设函数f(x)=ax^3+3/2(2a-1)x^2-6x (a∈R),若函数f(x)在区间（-∞,-3）是增函数,求实数a的取值范围?
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f(x)=ax^3+(3/2)(2a-1)x^2-6x
则,f'(x)=3ax^2+3(2a-1)x-6=3[ax^2+(2a-1)x-2]=3(ax-1)(x+2)
(i)当a=0时,f'(x)=-3x-6
则,x=-2时,f'(x)=0
当x＞-2时,f'(x)＜0,f(x)递减；当x＜-2时,f'(x)＞0,f(x)递增.
此时无法满足条件
(ii)若a＞0,则f'(x)=3(ax-1)(x+2)＞0时有：x＞1/a,或者x＜-2
即,f(x)在x＜-2时递增
则满足在区间(-∞,-3)上递增
若a＜0,则f'(x)=3(ax-1)(x+2)＞0时,===> (-ax+1)(x+2)＜0
则无论是在区间(1/a,-2),还是区间(-2,1/a)内,f(x)递增
此时都无法满足在区间(-∞,-3)上递增
综上：a＞0