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 $$f'(c_1) = f'(c_2) = 0$$


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)?

  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.