已知函数f(x)= 1 3 x 3 + 1-a 2 x 2 -ax-a,x∈R,其中a>0. (1)求函数f(x)的单调区间; (2)若函数f(x)在区间(-2,0)内恰有两个零点,求a的取值范围; (3)当a=1时,设函数f(x)在区间[t,t+3]上的最大值为M(t),最小值为m(t).记g(t)=M(t)-m(t),求函数g(t)在区间[-3,-1]上的最小值.
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