# Applying Rolle's theorem

This is the first time that I'm doing a proof-problem on my own and I'm not really sure how to check if my answer is correct, so I was wondering if someone could tell me with certainty if my proof is correct.

The problem states:

Let $f:\mathbb{R} \to \mathbb{R}$ be a twice differentiable function (on its entire domain) such that $f(0) = f(1) = f(2)$. Prove that there exists some point $x_0 \in (0, 2)$ such that $f''(x_0) = 0$.

So I developed my proof based on Rolle's theorem that states that if a function $f$ is continous on a closed interval $[a, b]$ and differentiable on an open interval $(a, b)$ and if $f(a) = f(b)$ then there exists some point $c \in (a, b)$ such that $f'(c) = 0$.

Proof:

1. $f$ is differentiable $\forall x \in \mathbb{R}$, meaning $f$ must be differentiable on $(0, 1)$ and $(1, 2)$

2. $f$ is differentiable $\forall x\in\mathbb{R}$ meaning $f$ must be continuous $\forall x \in \mathbb{R}$ meaning $f$ must be continuous on $[0, 1]$ and $[1, 2]$.

3. $f(0) = f(1)$ and $f(1) = f(2)$.

From 1., 2. and 3. we conclude that $f$ satisfies the conditions of Rolle's theorem for both of these intervals, meaning: $\exists c_1 \in (0, 1) : f'(c_1) = 0$ and $\exists c_2 \in (1, 2) : f'(c_2) = 0$

Let $F(x) = f'(x)$. We know that the original function is twice differentiable $\forall x \in \mathbb{R}$, therefore $F(x)$ is differentiable on $\mathbb{R}$ meaning it must be differentiable on $(c_1, c_2)$ also meaning it must be continuous on $[c_1, c_2]$.

Now, since $f'(c_1) = f'(c_2) = 0$, the function $F(x)$ satisfies the conditions of Rolle's theorem (on the interval $(c_1, c_2)$) meaning :

$\exists x_0 \in (c_1, c_2) \subset(0, 2): F'(x_0) = f''(x_0) = 0$.

• Hello there. Please do not type all the proof in math mode. This makes the post hard to read. Just type those symbols in mathjax. . Thanks.
– xbh
Sep 16 '18 at 15:31
• Usually, new users have to be encouraged to use mathjax. In your case, I'd say you that you are using mathjax too much. Please, write text without mathjax. Sep 16 '18 at 15:32
• After a cursory reading, i would say the proof seems fine. Although it can be neater.
– xbh
Sep 16 '18 at 15:36
• I've edited your answer, now it's more legible. I would say your proof is right Sep 16 '18 at 15:36
• Your proof is solid. Sep 16 '18 at 15:47