設(shè)函數(shù)f(x)=(ax-2)ex,a∈R,(e為自然對數(shù)的底數(shù)).
(Ⅰ)若x=1是函數(shù)f(x)的一個極值點,求a的值;
(Ⅱ)討論函數(shù)f(x)的單調(diào)性;
(Ⅲ)若a=1,t1,t2∈[0,1]時,證明:f(t1)-f(t2)≤e-2.
【答案】
分析:(I)先求出函數(shù)f(x)的導(dǎo)函數(shù),然后根據(jù)在極值點處的導(dǎo)數(shù)等于0,建立等式關(guān)系,求出a即可;
(II)分別討論a與0的大小,根據(jù)導(dǎo)函數(shù)的符號進行判斷函數(shù)f(x)的單調(diào)性,使f'(x)>0成立的是單調(diào)增區(qū)間,使f'(x)<0成立的是單調(diào)減區(qū)間;
(III)a=1,當(dāng)x∈[0,1]時,f'(x)=(x-1)e
x≤0,則f(x)單調(diào)減函數(shù),從而f(t
1)-f(t
2)≤f
max(x)-f
min(x)=f(0)-f(1)=e-2,得到結(jié)論.
解答:解:(Ⅰ)由已知f'(x)=(ax+a-2)e
x,f'(1)=0,∴a=1.
(Ⅱ)①當(dāng)a=0時,f'(x)<0,∴f(x)在R上是減函數(shù).
②當(dāng)a>0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181359467071627/SYS201310241813594670716017_DA/0.png)
時,f'(x)>0;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181359467071627/SYS201310241813594670716017_DA/1.png)
時,f'(x)<0,
∴f(x)的單調(diào)增、減區(qū)間分別是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181359467071627/SYS201310241813594670716017_DA/2.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181359467071627/SYS201310241813594670716017_DA/3.png)
.
③當(dāng)a<0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181359467071627/SYS201310241813594670716017_DA/4.png)
時,f'(x)<0;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181359467071627/SYS201310241813594670716017_DA/5.png)
時,f'(x)>0,
∴f(x)的單調(diào)減、增區(qū)間分別是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181359467071627/SYS201310241813594670716017_DA/6.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181359467071627/SYS201310241813594670716017_DA/7.png)
.
(Ⅲ)∵a=1,當(dāng)x∈[0,1]時,f'(x)=(x-1)e
x≤0,
∴f(x)單調(diào)減函數(shù),
∴f(t
1)-f(t
2)≤f
max(x)-f
min(x)=f(0)-f(1)=e-2.
點評:本題綜合考查函數(shù)的極值以及利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,同時考查函數(shù)的最值的求解,是一道綜合題.