Original question:

$f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Show that for any $c \in (a,b)$ that is not a point of maximum or minimum for $f'$, there exist $x_1, x_2 \in (a,b)$ such that $f'(c)=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$

What I have done:

We argue by contradiction. Suppose there doesn't exist such $x_1$ and $x_2$, then $\forall x_1,x_2 \in (a,b)$, we have $f'(c)=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. By assumption, since $f$ is differentiable on $(a,b)$, then according to Mean Value Theorem, $\exists k \in (a,b)$ such that $f'(k)=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Since $f'(c) \ne f'(k)$, we must have either $f'(c) < f'(k)$ or $f'(c) > f'(x)$. But we know that $c$ is not a point for maximum or minimum for $f'$, so this is a contradiction.

My concern:

Is it possible that we are missing some of the $k$ in $(a,b)$? Because if that's the case, then we wouldn't end up with a contradiction any more. I'm thinking since the choise of $x_1, x_2$ are arbitrary and since $k$ must lie between them, so we should be able to capture all the $k$'s, but how should I formalize that?

  • $\begingroup$ Some $k$ can be missed - for example, $f(x)=\sin x$ over $[0,3\pi]$ and $c=\pi$, $k=2\pi$. However I think if there is such $k$, then $f'(k)$ must be maximum of minimum of $f'$ (I try the proof but I have no time :( ) $\endgroup$ – Hanul Jeon May 29 '14 at 23:13

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