0
$\begingroup$

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

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?

$\endgroup$
3
  • 1
    $\begingroup$ 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 $e^x -x-1$ $\endgroup$
    – CSch of x
    Sep 1, 2020 at 23:32
  • 1
    $\begingroup$ 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 $\endgroup$ Sep 1, 2020 at 23:34
  • $\begingroup$ There is no maximum. But there is a minimum at $x=0$. $\endgroup$
    – Crostul
    Sep 1, 2020 at 23:46

3 Answers 3

4
$\begingroup$

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 ?

$\endgroup$
3
$\begingroup$

$\frac {xe^{x}} {e^{x}-1} =\frac x {1-e^{-x}} \to \infty $ as $x \to \infty$. So there is no maximum.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.