每日导数81

简单的同构

已知函数\(f(x)=(2-a-ax)e^x\)

(1)求\(f(x)\)的单调区间

(2)若\(a=1\)，证明:\(f(x)+e^x\ln(x+1)\leq x+1\)

解
(1)\(f^{\prime}(x)=(2-ax-2a)e^x\)

Case1 当\(a=0\)时，\(f(x)\)单调递增

Case2 当\(a<0\)时，\(f^{\prime}(x)=0\)得\(x_0=\dfrac{2-2a}{a}=\dfrac{2}{a}-2\)

则\(f(x)\)在\(\left(-\infty,\dfrac{2}{a}-2\right)\)上单调递增，在\(\left(\dfrac{2}{a}-2,+\infty\right)\)上单调递减

Case3 当\(a>0\)时，同上

则\(f(x)\)在\(\left(-\infty,\dfrac{2}{a}-2\right)\)上单调递减少，在\(\left(\dfrac{2}{a}-2,+\infty\right)\)上单调递增

(2)当\(a=1\)，\(f(x)=(1-x)e^x\)，原不等式转为

\((1-x)e^x+e^x\ln (x+1)\leq x+1\)

即\(e^x\left[1-x+\ln(x+1)\right]\leq x+1\)

即\(1-x+\ln(x+1)\leq (x+1)e^{-x}\)

即\(1+\ln(x+1)-\ln e^x\leq (x+1)e^{-x}\)

即\(1+\ln e^{-x}(x+1)\leq (x+1)e^{-x}\)

记\(e^{-x}(x+1)=t,t\in(0,1]\)

即证:\(1+\ln t\leq t\)

这是经典放缩，证明略

得证!