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

calculusrolles-theorem

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)?

Best 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).

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