# How do you solve for x?: e^(-x)+x = 1/3?

This equation has no real solutions.

#### Explanation:

Let's rewrite the equation as
${e}^{- x} + x - \frac{1}{3} = 0$

We can show that the function $f \left(x\right) = {e}^{- x} + x - \frac{1}{3}$ has a global minimum whose value is positive and so the former equation $f \left(x\right) = 0$ has no real solutions.
To find the extrema of the function we look at the boundary of the domain (in this case, the reals $\mathbb{R}$) and to the stationary points.

$f$ is continuous on $\mathbb{R}$ and ${\lim}_{x \to \pm \infty} f \left(x\right) = + \infty$. This means that $f$ has no global maximum and at least one global minimum.
$f ' \left(x\right) = - {e}^{- x} + 1$ is null when ${e}^{- x} = 1$, i.e. there's only one stationary point in $x = 0$. This has to be the global minimum, because the extrema are always stationary points and we already guaranteed the existence of a global minimum.

${\min}_{x \in \mathbb{R}} f \left(x\right) = f \left(0\right) = {e}^{0} + 0 - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3} > 0$, so the function $f$ is always positive and never crosses the $x$-axis.
We can conclude that $f \left(x\right) = 0$ doesn't have real solutions.