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Rolle's Theorem

Let $f(x):[a,b]\to\mathbb{R}$ where $f$ is differentiable at $(a,b)$ and continuous at $[a,b]$, with $f(a) = f(b)$.

We know from Rolle's theorem that $\exists$ at least one $x_o: f'(x_0)=0$


The problem

Let $x_1,x_2,x_3$ be the successive solutions of $f$.

  1. Prove $f''(x)$ has at least one solution

Solution Attempt

From Rolle's theorem is is obvious that $f'(x)$ has at least two solutions $$f'(c_1) = f'(c_2) = 0$$

$$x_1<c_1<x_2<c_2<x_3$$

Therefore, if we could prove $f'(x)$ is differentiable, then $f'(x)$ would also satisfy Rolle's theorem. Thus, we will be able to prove that $f''(x):(c_1,c2)\to\mathbb{R}$ has also at least one solution.


The Question

How to prove that $f'(x)$ is differentiable (given the fact $f$ satisfies Rolle's conditions)?

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    $\begingroup$ You won’t, unfortunately. $\endgroup$ – Michael Hoppe Jul 12 at 5:58
  • $\begingroup$ @MichaelHoppe How can we prove that its not then? $\endgroup$ – Dimitris Jul 12 at 5:59
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    $\begingroup$ I dont see any reason for $f'$ to be differentiable (or even continuous) as Rolle's theorem does not have this cosequence. You will need to find a counter exmaple. $\endgroup$ – eminem Jul 12 at 6:00
  • $\begingroup$ It makes sense, Slim Shady $\endgroup$ – Dimitris Jul 12 at 6:01
  • $\begingroup$ If any of you wants to post an answer, I'll accept it. I was just wondering if I am missing any Rolle's theorem consequence. $\endgroup$ – Dimitris Jul 12 at 6:09
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You will have to assume from the beginning that both $f$ and $f'$ satisfy the hypothesis of Rolle's theorem. For a counter-example, think of a function $g$ which is continuous, but not differentiable (like $|x|$), but modify it slightly (think of "piecing together" a few absolute value functions, so that the graph looks like a bunch of letter "W" stuck side-by-side, like a jagged sine graph). Then, integrate take $f(x):= \int_0^x g(t)\, dt$.

Then, $f$ is not twice differentiable on $\Bbb{R}$, but wherever $f''(x)$ exists, it is always $\pm 1$ (in particular non-zero).

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