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I have a question concerning the Mean Value Theorem (and maybe Rolle's Theorem).

In my calc book by Stewart, the concept of both theorems seemed to be thrown out of nowhere with a bunch of conditions and statements like.

"If $f$ is differentiable on an open interval $(a,b)$, then..."

So here is what i don't understand, why can't the interval be closed for differentiability? And what was the motivation behind creating the two theorems? I don't see how anyone, one day, could sit down and just write down a bunch of rules and conclude a formula and give a name to it.

I won't take the equation as face value and accept it like the rest of the mindless sheeps in my university

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  • $\begingroup$ What does "differentiability at a point" mean? $\endgroup$
    – JohnD
    Dec 19, 2012 at 6:14
  • $\begingroup$ It means f'(at a point) exists or in some books, they say the function is approximately linear near that point $\endgroup$
    – Lemon
    Dec 19, 2012 at 6:19
  • $\begingroup$ The function can certainly be differentiable on the closed interval, and more. But the point is that the theorem even holds when we don't have differentiability at an end point. That makes the theorem applicable in a wider variety of situations. For example, the commonly occurring function $f(x)=\sqrt{x}$ is not differentiable at $0$, but the Mean Value Theorem can be used on this function to conclude that there is a $c$ between $0$ and $x$ such that $f'(c)=\frac{\sqrt{x}-0}{x-0}$. $\endgroup$
    – André Nicolas
    Dec 19, 2012 at 6:20
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    $\begingroup$ Mine was a rhetorical comment... meant to lead him to the crux of what he's asking about. $\endgroup$
    – JohnD
    Dec 19, 2012 at 6:23
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    $\begingroup$ @sizz: When we say differentiable on an open interval, we do not commit ourselves to what might be happening beyond, so we are saying less than when we say differentiable on a closed interval. $\endgroup$
    – André Nicolas
    Dec 19, 2012 at 6:28

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The intuitive motivation for Rolle’s theorem is pretty simple. If $f(a)=0=f(b)$, then either $f$ is constant, or its value moves away from and then back to $0$. If $f$ is continuous, its value can’t jump instantaneously back to $0$: it has to turn around. If in addition $f$ has a derivative at every point between $a$ and $b$, its value can’t make a sharp turn: it has to turn gradually. And if it turns gradually, intuition says that at some point during the turn it has to be ‘moving’ horizontally $-$ which is exactly what Rolle’s theorem says in a rigorous way.

This also explains (at the intuitive level) why we don’t care about differentiability at the endpoint: the turnaround has to be somewhere in $(a,b)$, not at an endpoint of the interval.

The intuition behind the Mean Value Theorem is pretty much the same, except that horizontal is replaced by in the direction from $\langle a,f(a)\rangle$ to $\langle b,f(b)\rangle$.

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There seem to be several aspects to your question, I will try to address them as I understand what you are asking.

You seem to wonder about why the condition is differentiability on an open interval and not a closed one. Well, if a function is differentiable on a closed interval then certainly it is differentiable on an open one. Now, if a theorem can be proved from a weak condition then there is no need to strengthen the condition. In general one wants to impose the weakest possible conditions for a result. This is done so that the theory applicable to more situations.

The motivation for the creation/discovery of these and similar theorems from calculus is mainly to understand the relationship between a function and its derivative. In the initial stages of development of calculus there have been many subtleties that needed to be worked out and it took quite some time before a rigorous approach was established (by Cauchy and Weierstrass).

There is a lot of geometric intuition related to the study of functions. This intuition is sometimes helpful, guiding to correct theorems, and sometimes wrong, as witnessed by that or other unintuitive counter example (a famous case is the existence of a nowhere differentiable continuous function (first constructed by Bolzano (unpublished) and most famously by Weierstrass).

Historically, a lot of research and debate went into clarifying these issues and fine-tuning both the theory and ones geometric intuition. This is done by takings ones intuition, formulating a conjecture and then trying to prove it. The mean value theorem is very intuitive geometrically and indeed can be proved.

If you can get your hands on the book Adventures in Formalism by Criag Smorynski, I believe you fill find very interesting examples there from calculus that may help you understand how things work.

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  • $\begingroup$ Well, if a function is differentiable on a closed interval then certainly it is differentiable on an open one I am kinda stupid here. But is it true the other way around? If $f$ is diff on an open interval, then it is also diff on a closed one? $\endgroup$
    – Lemon
    Dec 24, 2012 at 1:50
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    $\begingroup$ no:f(x)=sin(1/x) is diff on (0,1) but cant be extended to a diff function on [0,1]. $\endgroup$
    – Ittay Weiss
    Dec 24, 2012 at 1:53
  • $\begingroup$ So that's why it was stated it should be diff on an open interval. Because the implication here is that it can be also done on a closed interval. If I do it the other way, I would restrict it to only a class of functions. Another dumb question, what about continuity? Can something be continuos on an open interval? And does that imply continuity on a closed interval? Could I see some examples? $\endgroup$
    – Lemon
    Dec 24, 2012 at 16:33
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    $\begingroup$ same function as above is also cont on (0,1) but can't be made cont on [0,1]. $\endgroup$
    – Ittay Weiss
    Dec 24, 2012 at 18:35
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    $\begingroup$ because then Rolle's thoerem does not apply: take f(x)=x for all 0<x<1 and f(0)=f(1)=7. Then f(0)=f(1), f is cont and diff on (0,1) yet for no value of c will you have f'(c)=0. $\endgroup$
    – Ittay Weiss
    Dec 24, 2012 at 19:47
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As a matter of fact, in Rolle's time (the late 17th and early 18th centuries) the modern concept of differentiability was unknown. Moreover, Rolle himself was a vocal opponent of the calculus (see Rolle's biography at http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Rolle.html ). I don't know how Rolle himself stated his result.

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The way that mathematics is actually done and the way that mathematics books are written are completely different. No mathematician would ever just sit down and guess a formula to be proven later. For example, if you wanted to prove the quadratic formula, it would be madness to guess how the formula worked first. Rather you would start with the equation $ax^2 + bx + c = 0$, and derive the quadratic equation from there. However, mathematics research is messy and complicated. When you write down your discoveries, you tend to organize them to make it as easy as possible for the reader to digest the results. In doing this organizing and polishing, you lose the feel of research mathematics, and results tend to be nicely polished but sometimes not very well motivated. This is likely how all calculus textbooks deal with Rolle's Theorem and the Mean Value Theorem. They are results which are crucial for the foundations of calculus, but to the first time reader they do not appear to be well motivated.

In regards to why we want differentiability on an open interval $(a,b)$ rather than a closed interval $[a,b]$, think about how the derivative is defined: $$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$$ In order for $f(x+h)$ to make sense, $x+h$ must be in the domain of $f$, and hence $x+h \in (a,b)$. If we are on an endpoint, moving $h$ units in either direction might take us out of the domain of $f$.

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  • $\begingroup$ In order for f(x+h) to make sense, x+h must be in the domain of f, and hence x+h∈(a,b). If we are on an endpoint, moving h units in either direction might take us out of the domain of f. Are we assuming $x \in (a,b)$ in the first place? $\endgroup$
    – Lemon
    Dec 24, 2012 at 1:46
  • $\begingroup$ Say $f(x)$ is only defined when $x\in[a,b]$. That is, for every $\epsilon > 0, f(a - \epsilon)$ is undefined. Then there would be no way to interpret $$\lim_{h \to 0^-} \frac{f(x+h) - f(x)}{h}$$ and thus no way to interpret the two-sided limit as having negative values of $h$ would take us out of the domain of $f$. Even if the values of $h$ were arbitrarily small (which would be permissible as taking a limit only involves the value near $h$), $f(a+h)$ would still be undefined. On the other hand, if we take $x \in (a,b)$, then $|x - a| > 0$ so that taking a derivative is well defined. $\endgroup$
    – JavaMan
    Dec 24, 2012 at 5:26

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