再来点简单的
已知函数\(f(x)=e^x\cos x\)
\((1)\)求\(f(x)\)的单调区间
\((2)\) \(F(x)=-f^{\prime}(x)-ax\)在\(\left(\dfrac{\pi}{2},\pi\right)\)上有两个极值点，求实数\(a\)的取值范围.
解
\((1)\) \(f^{\prime}(x)=e^x(\cos x-\sin x)=\sqrt{2}e^x\cos\left(x+\dfrac{\pi}{4}\right)\)
\(f(x)\)的单调递增区间\(\left(2k\pi-\dfrac{3\pi}{4},2k\pi+\dfrac{\pi}{4}\right)\)
\(f(x)\)的单调递减区间\(\left(2k\pi+\dfrac{\pi}{4},2k\pi+\dfrac{5\pi}{4}\right)\)
\((2)\) \(F(x)=e^x(\sin x-\cos x)-ax,F^{\prime}(x)=2\sin xe^{x}-a=0\)有两解
即\(a=2\sin xe^x\)有两解
即\(y=a\)与\(y=2e^x\sin x\)有两交点
记\(G(x)=2\sin xe^x,G^{\prime}(x)=\sqrt{2}e^x\sin\left(x+\dfrac{\pi}{4}\right)\)
不难得到\(x\in\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right)\)单调递增，\(x\in \left(\dfrac{3\pi}{4},\pi\right)\)上单调递减
作出草图不难得到
\(2e^{\frac{\pi}{2}}<a<\sqrt{2}e^{\frac{3\pi}{4}}\)