\(\ln x<x-1\)放缩应用
已知函数\(f(x)=mx-\ln x-1\)
\((1)\) 讨论函数的单调性
\((2)\) 若不等式\(e^{x-1}+a\ln x-(a+1)x+a\geq 0\)恒成立,求\(a\)的取值范围
解
\((1)\) \(f^{\prime}(x)=m-\dfrac{1}{x}\)
若\(m\leq 0\)从而\(f^{\prime}(x)<0\)从而\(f(x)\)单调递减
若\(m>0\)不难得到\(f(x)\)在\(\left(0,\dfrac{1}{m}\right)\)上单调递减,在\(\left(\dfrac{1}{m},+\infty\right)\)上单调递增
\((2)\) 由\((1)\)得\(m=1\)有\(\ln x\leq x-1\)
则有\(e^x-1\geq x\)
从而当\(a\leq 0\)时
\(e^{x-1}+a\ln x-(a+1)x+a=e^{x-1}+a(\ln x_1+1)-(a+1)x\geq x+ax-(a+1)x=0\)
当\(a>0\)时,记\(g(x)=e^{x-1}+a\ln x-(a+1)x+a\)
\(g(x)<e^{x-1}+a(x-1)-(a+1)x+a=e^{x-1}-x\)
而在\(x\in(0,1)\)上\(e^{x-1}-x<0\)
从而在\(x\in(0,1)\)上\(g(x)<0\)
不合题!
综上\(a\leq 0\).