【答案】
分析:(1)求出函數(shù)f(x)的導(dǎo)數(shù),因為e
=1,所以根據(jù)參數(shù)t是否大于1來討論函數(shù)f(x)在區(qū)間[0,+∞)上的單調(diào)性,從而得到函數(shù)f(x)在區(qū)間[0,+∞)上的最小值;
(2)求函數(shù)g(x)的導(dǎo)數(shù)g'(x),得出函數(shù)g'(x)在區(qū)間[0,+∞)上是增函數(shù),從而g'(x)≥g'(0)=2>0,根據(jù)g'(x)恒正得出函數(shù)g(x)在區(qū)間[0,+∞)上為增函數(shù),從而g(x)≥g(0)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/0.png)
;
(3)構(gòu)造函數(shù)h(x)=f(x)-g(x),再將所得函數(shù)h(x)進行配方,得到恒比h(x)小的一個函數(shù),再通過討論這個函數(shù)的最小值為非負,從而得出h(x)≥0,命題得理證.
解答:解:(1)f'(x)=2e
2x-2t=2(e
2x-t)
①若t≤1
∵x≥0,則e
2x≥1,∴e
2x-t≥0,即f'(x)≥0.
∴f(x)在區(qū)間[0,+∞)是增函數(shù),故f(x)在區(qū)間[0,+∞)的最小值是f(0)=1
②若t>1
令f'(x)=0,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/1.png)
.
又當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/2.png)
時,f'(x)<0;當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/3.png)
時,f'(x)>0,
∴f(x)在區(qū)間[0,+∞)的最小值是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/4.png)
(2)證明:當t=1時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/5.png)
,則g'(x)=-2x+2e
x=2(e
x-x),
∴[g'(x)]'=2(e
x-1),
當x∈[0,+∞)時,有[g'(x)]'≥0,∴g'(x)在[0,+∞)內(nèi)是增函數(shù),
∴g'(x)≥g'(0)=2>0,
∴g(x)在[0,+∞)內(nèi)是增函數(shù),
∴對于任意的x∈[0,+∞),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/6.png)
恒成立
(3)證明:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/8.png)
,
令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/9.png)
則當t∈R時,h(t)≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/11.png)
,
令F(x)=e
x-x,則F'(x)=e
x-1,
當x=0時,F(xiàn)'(x)=0;當x>0時,F(xiàn)'(x)>0;當x<0時,F(xiàn)'(x)<0,
則F(x)=e
x-x在(-∞,0]是減函數(shù),在(0,+∞)是增函數(shù),
∴F(x)=e
x-x≥F(0)=1,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181309968460885/SYS201310241813099684608025_DA/12.png)
,
∴h(t)≥0,即不等式f(x)≥g(x)對于任意的x∈R恒成立
點評:利用導(dǎo)數(shù)工具討論函數(shù)的單調(diào)性,是求函數(shù)的值域和最值的常用方法,考查了分類討論的思想與轉(zhuǎn)化的思想.解決本題同時應(yīng)注意研究導(dǎo)函數(shù)的單調(diào)性得出導(dǎo)數(shù)的正負,從而得出原函數(shù)的單調(diào)性的技巧.