每日导数94

经典找点问题，如果利用保号性是显然的

已知函数\(f(x)=(m+1-x)e^x-\dfrac{1}{2}me^{2x}-2\)

(1)当\(m=2\)，讨论\(f(x)\)的单调性

(2)若\(x=0\)，是\(f(x)\)的极小值点，求\(m\)的取值范围

解

(1)\(m=2,f(x)=(3-x)e^x-e^{2x}-2\)，\(f^{\prime}(x)=e^{x}(2-x)-2e^{2x}=e^x(2-x-2e^x)\)

因\(2-x-2e^x\)单调递减，并且\(f^{\prime}(0)=0\)
则\(f(x)\)在\((-\infty,0)\)上增，\([0,+\infty)\)上减

(2)
\(f^{\prime}(x)=e^x(m-x)-me^{2x}=e^x(m-x-me^x)\)

记\(\varphi(x)=m-x-me^x,\varphi(0)=0\)

\(\varphi^{\prime}(x)=-1-me^x,\varphi^{\prime}(0)=-1-m,\varphi^{\prime}\left(\ln-\dfrac{1}{m}\right)=-1-m\cdot\left(-\dfrac{1}{m}\right)=0\)

Case1 当\(m=-1\)时，\(f^{\prime}(x)=e^x(-1-x+e^x)\geq 0\)

此时没有极值点，舍

Case2 当\(m<-1\)时\(\ln-\dfrac{1}{m}<0\)，此时\(\varphi^{\prime}(x)\)单调递增，

则\(\varphi^{\prime}(x)\)在\(\left(-\infty,\ln-\dfrac{1}{m}\right)\)上为负，\(\left(\ln-\dfrac{1}{m},+\infty\right)\)上为正

\(\varphi(x)\)在\(\left(-\infty,\ln-\dfrac{1}{m}\right)\)上为减，\(\left(\ln-\dfrac{1}{m},+\infty\right)\)上增

\(\varphi\left(\ln-\dfrac{1}{m}\right)=m-\ln-\dfrac{1}{m}+1<m+1-\left(1-\dfrac{1}{-\dfrac{1}{m}}\right)=0\)

而\(\varphi(0)=0\)，则在\(\left(\ln-\dfrac{1}{m},0\right)\)上负，\((0,+\infty)\)上增

则\(x=0\)是\(f(x)\)的极小值点，合题

Case3 当\(-1<m<0\)时，\(\varphi^{\prime}(0)=-1-m<0\)，此时\(\varphi^{\prime}(x)\)单调递增

此时\(\ln-\dfrac{1}{m}>0\)，同Case3，

则\(\varphi^{\prime}(x)\)在\(\left(-\infty,\ln-\dfrac{1}{m}\right)\)上为负，\(\left(\ln-\dfrac{1}{m},+\infty\right)\)上为正

\(\varphi(x)\)在\(\left(-\infty,\ln-\dfrac{1}{m}\right)\)上为减，\(\left(\ln-\dfrac{1}{m},+\infty\right)\)上增

从而\(\varphi(x)\)在\(\left(0,\ln-\dfrac{1}{m}\right)\)减，因\(\varphi(0)=0\)

则此时\(x=0\)是\(f(x)\)的极大值点，舍

Case4 当\(m>0\)时，\(\varphi^{\prime}(x)<0\)，则\(\varphi(x)\)单调递减

此时\(x=0\)是\(f(x)\)的极大值点，舍

综上\(m<-1\)