MT【350】隐零点两题

已知函数$f(x)=ae^x+\dfrac{a+1}{x}-2(a+1)\ge0,(a>0)$对任意的$x\in(0,+\infty)$恒成立,求$a$的范围.

分析:$f^{'}(x)=\dfrac{g(x)}{x^2},$其中$g(x)=ae^xx^2-(a+1),(a>0,x>0).$
$\because g(x)=ae^xx^2-(a+1)>ax^2-(a+1)$故$g(\sqrt{\dfrac{a+1}{a}})>0$又$g(0)=-(a+1)<0$
故存在$x_0\in(0,\infty),g(x_0)=0$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$x_0\in(0,1]$,由于$g(x)$显然在$(0,\infty)$上单调递增,故$g(1)\ge g(x_0)=0$即$a\ge\dfrac{1}{e-1}$

 

练习：

已知函数$f(x)=e^x+e^{-x}-2\ln x $的极值点$x_0\in\left(\dfrac{1}{2},1\right)$,若$f(x)\ge b,b\in 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

$f(x)_{min}=f(x_0)=e^{x_0}+e^{-x_0}-2\ln x_0\ge1+x_0+\dfrac{x_0^2}{2}+1-x_0+\dfrac{x_0^2}{2}-2\ln x_0=2+x_0^2-2\ln x_0\ge3,$
又$f(x_0)<f(1)=e+\dfrac{1}{e}\in(3,4)$故$f(x)_{min}\in(3,4)$结合$b\in Z$ 知道$b_{max}=3$