All possible values of $x$ for which the function, $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x} {\log }_{\mathrm{e}}\mathrm{x}-\mathrm{x}+1$is positive is

We have $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{xlog}}_{\mathrm{e}}\mathrm{x}-\mathrm{x}+1,\mathrm{x}>0$
$$\begin{gathered} \begin{array}{l}\therefore \mathrm{f}\left(1\right)=0\\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=1+{\log }_{\mathrm{e}}\mathrm{x}-1={\log }_{\mathrm{e}}\mathrm{x}\end{array} \\ \begin{array}{l}\therefore {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)<0\text{ if }0<\mathrm{x}<1\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)>0\text{ if }\mathrm{x}>1\end{array} \end{gathered}$$
Also $\underset{\mathrm{x}\to 0^{+}}{lim} \left({\mathrm{xlog}}_{\mathrm{e}}\mathrm{x}-\mathrm{x}+1\right)=1$
So, in $\left(0,1\right)$,$f\left(x\right)$ decreases from 1 to 0 and in $(1,\mathrm{\infty })$increases from 0 to $\infty$.
Hence,  $\mathrm{f}\left(\mathrm{x}\right)>0\text{ for }\mathrm{x}\in (0,1)\cup (1,\mathrm{\infty })$