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Show that the equation $x+e^x=2$ has only one solution in $\mathbb{R}$. Which solution is this?

I don't know how to show this formally but if we take $e^x=2-x$, than the left part ist increasing and the right part decreasing. This implies that we have only one solution in $\mathbb{R}$. Right? How to show this formally and how to find this solution?

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    The solution is between $0$ and $1$ and I don't know if you can make it explicit. Else you can see that $x + e^x - 2$ is strictly increasing and negative for e.g $x = 0$ and positive for $x = 1$ so it has a unique zero. – 2017-01-15
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    Maybe it will not be so helpful, but I can give you some bounds. The solution is srictly less than $0,5$ and strictly greater than $0,4$. – 2017-01-15

4 Answers 4

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Showing that this equation has a unique real solution is easy : the map $$f:\mathbb{R}\to\mathbb{R},x\mapsto x+e^x$$ is continuous, strictly increasing and verifies $\lim_{-\infty}f=-\infty$, $\lim_{+\infty}f=+\infty.$

So $f$ is a bijection : for each real number $y$ there exists a unique $x\in\mathbb{R}$ such that $f(x)=y$.

Now finding a closed form for $f^{-1}(2)$ seems to me unreachable.

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Use Rolle's Theorem and the Intermediate Value Theorem. Consider $f(x)=e^x+x-2$. Show that it has one root using IVT. Then assume it has at least two, so by Rolle's theorem there must exist $c \in \mathbb{R}$ such that $f'(c)=0$. Show that leads to a contradiction.

Let's say we have two functions $f(x)$ and $g(x)$ where $f'(x)>0$ and $g'(x)<0$. Define $h(x)=f(x)-g(x)$ assume we know $h$ has at least one root. Then assume it has two roots, by Rolle's Theorem we must have $h'(c)=0$ for some $c$. But $f'(x) > g'(x) \implies f'(x)-g'(x)=h'(x)>0$. Hence there cannot be two roots. So your intuition is correct.

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Consider the function $f(x):=e^x+x-2$.

We calculate two limits:

$$\lim\limits_{x\rightarrow -\infty}f(x) = -\infty$$

$$\lim\limits_{x\rightarrow\infty}f(x) = \infty$$

As $f$ is the sum of continuous functions and thus continuous, there must exist a $c\in\mathbb{R}$ such that $f(c) = 0$.

Now we calculate the first derivative.

$$f'(x) = e^{x}+1$$

Since $f'(x)>0$ for any $x\in\mathbb{R}$, $f(x)$ is a strictly increasing function. Thus it has at most one zero.

Combining these two facts gives you that $f(x)$ has precisely one zero.

Writing this solution down explicitly is probably not possible. Numerical methods can help you approximate it fairly easily, though.

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It is easy to see that $f(x)=x+e^x$ is a strictly increasing function. Indeed, $f'(x)=1+e^x>0$. Next, $f(0)=1<2$ while $f(1)=1+e>2$, By the intermediate value theorem (see the @Ahmed S. Attaalla's answer) $f$ admits the value $2$. The solution is unique because $f$ is strictly increasing.