设f(x)=x^3-3/2(a+1)x^2+3ax+1,若函数f(x)在区间(1,4)内单调递减,求a的取值范围?
问题描述:
设f(x)=x^3-3/2(a+1)x^2+3ax+1,若函数f(x)在区间(1,4)内单调递减,求a的取值范围?
答
f '(x)=3x²-3(a+1)x+3a,
因为函数f(x)在区间(1,4)内单调递减,
所以,f '(x)即当x=2时,f '(2)=3×2²-6(a+1)+3a算得:a>2