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 $f'(c) = 0$.


  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 :

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

  • 1
    $\begingroup$ 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. $\endgroup$
    – xbh
    Sep 16 '18 at 15:31
  • 5
    $\begingroup$ 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. $\endgroup$
    – ajotatxe
    Sep 16 '18 at 15:32
  • 1
    $\begingroup$ After a cursory reading, i would say the proof seems fine. Although it can be neater. $\endgroup$
    – xbh
    Sep 16 '18 at 15:36
  • 1
    $\begingroup$ I've edited your answer, now it's more legible. I would say your proof is right $\endgroup$ Sep 16 '18 at 15:36
  • 1
    $\begingroup$ Your proof is solid. $\endgroup$ Sep 16 '18 at 15:47

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