How to solve for x: $e*x + e^{-x} = 0$

Question

I already know the answer is supposed to be $x=-1$, but I have no idea how to get to that. What I've done so far is:

$ln(e^{-x}) = ln(e)+ln(-x)$
$-x = 1 + ln(-x)$
$ln(-x) + x = -1$ or $ln(x * -e^x) = -1$

I just don't know what to do from here. What can I use to now conclude that indeed $x=-1$?

Answer 1

Let $f(x)= ex+e^{-x}$. Then $$f'(x)=e-e^{-x}$$

It is easy to prove that $f'(x) <0$ on $(-\infty,-1)$ and $f'(x) >0$ on $(-1, \infty)$. This implies that $f$ is strictly decreasing on $(- \infty, -1]$ and strictly increasing on $[-1, \infty)$.

Answer 2

Set $x={-y}-1$, then it becomes $$e^y=1+y$$ But $e^y\ge1+y$ with equality only when $y=0$ because $e^y$ is a convex function and $1+y$ is its tangent line at $y=0$.

Answer 3

Consider $f(x)=ex+e^{-x}$ then we have $f'(x)=e-e^{-x}$ and $f''(x)=e^{-x}>0$ thus the equation $f'(x)=0$ has only the solution $x=-1$ and this gives us a local minimum and $f(-1)=0$.