# Question #53368

Apr 2, 2017

#### Explanation:

The equation $x = {e}^{-} x$ has a real root in $\left[0 , 1\right]$ if and only if the function
$f \left(x\right) = {e}^{-} x - x$
has a zero in $\left[0 , 1\right]$.
$f \left(0\right) = {e}^{- 0} - 0 = 1$
and
$f \left(1\right) = {e}^{- 1} - 1 = \frac{1}{e} - 1 < 1 - 1 = 0$,
so $f \left(x\right)$ undergoes at least one sign change in $\left[0 , 1\right]$. This is a consequence of the continuity of $f \left(x\right)$ and the Intermediate value theorem , which in this case says the following:

Since $f \left(0\right) > 0$ and $f \left(1\right) < 0$, there exists some real number ${x}_{m}$ in $\left(0 , 1\right)$ such that $f \left({x}_{m}\right) = 0$, and we have proved the existence of a real zero of $f \left(x\right)$ in $\left[0 , 1\right]$.

Due to the relation between the zeros of $f \left(x\right)$ and the roots of $x = {e}^{-} x$, we can conclude that $x = {e}^{-} x$ has a real root in $\left[0 , 1\right]$.

EDIT: As pointed out by George, the function $f \left(x\right)$ must be continuous and defined on the entire interval of interest. You can, as an exercise, show that this is in fact the case. One shortcut is to use that $f \left(x\right)$ is a sum of two analytic functions in combination with (https://proofwiki.org/wiki/Combination_Theorem_for_Continuous_Functions/Sum_Rule) .