# $x(e^x - e^{-x})-e^x = 0$ has two reals solutions but am I supposed to use the intermediate value theorem?

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.

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

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 $$x>0$$.