观察放缩
已知函数\(f(x)=\dfrac{\sin x}{e^x}\)
\((1)\) 求函数\(f(x)\)在\((0,3)\)上的单调区间
\((2)\) 若\(x>0\)时，\(f(x)\leq a\ln (x+1)\)，求实数\(a\)的取值范围
解
\((1)\) \(f^{\prime}(x)=\dfrac{\cos x-\sin x}{e^x}=\dfrac{\sqrt{2}\sin\left(\dfrac{\pi}{4}-x\right)}{e^x}=0\)得\(x_1=\dfrac{\pi}{4}+k\pi\)
又\(x\in(0,3)\)从而\(f(x)\)在\(\left(0,\dfrac{\pi}{4}\right)\)单调递增，\(\left(\dfrac{\pi}{4},3\right)\)单调递减.
\((2)\)原不等式为\(\sin x-ae^x\ln (x+1)\leq 0\)
记\(G(x)=\sin x-ae^x\ln (x+1)\)，不难发现\(G(0)=0\)
\(G^{\prime}(x)=\cos x-ae^x\ln(x+1)-\dfrac{ae^x}{x+1}\)
\(G^{\prime}(0)=1-a\)
若\(1-a> 0\)即\(a<1\)时
由保号性，一定存在\((0,x_0)\)，使得\(x\in(0,x_0)\)，\(G^{\prime}(x)>0\)
从而\(G(x)\)在\((0,x_0)\)上单调递增，进而\(G(x)>G(0)=0\)不合题，舍.
若\(a\geq 1\)，则\(G^{\prime}(x)=\cos x-ae^x\left[\ln(x+1)+\dfrac{1}{x+1}\right]\)
考虑\(\varphi(x)=\ln(x+1)+\dfrac{1}{x+1},\varphi^{\prime}(x)=\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}>0\)
从而\(\varphi(x)\)单调递增，进而\(\varphi(x)>\varphi(0)=1\)
所以\(\cos x-ae^x\left[\ln(x+1)+\dfrac{1}{x+1}\right]\leq \cos x-e^x\left[\ln(x+1)+\dfrac{1}{x+1}\right]\)
又因\(e^x>e^0=1\)
从而\(e^x\left[\ln(x+1)+\dfrac{1}{x+1}\right]>1\)
从而\(\cos x-e^x\left[\ln(x+1)+\dfrac{1}{x+1}\right]<0\)
从而\(G^{\prime}(x)\)恒为负，进而\(G(x)\)单调递减
则有\(G(x)<G(0)=0\) 合题.
综上，\(a\geq 1\).