Deconstructing the Mean Value Theorem, Part 1

Visual interpretation of the Mean Value Theorem. — The Mean Value Theorem says there’s at least one point where the slope of the tangent line on a smooth curve equals the slope of the secant line between the endpoints.

Many mathematicians start chanting “MVT! MVT! MVT!” when they see the Mean Value Theorem in action. This is indeed because it is so valuable.

What does it mean for a theorem to be valuable to a mathematician? It means the theorem can be used to prove a lot of other theorems!

Of course, that’s not what most people think of when they hear that something in mathematics is valuable. Most people think it must be directly useful for daily life, business, engineering, science, health care, or public service.

But pure mathematicians appreciate mathematics for its own sake. They might also appreciate “real world” applications (though a pure mathematician here or there might not). But the “cold and austere” beauty of mathematics, as well as the fact that they find it “fun“, are big parts of what compel them to continue studying the subject.

How then, is the Mean Value Theorem valuable to a mathematician? What other theorems can it be used to prove?

A few theorems from calculus stand out immediately. They are the Monotonicity Theorem (Increasing/Decreasing/Constant Function Theorem), the First Derivative Test, and the Fundamental Theorem of Calculus. Moreover, as is often the case in mathematics, these pure mathematical theorems actually do have “real world” applications.

In this article, we will not consider these real world applications. Instead, we will explore many Mean Value Theorem examples.

Goals of this Series of Articles

In this post and others in this series, I “deconstruct” the Mean Value Theorem (MVT). I use the word “deconstruct” here in two senses. 1) to break down the theorem into its parts or pieces. And 2) to trace the deductive logic backwards to find the core reasons why the theorem is true. These tasks are accomplished in the first two posts on the MVT. The third post described below is the icing on the cake.

Here is an outline of the series:

In Part 1, the goals are to understand the statement of the MVT, see why its hypotheses (premises) are necessary, and see the theorem illustrated for some examples.
In Part 2, the goals are to understand why the MVT is true, both from an intuitive physical perspective and from a proof-oriented mathematical perspective.
And in Part 3, the goals are to see how the MVT is useful in proving other theorems and to describe how this is useful for the “real world”.

As you work through this series, you may find parts of Videos #1, 17, 18, 19, and 20 from my YouTube playlist on Introductory Real Analysis to be helpful.

Statement of the Mean Value Theorem

In mathematics, theorems are often stated as “if…, then…” clauses. The part of the statement right after the word “if” is called the premise or hypothesis of the theorem. The part that comes immediately after the word “then” is called the conclusion of the theorem. The hypothesis and conclusion can each be comprised of more than one sub-clause. The plural forms of these words are “hypotheses” and “conclusions”.

When mathematicians call such a statement a “theorem”, “proposition”, “lemma”, or “corollary”, they are saying that the statement is “true“. This means that if the premise is assumed to be true, the conclusion can be proved to be true using deductive reasoning. This involves the use of the rules of deductive logic, such as modus ponens. Note: the words “theorem”, “proposition”, “lemma”, and “corollary” are not exact synonyms. There are subtle differences in meaning between them.

Here at last is a precise statement of the MVT.

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 $f'(c)=\frac{f(b)-f(a)}{b-a}$ .

Alternative Description

When $f$ satisfies these hypotheses, it has a “smooth” graph. It that has tangent lines of finite slope at every point $x$ in the open interval $(a,b)$ . Now the derivative $f'(c)$ is the slope of the tangent line to $f$ at $x=c$ and $\frac{f(b)-f(a)}{b-a}$ is the slope of the secant line connecting the points $(a,f(a))$ and $(b,f(b))$ . Thus, the MVT asserts that a smooth function $f$ has at least one tangent line with the same slope as the secant line between the points that are “endpoints” of the graph of $f$ .

Deciphering the Mean Value Theorem

Let’s start by deconstructing the hypotheses of the theorem.

We are given a real-valued function $f$ defined on a closed interval $[a,b]$ , consisting of the real numbers $x$ with the property that $a\leq x\leq b$ (it is implicit here that $a$ and $b$ are real numbers making the open interval $(a,b)=\{x\in \Bbb R| a<x<b\}$ nonempty, so that $a<b$ .) Moreover, the interval $[a,b]$ is the domain of this function, and its codomain, or “target set”, is the collection of all real numbers ${\Bbb R}$ . Note however, that $f$ is not necessarily an onto function.
The function $f$ is continuous on the closed interval $[a,b]$ . Intuitively, this means that the graph of $f$ can be hand-drawn without picking up a writing utensil off the page. To be rigorous, this concept must be defined in a precise way. This involves using the concept of a limit of a function.
The function $f$ is differentiable on the open interval $(a,b)$ . This means that the derivative $f'(x)$ must exist for each $x$ in the interval $(a,b)$ . This is also based on the concept of a limit of a function.

Hypothesis 1 just sets the stage for our discussion. The hypotheses of most interest are 2 and 3.

What Happens when Hypothesis 2 is Not Satisfied?

What happens if we consider situations where Hypothesis 2 is not valid (not true)? The answer is: it depends. The conclusion of the theorem may be satisfied (true) or it may not be satisfied (not true). The fact that this second case occurs for at least one example is what renders Hypothesis 2 to be “necessary“. In other words, the statement of the MVT is not true without this hypothesis.

These claims can be verified with many Mean Value Theorem examples.

Example 1: a discontinuous function that satisfies the conclusion of the MVT

Let $f$ be a piecewise-defined function, defined on the closed interval $[0,1]$ by the formula $f(x)=\begin{cases} x & \mbox{if } 0\leq x\leq \frac{1}{2} \\ x^{2} & \mbox{if } \frac{1}{2}<x\leq 1 \\ \end{cases}.$

The graph of $f$ is shown below. It is clearly discontinuous at $x=\frac{1}{2}$ and therefore discontinuous on $[0,1]$ . However, for every point $c\in (0,\frac{1}{2})$ , we have $f'(c)=1=\frac{f(1)-f(0)}{1-0}$ .

The conclusion of the mean value theorem can still be true, even when the hypothesis that the function be continuous does not hold. — This function is discontinuous on [0,1], but there are still tangent lines whose slope is 1, which is the slope of the secant line between (0,0) and (1,1).

Example 2: a discontinuous function that does not satisfy the conclusion of the MVT

Let $f$ be a piecewise-defined function, defined on the closed interval $[0,1]$ by the formula $f(x)=\begin{cases} 2x & \mbox{if } 0\leq x\leq \frac{1}{2} \\ 2x-1 & \mbox{if } \frac{1}{2}<x\leq 1 \\ \end{cases}.$

The graph of $f$ is shown below. It is clearly discontinuous at $x=\frac{1}{2}$ and therefore discontinuous on $[0,1]$ . Also, at no point is there a tangent line whose slope equals $\frac{f(1)-f(0)}{1-0}=1$ . This is because all tangent lines have slope equal to 2.

The conclusion of the mean value theorem can be false when the function is not continuous. — *This function is discontinuous on [0,1], and there are no tangent lines whose slope is 1, which is the slope of the secant line between (0,0) and (1,1).*

Example 2 thus demonstrates that Hypothesis 2 is necessary.

Can a Function Satisfy the Hypotheses as Written?

It is also interesting to consider an example showing why it is sufficient for $f$ to be differentiable on just the open interval $(a,b)$ rather than the entire closed interval $[a,b]$ (assuming we are only talking about one-sided derivatives at the endpoints). Before looking at this example, we note that there is a theorem that differentiability implies continuity, and therefore there is a version of the MVT which just assumes that the function is defined and differentiable on the closed interval $[a,b]$ and makes no other assumptions. In writing the statement of the MVT in the way we have done (and the way that is done in all textbooks), we have made the hypothesis “weaker”. We have thereby made the MVT a “stronger” theorem, because it is more widely applicable.

Example 3: a function that is continuous on [a,b], differentiable on (a,b), but not differentiable on [a,b]

Let $f$ be defined by $f(x)=\arcsin(x)=\sin^{-1}(x)$ over the interval $[-1,1]$ . Then $f$ is continuous on $[-1,1]$ , differentiable on $(-1,1)$ , but not differentiable at $x=\pm 1$ and therefore not differentiable on $[-1,1]$ . The graph of $f$ has vertical one-sided tangents at $x=\pm 1$ .

However, $f$ still satisfies the hypotheses of the (strong version of) the MVT, so the conclusion of the MVT still holds. The slope of the secant line between the endpoints of the graph is $\frac{f(1)-f(-1)}{1-(-1)}=\frac{\pi/2+\pi/2}{2}=\frac{\pi}{2}$ . Moreover, since $f'(x)=\frac{1}{\sqrt{1-x^{2}}}$ , we see that $f'(c)=\frac{\pi}{2}$ if and only if $c=\pm \frac{\sqrt{\pi^{2}-4}}{\pi}\approx 0.771$ . Exercise: Check this algebra!

The Graph

A graph illustrating all this is shown in the picture below.

An example of a function that is continuous on a closed interval but only differentiable on the interior of that closed interval. — The graph of the arcsine (inverse sine) function is differentiable on the open interval (-1,1) but not the closed interval [-1,1] (it has one-sided tangent lines with infinite slopes at the endpoints). It still satisfies the hypothesis, and therefore the conclusion, of the MVT.

How about Hypothesis 3? Just as before, the conclusion of the theorem may be satisfied (true) or it may not be satisfied (not true). The fact that this second case occurs for at least one example is what renders Hypothesis 3 to be “necessary“. In other words, the statement is not true without this hypothesis.

We consider two examples to illustrate these claims.

Example 4: a non-differentiable function that satisfies the conclusion of the MVT

The example we discuss here is a function whose graph is “smooth” in that it has no “breaks” or “corners”. However, it is not differentiable everywhere in the open interval because there is one point whose tangent line is vertical (infinite slope). The example is the continuous function $f$ defined, say, on the interval $[-1,1]$ , by the formula $f(x)=\sqrt[3]{x}=x^{1/3}$ .

When $x\not=0$ , the Power Rule from calculus implies that $f'(x)=\frac{1}{3}x^{-2/3}=\frac{1}{3\sqrt[3]{x^{2}}}.$ The fact that $\displaystyle\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}=\displaystyle\lim_{h\rightarrow 0}\frac{1}{h^{2/3}}$ does not exist implies that $f'(0)$ does not exist and that $f$ fails to be differentiable at $x=0$ and therefore $f$ fails to be differentiable on $(-1,1)$ .

The secant line between the “endpoints” of the graph has slope $\frac{f(1)-f(-1)}{1-(-1)}=\frac{1+1}{1+1}=1$ . Furthermore, $f'(c)=\frac{1}{3\sqrt[3]{c^{2}}}=1$ if and only if $c=\pm \frac{1}{3\sqrt{3}}\approx \pm 0.192\in (-1,1)$ .

The Graph

Therefore, the conclusion of the MVT is satisfied for this example. The graph shown below illustrates these calculations.

The conclusion of the mean value theorem can still be true, even when the hypothesis that the function be differentiable does not hold. — *This function is not differentiable on [-1,1], but there are still tangent lines whose slope is 1, which is the slope of the secant line between (-1,-1) and (1,1).*

Example 5: a non-differentiable function that does not satisfy the conclusion of the MVT

The standard example to use here is $f(x)=|x|$ over any interval $[a,b]$ where $a<0$ and $b>0$ . Then $f$ fails to be differentiable on $(a,b)$ since $f'(0)$ does not exist. Furthermore, $f'(x)=-1$ when $x<0$ and $f'(x)=1$ when $x>0$ . But $\frac{f(b)-f(a)}{b-a}=\frac{b+a}{b-a}$ , which does not equal 1 or $-1$ . You should make a graph on your own to confirm this visually. It looks like the letter “V”, with the “corner” of the graph occurring at the origin $(0,0).$

Believing the MVT

If you make drawings of “random” smooth (and necessarily continuous) functions without vertical tangent lines, you will see that the truth of the MVT seems eminently reasonable. However, as your mathematics teachers should always emphasize, a picture is not a proof.

Is there anything else we can do to convince ourselves that the MVT is “probably” true? That is, without having to construct a complete proof? Yes, we can consider more examples – examples where the value of $c$ can actually be solved for algebraically.

Such examples definitely do not prove the theorem. They just help us to believe it. However, it is possible in such a situation that a purported “theorem” is actually false (not for the MVT, however), and perhaps someone can find a “counterexample” to demonstrate this.

We end Part 1 of this series of posts with a couple examples to help us believe the MVT. These examples also have the benefit of leading to interesting facts in their own right.

Example 6: finding c for a general quadratic function

Let $f(x)=\alpha x^{2}+\beta x+\gamma$ over the interval $[a,b]$ . Then $f'(x)=2\alpha x+\beta$ and $\frac{f(b)-f(a)}{b-a}=\frac{\alpha(b^{2}-a^{2})+\beta(b-a)}{b-a}=\frac{(b-a)(\alpha(b+a)+\beta)}{b-a}=\alpha(b+a)+\beta$ .

From these equations we see that $f'(c)=\frac{f(b)-f(a)}{b-a}$ if and only if $c=\frac{a+b}{2}$ . In other words, for a quadratic function defined on $[a,b]$ , the value of $x$ where the tangent line to $f$ has the same slope as the secant line between $(a,f(a))$ and $(b,f(b))$ is halfway between $a$ and $b$ (at their arithmetic mean, or average).

Example 7: finding c for a general cubic function (this involves calculations you should definitely check on your own)

Let $f(x)=\alpha x^{3}+\beta x^{2}+\gamma x+\delta$ over the interval $[a,b]$ . Then $f'(x)=3\alpha x^{2}+2\beta x+\gamma$ and

$\\frac{f(b)-f(a)}{b-a}=\frac{\alpha(b^{3}-a^{3})+\beta(b^{2}-a^{2})+\gamma(b-a)}{b-a}$ $=\frac{(b-a)(\alpha(b^{2}+ab+a^{2})+\beta(b+a)+\gamma)}{b-a}=\alpha(b^{2}+ab+a^{2})+\beta(b+a)+\gamma.$

You should check on your own that $f'(c)=\frac{f(b)-f(a)}{b-a}$ in this situation when

$c=\frac{-\beta\pm \sqrt{3\alpha^{2}(a^{2}+ab+b^{2})+3\alpha\beta(a+b)+\beta^{2}}}{3\alpha}.$

It would take proof, but evidently these values are always real numbers. Futhermore, at least one of these two values is always between $a$ and $b$ . This is true no matter what the values of $\alpha, \beta, \gamma,$ and $\delta$ are.

Graphs To Illustrate These Two Cases

Here are graphs that illustrate these two cases.

Applying the mean value theorem to a cubic over an interval containing its inflection point. — The two values of $c$ are both between $a=-2$ and $b=2$ for $f(x)=x^{3}-x^{2}-x+2$ .

The two values of $c$ are both between $a=-2$ and $b=2$ for $f(x)=x^{3}-x^{2}-x+2$ .

Applying the mean value theorem to a cubic over an interval not containing its inflection point. — There is only one value of $c$ that is between $a=1$ and $b=3$ for $f(x)=x^{3}-x^{2}-x+2$ .

There is only one value of $c$ that is between $a=1$ and $b=3$ for $f(x)=x^{3}-x^{2}-x+2$ .

The two values of $c$ would both be between the endpoints when $a$ and $b$ are on opposite sides of the inflection point of the cubic at $x=-\frac{\beta}{3\alpha}$ . Otherwise, only one of these values would be between $a$ and $b.$

Video to Illustrate This Fact

The following video illustrates this fact for fixed $a$ and $b$ while $\alpha$ and $\beta$ vary. It is difficult to interpret. You have to think very carefully about it to understand it – but it is worth the effort! It is showing the graphs of the $c$ values above, as functions of $\beta$ (in red and bright pink), when we set $a=-3$ (horizontal line in blue) and $b=5$ (horizontal line in green). Note that at least one of the red and bright pink curves are between the blue and green lines, no matter what $\alpha$ and $\beta$ are (note: both $\alpha$ and $\beta$ increase from $-5$ to 5). This is most interesting when $\alpha$ is near zero because that’s where only one of the curves is between the lines for many values of $\beta$ .

The outputs of the red and bright pink curves show the c-values guaranteed to exist by the MVT for a cubic $\alpha x^{3}+\beta x^{2}+\gamma x+\delta$ , as functions of $\beta$ , for various values of $\alpha$ (which is the animation parameter). We have set $a=-3$ (blue) and $b=5$ (green). Note that, no matter the value of $\alpha$ , at least one of the red or pink curves is always between the blue and green lines for any given value of $\beta$ .

That is the end of our exploration of Mean Value Theorem examples. We end by including the Mathematica code for the animation above.

Mathematica Code

The Mathematica code used to generate the animation is shown in the figure below.

Many mathematicians start chanting “MVT! MVT! MVT!” when they see the Mean Value Theorem in action. This is indeed because it is so valuable.

Goals of this Series of Articles

Statement of the Mean Value Theorem

Alternative Description

Deciphering the Mean Value Theorem

What Happens when Hypothesis 2 is Not Satisfied?

Example 1: a discontinuous function that satisfies the conclusion of the MVT

Example 2: a discontinuous function that does not satisfy the conclusion of the MVT

Can a Function Satisfy the Hypotheses as Written?

Example 3: a function that is continuous on [a,b], differentiable on (a,b), but not differentiable on [a,b]

The Graph

Example 4: a non-differentiable function that satisfies the conclusion of the MVT

The Graph

Example 5: a non-differentiable function that does not satisfy the conclusion of the MVT

Believing the MVT

Example 6: finding c for a general quadratic function

Example 7: finding c for a general cubic function (this involves calculations you should definitely check on your own)

Graphs To Illustrate These Two Cases

Video to Illustrate This Fact

Mathematica Code

Next: Deconstructing the Mean Value Theorem, Part 2