每日导数68

切线放缩

已知函数\(f(x)=\dfrac{1}{2}x^2+(a-m-1)x-ax\ln x\)

(1)若\(m=-1\)时，\(y=f(x)\)不是单调函数，求\(a\)范围

(2)若\(a=2,m<0\)时，\(f(x)\)存在两个极值点\(x_1,x_2(x_1<x_2)\)，证明：\(x_2-x_1<3(m+1)\)

解

(1)\(f(x)=\dfrac{1}{2}x^2+ax-ax\ln x,f^{\prime}(x)=x-a\ln x=a\left(\dfrac{x}{a}-\ln x\right)\)

因\(\dfrac{x}{e}\geq \ln x\)，从而

当\(a>e\)合题，\(a<0\)合题，\(a=0\)不合

(2)\(f(x)=\dfrac{1}{2}x^2+(1-m)x-2x\ln x,f^{\prime}(x)=x-2\ln x-m-1=0\)

即\(x-2\ln x-1=m\)

记\(g(x)=x-2\ln x-1,y=m\)（因为发现这样分离后\(g(1)=0\)，很特殊)

变成了\(y=g(x),y=m\)有两个交点

不难得到\(g(x)\)在\((0,2)\)减，\((2,+\infty)\)上增

又因\(m<0\)，从而得到\(1<x_1<2<x_2\)

因\(g(1)=0\)，设在\(x>2\)上另外一个零点为\(x_0\)即\(g(x_0)=0\)

因\(g(3)=2-2\ln 3<0,g(4)=3-4\ln 2>0\)，从而\(x_0\in(3,4)\)

\(g(x)\)在\(x=1\)上的切线\(l_1:y=1-x\)，在\(x=x_0\)上切线:\(l_2=y=\left(1-\dfrac{2}{x_0}\right)(x-x_0)\)

设\(l_1,l_2\)与\(y=m\)的交点为\(x_1^{'},x_2^{''}\)

从而\(x_1^{'}=\dfrac{mx_0}{x_0-2}+x_0,x_2^{'}=1-m\)

则\(x_2-x_1<x_2^{'}-x_1^{'}=\dfrac{mx_0}{x_0-2}+x_0-(1-m)=\dfrac{m(x_0-2)+2m}{x_0-2}+x_0+m-1\)

\(=\dfrac{2m}{x_0-2}+x_0+2m-1=\dfrac{2m}{x_0-2}+x_0-2+2m+1\)

\(<\dfrac{2m}{2}+2+2m+1=3(m+1)\)