- #1

evinda

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Let $I$ an interval and $f: I \to \mathbb{R}$ a differentiable (as many times as we want) function.

If $\xi \in I$ with $f''(\xi) \neq 0$, then there are $a,b \in I: a< \xi <b$ and $\frac{f(b)-f(a)}{b-a}=f'(\xi)$.

Hint: First suppose that $f'(\xi)=0$ and show that at $\xi$ we have a local extremum. Then find (for example with the intermediate value theorem) $a,b$ with $a< \xi<b$ and $f(a)=f(b)$, therefore... etc.

For $f'(\xi) \neq 0$, consider the function $g(x)=f(x)-f'(\xi)x$ to get reduced to the previous case.So we suppose that $f'(\xi)=0$.

How do we show that $f$ has a local extremum at $\xi$ ?

I have thought to calculate the Taylor series of second order.

Then, $f(x)=f(\xi)+f'(\xi) (x-\xi)+\frac{f''(\xi)}{2!}(x-\xi)^2=f(\xi)+\frac{f''(\xi)}{2}(x-\xi)^2$.

From this we get that $f(x)-f(\xi) \neq 0$.

But this does not help somehow, does it?

How else can we start the proof? (Thinking)