開啟絕對值，隱零點
已知函式\(f(x)=ae^x-\dfrac{1}{x}\)
(1)討論\(f(x)\)零點個數
(2)當\(a>0\)時，\(|f(x)|>1+\ln x\)，求\(a\)的取值範圍
解
(1)\(f(x)=ae^x-\dfrac{1}{x}=0\)
即\(ax=e^{-x}\)，因\(y=-ex\)與\(y=e^{-x}\)相切
則有\(a\in\begin{cases} \{-e \}\cup(0,+\infty),&\text{一個零點}\\ (-\infty,-e),&\text{兩個零點}\\ \{0\},&\text{零個零點} \end{cases}\)
(2)當\(a>0\)時，\(f(x)\)單調遞增
因\(x\to 0,f(x)\to +\infty,f(x)\to-\infty,x\to+\infty,f(x)\to+\infty\)
則一定存在\(x_0\)使得\(f(x_0)=0\)
即\(ae^{x_0}-\dfrac{1}{x_0}=0\)
即\(\ln a=-\ln x_0-x_0\)
因\(f\left(\dfrac{1}{a}\right)=ae^{\frac{1}{a}}-a=a\left(e^{\frac{1}{a}}-1\right)>0\)，則\(x_0<\dfrac{1}{a}\)
當\(x\in(0,x_0)\)時，\(|f(x)|=\dfrac{1}{x}-ae^x\)
則原不等式為\(\dfrac{1}{x}-ae^x-1-\ln x>0\)恆成立
記\(\varphi(x)=\dfrac{1}{x}-ae^x-1-\ln x\)，\(\varphi^{\prime}(x)=-\dfrac{1}{x^2}-ae^x-\dfrac{1}{x}<0\)
則\(\varphi(x)\)單調遞減
從而有\(\varphi(x_0)>0\)
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\(a>e^{1-\frac{1}{e}}\)
當\(x>x_0\)時，原不等式為\(ae^x-\dfrac{1}{x}-1-\ln x>0\)
記\(\gamma(x)=ae^x-\dfrac{1}{x}-1-\ln x,\gamma^{\prime}(x)=ae^x+\dfrac{1}{x^2}-\dfrac{1}{x}=\dfrac{1}{x^2}+f(x)>0\)
則\(\gamma(x)>\gamma(x_0)\)，即\(-1-\ln x_0>0\)
得\(x_0<\dfrac{1}{e}\)，同上得到\(a>e^{1-\frac{1}{e}}\)
綜上\(a>e^{1-\frac{1}{e}}\)