Applying Rolle's Theorem, prove that the given equation has only one root:
$$e^x=1+x$$
By inspection, we can say that $x=0$ is one root of the equation. But how can we use Rolle's theorem to prove this root is unique?
[Math] Using Rolle’s theorem to show $e^x=1+x$ has only one real root
applicationscalculusrolles-theorem
Related Question
- Real Analysis – Proving an Equation Has Exactly One Real Solution Using Rolle’s Theorem
- [Math] prove to have at least one real root by Rolle’s theorem
- [Math] Using Rolle’s theorem and IVT, show that $x^4-7x^3+9=0$ has exactly $2$ roots.
- Using Rolle’s Theorem to prove that there are two roots to a function.
Best Answer
Let $f(x) = e^x - 1 - x$, and we observe that $f(0)=0$. $f$ is also obviously continuous and differentiable over the real numbers (if you wish to verify that in detail, you can do that separately).
Suppose there exists a second root $b \neq 0$ such that $f(0) = f(b) = 0$. Then there exists some $c \in (0,b)$ (or $(b,0)$ if $b<0$) such that $f'(c) = 0$ by Rolle's theorem.
$f'(x) = e^x - 1$, however, which satisfies $f'(x) = 0$ only when $x=0$, which is not in any interval $(0,b)$ (or $(b,0)$).
Thus, since no satisfactory $c$ exists, we conclude the equation only has one real root.