簡單的參變分離
已知函式$ f(x)=\dfrac{\ln x}{x^2}$
(1)討論\(f(x)\)的最值；
(2)若函式\(g(x)=e^x+x^4f(x)-x^2-ax\)有兩個零點，求\(a\)的範圍
解
(1)\(f(x)=\dfrac{1-2\ln x}{x^3}\)
得\(f^{\prime}\left(\sqrt{e}\right)=0\)
則\(f(x)\)在\(\left(0,\sqrt{e}\right)\)上增，在\(\left(\sqrt{e},+\infty\right)\)上減
從而\(f(x)_{\max}=f(\sqrt{e})=\dfrac{1}{2e}\)
(2)\(g(x)=e^x+x^2\ln x-x^2-ax=e^x-ax+x^2(\ln x-1)=0\)
即\(a=\dfrac{e^x}{x}+x(\ln x-1)\)
記\(\varphi(x)=\dfrac{e^x}{x}+x(\ln x-1),\varphi^{\prime}(x)=\dfrac{e^x(x-1)}{x^2}+\ln x\)
不難發現\(0<x<1\)時，\(\varphi^{\prime}(x)<0,x>0\varphi^{\prime}(x)>0,x=1,\varphi^{\prime}(1)=0\)
則\(\varphi(x)\)在\((0,1)\)上減，在\((1,+\infty)\)上增，\(\varphi(x)_{\min}=\varphi(1)=e+1\)
則使得\(y=a\)與\(y=\varphi(x)\)有兩個交點，要有\(a>e+1\)