已知函数$f(x)=ae^x+dfrac{a+1}{x}-2(a+1)ge0,(a>0)$对任意的$xin(0,+infty)$恒成立,求$a$的范围.
分析:$f^{'}(x)=dfrac{g(x)}{x^2},$其中$g(x)=ae^xx^2-(a+1),(a>0,x>0).$
$ecause g(x)=ae^xx^2-(a+1)>ax^2-(a+1)$故$g(sqrt{dfrac{a+1}{a}})>0$又$g(0)=-(a+1)<0$
故存在$x_0in(0,infty),g(x_0)=0$即$ae^{x_0}=dfrac{a+1}{x^2_0}$
易知$f(x)_{min}=f(x_0)=ae^{x_0}-dfrac{a+1}{x_0}-2(a+1)=dfrac{a+1}{x^2_0}-dfrac{a+1}{x_0}-2(a+1)ge0$
故$x_0in(0,1]$,由于$g(x)$显然在$(0,infty)$上单调递增,故$g(1)ge g(x_0)=0$即$agedfrac{1}{e-1}$
练习：
已知函数$f(x)=e^x+e^{-x}-2ln x $的极值点$x_0inleft(dfrac{1}{2},1 ight)$,若$f(x)ge b,bin Z$恒成立,求$b$的范围.
分析:导函数$f^{'}(x)=e^x-e^{-x}-dfrac{2}{x}$单调递增,
易知
$f(x)_{min}=f(x_0)=e^{x_0}+e^{-x_0}-2ln x_0ge1+x_0+dfrac{x_0^2}{2}+1-x_0+dfrac{x_0^2}{2}-2ln x_0=2+x_0^2-2ln x_0ge3,$
又$f(x_0)<f(1)=e+dfrac{1}{e}in(3,4)$故$f(x)_{min}in(3,4)$结合$bin Z$ 知道$b_{max}=3$