Answer
\[\begin{align}
& \text{Relative maximum at }x=\frac{1-\sqrt{5}}{2} \\
& \text{Relative minimum at }x=\frac{1+\sqrt{5}}{2} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{x{{e}^{x}}}{x-1} \\
& \text{Calculating the first derivative} \\
& f'\left( x \right)=\frac{d}{dx}\left[ \frac{x{{e}^{x}}}{x-1} \right] \\
& f'\left( x \right)=\frac{\left( x-1 \right)\frac{d}{dx}\left[ x{{e}^{x}} \right]-x{{e}^{x}}\frac{d}{dx}\left[ x-1 \right]}{{{\left( x-1 \right)}^{2}}} \\
& f'\left( x \right)=\frac{\left( x-1 \right)\left( x{{e}^{x}}+{{e}^{x}} \right)-x{{e}^{x}}}{{{\left( x-1 \right)}^{2}}} \\
& f'\left( x \right)=\frac{{{x}^{2}}{{e}^{x}}+x{{e}^{x}}-x{{e}^{x}}-{{e}^{x}}-x{{e}^{x}}}{{{\left( x-1 \right)}^{2}}} \\
& f'\left( x \right)=\frac{{{x}^{2}}{{e}^{x}}-x{{e}^{x}}-{{e}^{x}}}{{{\left( x-1 \right)}^{2}}} \\
& \text{Calculating the critical points, set }f'\left( x \right)=0 \\
& f'\left( x \right)={{x}^{2}}{{e}^{x}}-x{{e}^{x}}-{{e}^{x}} \\
& {{e}^{x}}\left( {{x}^{2}}-x-1 \right)=0 \\
& {{x}^{2}}-x-1=0 \\
& \text{By the quadratic formula} \\
& {{x}_{1}}=\frac{1+\sqrt{5}}{2},\text{ }{{x}_{2}}=\frac{1-\sqrt{5}}{2} \\
& \text{And the derivative is not defined for }x=1 \\
& \text{Set the intervals }\left( -\infty ,\frac{1-\sqrt{5}}{2} \right),\left( \frac{1-\sqrt{5}}{2},1 \right) \\
& \text{and}\left( 1,\frac{1+\sqrt{5}}{2} \right),\left( \frac{1+\sqrt{5}}{2},\infty \right) \\
& \text{Making a table of values} \\
& \begin{matrix}
\text{Interval} & \text{Test Value} & \text{Sign of }f'\left( x \right) & \text{Conclusion} \\
\left( -\infty ,\frac{1-\sqrt{5}}{2} \right) & x=-2 & + & \text{Increasing} \\
\left( \frac{1-\sqrt{5}}{2},1 \right) & x=0 & - & \text{Decreasing} \\
\left( 1,\frac{1+\sqrt{5}}{2} \right) & x=\frac{3}{2} & - & \text{Decreasing} \\
\left( \frac{1+\sqrt{5}}{2},\infty \right) & x=2 & + & \text{Increasing} \\
\end{matrix} \\
& \text{Therefore,} \\
& \text{Relative maximum at }x=\frac{1-\sqrt{5}}{2} \\
& \text{Relative minimum at }x=\frac{1+\sqrt{5}}{2} \\
\end{align}\]
