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I find the proof of the Mean Value Theorem not intuitive because it uses Rolle's theroem on an auxiliary function.I am sure that there must be another proof which is longer and intuitive but I can't find it in any calculus or analysis book.I wonder if anyone can show me such a proof or perhaps tell me where I can find one that doesn't use Rolle's theorem or at least is more intuitive (perhaps the original proof).

Narek Margaryan
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  • Well, Rolle's theorem is a particular case of Lagrange's mean value theorem. Hence the two statements are actually equivalent. – Siminore Aug 04 '13 at 11:11
  • Would the following argument suit you: Given $f$ on $[a,b]$ there has to be a point $c\in\ ]a,b[\ $ where $f$ deviates most from its linear interpolation between $a$ and $b$. – Christian Blatter Aug 04 '13 at 11:24
  • The author of the popular proof of the MVT is Ossian Bonnet. His proof is shown for the first time by J.A.Serret in his Cours de Calcul Différentiel et Intégral (1868) quoting Bonnet but without producing a work of his (in fact this doesn't exist). – Tony Piccolo Aug 04 '13 at 16:38
  • It is less known that E.Galois (!) used the same idea of auxiliary function in a writing published in Gergonne's Annales in December 1830 link pp. 182-183: are there sufficient grounds for a quarrel about priority ? – Tony Piccolo Aug 05 '13 at 23:13

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Lagrange's Mean Value Theorem is nothing but a tilted version of Rolle's Theorem. If the proof you have does not include a drawing, make some drawing for yourself and then it should be clear where the auxiliary function is coming from.

As for the original proofs, it is highly unlikely that the original will be easy to read. The notation will be very different than what you are used to (not to mention that it will be in French) and the standards of rigor completely different. There is a lot to learn from the original work of the great masters of the calculus but I don't think that is what you are looking for right now.

Ittay Weiss
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Let $f$ satisfy the hypotheses of the MVT on the interval $[a,b]$. Define $g(x):=f(x)-\frac{f(a)-f(b)}{a-b}x$

Case 1: $g$ attains its maximum and minimum at $a$ and $b$. Then $g$ is constant. In this case we get $0=g'(x)=f'(x)-\frac{f(a)-f(b)}{a-b}$ at every point in $[a,b]$.

Case 2 If the maximum or the minimum is attained at an interior point $c$, then $g'(c)=0$. Therefore $f'(c)=\frac{f(a)-f(b)}{a-b}$.

OR.
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