Maximize $\frac{xe^x}{e^x-1}$

$$\cfrac{\mathrm d}{\mathrm dx} \cfrac{xe^x}{e^x-1} = \cfrac{e^{2x}-(x+1)e^x}{(e^x-1)^2} = 0$$

From L'Hospital's we get $$x=0$$ as a root, but to find the other roots I have to solve $$e^x-x-1=0$$. How can I do this, or rather, how can I decide I can't?

• There are no roots other than what you just found, you can prove this function is always positive, by proving it is decreasing towards $(0,0)$ and increasing from $(0,0)$ - I am taking about $e^x -x-1$ Sep 1, 2020 at 23:32
• Look at the graphs of $y= e^x$ and $y= x+1$. You want to show they don't intersect when $x$ is strictly positive. Consider their slopes Sep 1, 2020 at 23:34
• There is no maximum. But there is a minimum at $x=0$. Sep 1, 2020 at 23:46

From expression you obtain for derivative follows that function is increasing. Also $$\lim\limits_{x \to \infty}{\cfrac{xe^x}{e^x-1}}=\infty$$ So ?
$$\frac {xe^{x}} {e^{x}-1} =\frac x {1-e^{-x}} \to \infty$$ as $$x \to \infty$$. So there is no maximum.
Hint: The tangent to the curve $$f(x)=\exp(x)$$ at $$x=0$$ is the line $$y=x+1$$.
(slope of tangent to $$f(x)$$ at $$x=c$$ is $$f'(c)$$)
$$0$$ is the only solution. However, in case you have equations like that involve both $$\exp(x)$$ and $$x$$, always pay a visit to the Lambert W function.