37
$\begingroup$

I'm going through my first year of teaching AP Calculus. One of the things I like to do is to impress upon my students why the topics I introduce are interesting and relevant to the big picture of understanding the nature of change.

That being said, while I know that the Mean Value Theorem is one of the central facts in the study of calculus, I'm not really clear on why. I feel that it's a bit like IVT for the derivatives of a continuous and differentiable function. But I feel that the only thing I did with it when I studied Calc is to identify the point where the tangent was parallel to the secant of the endpoints. If the class had been a little smaller or I had been a little bolder, I might have raised my hand and asked the professor "So what?" But I didn't, so here we are.

To be clear, I am not at all arguing that MVT is not critical, so I don't plan on the answers being opinion-based. But can you discuss some uses of MVT that justify the lofty place it has in the curriculum?

$\endgroup$
  • 2
    $\begingroup$ In the title you ask about motivation, but in the body you talk about uses. It can be used to get pointwise information of the function from global information about its derivative. This can serve to prove inequalities, compute limits (which are just inequalitites). As you observed, its direct use gives existence of solutions of certain equations. So, it can be used to locate or count zeros. $\endgroup$ – conditionalMethod Oct 21 at 17:03
  • $\begingroup$ What is mvt that mean value is greater then min and less than max or equal. Well mean velocity is the original idea of approx instant acceleration. So you have mvt, which kind of defines mean value, then you take smaller and smaller steps to get derivative as limit. $\endgroup$ – marshal craft Oct 21 at 17:04
  • 1
    $\begingroup$ What about Taylor's expansion? $\endgroup$ – Vasya Oct 21 at 17:07
  • 2
    $\begingroup$ Körner's book "Calculus for the Ambitious" says (p47) "the mean value inequality looks unimpressive, but turns out to be useful in all sorts of circumstances." $\endgroup$ – Mark Bennet Oct 21 at 17:31
  • 2
    $\begingroup$ Maybe useful : Craig Smorynski, MVT : A Most Valuable Theorem (Springer, 2017). $\endgroup$ – Mauro ALLEGRANZA Oct 23 at 7:00
72
$\begingroup$

When I teach the MVT in a Calculus class, I do three things:

a) Show the one real-world example I know and which everyone gets: Police has two radar controls at a highway, say at kilometre $11$ and at kilometre $20$. Speed limit is $70$ km/h. They measure a truck going through the first control, at 11.11am, at $65$ km/h, and going through the second control at 11.17am, at $67$ km/h. They issue a speeding ticket. Why?

Let the class think about this. Every time I've taught this, someone realised after a short while that the truck passed $9$ km in $6$ minutes, so its average speed was $90$ km/h. Then someone says something like: you cannot go at an average speed of $90$ km/h without ever going at a speed of $90$ km/h (and certainly not without ever going more than $70$ km/h). This is totally common sense, but also it is exactly the MVT. Draw a graph of the position function, realise that the numbers $65$ and $67$ were just red herrings (tangent slopes at the endpoints, irrelevant to the argument), discuss whether there is some way out: Can the function have discontinuites? Well, a jump discontinuity would be a wormhole the truck fell through, or more realistically some shortcut off-highway which would be illegal too. Points where the derivative does not exist? Actually yes, if the truck braked somewhere, but it cannot have done that more than finitely many times, and then we break down the problem into subintervals. Turns out: No, even sharp braking cannot create a "sharp turn" of the function under standard assumptions of physics, see comments by users @leftaroundabout and @llama.

b) Mention that aside from that, it is a "workhorse theorem" which we never see but which makes the entire curve sketching routine work. How do you prove Positive derivative means increasing function: with the MVT. How do you prove Derivative $0$ on an interval means constant: with the MVT. Of course we never think of the proofs of those, we just use them as "well-known", but without MVT, they would not be there.

c) Related to b, it comes up crucially in the Fundamental Theorem later, compare Arturo Magidin's answer. I point it out when I'm there.

Added: As this answer seems to get a lot of attention, I want to put in one more thing part of which I try to get across in class when the MVT is up.

d) The derivative is a cool thing because it carries a lot of information about the original function, but in a subtle way. To the non-initiated, the graphs of an $f$ and $f'$ would most often look totally unrelated. But the initiated, i.e. your calculus class, at this point should already "get" intuitively "hmm, $f'$ is very negative around here, so $f$ should decrease with a steep slope in this neighbourhood". Now the MVT is the one theorem which attaches actual numbers to this intuition, it is the first result which gives an explicit (albeit subtle) relation between values of $f$ and values of $f'$. That is why it underlies the proofs of all the fancy machinery that, later, gives seemingly much stronger relations between $f$ and $f'$, like Curve Sketching, the Fundamental Theorem, Taylor Series, and even L'Hôpital's rules (thanks @JavaMan for pointing out this one). They get all the limelight, but in a way, they all are refined versions of repeated applications of the MVT plus special conditions.


Further update: Since the "speeding" application of the MVT keeps getting mentioned everywhere (and of course I don't even remember where I got it from originally), I googled a little and see that it's been around for quite a while. This educational video of the MAA's from 1966 is almost of historical value (although I hardly understand the voice-over due to its very American accent). As for the question whether this is actually done, thanks to User Bracco23 for providing one source from Italy in a comment. Here is another one from Scotland: http://news.bbc.co.uk/2/hi/uk_news/scotland/4681507.stm The internet has more hearsay and debates: 1 2 3.

$\endgroup$
  • 11
    $\begingroup$ “Points where the derivative does not exist? Actually yes, if the truck braked somewhere” – with infinite braking force... $\endgroup$ – leftaroundabout Oct 22 at 9:45
  • 5
    $\begingroup$ It's a nitpick, but to make more explicit @leftaroundabout's point: braking doesn't imply the derivative doesn't exist. The derivative of velocity is acceleration, and you can't get discontinuous acceleration without discontinuous force (since $F = ma$), which is impossible. Crashing is about the closest thing you can get, but that wouldn't help the driver's case at all. $\endgroup$ – llama Oct 22 at 15:43
  • 3
    $\begingroup$ @leftaroundabout, llama: Thanks for correcting that! I will edit when I have more time later. -- eric_kernfeld: Indeed, I close that lesson by saying: If you face police who know calculus, you are screwed. $\endgroup$ – Torsten Schoeneberg Oct 22 at 15:55
  • 3
    $\begingroup$ The driver says he has an amazing truck and is following a Cantor function, and therefore never went in a wormhole nor sped. Now what? $\endgroup$ – Passer By Oct 23 at 5:43
  • 1
    $\begingroup$ @PasserBy : Now that we're following ‘standard assumptions of physics’, we're just assuming that the truck always has a velocity. Even assuming that it has an instantaneous position is already a kludge (since it is, in fact, spread out in space), so this isn't really worse. $\endgroup$ – Toby Bartels Oct 23 at 21:23
31
$\begingroup$

In terms of why students are taught the Mean Value Theorem in Calculus 1, I think it is often obscured because they are often not shown a proof of the Fundamental Theorem of Calculus. But the Mean Value Theorem is a key step in the proof of the first part of the Fundamental Theorem of Calculus. If you want to prove the first part of the Fundamental Theorem of Calculus, the simplest way is to use the MVT:

Namely, to calculate the integral $\int_a^b f'(x)\,dx$, pick a partition of the interval $[a,b]$, $a=x_0\lt x_1\lt\cdots\lt x_n=b$. We want to select points $x_i^*$, $x_{i-1}\leq x_i^*\leq x_i$ to do the Riemann sum $$\sum_{i=1}^{n}f'(x_i^*)(x_i-x_{i-1}).$$ By the Mean Value Theorem, there exists a point $x_i^*$ in $[x_{i-1},x_i]$ such that $f'(x_i^*)(x_{i}-x_i) = f(x_i)-f(x_{i-1})$. So pick those points so that the Riemann sum becomes $$\sum_{i=1}^n f'(x_i^*)(x_i-x_{i-1}) = \sum_{i=1}^n (f(x_i)-f(x_{i-1}))$$ which is a telescoping sum that equals $f(x_n)-f(x_0) = f(b)-f(a)$.

Since every Riemann sum can be selected to equal $f(b)-f(a)$, the limit as the mesh size goes to zero is $f(b)-f(a)$, proving that $$\int_a^b f'(x)\,dx = f(b)-f(a),$$ the first part of the FTC.

(There are other ways of proving the first part of the FTC, e.g., using the second part, but this is still a major highway to the FTC.)

In terms of "motivation", there is the classic one: if you drive at an average speed of 60 mph, is there necessarily an instant where your instantaneous speed is exactly 60 mph? In other words, is this notion of "instantaneous rate of change" reasonable when compared to our much more sensible notion of "average rate of change"?

All in all, if one is not going to prove the FTC, then there is an argument for not emphasizing (or going over) the MVT; though I confess that not having taken the relevant physics course, I am not aware if there is a need for the MVT somewhere in the parallel physics curriculum of dynamics.

$\endgroup$
  • $\begingroup$ I have taken plenty of physics courses and the MVT has never been brought up. I think it is only important for the proof of the fundamental theorem of calculus. $\endgroup$ – Display Name Oct 24 at 15:07
  • 2
    $\begingroup$ @DisplayName: It is also needed to prove the Increasing, Decreasing, and Constant Function theorems, which are used all the time. $\endgroup$ – Arturo Magidin Oct 24 at 16:14
  • 1
    $\begingroup$ @DisplayName: I mention physics because I am told, for example, that Stokes's and Green's Theorem are less "magical" and make more sense (as does Gauss's Divergence Theorem) if you are looking at the physics of electrostatics and fluid dynamics. $\endgroup$ – Arturo Magidin Oct 24 at 16:15
12
$\begingroup$

I speak as a person that worked with maths with the typical rule-of-thumb approach of engineers for many years, before deciding to get back to maths and study it thoroughly by myself. What was illutimating to me was actually the following "chain"

  1. Axiom of Completeness
  2. Extreme Value Theorem
  3. Rolle's Theorem

I sort of "see" the Mean Value Theorem as a direct consequence of Rolle's Theorem, in the sense that if $f(x)$ satisfies MVT's hypotheses in $[a,b]$, then $g(x)=f(b) -f(x)-\frac{f(b)-f(a)}{b-a}(b-x)$ satisfies Rolle's Theorem hypotheses in $[a,b]$, and hence its demonstration. Use instead $h(x) = f(b)-f(x)-\frac{f(b)-f(a)}{g(b)-g(a)}(g(b)-g(x))$, and you'll demonstrate Cauchy's generalization of MVT.

As in other comments and answers, it is fascinating that a repeated use of MVT brings to such powerful an instrument as Taylor polynomial.

$\endgroup$
8
$\begingroup$

Suppose $f$ is a constant function. Then its derivative is zero. This follows directly from the definition.

Suppose $f \colon \mathbb{R} \to \mathbb{R}$ satisfies $f'(x) = 0$ for all $x$. Must $f$ be a constant function?

The answer is yes. Because for any $x \in \mathbb{R}$ there exists $t$ between $0$ and $x$ such that $$ f(x) = f(0) + x f'(t) $$ but $f'(t) = 0$, so $f(x) = f(0)$.

The Mean Value Theorem (or Rolle's Theorem, but MVT is more flexible) is the fundamental theorem which connects information about the derivative of a function back to the original function.

$\endgroup$
7
$\begingroup$

Staying inside the classroom: Looking ahead to its use in a proof of Taylor's theorem is probably the best answer for a first year calculus student, given how important that theorem is. It also comes up all the time in deriving various formulas in both single variable and multivariable calculus, at least the heuristic that $f(x+h)-f(x)\approx hf'(x)$, although you don't need the mean value theorem for this expression.

But real life examples abound: one I have had success with in teaching is a method for handing out speeding tickets in Italy (and probably elsewhere). One's car is photographed at point $a$ at time $t_0$, and then again at time $t_f$ at point $b$. From this information, and the mean value theorem, they are able to detect those who sped. Of course, this doesn't account for the 20 minute espresso stops some speeders take.

$\endgroup$
  • 1
    $\begingroup$ And one can do the reformulation of the MVT, $f(x+h) = f(x) + h\cdot f'(\xi)$ for some $\xi$ between $x$ and $x+h$, immediately. $\endgroup$ – Carsten S Oct 22 at 14:56
2
$\begingroup$

One important application of MVT is in proofs of what's called Uniform Continuity, a global property of an interval rather than a local continuity property at a point.

We can't say $\sum_{n=0}^\infty \int_a^bf_n(x)dx=\int_a^b\sum_n^\infty f_n(x) dx$ even if all $f_n$'s are continuous, we need Uniform Continuity. This in turn helps establish various approximation techniques like Fourier Series. Sines and cosines are uniformly continuous on their entire domain.

Such methods are the basis for analysis using Hilbert Spaces, essential to solving problems in Quantum Mechanics.

$\endgroup$
1
$\begingroup$

The MVT is one of the things where the first approach to it makes you think, "Where's the point?" because it seems so obvious that it can be hard to understand that this is worth a proof.

You notice its value when you apply it to a problem where it isn't really obvious that it can be applied. A nice example was shown on Numberphile https://youtu.be/OuF-WB7mD6k: Given is a round table with four legs on uneven (but continuous!) ground which is wobbly because not all of its four legs reach the ground. Then you can turn this table, and before you have turned it by 90 degrees you must have found a place where all its legs touch the ground at the same time, so that it isn't wobbly anymore.

This application of the MVT not necessarily is obvious, and it gives us a hint on that sometimes also the most profound appearing theorem can have applications which aren't obvious.

In pure mathematics you sometimes drop back on the MVT (and the way its formal proof works) to approach a problem in higher dimensions where intuition quickly leaves us.

$\endgroup$
  • $\begingroup$ Sorry, but I think that video is actually about the Intermediate Value Theorem (IVT). I guess the professor mistranslated (in German, IVT = Zwischenwertsatz, MVT = Mittelwertsatz). $\endgroup$ – Torsten Schoeneberg Oct 25 at 4:06
  • $\begingroup$ @TorstenSchoeneberg Right, indeed. But I always felt like these two are just the same as the basic idea. At least understanding the IVT made me at once also see that the MVT must be true. While that sudden feeling of intuitive insight can be dangerous (because it can be misleading in other cases and thus shouldn't be relied on), I still think the two are mostly alike in nature. $\endgroup$ – Alfe Nov 6 at 13:32
0
$\begingroup$

In the paper Berrone, Lucio R.; Moro, Julio, Lagrangian means, Aequationes Math. 55, No. 3, 217-226 (1998), there is an extensive study of the means generated by the MVT. There are also related MVT-s, see Flett's MVT and Pawlikowska studies on it. It seems to me that there could be some MVT's apology.

$\endgroup$
  • 3
    $\begingroup$ You should probably at least cite the main points here so that the answer is self contained. $\endgroup$ – YiFan Oct 23 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.