【答案】
分析:(1)當(dāng)a=-2時(shí),求函數(shù)f(x)=x
3+3ax-1的導(dǎo)函數(shù)為f
′(x),令f
′(x)>0,求出單調(diào)增區(qū)間;令f
′(x)<0求出單調(diào)減區(qū)間;
(2)若對(duì)滿足-1≤a≤1的一切a的值,都有g(shù)(x)<0,變更主元,轉(zhuǎn)化為關(guān)于a的一次函數(shù),求出實(shí)數(shù)x的取值范圍;
(3)依題意,x•g
′(x)+lnx>0對(duì)一切x≥2恒成立,采取分離參數(shù)的方法,轉(zhuǎn)化為求函數(shù)的最值問題.
解答:解:(1)當(dāng)a=-2時(shí),f′(x)=3x
2-6.令f′(x)=0得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/0.png)
,
故當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/1.png)
或x>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/2.png)
時(shí)f′(x)>0,f′(x)單調(diào)遞增;
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/3.png)
時(shí)f
′(x)<0,f(x)單調(diào)遞減.
所以函數(shù)f′(x)的單調(diào)遞增區(qū)間為(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/4.png)
,[
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/5.png)
);單調(diào)遞減區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/6.png)
;
(2)因f′(x)=3a
2+3a,故g(x)=3x
2-ax+3a-3.
令g(x)=h(a)=a(3-x)+3x
2-3,要使h(a)<0對(duì)滿足-1≤a≤1的一切a成立,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/7.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/8.png)
;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/9.png)
.
(3)因?yàn)間(x
′)=6x-a,
所以X(6x-a)+lnx>0
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/10.png)
對(duì)一切x≥2恒成立.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/11.png)
,
令6x
2+1-lnx=φ(x),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/12.png)
.
因?yàn)閤≥2,所以φ
′(x)>0,
故φ(x)在[2,+∞)單調(diào)遞增,有φ(x)≥φ(2)=25-ln2>0.
因此h
′(x)>0,從而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/13.png)
.
所以a
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058821344/SYS201310232135180588213019_DA/14.png)
.
點(diǎn)評(píng):考查利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性和最值問題,特別是恒成立問題,(2)若對(duì)滿足-1≤a≤1的一切a的值,都有g(shù)(x)<0,變更主元,轉(zhuǎn)化為關(guān)于a的一次函數(shù),求出實(shí)數(shù)x的取值范圍;(3)x•g
′(x)+lnx>0對(duì)一切x≥2恒成立,采取分離參數(shù)的方法,轉(zhuǎn)化為求函數(shù)的最值問題體現(xiàn)了轉(zhuǎn)化的思想方法,屬難題.