设f(x)=x3-3/2(a+1)x2+3ax+1. (Ⅰ)若函数f(x)在区间(1,4)内单调递减,求a的取值范围; (Ⅱ)若函数f(x)在x=a处取得极小值是1,求a的值,并说明在区间(1,4)内函数f(x)的单调性.

问题描述:

f(x)=x3-

3
2
(a+1)x2+3ax+1.
(Ⅰ)若函数f(x)在区间(1,4)内单调递减,求a的取值范围;
(Ⅱ)若函数f(x)在x=a处取得极小值是1,求a的值,并说明在区间(1,4)内函数f(x)的单调性.

f'(x)=3x2-3(a+1)x+3a=3(x-1)(x-a)(2分)(1)∵函数f(x)在区间(1,4)内单调递减,∴f'(4)≤0,∴a∈[4,+∞);(5分)(2)∵函数f(x)在x=a处有极值是1,∴f(a)=1,即a3-32(a+1)a2+3a2+1=-12a3...