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


*

*Intermediate Value Theorem

*Rolle's Theorem

*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)
 A: 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?)
A: Let $f(x)=x+\sin x + e^x$, then $f'(x)=1-\cos x+ e^x>0$ implies that $f(x)$ is monotonicaly increasing so $f(x)=0$ has almost one real root. Since $f(-\infty)=-\infty$  and $f(\infty)=+\infty$, so by IVT this eq. has one real root. Hence the eq. Has exactly one real root.
A: The Intermediate Value Theorem give you existence of one $x$ satisfying their statements, for the unicity you have to show the bijection of $f(x)=1+\cos x+ e^x$
Consider this fonction $$f(x)=x+\cos x+e^x$$
For all $ x\in \mathbb{R}, f'(x)=1-\sin(x) + e^{x}$ and $|sinx|\le 1$ then  $1-\sin x\ge0\Longrightarrow f'(x)\ge0$.
So, $\displaystyle{\lim _{x\to +\infty} f(x)=\lim _{x\to +\infty}x\left(\dfrac{1}{x}+\dfrac{\cos x}{x}+\dfrac{e^x}{x}\right)=+\infty}$ and $\displaystyle{\lim _{x\to -\infty} f(x)=\lim _{x\to -\infty}x\left(\dfrac{1}{x}+\dfrac{\cos x}{x}+\dfrac{e^x}{x}\right)=-\infty}$
Finally $f$ increasing and $f(\mathbb{R})=\mathbb{R}$, hence they are only one root for $f(x)=0$ i.e $1+\cos x +e^x=0.$
A: 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
A: The given equation can be written as $$x+\cos x=-e^x$$
$x+\cos x$ is a familiar increasing graph.
$-e^x$ is also easy to draw.
They intersect just once.
