【答案】
分析:(1)對函數(shù)f(x)進行求導然后整理成f′(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/0.png)
e
-x(-ax
2+2ax-a-1)的形式,因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/1.png)
e
-x>0,根據(jù)導函數(shù)大于0原函數(shù)單調(diào)遞增,導函數(shù)小于0原函數(shù)單調(diào)遞減通過討論函數(shù)g(x)=-ax
2+2ax-a-1值的情況來確定原函數(shù)的單調(diào)性.
(2)先根據(jù)a的范圍確定導函數(shù)等于0的兩根的范圍,進而可判斷函數(shù)在區(qū)間[1,2]上的單調(diào)性,最后可得到最小值.
解答:解:(1)由已知f′(x)=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/2.png)
e
-x(ax
2+a+1)+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/3.png)
e
-x•2ax
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/4.png)
e
-x(-ax
2+2ax-a-1).
因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/5.png)
e
-x>0,以下討論函數(shù)g(x)=-ax
2+2ax-a-1值的情況:
當a=0時,g(x)=-1<0,即f′(x)<0,所以f(x)在R上是減函數(shù).
當a>0時,g(x)=0的判別式△=4a
2-4(a
2+a)=-4a<0,所以g(x)<0,
即f′(x)<0,所以f(x)在R上是減函數(shù).
當a<0時,g(x)=0有兩個根x
1,2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/6.png)
,并且
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/7.png)
<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/8.png)
,
所以在區(qū)間(-∞,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/9.png)
)上,g(x)>0,即f'(x)>0,f(x)在此區(qū)間上是增函數(shù);
在區(qū)間(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/11.png)
)上,g(x)<0,即f′(x)<0,f(x)在此區(qū)間上是減函數(shù).
在區(qū)間(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/12.png)
,+∞)上,g(x)>0,即f′(x)>0,f(x)在此區(qū)間上是增函數(shù).
綜上,當a≥0時,f(x)在R上是減函數(shù);
當a<0時,f(x)在(-∞,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/13.png)
)上單調(diào)遞增,在(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/15.png)
)上單調(diào)遞減,
在(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/16.png)
,+∞)上單調(diào)遞增.
(2)當-1<a<0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/17.png)
=1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/18.png)
<1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/19.png)
=1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/20.png)
>2,
所以在區(qū)間[1,2]上,函數(shù)f(x)單調(diào)遞減.
所以函數(shù)f(x)在區(qū)間[1,2]上的最小值為f(2)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213518058818914/SYS201310232135180588189017_DA/21.png)
.
點評:本題主要考查函數(shù)的單調(diào)性與其導函數(shù)的正負之間的關(guān)系、函數(shù)的最值問題.函數(shù)的最值和函數(shù)的單調(diào)性有緊密聯(lián)系.判斷較復雜函數(shù)的單調(diào)性,利用導函數(shù)的符號是基本方法.