經典找點問題，如果利用保號性是顯然的
已知函式\(f(x)=(m+1-x)e^x-\dfrac{1}{2}me^{2x}-2\)
(1)當\(m=2\)，討論\(f(x)\)的單調性
(2)若\(x=0\)，是\(f(x)\)的極小值點，求\(m\)的取值範圍
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\(2-x-2e^x\)單調遞減，並且\(f^{\prime}(0)=0\)
則\(f(x)\)在\((-\infty,0)\)上增，\([0,+\infty)\)上減
(2)
\(f^{\prime}(x)=e^x(m-x)-me^{2x}=e^x(m-x-me^x)\)
記\(\varphi(x)=m-x-me^x,\varphi(0)=0\)
\(\varphi^{\prime}(x)=-1-me^x,\varphi^{\prime}(0)=-1-m,\varphi^{\prime}\left(\ln-\dfrac{1}{m}\right)=-1-m\cdot\left(-\dfrac{1}{m}\right)=0\)
Case1 當\(m=-1\)時，\(f^{\prime}(x)=e^x(-1-x+e^x)\geq 0\)
此時沒有極值點，舍
Case2 當\(m<-1\)時\(\ln-\dfrac{1}{m}<0\)，此時\(\varphi^{\prime}(x)\)單調遞增，
則\(\varphi^{\prime}(x)\)在\(\left(-\infty,\ln-\dfrac{1}{m}\right)\)上為負，\(\left(\ln-\dfrac{1}{m},+\infty\right)\)上為正
\(\varphi(x)\)在\(\left(-\infty,\ln-\dfrac{1}{m}\right)\)上為減，\(\left(\ln-\dfrac{1}{m},+\infty\right)\)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\(\varphi(0)=0\)，則在\(\left(\ln-\dfrac{1}{m},0\right)\)上負，\((0,+\infty)\)上增
則\(x=0\)是\(f(x)\)的極小值點，合題
Case3 當\(-1<m<0\)時，\(\varphi^{\prime}(0)=-1-m<0\)，此時\(\varphi^{\prime}(x)\)單調遞增
此時\(\ln-\dfrac{1}{m}>0\)，同Case3，
則\(\varphi^{\prime}(x)\)在\(\left(-\infty,\ln-\dfrac{1}{m}\right)\)上為負，\(\left(\ln-\dfrac{1}{m},+\infty\right)\)上為正
\(\varphi(x)\)在\(\left(-\infty,\ln-\dfrac{1}{m}\right)\)上為減，\(\left(\ln-\dfrac{1}{m},+\infty\right)\)上增
從而\(\varphi(x)\)在\(\left(0,\ln-\dfrac{1}{m}\right)\)減，因\(\varphi(0)=0\)
則此時\(x=0\)是\(f(x)\)的極大值點，舍
Case4 當\(m>0\)時，\(\varphi^{\prime}(x)<0\)，則\(\varphi(x)\)單調遞減
此時\(x=0\)是\(f(x)\)的極大值點，舍
綜上\(m<-1\)