端点效应难在放缩语言的叙述

已知函数\(f(x)=\ln x+ax^2-x+a+1\)，若\(f(x)\leq e^x\)，求\(a\)取值范围

解

\(\ln x+ax^2-x+a+1-e^x\leq 0\)

记\(g(x)=\ln x+ax^2-x+a+1-e^x,g(1)=2a-e\leq 0\)

则一定有\(a\leq \dfrac{e}{2}\)

现说明\(a\leq \dfrac{e}{2}\)是合题的

\(g(x)=\ln x+ax^2-x+a+1-e^x=\ln x+a(x^2+1)-x+1-e^x\leq \ln x+\dfrac{e}{2}(x^2+1)-x+1-e^x\)

记\(\varphi(x)=\ln x+\dfrac{e}{2}(x^2+1)-x+1-e^x,\varphi^{\prime}(x)=\dfrac{1}{x}+ex-1-e^x\)

当\(x\geq 1\)时，\(\varphi^{\prime}(x)\leq \dfrac{1}{x}+ex-1-ex=\dfrac{1}{x}-1\leq 0\)

当\(x\in(0,1)\)时,\(\varphi^{\prime\prime}(x)=-\dfrac{1}{x^2}+e-e^x<-\dfrac{1}{x^2}-ex+e<-2\sqrt{\dfrac{e}{x}}+e<-2\sqrt{e}+e<0\)

从而\(\varphi^{\prime}(x)\)单调递减，而\(\varphi^{\prime}(1)=0\),

则当\(x\in(0,1)\)，\(\varphi(x)>0\)

从而\(\varphi(x)\leq \varphi(1)=0\)

则\(g(x)\leq \varphi(x)\leq \varphi(1)=0\)

得证.