每日导数67

有点硬凑的找点问题

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

(1)讨论\(f^{\prime}(x)\)单调性

(2)若\(x_1,x_2\)是\(f(x)\)的极值点，证明:\(x_2-x_1<\ln(3-a)-\ln a+\dfrac{2}{a}-1\)

解

(1)\(f^{\prime}(x)=ae^{2x}+(a-2)e^x-x\)

\(f^{\prime\prime}(x)=2ae^{2x}+(a-2)e^x-1=(2e^x+1)(ae^x-1)\)

当\(a\leq0\) 时，\(f^{\prime\prime}(x)\leq 0\)，则\(f^{\prime}(x)\)单调递减

当\(a>0\)时，\(f^{\prime\prime}(-\ln a)=0\)

则\(f^{\prime}(x)\)在\((-\infty,-\ln a)\)递减，在\([-\ln a,+\infty)\)上增

(2) 由(1)要使\(f(x)\)有两个极值点，则要有\(a>0\)，并且\(f^{\prime}(-\ln a)<0\)

则\(1-\dfrac{1}{a}-\ln\dfrac{1}{a}<0\)得\(a\in(0,1)\)

不妨设\(x_1<-\ln a<x_2\)

处理\(\ln(3-a)-\ln a+\dfrac{2}{a}-1\)为\(\ln\left(\dfrac{3}{a}-1\right)-\left(1-\dfrac{2}{a}\right)\)

①先说明\(-\ln a<x_2<\ln\left(\dfrac{3}{a}-1\right)\)

因\(f^{\prime}\left[\ln\left(\dfrac{3}{a}-1\right)\right]=a\left(\dfrac{3}{a}-1\right)^2+(a-2)\left(\dfrac{3}{a}-1\right)-\ln\left(\dfrac{3}{a}-1\right)=\left(\dfrac{3}{a}-1\right)-1-\ln\left(\dfrac{3}{a}-1\right)\)

因\(a\in(0,1)\)，从而\(\dfrac{3}{a}-1\neq 1\)

则\(\left(\dfrac{3}{a}-1\right)-1-\ln\left(\dfrac{3}{a}-1\right)>\left(\dfrac{3}{a}-1\right)-1-\left[\left(\dfrac{3}{a}-1\right)-1\right]=0\)

由零点存在定理

\(-\ln a<x_2<\ln\left(\dfrac{3}{a}-1\right)\)，并且是唯一的.

②再说明\(1-\dfrac{2}{a}<x_1<-\ln a\)

因\(f^{\prime}\left(1-\dfrac{2}{a}\right)=1-\dfrac{1}{a}-\ln\dfrac{1}{a}\)

因\(\ln x\leq x-1\)

从而\(1-\dfrac{1}{a}-\ln\dfrac{1}{a}>1-\dfrac{1}{a}-\left(1-\dfrac{1}{a}\right)>0\)

由零点存在定理\(1-\dfrac{2}{a}<x_1<-\ln a\)，并且是唯一的.

从而\(x_2-x_1<\ln(3-a)-\ln a+\dfrac{2}{a}-1\)