# Using Rolle's theorem to show $e^x=1+x$ has only one real root

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?

• It is $$\exp(x)\geq 1+x$$ for all real $x$ – Dr. Sonnhard Graubner Mar 31 '19 at 6:31

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.

• I don't understand the second para. – pi-π Mar 31 '19 at 6:07
• We want to show that there exists no second (unique) root, so we seek a contradiction by supposing it exists. Okay, so if the second root is not unique, it is some real number $b$ that is not equal to our first root, $0$. If $b$ is a root, then we are ensured $f(b) =0$. Coincidentally, $f(b) = f(0)$, which gives us a situation in which Rolle's theorem applies. Then, there exists some point $c$ between $b$ and $0$ such that the derivative of $f$ is equal to zero. – Eevee Trainer Mar 31 '19 at 6:09
• Do we not need to check for continuity and differentiability of $f(x)$ in $[0,b]$ and $(0,b)$ respectively before applying Rolle's Theorem? – pi-π Mar 31 '19 at 6:14
• Yeah, technically you do if you want to be rigorous (and that's a fair point to bring up). Though in this case it's one of those cases where it's "obvious" in the sense that $f$ is obviously continuous and differentiable over $\Bbb R$. I suppose whether you want to prove that, or just state it as an obvious thing, depends on the rigor expected of you in your course. – Eevee Trainer Mar 31 '19 at 6:25
• With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it. – pi-π Mar 31 '19 at 6:28