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Suppose that $a<b<c$ for real numbers $a, b, c$. Suppose that $f(x)$ is continuous on $[a, c]$ and differentiable on $(a, c)$. Then prove that there exists $a<\alpha<\beta<c$ such that $$\frac{f(b)-f(a)}{b-a}=f'(\alpha), \frac{f(c)-f(a)}{c-a}=f'(\beta)$$

I understand that there exists $\alpha$, $\beta$, but how would I prove that $\alpha<\beta$? Do I have to divide into cases? Can anyone give me any hints?

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  • $\begingroup$ Suppose $f(x)=1$ on $[a,b]$ and $f(x)=2$ for $x>b$. Then $\frac{f(c)-f(a)}{c-a}=\frac{1}{c-a}\ne0$, but $f'(x)=0$ everywhere it is defined- and you are not requiring it to be defined at $x=b$. So the required $\beta$ cannot exist. $\endgroup$
    – almagest
    May 28, 2016 at 6:32
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    $\begingroup$ Should the function be defined on $[a, c]$ and differentiable on $(a, c)$ instead? For all we know, $f(c)$ need not be even be defined. $\endgroup$
    – Axoren
    May 28, 2016 at 6:39
  • $\begingroup$ @Axoren: Thanks for pointing it out. Edited it. $\endgroup$
    – zxcvber
    May 28, 2016 at 6:56
  • $\begingroup$ Hint: derivatives have the intermediate value property. $\endgroup$ May 28, 2016 at 7:46
  • $\begingroup$ @DavidSchneider-Joseph I don't the using Darboux's Theorem would be fair here... Anyway, OP didn't ask for an elementary proof so it might be eligible. $\endgroup$
    – BigbearZzz
    May 28, 2016 at 7:48

3 Answers 3

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Define $S(x, y) = \frac {f(y) - f(x)} {y - x}$.

We have $\alpha \in (a, b), \gamma \in (b, c)$ such that $f'(\alpha) = S(a, b)$ and $f'(\gamma) = S(b, c)$.

$S(a, c)$ lies between $S(a, b)$ and $S(b, c)$, so the intermediate value property for derivatives gives us $\beta \in (\alpha, \gamma)$ such that $f'(\beta) = S(a, c)$.

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  • $\begingroup$ Does the converse of Mean Value Theorem always hold? There exists $\beta \in (\alpha, \gamma)$ but why does $f'(\beta)$ have to be $S(a, c)$? $\endgroup$
    – zxcvber
    May 28, 2016 at 8:10
  • $\begingroup$ I'm not using the converse of anything. I'm using the intermediate value property of derivatives (given by Darboux's theorem) and the fact that $S(a, c)$ lies between $f’(\alpha)$ and $f’(\gamma)$. $\endgroup$ May 28, 2016 at 8:27
  • $\begingroup$ Then does $\alpha < \gamma$ always hold in your proof? $\endgroup$
    – zxcvber
    May 28, 2016 at 8:52
  • $\begingroup$ Yes, because $\alpha$ was given by MVT in $(a, b)$, and $\gamma$ by MVT in $(b, c)$. $\endgroup$ May 28, 2016 at 9:10
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    $\begingroup$ @Zack Ni some simple algebra can show it's a convex combination of the two. (Intuitive explanation: if you break a trip up into two legs, the average speed over the whole trip can't be greater than the average speed of both legs.) $\endgroup$ May 28, 2016 at 17:00
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Is your function convex or concave? For convex function it's proved easily using the second order derivative. But for concave,I don't know

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  • $\begingroup$ There weren't any further conditions for the function $f$. I was thinking about the solution for functions that can have both concave and convex intervals in $[a, c]$. $\endgroup$
    – zxcvber
    May 28, 2016 at 6:32
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The fact that $\alpha<\beta$ should follow directly from the mean value theorem.

You said that you can easily observe that such $\alpha$ and $\beta$ exist. I am assuming that you are doing so by considering $f$ restricted to the intervals $[a,b]$ and $[b,c]$. In which case, we get that $\alpha\in(a,b)$ and $\beta\in(b,c)$. And so $\alpha<b$ and $b<\beta$, which together imply $\alpha<\beta$.

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  • $\begingroup$ How does one know that $\beta$ is in $[b, c]$? $\endgroup$
    – zxcvber
    May 28, 2016 at 7:17
  • $\begingroup$ I am sorry, I misread the question. $\endgroup$ May 28, 2016 at 7:22

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