端点效应与放缩
已知函数\(f(x)=(ax^2+x+a)e^{-x}(a\in\mathbb{R})\)
(1)若\(a\geq 0\),函数\(f(x)\)的极大值为\(\dfrac{3}{e}\),求\(a\)
(2)对任意的\(a\leq 0\),\(f(x)\leq b\ln(x+1)\)在\((0,+\infty)\)上恒成立,求\(b\)的取值范围
解
(1)\(f^{\prime}(x)=e^{-x}(-ax^2-x-a+2ax+1)=e^{-x}\left[-ax+(1-a)\right](x-1)=0\)
得\(x_1=1,x_2=1-\dfrac{1}{a}\)
Case1 当\(a=0\)时,\(f^{\prime}(x)=e^{-x}(1-x)\),得\(f(x)\)的极大值为\(f(1)=e^{-1}\)舍
Case2 当\(a>0\)时,\(1-\dfrac{1}{a}<1\)
,则不难得到\(f(x)\)的极大值为\(f(1)=3e^{-1}\)得\(a=1\)
综上\(a=1\)
(2)记\(F(a)=(ax^2+x+a)e^{-x}=a(x^2+1)e^{-x}+xe^{-x}\)关于\(a\)是单调递增的
则\(f(x)=F(a)\leq xe^x{-x}\)
要证:\((ax^2+x+a)e^{-x}\leq b\ln(x+1)\)
即证:$ xe^{-x}\leq b\ln(x+1)$
记\(\varphi(x)=xe^{-x}-b\ln (x+1)\),\(\varphi(0)=0\)
\(\varphi^{\prime}(x)=e^{-x}(1-x)-\dfrac{b}{x+1},\varphi^{\prime}(0)=1-b\)
Case1 当\(b\geq 1\)时,\(\varphi^{\prime}(0)\leq 0\)
\(\varphi(x)\leq xe^{-x}-\ln(x+1)\)
记\(\gamma(x)=xe^{-x}-\ln(x+1)\)
\(\gamma^{\prime}(x)=e^{-x}(-x+1)-\dfrac{1}{x+1}=\dfrac{1-x^2-e^x}{(x+1)e^x}<\dfrac{1-x^2-1-x-\dfrac{x^2}{2}}{(x+1)e^x}=\dfrac{-x-\dfrac{3x^2}{2}}{(x+1)e^x}<0\)
则\(\gamma(x)\)单调递减,即\(\gamma(x)\leq \gamma(0)=0\)
合题
Case2 当\(0<b<1\)时,\(\varphi^{\prime}(0)=1-b>0,\varphi^{\prime}(1)=-\dfrac{b}{2}<0\)
则由零点存在定理,存在\(x_0\)使得\(\varphi^{\prime}(x_0)=0\)
则在\((0,x_0)\)上,\(\varphi(x)\)单调递增,即\(\varphi(x)>\varphi(0)=0\)
矛盾!,不合题
Case3 当\(b\leq0\)时,左边极限是0,右边极限是\(-\infty\),矛盾!
综上\(b\geq 1\)
标签:prime,导数,ln,dfrac,每日,varphi,xe,leq,80 From: https://www.cnblogs.com/manxinwu/p/18063795