已知函数$ extit{f}( extit{x})= extit{a}ln extit{x}+ extit{x}$, $ extit{x}in(0$,$+infty) ext{且} extit{a}in extbf{R}$.
$(1)$讨论函数$ extit{f}( extit{x})$的单调性;
$(2)$ 若$ extit{x}^{hspace{0.021cm} extit{a}} ext{e}^{ extit{x}}-1geqslant (1-frac{hspace{0.05cm}1hspace{0.05cm}}{ ext{e}})hspace{0.08cm} extit{f}( extit{x})$恒成立,求实数$ extit{a}$的取值范围.
解:(1) (f(x)=aln x+xRightarrow f'(x)=frac{a}{x}+1=frac{x+a}{x})
当(ageqslant 0)时,(f'(x)>0 Rightarrow f(x))在((0,+infty))上单调递增,
当(a<0)时,(xin(0,-a),f'(x)<0),(xin(-a,+infty),f'(x)>0)
(Rightarrow f(x))在((0,-a))上单调递减,(f(x))在((-a,+infty))上单调递增,综上可知,当(ageqslant 0)时,函数(f(x))在((0,+infty))上单调递增,
当(a<0)时,函数(f(x))在((0,-a))上单调递减,在((-a,+infty))上单调递增;
(2) 由题可知,( ext{e}^{aln x+x}+(frac{1}{ ext{e}}-1)(aln x+x)-1geqslant 0), 设(h(x)= ext{e}^x+(frac{1}{ ext{e}}-1)x-1),则
(Rightarrow h'(x)= ext{e}^x+frac{1}{ ext{e}}-1Rightarrow h''(x)= ext{e}^x>0)
(Rightarrow h'(x))在( extbf{R})上单增且(h'(0)=frac{1}{ ext{e}}>0,h'(-1)=frac{2}{ ext{e}}-1<0)
(Rightarrow)存在(x_0)使(h'(x_0)=0)
(Rightarrow h(x))在((-infty,x_0))上单减,在((x_0,+infty))上单增,且(h(0)=0,h(-1)=0)
(Rightarrow)当(x<-1)时(h(x)>0),当(-1<x<0)时(h(x)<0,)当(x>0)时(h(x)>0)
(Rightarrow aln x+xleqslant -1)恒成立, 或(aln x+xgeqslant 0)恒成立,而由(f(x)=aln x+x)有$ f(1)=1>0$
(Rightarrow f(x)=aln x+xgeqslant 0)恒成立
(Rightarrow f'(x)=frac{a+x}{x})
ding{192}当(a>0)时(f(frac{a}{a+1})<0),不符合题意;
ding{193}当(a=0)时(f(x)=x > 0),符合题意;
ding{194}当(a<0)时(f(x))在((0,-a))上单调递减,在((-a,+infty))上单调递增(Rightarrow f(-a)geqslant 0RightarrowcdotsRightarrow - ext{e}leqslant a<0).
综上可知,(- ext{e}leqslant aleqslant0).