簡單的一道
已知\(f(x)=e^{x+a}-\dfrac{e^2}{2}a^2x(a>0)\)
(1)求\(f(x)\)極小值點的最大值
(2)證明:當\(x\geq 0\),\(f(x)>e^x\)恆成立
解
(1)\(f^{\prime}(x)=e^{x+1}-\dfrac{e^2}{2}a^2\)單調遞增，則有唯一的零點\(x_0\)，使得\(f^{\prime}(x_0)=0\)
並且\(x=x_0\)是\(f(x)\)的極小值點，即\(e^{x_0+a}=\dfrac{e^2}{2}a^2\)
即
$$x_0=2-\ln 2+2\ln a-a$$
記\(h(a)=2-\ln 2+2\ln a-a,h^{\prime}(a)=\dfrac{2}{a}-1\)
則\(h(a)\leq h\left(2\right)=2-\ln 2+2\ln 2-2=\ln 2\)
(2)\(e^{x+a}-\dfrac{e^2}{2}a^2x>e^x\)
即\(e^a-\dfrac{a^2x}{2}e^{2-x}-1>0\)
記\(\varphi(x)=e^a-\dfrac{a^2x}{2}e^{2-x}-1,\varphi^{\prime}(x)=-\dfrac{a^2}{2}e^{2-x}(1-x)\)
則有\(\varphi(x)_{\min}=\varphi(1)=e^a-\dfrac{a^2}{2}e-1\)
記\(\gamma(a)=e^a-\dfrac{a^2}{2}e-1,\gamma^{\prime}(a)=e^a-ae\geq ae-ae=0\)
則\(\varphi_{\min}=\varphi(1)=\gamma(a)\geq \gamma(0)=0\)
得證.