Question : Prove that the equation $x + \cos(x) + e^{x} = 0$ has exactly one root

This is what I thought of doing:

$$\text{Let} \ \ \ f(x) = x + \cos(x) + e^{x}$$ By using the Intermediate Value Theorem on the open interval $(-\infty, \infty)$, and then by showing that

$$\left(\lim_{x \to -\infty}f(x) < 0 < \lim_{x \to +\infty}f(x)\right) \lor \left(\lim_{x \to +\infty}f(x) < 0 < \lim_{x \to -\infty}f(x)\right)$$

I could show that $\exists\ x \in \mathbb{R} \ s.t.\ f(x) = 0$.

However this method, although it does show the existence of an $x$ such that $f(x)=0$, it doesn't show that there is only one $x$ that satisfies the statement $f(x)=0$.

The original question, suggest's using either Rolle's Theorem or the Mean Value Theorem, however we face the same problem with both theorems as both theorems prove the existence of at least a single $x$ (or any arbitrary number) satisfying their given statements, they don't prove the existence of only one $x$ satisfying their statements.

All three theorem's I've mentioned here :

  1. Intermediate Value Theorem
  2. Rolle's Theorem
  3. Mean Value Theorem

Can all be used to show that the equation $x + cos(x) + e^{x} = 0$ has at least one root, but they can't be used to show $x + cos(x) + e^{x} = 0$ has only one root. (Or can they?)

How can this problem be solved then, using either Rolle's Theorem or the Mean Value Theorem (or even the Intermediate Value Theorem)

  • $\begingroup$ Question: Does an equaltion have roots or solutions? I've always thought it was the latter. $\endgroup$
    – zhw.
    May 17, 2016 at 20:59

2 Answers 2


Hint: Consider the derivative: $$ 1-\sin(x)+e^x. $$ Since $1\geq \sin(x)$, $1-\sin(x)\geq0$. Therefore, the derivative is always positive. If there were two roots, then there would be a place where the derivative is zero. (why?)

  • $\begingroup$ Is there a solution working without derivatives (?) $\endgroup$
    – Imago
    May 17, 2016 at 19:53
  • $\begingroup$ @Imago Unlikely. $\endgroup$ May 17, 2016 at 19:57
  • $\begingroup$ How do you see that it is unlikely? I find it easily believeable that there is a root between $\frac{-\pi}{2} $ and $0$ and beyond $\frac{-\pi}{2} $ unlikely to be another one. But I think argumentation would go in a similar direction as using derivatives... mhm.. $\endgroup$
    – Imago
    May 17, 2016 at 20:02

Consider the first derivate of it: $f'(x)=1-\sin(x) + e^{x} \ge e^{x}$, for $x \in R$. So function is grow up , now you should use the Rolle's theorem

  • $\begingroup$ @MichaelBurr yes $\endgroup$
    – openspace
    May 17, 2016 at 19:45

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