What are the applications of the Mean Value Theorem?

Question

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?

Answer 1

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.

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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. Cf. also this answer.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

Having taught calculus a few times, I'm becoming less convinced of its importance in a first course. The important player has always been the Fundamental Theorem of Calculus, but as 19th-century mathematicians probed the nature of real numbers, and what exactly made calculus worked, they found that the mean value theorem — which seems obvious and could be "deduced" from FTC: $$f(b)-f(a) = \int_a^b f'(x)\,dx =f'(\xi)\,(b-a)$$ if $f'$ doesn't jump over its mean value — at the basis of FTC while not relying on the troublesome definition of the integral. I think that's what made MVT important: it's the key stepping stone between the completeness of real numbers and FTC.

Other theorems that could be deduced from the FTC (e.g., Taylor's theorem, L'Hopital's rule) are now seen as more direct consequences of the MVT — direct in the sense that it circumvents integrals. Also, it removes the condition that $f'$ be continuous, and hence more "fundamental". (Paradoxically, it's the desire to generalize and "simplify" the proofs that made things more difficult to understand.)

In my opinion, we shouldn't make the integral such a formidable beast. It is a very intuitive object (the method of exhaustion of Archimedes) for all the functions that the students will ever see in a first course, and in almost all applications. In fact, the order that a standard course follows, from limit, to derivative, to integral, is almost the complete opposite of when they were first discovered.

To drive the point further, it helps to take a computable/constructive perspective. Integrals can be defined over the rationals, or even the computable numbers, while the MVT does require the completeness of real numbers (or $f'$ could jump over its mean value). To prove the FTC over computable numbers would be a "good exercise", and you'll soon realize that real numbers make life much easier.

[Update] While browsing through the last chapter of MVP — A Most Valuable Theorem, which summarizes the decades worth of debate around MVP, I think I'm more inclined to say that the "mean value inequality" is of more importance, as that's my opinion of Taylor's theorem vs inequality, too.