已知函數(shù)f(x)=alnx+x2(a為實常數(shù)).
(Ⅰ)若a=-2,求證:函數(shù)f(x)在(1,+∞)上是增函數(shù);
(Ⅱ)求函數(shù)f(x)在[1,e]上的最小值及相應的x值.
【答案】
分析:(Ⅰ)將a=-2代入,然后求出導函數(shù)f'(x),欲證函數(shù)f(x)在(1,+∞)上是增函數(shù)只需證導函數(shù)在(1,+∞)上恒大于零即可;
(Ⅱ)先求出導函數(shù)f'(x),然后討論a研究函數(shù)在[1,e]上的單調性,將f(x)的各極值與其端點的函數(shù)值比較,其中最小的一個就是最小值.
解答:解:(Ⅰ)當a=-2時,f(x)=x
2-2lnx,當x∈(1,+∞),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/0.png)
,
故函數(shù)f(x)在(1,+∞)上是增函數(shù).
(Ⅱ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/1.png)
,當x∈[1,e],2x
2+a∈[a+2,a+2e
2].
若a≥-2,f'(x)在[1,e]上非負(僅當a=-2,x=1時,f'(x)=0),
故函數(shù)f(x)在[1,e]上是增函數(shù),此時[f(x)]
min=f(1)=1.
若-2e
2<a<-2,當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/2.png)
時,f'(x)=0;當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/3.png)
時,f'(x)<0,
此時f(x)是減函數(shù);當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/4.png)
時,f'(x)>0,此時f(x)是增函數(shù).
故[f(x)]
min=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/6.png)
若a≤-2e
2,f'(x)在[1,e]上非正(僅當a=-2e
2,x=e時,f'(x)=0),
故函數(shù)f(x)在[1,e]上是減函數(shù),此時[f(x)]
min=f(e)=a+e
2.
綜上可知,當a≥-2時,f(x)的最小值為1,相應的x值為1;
當-2e
2<a<-2時,f(x)的最小值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/7.png)
,相應的x值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816276068/SYS201310251245028162760021_DA/8.png)
;
當a≤-2e
2時,f(x)的最小值為a+e
2,相應的x值為e
點評:本題主要考查了利用導數(shù)研究函數(shù)的單調性,以及利用導數(shù)求閉區(qū)間上函數(shù)的最值,屬于中檔題.