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.