多变量问题转化成单变量问题
设\(a\in \mathbb{R}\)，函数\(f(x)=x^2e^{1-x}-a(x-1)\)
\((1)\)当\(a=1\)时，求\(f(x)\)在\(\left(\dfrac{3}{4},2\right)\)内的极值
\((2)\)设函数\(g(x)=f(x)+a(x-1-e^{1-x})\)，当\(g(x)\)有两个极值点\(x_1,x_2(x_1<x_2)\)时，总有\(x_2g(x_1)\leq \lambda f^{\prime}(x_1)\)，求实数\(\lambda\)的值.
解
\((1)\) \(a=1,f(x)=x^2e^{1-x}-x+1\)
\(f^{\prime}(x)=2xe^{1-x}-x^2e^{1-x}-1=e^{1-x}(2x-x^2)-1\)
\(f^{\prime\prime}(x)=e^{1-x}(2-2x-2x+x^2)=e^{1-x}(x^2-4x+2)\)
因\(x^2-4x+2\)在\(\left(\dfrac{3}{4},2\right)\)单调递减
从而\(x^2-4x+2<0\) 进而\(f^{\prime}(x)单调递减\)
不难发现\(f^{\prime}(1)=0\)
从而\(x=1\)是极大值点.
\((2)\) \(g(x)=x^2e^{1-x}-a(x-1)+a(x-1-e^{1-x})=x^2e^{1-x}-ae^{1-x}=e^{1-x}(x^2-a)\)
\(g^{\prime}(x)=e^{1-x}(-x^2+2x+a),f^{\prime}(x)=e^{1-x}(2x-x^2)-a\)
若\(g(x)\)有两个极值点则\(x_1+x_2=2,x_1x_2=-a\)
则\(x_2g(x_1)=x_2e^{1-x_1}(x_1^2-a),f^{\prime}(x_1)=e^{1-x_1}(2x_1-x_1^2)-a\)
则原不等式为
因\(x_1+x_2=2,x_1x_2=-a\)从而\(x_2=2-x_1,a=(x_1-2)x_1\)代回不等式有
即
当\(x_1\in(0,1)\)时
即
即
当\(x_1\in(-\infty,0)\)时，
综上\(\lambda=\dfrac{2e}{e+1}\)