8.(2020秋•开福区校级期末)已知函数f(x)=−(a+1)lnx+ax−1x(a∈R)f(x)=-(a+1)lnx+ax-\frac{1}{x}(a\in R)f(x)=−(a+1)lnx+ax−x1(a∈R).(1)求函数f(x)f(x)f(x)的单调区间;(2)当a=−2a=-2a=−2时,g(x)=f(x)+(x−2)ex+x+1xg(x)=f(x)+(x-2)e^{x}+x+\frac{1}{x}g(x)=f(x)+(x−2)ex+x+x1,记函数y=g(x)y=g(x)y=g(x)在[14,1][\frac{1}{4},1][41,1]上的最大值为mmm,证明:−4<m<−3-4<m<-3−4<m<−3.