(2014•呼伦贝尔一模)若函数f(x)=13x3-12ax2+(a-1)x+1在区间(1,4)内为减函数,在区间(6,+∞)为增函数,则实数a的取值范围是( ) A.(-∞,2] B.[5,7] C.[4,6] D.(-∞,5]∪[7,
问题描述:
(2014•呼伦贝尔一模)若函数f(x)=
x3-1 3
ax2+(a-1)x+1在区间(1,4)内为减函数,在区间(6,+∞)为增函数,则实数a的取值范围是( )1 2
A. (-∞,2]
B. [5,7]
C. [4,6]
D. (-∞,5]∪[7,+∞)
答
由函数f(x)=13x3−12ax2+(a−1)x+1,得f′(x)=x2-ax+a-1.令f′(x)=0,解得x=1或x=a-1.当a-1≤1,即a≤2时,f′(x)在(1,+∞)上大于0,函数f(x)在(1,+∞)上为增函数,不合题意;当a-1>1,即a>2时,...