Recently I've stumbled across a problem asking how many real solutions there was to $x(e^x-e^{-x})-e^x = 0 $, the answer is at least two, and it's easy to verify when plotting the function. However, in exams conditions, I would not have the opportunity to plot and was wondering if there is an "easy" way to find it?

Personally, I did use the intermediate values theorem. But here's the problem. I found that there was a $x_0 \in \mathbb{R}$ where $f(x_0)$ was negative and two values such as $ a < x_0 < b$ where $f(a) > 0$ and $f(b) > 0$

I think it's okay but I would be glad if there was a better way, not relying on roughly knowing how much $e^2$, etc is equal to. In this case it was fairly easy to find the values but it's always stressing me out to think I could've not think of a specific value.

  • $\begingroup$ It could be a good idea to divide the equality by $e^x$ first. This can be done, since $e^x$ cannot be $0$ $\endgroup$
    – Peter
    Jan 4, 2022 at 14:53
  • $\begingroup$ Yeah that's a nice idea to simplify it at first thanks you $\endgroup$ Jan 4, 2022 at 15:02

1 Answer 1


Your problem is equivalent than showing that $$f(x):=x(1-e^{-2x})=1,$$ has several solutions. It's rather clear that $f(0)=0$ and $f(x)\to +\infty$ whenever $x\to \pm \infty $. So, oblviously, $f(x)=1$ has at least two solutions. It's also quite clear that $f(x)=1$ has exactly two solutions since $f'(x)<0$ whenever $x<0$ and $f(x)>0$ whenever $x>0$.

  • 1
    $\begingroup$ Way more clear that way thanks you, I didn't not thought of making the function easier to visualize ! $\endgroup$ Jan 4, 2022 at 15:03

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