Solve $e^x+x=1$

This seems to have stumped even my TA, so I'm asking it here.

Given $e^x + x = 1$, solve for $x$.

I already know that the answer is zero, but have no idea how to get there.

Solution 1:

Using the series expansion we have:

$$1+x+\frac{x^2}{2!}+ \dots + \frac{x^n}{n!}+\dots = 1-x$$

If $x$ is positive it is immediately obvious that there can be no equality.

If $x<0$ then the RHS is greater than 1 and $e^{x}<1$.

This is not strictly an "algebraic" solution, but with the term in $e^x$ we do not expect anything purely algebraic.

Solution 2:

"Lambert W" is a hint for "algebraic solution".
The solution for $\mathrm{e}^x + x = 1$ is $1-\mathrm W(\mathrm{e})$,
to find ALL complex solutions, use all branches of the Lambert W ...

$$ \begin{align*} &\dots \\ 1 - \mathrm{W}_{-4}(\mathrm{e}) &= 3.159947300 + 23.47017395 i \\ 1 - \mathrm{W}_{-3}(\mathrm{e}) &= 2.849014724 + 17.17149358 i \\ 1 - \mathrm{W}_{-2}(\mathrm{e}) &= 2.393982241 + 10.86800606 i \\ 1 - \mathrm{W}_{-1}(\mathrm{e}) &= 1.532092122 + 4.597158013 i \\ 1 - \mathrm{W}_{0}(\mathrm{e}) &= 0.000000000 \\ 1 - \mathrm{W}_{1}(\mathrm{e}) &= 1.532092122 - 4.597158013 i \\ 1 - \mathrm{W}_{2}(\mathrm{e}) &= 2.393982241 - 10.86800606 i \\ 1 - \mathrm{W}_{3}(\mathrm{e}) &= 2.849014724 - 17.17149358 i \\ 1 - \mathrm{W}_{4}(\mathrm{e}) &= 3.159947300 - 23.47017395 i \\ 1 - \mathrm{W}_{5}(\mathrm{e}) &= 3.396557044 - 29.76478701 i \\ &\dots \end{align*} $$

explanation

$\mathrm{e}^x+x=1$
$\mathrm{e}^x=1-x$
$\mathrm{e} = (1-x)\mathrm{e}^{1-x}$
$\mathrm{W}(\mathrm{e}) = 1-x$
$x = 1-\mathrm W(\mathrm{e})$

Solution 3:

You can see this very easily graphically. The equation is $$e^x=1-x$$ and the two sides of the equation are plotted here (from Wolfram Alpha):

The intuition for a formal proof also follows directly from the picture (the functions are both monotonic but in opposite directions), if that's your aim.

Solution 4:

Let $f(x) = e^x + x - 1$. Then, for any given $x$, $f(x) = 0$ if and only if $e^x + x = 1$.

You have already noticed that $f(0) = 1 + 0 - 1 = 0$, so it is a solution.

Now, we turn to calculus, not algebra. We have $f'(x) = e^x + 1$. Since $e^x > 0$ for all $x$, we know that $e^x + 1 > 0$ as well. In other words, $f'(x)$ is positive for all $x$ which tells us that $f(x)$ is an increasing function on the entire real line. Therefore, it could only possibly be 0 at one point, and you already found that point.

Now, if you haven't had calculus, you could still get the same basic idea. For example, you know $y = x$ is increasing. That is something you should know. Perhaps you have learned that $y = e^x$ is always increasing as well, because even in an algebra class, they would probably give you a bunch of properties of $y = e^x$ when they introduce it. Add these two functions together, and it's still increasing. Subtract 1, and the function is simply translated downward 1 unit, so it's still increasing everywhere. Again, the conclusion is the same.

Solution 5:

Just to add weight to @Jonathan’s response: If $f$ and $g$ are an increasing and a decreasing function on $\mathbb{R}$ respectively, their graphs can cross at only one point. Inspection finds that point to be $(0,1)$, and you’re done.