【答案】
分析:(Ⅰ)求出導函數(shù)的根,判斷導函數(shù)左右兩邊的符號,得函數(shù)的單調性,據(jù)極值的定義求出極值.
(Ⅱ)求出導函數(shù)的根,討論根在不在定義域內;若根在定義域內,討論兩根的大小;判斷根左右兩邊導函數(shù)的符號,據(jù)單調性與導函數(shù)的關系求出單調性.
解答:解:(Ⅰ)函數(shù)f(x)=x
2+x-lnx,則f′(x)=2x+1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/0.png)
,
令f′(x)=0,得x=-1(舍去),x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/1.png)
.
當0<x<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/2.png)
時,f′(x)<0,函數(shù)單調遞減;
當x>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/3.png)
時,f′(x)>0,函數(shù)單調遞增;
∴f(x)在x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/4.png)
處取得極小值
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/5.png)
+ln2.
(Ⅱ)由于a+b=-2,則a=-2-b,從而f(x)=x
2-(2+b)x+blnx,
則f′(x)=2x-(2+b)+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/7.png)
令f′(x),得x
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/8.png)
,x
2=1.
1、當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/9.png)
≤0,即b<0時,函數(shù)f(x)的單調遞減區(qū)間為(0,1),
單調遞增區(qū)間為(1,+∞);
2、當0<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/10.png)
<1,即0<b<2時,列表如下:
所以,函數(shù)f(x)的單調遞增區(qū)間為(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/11.png)
),(1,+∞),
單調遞減區(qū)間為(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/12.png)
,1);
3、當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/13.png)
=1,即b=2時,函數(shù)f(x)的單調遞增區(qū)間為(0,+∞);
4、當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/14.png)
>1,即b>2時,列表如下:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/images15.png)
所以函數(shù)f(x)的單調遞增區(qū)間為(0,1),(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/15.png)
,+∞),
單調遞減區(qū)間為(1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/16.png)
);
綜上:當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/17.png)
≤0,即b<0時,
函數(shù)f(x)的單調遞減區(qū)間為(0,1),
單調遞增區(qū)間為(1,+∞);
當0<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/18.png)
<1,即0<b<2時,
函數(shù)f(x)的單調遞增區(qū)間為(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/19.png)
),(1,+∞),
單調遞減區(qū)間為(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/20.png)
,1);
當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/21.png)
=1,即b=2時,函數(shù)f(x)的單調遞增區(qū)間為(0,+∞);
當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/22.png)
>1,即b>2時,函數(shù)f(x)的單調遞增區(qū)間為(0,1),(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/23.png)
+∞),
單調遞減區(qū)間為(1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184917694/SYS201310241821531849176018_DA/24.png)
).
點評:本題考查利用導數(shù)研究函數(shù)的性質:求極值,求單調區(qū)間.考查分類討論時注意分類的起點.