每日导数87

打开绝对值，隐零点

已知函数\(f(x)=ae^x-\dfrac{1}{x}\)

(1)讨论\(f(x)\)零点个数

(2)当\(a>0\)时，\(|f(x)|>1+\ln x\)，求\(a\)的取值范围

解
(1)\(f(x)=ae^x-\dfrac{1}{x}=0\)

即\(ax=e^{-x}\)，因\(y=-ex\)与\(y=e^{-x}\)相切

则有\(a\in\begin{cases} \{-e \}\cup(0,+\infty),&\text{一个零点}\\ (-\infty,-e),&\text{两个零点}\\ \{0\},&\text{零个零点} \end{cases}\)

(2)当\(a>0\)时，\(f(x)\)单调递增

因\(x\to 0,f(x)\to +\infty,f(x)\to-\infty,x\to+\infty,f(x)\to+\infty\)

则一定存在\(x_0\)使得\(f(x_0)=0\)

即\(ae^{x_0}-\dfrac{1}{x_0}=0\)

即\(\ln a=-\ln x_0-x_0\)

因\(f\left(\dfrac{1}{a}\right)=ae^{\frac{1}{a}}-a=a\left(e^{\frac{1}{a}}-1\right)>0\)，则\(x_0<\dfrac{1}{a}\)

当\(x\in(0,x_0)\)时，\(|f(x)|=\dfrac{1}{x}-ae^x\)

则原不等式为\(\dfrac{1}{x}-ae^x-1-\ln x>0\)恒成立

记\(\varphi(x)=\dfrac{1}{x}-ae^x-1-\ln x\)，\(\varphi^{\prime}(x)=-\dfrac{1}{x^2}-ae^x-\dfrac{1}{x}<0\)

则\(\varphi(x)\)单调递减

从而有\(\varphi(x_0)>0\)

即\(-1-\ln x_0>0\)，得\(x_0<\dfrac{1}{e}\)

而\(\ln a=-\ln x_0-x_0\)

得\(\ln a>-\ln\dfrac{1}{e}-\dfrac{1}{e}=1-\dfrac{1}{e}\)

即\(a>e^{1-\frac{1}{e}}\)

当\(x>x_0\)时，原不等式为\(ae^x-\dfrac{1}{x}-1-\ln x>0\)

记\(\gamma(x)=ae^x-\dfrac{1}{x}-1-\ln x,\gamma^{\prime}(x)=ae^x+\dfrac{1}{x^2}-\dfrac{1}{x}=\dfrac{1}{x^2}+f(x)>0\)

则\(\gamma(x)>\gamma(x_0)\)，即\(-1-\ln x_0>0\)

得\(x_0<\dfrac{1}{e}\)，同上得到\(a>e^{1-\frac{1}{e}}\)

综上\(a>e^{1-\frac{1}{e}}\)