APEX The Mean Value Theorem

Section 3.2 The Mean Value Theorem

We motivate this section with the following question: Suppose you leave your house and drive to your friend's house in a city \(100\) miles away, completing the trip in two hours. At any point during the trip do you necessarily have to be going \(50\) miles per hour?

In answering this question, it is clear that the average speed for the entire trip is 50 mph (i.e. \(100\) miles in \(2\) hours), but the question is whether or not your instantaneous speed is ever exactly 50 mph. More simply, does your speedometer ever read exactly 50 mph?. The answer, under some very reasonable assumptions, is “yes.”

Figure 3.2.1. Video introduction to Section 3.2

Let's now see why this situation is in a calculus text by translating it into mathematical symbols.

First assume that the function \(y = f(t)\) gives the distance (in miles) traveled from your home at time \(t\) (in hours) where \(0\le t\le 2\text{.}\) In particular, this gives \(f(0)=0\) and \(f(2)=100\text{.}\) The slope of the secant line connecting the starting and ending points \((0,f(0))\) and \((2,f(2))\) is therefore

\begin{align*} \frac{\Delta f}{\Delta t} \amp = \frac{f(2)-f(0)}{2-0}\\ \amp = \frac{100-0}{2}\\ \amp = 50 \text{ mph}\text{.} \end{align*}

The slope at any point on the graph itself is given by the derivative \(\fp(t)\text{.}\) So, since the answer to the question above is “yes,” this means that at some time during the trip, the derivative takes on the value of 50 mph. Symbolically,

\begin{equation*} \fp(c) = \frac{f(2)-f(0)}{2-0} = 50 \end{equation*}

for some time \(0\le c \le 2\text{.}\)

How about more generally? Given any function \(y=f(x)\) and a range \(a\le x\le b\) does the value of the derivative at some point between \(a\) and \(b\) have to match the slope of the secant line connecting the points \((a,f(a))\) and \((b,f(b))\text{?}\) Or equivalently, does the equation \(\fp(c) = \frac{f(b)-f(a)}{b-a}\) have to hold for some \(a \lt c \lt b\text{?}\)

Let's look at two functions in an example.

Example 3.2.2. Comparing average and instantaneous rates of change.

Consider functions

\begin{align*} f_1(x)\amp=\frac{1}{x^2}\amp f_2(x)\amp= \abs{x} \end{align*}

with \(a=-1\) and \(b=1\) as shown in Figure 3.2.3. Both functions have a value of \(1\) at \(a\) and \(b\text{.}\) Therefore the slope of the secant line connecting the end points is \(0\) in each case. But if you look at the plots of each, you can see that there are no points on either graph where the tangent lines have slope zero. Therefore we have found that there is no \(c\) in \([-1,1]\) such that

\begin{equation*} \fp(c) = \frac{f(1)-f(-1)}{1-(-1)} = 0\text{.} \end{equation*}

So what went “wrong”? It may not be surprising to find that the discontinuity of \(f_1\) and the corner of \(f_2\) play a role. If our functions had been continuous and differentiable, would we have been able to find that special value \(c\text{?}\) This is our motivation for the following theorem.

Theorem 3.2.4. The Mean Value Theorem of Differentiation.

Let \(y=f(x)\) be a continuous function on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\text{.}\) There exists a value \(c\text{,}\) \(a \lt c \lt b\text{,}\) such that

\begin{equation*} \fp(c) = \frac{f(b)-f(a)}{b-a}\text{.} \end{equation*}

That is, there is a value \(c\) in \((a,b)\) where the instantaneous rate of change of \(f\) at \(c\) is equal to the average rate of change of \(f\) on \([a,b]\text{.}\)

Note that the reasons that the functions in Example 3.2.2 fail are indeed that \(f_1\) has a discontinuity on the interval \([-1,1]\) and \(f_2\) is not differentiable at the origin.

We will give a proof of the Mean Value Theorem below. To do so, we use a fact, called Rolle's Theorem, stated here.

Theorem 3.2.5. Rolle's Theorem.

Let \(f\) be continuous on \([a,b]\) and differentiable on \((a,b)\text{,}\) where \(f(a) = f(b)\text{.}\) There is some \(c\) in \((a,b)\) such that \(\fp(c) = 0\text{.}\)

Figure 3.2.6. Video presentation of Theorem 3.2.5

Consider Figure 3.2.7 where the graph of a function \(f\) is given, where \(f(a) = f(b)\text{.}\) It should make intuitive sense that if \(f\) is differentiable (and hence, continuous) that there would be a value \(c\) in \((a,b)\) where \(\fp(c)=0\text{;}\) that is, there would be a relative maximum or minimum of \(f\) in \((a,b)\text{.}\) Rolle's Theorem guarantees at least one; there may be more.

Figure 3.2.7. A graph of \(f(x) = x^3-5x^2+3x+5\text{,}\) where \(f(a) = f(b)\text{.}\) Note the existence of \(c\text{,}\) where \(a\lt c\lt b\text{,}\) where \(\fp(c)=0\text{.}\)

Rolle's Theorem is presented here as a stepping stone toward the Mean Value Theorem, but it's a useful result in its own right. It often turns up as a tool in mathematical problem solving. The following video illustrates one such use of Rolle's Theorem.

Figure 3.2.8. Using Rolle's Theorem to show a polynomial has at most one real root

Rolle's Theorem is really just a special case of the Mean Value Theorem. If \(f(a) = f(b)\text{,}\) then the average rate of change on \((a,b)\) is \(0\text{,}\) and the theorem guarantees some \(c\) where \(\fp(c)=0\text{.}\) We will prove Rolle's Theorem, then use it to prove the Mean Value Theorem.

Proof of Rolle's Theorem.

Let \(f\) be differentiable on \((a,b)\) where \(f(a)=f(b)\text{.}\) We consider two cases.

Case.

Consider the case when \(f\) is constant on \([a,b]\text{;}\) that is, \(f(x) = f(a) = f(b)\) for all \(x\) in \([a,b]\text{.}\) Then \(\fp(x) = 0\) for all \(x\) in \([a,b]\text{,}\) showing there is at least one value \(c\) in \((a,b)\) where \(\fp(c)=0\text{.}\)

Case.

Now assume that \(f\) is not constant on \([a,b]\text{.}\) The Extreme Value Theorem (Theorem 3.1.4) guarantees that \(f\) has a maximal and minimal value on \([a,b]\text{,}\) found either at the endpoints or at a critical value in \((a,b)\text{.}\) Since \(f(a)=f(b)\) and \(f\) is not constant, it is clear that the maximum and minimum cannot both be found at the endpoints. Assume, without loss of generality, that the maximum of \(f\) is not found at the endpoints. Therefore there is a \(c\) in \((a,b)\) such that \(f(c)\) is the maximum value of \(f\text{.}\) By Theorem 3.1.15, \(c\) must be a critical number of \(f\text{;}\) since \(f\) is differentiable, we have that \(\fp(c) = 0\text{,}\) completing the proof of the theorem.

We can now prove the Mean Value Theorem.

Proof of the Mean Value Theorem.

Define the function

\begin{equation*} g(x) = f(x) - \frac{f(b)-f(a)}{b-a}x\text{.} \end{equation*}

We know \(g\) is differentiable on \((a,b)\) and continuous on \([a,b]\) since \(f\) is. We can show \(g(a)=g(b)\) (it is actually easier to show \(g(b)-g(a)=0\text{,}\) which suffices). We can then apply Rolle's theorem to guarantee the existence of \(c\) in \((a,b)\) such that \(g'(c) = 0\text{.}\) But note that

\begin{equation*} 0= g'(c) = \fp(c) - \frac{f(b)-f(a)}{b-a}\text{;} \end{equation*}

hence

\begin{equation*} \fp(c) = \frac{f(b)-f(a)}{b-a}\text{,} \end{equation*}

which is what we sought to prove.

Going back to the very beginning of the section, we see that the only assumption we would need about our distance function \(f(t)\) is that it be continuous and differentiable for \(t\) from \(0\) to \(2\) hours (both reasonable assumptions). By the Theorem 3.2.4, we are guaranteed a time during the trip where our instantaneous speed is 50 mph. This fact is used in practice. Some law enforcement agencies monitor traffic speeds while in aircraft. They do not measure speed with radar, but rather by timing individual cars as they pass over lines painted on the highway whose distances apart are known. The officer is able to measure the average speed of a car between the painted lines; if that average speed is greater than the posted speed limit, the officer is assured that the driver exceeded the speed limit at some time.

Note that the Theorem 3.2.4 is an existence theorem. It states that a special value \(c\) exists, but it does not give any indication about how to find it. It turns out that when we need the Theorem 3.2.4, existence is all we need.

Example 3.2.9. Using the Mean Value Theorem.

Consider \(f(x) = x^3+5x+5\) on \([-3,3]\text{.}\) Find \(c\) in \([-3,3]\) that satisfies the Theorem 3.2.4.

Solution.

The average rate of change of \(f\) on \([-3,3]\) is:

\begin{align*} \frac{f(3)-f(-3)}{3-(-3)} \amp =\frac{47-(-37)}{6}\\ \amp =\frac{84}{6}\\ \amp = 14\text{.} \end{align*}

We want to find \(c\) such that \(\fp(c) = 14\text{.}\) We find \(\fp(x) = 3x^2+5\text{.}\) We set this equal to \(14\) and solve for \(x\text{.}\)

\begin{align*} \fp(x) \amp = 14\\ 3x^2 +5 \amp = 14\\ x^2 \amp = 3\\ x \amp = \pm \sqrt{3} \approx \pm 1.732 \end{align*}

We have found two values \(c\) in \([-3,3]\) where the instantaneous rate of change is equal to the average rate of change; the Theorem 3.2.4 guaranteed at least one. In Figure 3.2.10, \(f\) is graphed with a line representing the average rate of change; the lines tangent to \(f\) at \(x=\pm \sqrt{3}\) are also given. Note how these lines are parallel (i.e., have the same slope) to the secant line.

Figure 3.2.10. Demonstrating the Mean Value Theorem in Example 3.2.9

While the Theorem 3.2.4 has practical use (for instance, the speed monitoring application mentioned before), it is mostly used to advance other theory. We will use it in the next section to relate the shape of a graph to its derivative.

Before ending this section, we give two important theoretical consequences of the Mean Value Theorem.

First, we recall that the derivative of any constant function is zero. Is the converse true? That is, are constant functions the only ones whose derivative is zero? The Mean Value Theorem says yes. Second, if two functions have the same derivative, does this tell us anything about the original functions? The Mean Value Theorem implies that the only way the functions can differ is the addition of a constant value.

Theorem 3.2.11. Corollaries of the Mean Value Theorem.

Let \(f\text{,}\) \(g\text{,}\) and \(h\) be differentiable (and therefore continuous) functions on an interval \(I\text{.}\)

If \(f'(x)=0\) for every point in the interval \(I\text{,}\) then \(f\) is a constant function on \(I\text{.}\)
If \(g'(x)=h'(x)\) for every point in \(I\text{,}\) then there is a constant \(C\) such that \(g(x)=h(x)+C\) for every point in \(I\text{.}\)

Proof of Corollary 1.

Figure 3.2.12. Showing that a function with zero derivative is constant

To prove that \(f\) is a constant function on \(I\text{,}\) we must prove that the outputs \(f(x)\) are the same for every point in \(I\text{.}\) In other words, given two points \(a, b\) in \(I\) with \(a \lt b\text{,}\) we must show that \(f(a)=f(b)\text{.}\)

Suppose \(a, b\) are in \(I\) with \(a \lt b\text{.}\) We will apply the Mean Value Theorem to the function \(f\) on \([a,b]\text{.}\) First, we check that \(f\) on \([a,b]\) satisfies all the assumptions of the Mean Value Theorem. Since \(f'(x)=0\) for every point in \(I\text{,}\) \(f\) is differentiable on \(I\) and hence \(f\) continuous on \(I\text{.}\) Therefore, \(f\) is continuous on \([a,b]\) because each point in \([a,b]\) is a point in \(I\text{.}\) Now, \(f\) is differentiable on \((a,b)\) because the assumption says that \(f'(x)=0\) for every point in \(I\text{.}\) Now that we have shown the assumptions of the Mean Value Theorem holds for \(f\) on \([a,b]\text{,}\) the conclusion of the Mean Value Theorem gives a number \(c\) in the interval \((a,b)\) such that

\begin{equation*} \frac{f(b)-f(a)}{b-a} = f'(c)\text{.} \end{equation*}

Since \(c\) is in the interval \((a,b)\text{,}\) \(c\) is in the interval \(I\text{.}\) Therefore, \(f'(c)=0\) which gives that

\begin{equation*} \frac{f(b)-f(a)}{b-a} = f'(c)=0\text{.} \end{equation*}

Hence, \(f(b)-f(a)=0\) which implies that \(f(b)=f(a)\text{.}\) Consequently, \(f\) is constant on \(I\text{.}\)

Proof of Corollary 2.

Figure 3.2.13. Showing that two functions with the equal derivatives differ by a constant

Note that the second part of theorem is equivalent to saying that \(g(x)-h(x)\) is a constant function on \(I\text{.}\) First we note that \(g'(x)=h'(x)\) is equivalent to \(g'(x)-h'(x)=0\text{.}\) Therefore,

\begin{equation*} \frac{d}{dx}(g(x)-h(x))=g'(x)-h'(x)=0 \end{equation*}

for every point in \(I\text{,}\) and by part 1 of this theorem, we know that \(g(x)-h(x)\) is a constant function on \(I\text{.}\) Thus, the functions \(g\) and \(h\) satisfy \(g(x)=h(x)+C\) for some constant \(C\text{.}\)

Example 3.2.14. Applying a Mean Value Theorem Corollary.

At all times \(t\ge 0\text{,}\) the positions of two objects \(A\) and \(B\) along a straight line are described by the functions \(g(t)=\tan^{-1}(t+1)\) (for object \(A\)) and \(h(t)=-\tan^{-1}(\frac{1}{t+1})\) (for object \(B\)). Show that the objects are always \(\dfrac{\pi}{2}\) units apart.

Solution.

By Theorem 3.2.11, if \(g'(t) = h'(t)\) for \(t \geq 0 \text{,}\) then there is a constant \(C\) such that \(g(t) = h(t) + C\) for all \(t \geq 0\text{.}\) In other words, if the velocity functions are the same, then the difference in position is constant, which is what we want to show.

First we check that the two functions are differentiable on the domain \([0, \infty)\text{.}\) This is accomplished by applying the Chain Rule (Theorem 2.5.4). First we note that Theorem 2.7.14 tells us that the inverse tangent function is differentiable everywhere. In the case of \(g\text{,}\) the inner function in the composition is a polynomial, and is therefore also differentiable everywhere. The Chain Rule then tells us that the composed function \(g\) is also differentiable everywhere.

Showing \(h\) is differentiable is slightly more work. In this case, the Quotient Rule (Theorem 2.4.11) tells us that \(\frac{1}{t+1}\) is differentiable on any interval where \(t+1\) is never zero, which only occurs at \(t=-1\text{.}\) Therefore \(\frac{1}{t+1}\) is differentiable on the interval \([0,\infty)\text{,}\) and the Chain Rule implies that the composition \(\tan^{-1}(\frac{1}{t+1})\) is differentiable on the interval as well. Finally the Constant Multiple Rule (or the Product Rule) allow us to conclude that \(h\) itself is differentiable on the interval of interest.

Now we compute derivatives and show that they are the same.

\begin{align*} g'(t)\amp= \dfrac{1}{1+(t+1)^2}\\ h'(t) \amp = -\dfrac{1}{1+\left( \frac{1}{t+1}\right)^2} \left(-(t+1)^{-2}\right)\\ \amp= \dfrac{1}{ 1 + \frac{1}{(t+1)^2}} \dfrac{1}{ (t+1)^2} \\ \amp= \dfrac{1}{ \left(1 + \frac{1}{(t+1)^2} \right) (t+1)^2} \\ \amp= \dfrac{1}{ (t+1)^2+1} \end{align*}

Therefore, \(g'(t)=h'(t)\) for every \(t \geq 0\text{.}\) Hence, by the second corollary to the Mean Value Theorem, there is a constant \(C\) such that \(g(t)=h(t)+C\) for every \(t \geq 0\text{.}\)

To find \(C\text{,}\) set \(t=0\text{,}\) then

\begin{align*} \dfrac{\pi}{4} \amp = \tan^{-1}(1) \\ \amp= g(0) \\ \amp = h(0) + C\\ \amp = -\tan^{-1}\left( \frac{1}{0+1}\right) + C \\ \amp = -\tan^{-1}(1)+C\\ \amp = - \frac{\pi}{4}+C \end{align*}

Solving for \(C\) in the equation \(\dfrac{\pi}{4} = - \dfrac{\pi}{4} + C \text{,}\) we get \(C = \dfrac{\pi}{2}\text{.}\) We conclude that the objects are always \(\dfrac{\pi}{2}\) units apart.

We end this section with one more proof involving the Mean Value Theorem; this time, establishing a property of the sine function.

Figure 3.2.15. Demonstrating a property of the sine function

Exercises Exercises

Terms and Concepts

1.

Explain in your own words what the Mean Value Theorem states.

2.

Explain in your own words what Rolle’s Theorem states.

Problems

Exercise Group.

A function \(f(x)\) and interval \([a,b]\) are given. Check if Rolle's Theorem can be applied to \(f\) on \([a,b]\text{;}\) if so, find \(c\) in \((a,b)\) such that \(\fp(c)=0\text{.}\)

3.

\(f(x) = 6\) on \([-1,1]\)

Explain why Rolle's Theorem can or cannot be applied.

4.

\(f(x) = 6x\) on \([-1,1]\)

Explain why Rolle's Theorem can or cannot be applied.

5.

\(f(x) = x^2+x-6\) on \([-3,2]\)

Explain why Rolle's Theorem can or cannot be applied.

6.

\(f(x) = x^2+x-2\) on \([-3,2]\)

Explain why Rolle's Theorem can or cannot be applied.

7.

\(f(x) = x^2+x\) on \([-2,2]\)

Explain why Rolle's Theorem can or cannot be applied.

8.

\(f(x) = \sin(x)\) on \([\pi/6,5\pi/6]\)

Explain why Rolle's Theorem can or cannot be applied.

9.

\(f(x) = \cos(x)\) on \([0,\pi]\)

Explain why Rolle's Theorem can or cannot be applied.

10.

\(f(x) = \frac{1}{x^2-2x+1}\) on \([0,2]\)

Explain why Rolle's Theorem can or cannot be applied.

Exercise Group.

A function \(f(x)\) and interval \([a,b]\) are given. Check if The Mean Value Theorem of Differentiation can be applied to \(f\) on \([a,b]\text{;}\) if so, find \(c\) in \((a,b)\) guaranteed by the Mean Value Theorem.