已知函數(shù)f(x)=ln(3-x)+ax+1.
(1)若函數(shù)f(x)在[0,2]上是單調(diào)遞增函數(shù),求實數(shù)a的取值范圍;
(2)求函數(shù)f(x)在[0,2]上的最大值.
【答案】
分析:(1)對函數(shù)進行求導,根據(jù)導函數(shù)大于等于0在[0,2]上恒成立可得答案.
(2)先得出當x∈[0,2]時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/0.png)
∈[-1,-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/1.png)
]下面對a進行分類討論:①當a≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/2.png)
時,②當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/3.png)
<a<1時,③當a≥1時,分別求得函數(shù)f(x)在[0,2]上的最大值,最后在總結(jié)即可.
解答:解:f′(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/4.png)
+a
(1)只要在x∈[0,2]上f'(x)≥0恒成立,?a≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/5.png)
而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/6.png)
∈[
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/7.png)
,1],∴a≥1 (5分)
(2)∵當x∈[0,2]時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/8.png)
∈[-1,-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/9.png)
]
∴①當a≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/10.png)
時,f′(x)≤0,這時f(x)在[0,2]上單調(diào)遞減,
f(x)≤f(0)=1+ln3(7分)
②當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/11.png)
<a<1時,令f′(x)=0,可解得x=3-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/12.png)
,
∵當x∈[0,3-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/13.png)
]時,有f′(x)>0
當x∈[3-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/14.png)
,2]時,有f′(x)<0,
∴x=3-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/15.png)
是f(x)在[0,2]上的唯一的極大值,
則f(x)≤f(3-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/16.png)
)=3a-lna (10分)
③當a≥1時,f'(x)≥0,這時f(x)在[0,2]上單調(diào)遞增,
f(x)≤f(2)=2a+1 (12分)
綜上所述:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/17.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125340987873877/SYS201310251253409878738019_DA/18.png)
(13分)
點評:本題主要考查利用導數(shù)求函數(shù)的單調(diào)性,考查分離參數(shù)法求恒成立問題.本題考查了函數(shù)單調(diào)性和導數(shù)的關(guān)系以及利用導數(shù)求出最值,第(2)要注意分情況求最值,屬于中檔題.