Deconstructing the Mean Value Theorem, Part 3

The Mean Value Theorem implies that there is at least one tangent line with the same slope as the secant line.

Pete Brady strolled out of the U.S. Postal Service office just off Weaver Lake Road in Maple Grove, Minnesota at exactly 10 am.

Coffee and important package in hand, he set off on his 161 mile (259 kilometer) journey to the northwest on Interstate 94 to the police department in Fergus Falls, Minnesota.

It was a cloudy and windy October day, with a hint of winter in the air. As he listened to his favorite tunes, the miles rolled by relatively unnoticed, as well as the readings on his car’s dashboard. He would stop for lunch after making the delivery to his uncle Robert Brady, who was head of public safety in Fergus Falls. But then he would get out of town as quickly as possible.

Robert had only recently made the transition from being a college mathematics professor for many years into his new role, but he brought his passion for mathematics with him. Pete was okay at math, but he didn’t like to talk about it outside of school, and spending too much time with Robert made his head hurt.

Robert was also a stickler about obeying the law and continually scrutinized other peoples’ lives. Pete wanted to make sure he didn’t reveal too much about his recent activities with his friends — the less time spent around Robert the better.

After arriving, Pete warily opened the door and sauntered in.

Robert saw him immediately and sprang up. “You are here too early.”

With a quizzical look on his face, Pete decided not to say anything, but just started to raise the package to hand it off. Shouldn’t Robert be pleased that he was there early — whatever he meant by it?

“You arrived at exactly 12:18 pm,” Robert continued. “You were speeding.”

Pete knew that he had a history of speeding, but he also had been in no rush to see uncle Robert. It didn’t seem possible. Maybe he hadn’t watched the speedometer enough?

“What do you mean?”

“I saw you leave the Maple Grove post office at exactly 10 am through our video feed.” Robert paused, then explained, “This package is too important, so I felt I had to watch the feed to make sure you got it.”

Pete was upset, but not completely surprised. “So?”

“So?” Robert sighed, “That means your overall average speed was 70 miles per hour!”

Now Pete was really perplexed. After about five seconds, he stammered, “How…how do you know? And why does it matter?” Pete instantly wished he had not asked.

“Simple mathematics — child’s play, really,” Robert proclaimed. “The distance between that post office and this police station is exactly 161 miles, and it took you exactly 2.3 hours, so your average speed was 161 miles divided by 2.3 hours, which is exactly 70 miles per hour. I’m going to give you a ticket for speeding. I will also have to consider contacting your parents. It is for your own good. I do care about you, you know.”

As Robert moved to a counter to start the process of issuing a citation, Pete thought he saw an out. “Wait a minute — isn’t the speed limit on the interstate highway 70 miles per hour? What you just said actually proves that I followed the speed limit!”

“Do not use the word ‘proves’ in such a casual manner young man! You had to speed up as you left the post office in Maple Grove and you had to slow down in Fergus Falls, so your average speed on the interstate must have been greater than 70 miles per hour. So you did indeed exceed the speed limit.”

Then Pete remember some physics from a few years before and decided to take on Robert’s scolding tone, “But that was just my average speed, not my instantaneous speed. You can’t prove that there was ever a point in time where I was going over the speed limit.”

Robert turned and smirked at his nephew’s attempt to mimic him. “Oh…yes…I…can. I can invoke the glorious Mean Value Theorem. It can be used to show that, at some point in time, your instantaneous speed equaled your average speed, which was higher than 70 miles per hour.”

Pete didn’t remember the Mean Value Theorem. There was no use in fighting. He was defeated. He gave his uncle the package in exchange for the ticket.

“You have 60 days to pay.”

“See ya later uncle Robert.”

Statement of the Mean Value Theorem

Yes, I went overboard on this story. The stereotypical portrayal of uncle Robert as a priggish mathematician was just for fun. Please forgive me.

Mathematicians do indeed enjoy mathematics to an extreme degree in many peoples’ eyes. But they are, in my personal experience, often kind and generous.

But was uncle Robert right? Did Pete break the speed limit? And what does the Mean Value Theorem (MVT) have to do with it? We should look at its precise statement, which was explored in Part 1 and Part 2 of this series.

Mean Value Theorem (MVT): If $f$ is a real-valued function defined and continuous on a closed interval $[a,b]$ and if $f$ is differentiable on the open interval $(a,b),$ then there exists a number $c\in (a,b)$ with the property that $f'(c)=\frac{f(b)-f(a)}{b-a}.$

Back to Pete’s Story

Let us first work on understanding the MVT in the context of Pete’s driving. Assume Pete left the Maple Grove post office at time 0 hours (10 am) and arrived at the police station in Fergus Falls at time 2.3 hours (12:18 pm). Then Pete’s average speed was indeed $\frac{161\mbox{ miles}}{2.3\mbox{ hours}}=70$ miles per hour.

Let $f(t)$ represent the distance, in miles, that Pete traveled from time 0 to time $t,$ in hours. If time $a>0$ was the time that Pete entered the interstate and time $b<2.3$ was the time that Pete exited the interstate (with $a<b$ ), then it is also true that $\frac{f(b)-f(a)}{b-a}$ was Pete’s average speed on the interstate. Because Pete had to speed up at the beginning and slow down at the end, we can say that $\frac{f(b)-f(a)}{b-a}=s,$ where $s>70$ miles per hour.

It is a reasonable physical assumption to say that $f(t)$ is continuous and differentiable over the the intervals $[a,b]$ and $(a,b),$ respectively. The MVT now implies that there is a time $c\in (a,b)$ when Pete’s instantaneous speed, the derivative $f'(c),$ equaled his average speed on the interstate $s>70$ . Pete did indeed break the speed limit of 70 miles per hour.

To some readers, this may seem intuitively obvious. Is the MVT really necessary to use? Isn’t it obviously true? If this is your situation, you should read Part 2 to see that its proof is not trivial. It rests on foundational principles (“axioms“), definitions, and facts in the subject real analysis.

A Return to Steady Eddie and Wild Kyle

The main goal of Part 1 of this series of posts was to “deconstruct” the MVT by understanding the parts of its statement. We also illustrated its implications and limitations through examples. The main goal of Part 2 was to “deconstruct” the MVT by working backwards from its statement toward first principles to understand why it is true — or, at least, to understand how it is proved within a certain axiomatic system.

Our main goal in this, Part 3, is not so much a “deconstruction”. Rather, we will explore why the MVT is so valuable by exploring more of its implications.

Let’s start by considering the physical example from Part 2: the “race” between Steady Eddie and Wild Kyle. Below is the animated graphic of their race. Steady Eddie is driving the constant-speed blue car, and Wild Kyle is driving the varying-speed red car.

The mean value theorem implies that Wild Kyle will sometimes travel at the same speed as Steady Eddie.

They start and end the race at the same times, so they have the same average speed. The MVT implies that there is at least one time where Kyle’s instantaneous speed equals Eddie’s constant speed. In fact, in this case, there is more than one such time.

Kyle Becomes More Wild

We can take this example even further. Suppose Kyle convinces Eddie to be a bit more “free-spirited” and not keep a constant speed. In addition, let’s say that Kyle decides to be even more wild and travel both forward (positive velocity) and backward (negative velocity) toward the finish line. Is there ever a time where their instantaneous velocities are equal?

The mean value theorem implies that Wild Kyle will sometimes travel at the same speed as Steady Eddie, this is true even when Wild Kyle sometimes has a negative velocity. — Now both cars have non-constant velocities and Wild Kyle even has times where his car has a negative velocity (backwards motion). Yet there is still at least one time where their instantaneous velocities are equal.

The Mean Value Theorem still implies that the answer is “yes”!

To verify this, let $f(t)$ be Steady Eddie’s position at time $t$ and let $g(t)$ be Wild Kyle’s position at time $t$ (all with respect to appropriate measurement units). The starting line is taken to be position 0 and the rightward direction is the positive direction. It is physically reasonable to assume that both $f$ and $g$ are continuous and differentiable on the interval $[0,b]$ , where $b$ is the ending time of the race.

Define a new function $h$ by the formula $h(t)=f(t)-g(t).$ Since $f$ and $g$ are continuous and differentiable on $[0,b],$ the same is true of $h$ . We also know that $h'(t)=f'(t)-g'(t),$ and that $h(0)=f(0)-g(0)=0$ and $h(b)=f(b)-g(b)=0.$

By the Mean Value Theorem (or Rolle’s Theorem), there exists a time $c\in (0,b)$ such that $h'(c)=\frac{h(b)-h(0)}{b-0}=\frac{0}{b}=0$ . But this means that $f'(c)=g'(c)$ . That does it. That means that they have the same instantaneous velocity (which could be positive or negative) at some point in time. At some point in time the slopes of tangent lines to the graphs of $f(t)$ and $g(t)$ are the same.

Monotonicity

The Mean Value Theorem can be used to prove the “Monotonicity Theorem”, which is sometimes split into three pieces: 1) the “Increasing Function Theorem”. 2) the “Decreasing Function Theorem”. And 3) the “Constant Function Theorem”. These are fundamental and useful facts from calculus related to monotonic functions. Here’s the full statement:

Monotonicity Theorem: Let $f$ be a real-value function defined on a closed interval $[a,b]$ . Suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ . Then the following three statements are true: 1) If $f'(x)>0$ for all $x\in (a,b)$ , then $f$ is (strictly) increasing on $[a,b]$ . 2) If $f'(x)<0$ for all $x\in (a,b)$ , then $f$ is (strictly) decreasing on $[a,b]$ . 3) If $f'(x)=0$ for all $x\in (a,b)$ , then $f$ is constant on $[a,b]$ .

Proof of Part 1 of the Monotonicity Theorem (using the MVT):

Let $x_{1},x_{2}\in [a,b]$ be arbitrary points with $x_{1}<x_{2}.$ Since these points are arbitrary, to prove that $f$ is increasing on $[a,b],$ we only need to show that $f(x_{1})<f(x_{2}).$

Our assumptions about $f$ allows us to apply the MVT to this function over the interval $[x_{1},x_{2}]$ to conclude that there exists a number $c\in (x_{1},x_{2})$ such that $f'(c)=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}.$ But this means that $f(x_{2})-f(x_{1})=f'(c)\cdot (x_{2}-x_{1}).$ However, this quantity is positive because $f'(c)>0$ and $x_{2}-x_{1}>0.$ We are finished. This means $f(x_{1})<f(x_{2})$ and therefore $f$ is (strictly) increasing on $[a,b].$

The proofs of Parts 2 and 3 of this theorem are similar.

A Possible Misinterpretation

Many students in calculus misinterpret this theorem to mean that if a differentiable function $f$ is strictly increasing on an interval, then its derivative must always be positive on that interval. However, this is technically a false statement (though a person might say be tempted to say that it is “usually true” in some vague sense). For example, the function $f(x)=x^{3}$ is strictly increasing on the interval $[-1,1],$ but its derivative is not always positive on $(-1,1)$ because $f'(0)=0.$ You might be tempted to say this example is a form of “cheating”, but it is really not. It illustrates an important subtlety to be aware of. There are many things in calculus and real analysis that sound “true”, and might indeed be “usually true”, but are technically “false” because of certain “counterexamples“.

The First Derivative Test

Very much related to the Monotonicity Theorem is the so-called First Derivative Test. This has important applications to calculus and its real-world consequences.

One of the main applications that calculus students tend to remember is optimization. This is the search for extreme points of functions (maxima and minima) in many real-life situations.

We first find the derivative of the function in question. Next, we find the critical points of the function. And then we draw a number line to help visualize the possible changes in the sign of the derivative as the independent variable increases through the critical points. This leads to the identification of local maximum and minimum points. However, some critical points are neither local maxima nor local minima — such as $x=0$ for $f(x)=x^{3}$ over any open interval containing zero.

A Real-Life Optimization

It’s a bit tough to tell, but in the following animation, the angle between the red lines is maximized when the boat is close to, but not too close to, the Statue of Liberty on the right. This is an example of a practical real-life maximization. The location where the angle between the red lines is largest is the “best” viewing location for viewing the “size” of the statue.

There is an optimal viewing angle at some location. — The location where the angle between the red lines is largest is the “best” viewing location for viewing the “size” of the statue. This occurs when the boat is close to, but not too close to, the statue.

Here’s the statement of one form of the First Derivative Test.

First Derivative Test: Suppose that $f$ is continuous on $(a,b)$ and differentiable on $(a,b)$ (it is actually sufficient to just assume the second fact here). Furthermore, suppose $c\in (a,b)$ is a critical point of $f$ (where, in this case, $f'(c)=0$ ). Then each of the following statements is true: 1) If $f'(x)>0$ for all $x\in (a,c)$ and $f'(x)<0$ for all $x\in (c,b)$ , then $f$ has a (local) maximum point at $x=c$ . 2) If $f'(x)<0$ for all $x\in (a,c)$ and $f'(x)>0$ for all $x\in (c,b)$ , then $f$ has a (local) minimum point at $x=c$ .

We can also prove this by using the Monotonicity Theorem above. It is therefore also a consequence of the Mean Value Theorem. We prove Part 1 here. The proof of Part 2 is similar.

Proof of Part 1 (using the Monotonicity Theorem):

The fact that $f'(x)>0$ for all $x\in (a,c)$ implies, by the Monotonicity Theorem, that $f$ is strictly increasing on $(a,c)$ (after all, $(a,c)\subseteq [a,b]$ ). In a similar manner, the fact that $f'(x)<0$ for all $x\in (c,b)$ implies, by the Monotonicity Theorem, that $f$ is strictly decreasing on $(c,b).$ Therefore, if $x\in (a,c)$ or if $x\in (c,b)$ , it must be the case that $f(x)<f(c),$ by what it means for a function to be strictly increasing and strictly decreasing. But this means that $f$ has a (local) maximum point at $x=c.$ This proves Part 1 of the First Derivative Test.

The Fundamental Theorem of Calculus (FTC)

The most important result in calculus is, as its name seems to imply, the Fundamental Theorem of Calculus (FTC). In fact, though more rudimentary ideas in calculus had been circulating before the time, it was the discovery and application of this theorem that marked the traditional “birth” of calculus by Isaac Newton and Gottfried Wilhelm Leibniz in the 1600’s.

Its statement is often written in two parts, though we can combine these under one theorem-heading. Take the term “Riemann integrable” as basically meaning that, if the graph of the function is always above the horizontal axis, then the area under its graph and above the interval $[a,b]$ is well-defined. We will not delve into a rigorous description of what this means.

Statement of the FTC

Fundamental Theorem of Calculus: Suppose the real-valued function $f$ is Riemann integrable on the interval $[a,b]$ . Then the following two statements are true: 1) If $f$ is continuous on $[a,b]$ and if $F$ is a function defined by the equation $F(x)=\displaystyle\int_{a}^{x}f(t)\, dt,$ then $F$ is differentiable on $[a,b]$ and $F'(x)=f(x)$ for all $x\in [a,b]$ (with appropriate one-sided derivatives at the endpoints). 2) If $F$ is an antiderivative of $f$ on $[a,b],$ so that $F'(x)=f(x)$ for all $x\in [a,b],$ and if $f$ is continuous on $[a,b],$ then $\displaystyle\int_{a}^{b}f(x)\, dx=F(b)-F(a).$

Note: the assumption of the continuity of $f$ on $[a,b]$ is actually not necessary for the truth of Part 2, but will make our proof using the MVT possible to do.

One way that Part 2 of the FTC can be proved is by using Part 1. However, Part 2 of the FTC can also be proved using the MVT. We will do this now, though the “proof” we give is really only an idea/sketch/outline of a proof because we have not defined what integration really means in terms of limits of Riemann sums. If you have done so in a previous course, however, this proof should make sense to you, though its very last part is very subtle.

Idea of Proof of Part 2 of the FTC (using the MVT):

For an arbitrary partition $a=x_{0}<x_{1}<x_{2}<\cdots<x_{n}$ of the interval $[a,b],$ our assumptions and the MVT allow us to conclude that there exist numbers $c_{1},c_{2},\ldots,c_{n},$ with $x_{i-1}<c_{i}<x_{i}$ for each $i,$ such that $f(c_{i})=F'(c_{i})=\frac{F(x_{i})-F(x_{i-1})}{x_{i}-x_{i-1}}.$

But this means that

$F(b)-F(a)=(F(x_{n})-F(x_{n-1}))+(F(x_{n-1})-F(x_{n-2}))+\cdots+(F(x_{2})-F(x_{1}))+(F(x_{1})-F(x_{0}))$

$=f(c_{n})(x_{n}-x_{n-1})+f(c_{n-1})(x_{n-1}-x_{n-2})+\cdots+f(c_{2})(x_{2}-x_{1})+f(c_{1})(x_{1}-x_{0}).$

But this last sum is an arbitrary Riemann sum for estimating the integral $\displaystyle\int_{a}^{b}f(x)\, dx,$ which exists because we are assuming $f$ is continuous. Since the “mesh” or “norm“, $\displaystyle\min_{1\leq i\leq n}(x_{i}-x_{i-1}),$ of such a Riemann sum can be made as small as we like, this implies that $\displaystyle\int_{a}^{b}f(x)\, dx$ must actually equal all of such Riemann sums, which all equal $F(b)-F(a).$

This last sentence is very subtle, but it is true. You should realize that other Riemann sums might not equal the integral. The key is that these particular Riemann sums, which arise from the truth of the MVT, do all equal the integral.

The FTC is key to many of the applications of integration. We will certainly explore the FTC in depth in the future, Lord willing.