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What is the best way to prove the mean value theorem.

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What exactly do you mean by 'best'? –  Aryabhata Jan 18 '11 at 0:53
    
I guess I meant easiest/most common –  DogDog Jan 18 '11 at 1:25
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Have you looked up any proofs yet? Are you looking for something easier than what you already have? –  Jonas Meyer Jan 18 '11 at 1:27
    
In English, "demonstrate" and "prove" are not synonyms. To "demonstrate the Mean Value Theorem" would be to show how it works (e.g., give a function that satisfies the hypothesis and show that the conclusion holds). To prove it is to provide a proof of the statement. –  Arturo Magidin Jan 18 '11 at 5:51

2 Answers 2

up vote 10 down vote accepted

The most common way to prove the Mean Value Theorem is to first prove Rolle's Theorem; then, given the arbitrary setting of the Mean Value Theorem, you just "tilt your head and apply Rolle's Theorem".

That is, you draw the line joining $(a,f(a))$ and $(b,f(b))$, and let $g(x)$ be the function whose value is the vertical distance between $f(x)$ and this line. Then $g(x)$ is differentiable, and has $g(a)=g(b)=0$; applying Rolle's Theorem to $g$ and then translating this into what it means for $f(x)$ yields the Mean Value Theorem.

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+1 "tilt your head and apply Rolle's Theorem" just made me laugh a lot. I'll try to see what effect this has on the students the next time I teach calculus. –  Adrián Barquero Jan 18 '11 at 4:48
    
Of course the natural next question is how you prove Rolle's theorem. –  Grumpy Parsnip Jan 18 '11 at 13:15
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@Jim Conant: Most calculus textbooks I'm familiar with use the Extreme Value Theorem; if $f(x)=f(a)=f(b)$ for all $x$ in $[a,b]$, there is nothing to do; otherwise, there is an absolute extreme that is not the value at $a$, hence must be a point where the derivative vanishes. This is also what Spivak does. –  Arturo Magidin Jan 18 '11 at 14:02
    
I agree. My point, not very well expressed, was that Rolle's Theorem and MVT are basically the same theorem. Proving Rolle's theorem is 95% of the way to the general MVT.In any event, your answer is helpful and correct. –  Grumpy Parsnip Jan 18 '11 at 14:41
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@Jim: Yes, they are "essentially the same theorem." Rolle's Theorem has the virtue of being slightly easier because one has a specific numerical "target" (a point where the derivative is $0$), but one could prove MVT directly by using the auxiliary function $g$ and invoking the Extreme Value Theorem for that one. –  Arturo Magidin Jan 18 '11 at 14:55

Have you seen the wikipedia article?

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