[例] (2018·高考全國卷Ⅰ)已知函數(shù)ƒ(x)=aex-ln x-1.
(1)設(shè)x=2是ƒ(x)的極值點(diǎn),求a,并求ƒ(x)的單調(diào)區(qū)間;
(2)證明:當(dāng)a≥時(shí),ƒ(x)≥0.
[解析] (1)ƒ(x)的定義域?yàn)?/span>(0,+∞),ƒ′(x)=aex-.
由題設(shè)知,ƒ′(2)=0,所以a=.
從而ƒ(x)=ex-ln x-1,ƒ′(x)=ex-.當(dāng)0<x<2時(shí),ƒ′(x)<0;當(dāng)x>2時(shí),ƒ′(x)>0.
所以ƒ(x)在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增.
(2)證明:當(dāng)a≥時(shí),ƒ(x)≥-ln x-1.設(shè)g(x)=-ln x-1,則g′(x)=-.當(dāng)0<x<1時(shí),g′(x)<0;當(dāng)x>1時(shí),g′(x)>0.
所以x=1是g(x)的最小值點(diǎn).
故當(dāng)x>0時(shí),g(x)≥g(1)=0.
因此,當(dāng)a≥時(shí),ƒ(x)≥0.
[破題技法] 利用導(dǎo)數(shù)證明不等式f(x)>g(x)的基本方法
(1)若f(x)與g(x)的最值易求出,可直接轉(zhuǎn)化為證明f(x)min>g(x)max;
(2)若f(x)與g(x)的最值不易求出,可構(gòu)造函數(shù)h(x)=f(x)-g(x),然后根據(jù)函數(shù)h(x)的單調(diào)性或最值,證明h(x)>0.