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)