设f(x)=x^3-3/2(a+1)X^2+3ax+1,(2)若函数f(x)在区间(1,4)内单调递减,求a的取值范

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设f(x)=x^3-3/2(a+1)X^2+3ax+1,(2)若函数f(x)在区间(1,4)内单调递减,求a的取值范

设f(x)=x^3-3/2(a+1)X^2+3ax+1,(2)若函数f(x)在区间(1,4)内单调递减,求a的取值范
设f(x)=x^3-3/2(a+1)X^2+3ax+1,(2)若函数f(x)在区间(1,4)内单调递减,求a的取值范

设f(x)=x^3-3/2(a+1)X^2+3ax+1,(2)若函数f(x)在区间(1,4)内单调递减,求a的取值范
f'(x)=3x^2-3(a+1)x+3a=3[x^2-(a+1)x+a]=3(x-a)(x-1)
令f'(x)=4

求导f'(x)=3[x^2-(a+1)x+a]=3(x-a)(x-1).这时a,1为极值点,由三次函数图像性质知,当a<1时,在区间(1,4)上是单调增加的,当a=1时,在R上是单增的,所以只有当a>1时是单调递减的。