+1-a2/3,对称轴x=-a/3函数在[-2/3,-1/3]上是减函数,即在[-2/3,-1/3]上f'(x)≤0(1)-a/3≤-2/3时,即a≥2时,函数单调递增,只要f'(-2/3)≤0f'(-2/3)=4/3-4a/3+1≤0解得a≥7/4 得a≥2(2)-a/3≥-1/3时,即a≤1时,函数单调递减,只要f'(-1/3)≤0f'(-1/3)=1/3-2a/3+1≤0 解得a≥2(舍去)(3)1=-1/3, x20 ==> a>√3 or a a>7/4f'(-1/3)...