Let $f(x)$ satisfy Rolle's theorem conditions and have three successive solutions $x_1, x_2, x_3$. How to prove that $f'(x)$ is differentiable?

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

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)?

• You won’t, unfortunately. – Michael Hoppe Jul 12 at 5:58
• @MichaelHoppe How can we prove that its not then? – Dimitris Jul 12 at 5:59
• 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. – eminem Jul 12 at 6:00
• It makes sense, Slim Shady – Dimitris Jul 12 at 6:01
• 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. – Dimitris Jul 12 at 6:09

1 Answer

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).