每日导数77

常规的零点分析

已知函数\(f(x)=\dfrac{ax}{e^x}+\dfrac{1}{2}x^2-x(a>0)\)

(1)若\(f(x)<\dfrac{1}{2}x^2-\ln x\)恒成立，求\(a\)的范围

(2)讨论\(f(x)\)的零点个数

解

(1)\(f(x)<\dfrac{1}{2}x^2-\ln x\)整理为

\(ae^{\ln x-x}<x-\ln x\)

记\(x-\ln x=t\)

即\(ae^t<t\)

因\(te^t\)在\(t>1\)上单调递增

则\(0<a\leq e\)

(2)\(f(x)=0\)整理得\(a=\left(1-\dfrac{1}{2}x\right)e^x\)

记\(\varphi(x)=\left(1-\dfrac{1}{2}x\right)e^x,F^{\prime}(x)=\dfrac{1}{2}e^{x}\left(1-x\right)\)

则\(\varphi(x)\)在\((-\infty,1)\)上减，在\((1,+\infty)\)上增，\(\varphi(x)_{\max}=\varphi(1)=\dfrac{e}{2}\)

又因\(x\to -\infty,\varphi(x)\to 0,x\to+\infty,\varphi(x)\to -\infty\)，

则有\(a\in\begin{cases} \left\{\dfrac{e}{2}\right\}\cup (-\infty,0],&1\text{个零点}\\ \left(0,\dfrac{e}{2}\right),&\text{两个零点}\\ \left(\dfrac{e}{2},+\infty\right),&0\text{个零点} \end{cases}\)