每日导数1

保号性应用

已知函数\(f(x)=e^x-mx\)

(1)讨论\(f(x)\)单调性

(2)若\(f(x)\geq (a-m)x-\sin x+1,\forall x>0\)恒成立，求\(a\)的取值范围

解.

\((1)\) 当\(m\leq 0\)时，\(f(x)\)为单调递增

当\(m>0\)时，\(f^{\prime}(x)=e^x-m=0\)得\(x_0=\ln m\)

则\(f(x)\)在\((0,x_0)\)单调递减，\((x_0,+\infty)\)单调递增.

\((2)\) 若\(f(x)\geq (a-m)x-\sin x+1,\forall x>0\)恒成立

则\(e^x-ax+\sin x-1\geq 0\)恒成立

记\(\varphi(x)=e^x-ax+\sin x-1,\varphi^{\prime}(x)=e^x-a+\cos x\)

不难发现\(\varphi(0)=0,\varphi^{\prime}(0)=2-a\)

若\(2-a< 0\)即 \(a> 2\)时

由保号性知存在\(m\)使得
\(x\in(0,m)\)有\(\varphi^{\prime}(x)<0\)

又因\(\varphi(0)=0\) 则在\(x\in(0,x_0)\)上\(\varphi(x)<0\)不合题！

当\(a\leq 2\)时，\(\varphi^{\prime\prime}(x)=e^x-\sin x>0\)

即\(\varphi^{\prime}(x)\)单调递增，则\(\varphi^{\prime}(x)>\varphi^{\prime}(0)=2-a>0\)

即\(\varphi(x)\)单调递增，即\(\varphi(x)\geq \varphi(0)=0 \square\)

综上\(a\leq 2\)