试题分析:解:(1)因
f(
x)=
a ln
x+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824005439027408.png)
+
x+1,
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824005439074985.png)
. (2分)
由于曲线y=f(x)在点(1,f(1))处的切线垂直于y轴,故该切线斜率为0,即f′(1)=0,从而a-+=0,解得a=-1. (4分)
(2)由(1)知
f(
x)=
-ln
x+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824005439027408.png)
+
x+1 (x>0),
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824005439136956.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824005439152684.png)
令f′(x)=0,解得x
1=1,x
2=-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824005439167327.png)
(因x
2=-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824005439167327.png)
不在定义域内,舍去).(6分)
当x∈(0,1)时,f′(x)<0,故f(x)在(0,1)上为减函数;
当x∈(1,+∞)时,f′(x)>0,故f(x)在(1,+∞)上为增函数.
故f(x)在x=1处取得极小值f(1)=3,无极大值. (10分)
点评:运用导数的符号判定函数的单调性,求解极值,属于基础题。