也来一道同构的导数题

已知函数$\textit{f}(\textit{x})=\textit{a}\ln\textit{x}+\textit{x}$, $\textit{x}\in(0$,$+\infty)\text{且}\textit{a}\in\textbf{R}$.

$(1)$讨论函数$\textit{f}(\textit{x})$的单调性；

$(2)$ 若$\textit{x}^{\hspace{0.021cm}\textit{a}}\text{e}^{\textit{x}}-1\geqslant (1-\frac{\hspace{0.05cm}1\hspace{0.05cm}}{\text{e}})\hspace{0.08cm}\textit{f}(\textit{x})$恒成立,求实数$\textit{a}$的取值范围.

解：(1) \(f(x)=a\ln x+x\Rightarrow f'(x)=\frac{a}{x}+1=\frac{x+a}{x}\)

当\(a\geqslant 0\)时,\(f'(x)>0 \Rightarrow f(x)\)在\((0,+\infty)\)上单调递增,

当\(a<0\)时,\(x\in(0,-a),f'(x)<0\),\(x\in(-a,+\infty),f'(x)>0\)
\(\Rightarrow f(x)\)在\((0,-a)\)上单调递减,\(f(x)\)在\((-a,+\infty)\)上单调递增,

综上可知,当\(a\geqslant 0\)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,\(\text{e}^{a\ln x+x}+(\frac{1}{\text{e}}-1)(a\ln x+x)-1\geqslant 0\), 设\(h(x)=\text{e}^x+(\frac{1}{\text{e}}-1)x-1\),则

\(\Rightarrow h'(x)=\text{e}^x+\frac{1}{\text{e}}-1\Rightarrow h''(x)=\text{e}^x>0\)

\(\Rightarrow h'(x)\)在\(\textbf{R}\)上单增且\(h'(0)=\frac{1}{\text{e}}>0,h'(-1)=\frac{2}{\text{e}}-1<0\)

\(\Rightarrow\)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\(h(x)>0\)

\(\Rightarrow a\ln x+x\leqslant -1\)恒成立, 或\(a\ln x+x\geqslant 0\)恒成立,而由\(f(x)=a\ln x+x\)有$ f(1)=1>0$

\(\Rightarrow f(x)=a\ln x+x\geqslant 0\)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\(\Rightarrow f(-a)\geqslant 0\Rightarrow\cdots\Rightarrow -\text{e}\leqslant a<0\).

综上可知,\(-\text{e}\leqslant a\leqslant0\).