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Section 5.6 The Mean Value Theorem

There are numerous applications of the derivative through its definition as rate of change and as the slope of the tangent line. In this section we shall look at some deeper reasons why the derivative turns out to be so useful. The simple answer is that the derivative of a function tells us a lot about the function. More important, “hard” questions about a function can sometimes be answered by solving a relatively simple problem about the derivative of the function.

The Mean Value Theorem tells us that there is an intimate connection between the net change of the value of any “sufficiently nice” function over an interval and the possible values of its derivative on that interval. Because of this connection, we can draw conclusions about the possible values of the derivative based on information about the values of the function, and conversely, we can draw conclusions about the values of the function based on information about the values of its derivative.

Let us illustrate the idea through the following two interesting questions involving derivatives:

  1. Suppose two different functions have the same derivative; what can you say about the relationship between the two functions?

  2. Suppose you drive a car from toll booth on a toll road to another toll booth at an average speed of 70 miles per hour. What can be concluded about your actual speed during the trip? In particular, did you exceed the 65 mile per hour speed limit?

While these sound very different, it turns out that the two problems are very closely related. We know that “speed” is really the derivative by a different name; let's start by translating the second question into something that may be easier to visualize. Suppose that the function \(f(t)\) gives the position of your car on the toll road at time \(t\text{.}\) Your change in position between one toll booth and the next is given by \(\ds f(t_1)-f(t_0)\text{,}\) assuming that at time \(\ds t_0\) you were at the first booth and at time \(\ds t_1\) you arrived at the second booth. Your average speed for the trip is \(\ds (f(t_1)-f(t_0))/(t_1-t_0)\text{.}\) If we think about the graph of \(f(t)\text{,}\) the average speed is the slope of the line that connects the two points \(\ds (t_0,f(t_0))\) and \(\ds (t_1,f(t_1))\text{.}\) Your speed at any particular time \(t\) between \(\ds t_0\) and \(\ds t_1\) is \(f'(t)\text{,}\) the slope of the curve. Now question (2) becomes a question about slope. In particular, if the slope between endpoints is 70, what can be said of the slopes at points between the endpoints?

As a general rule, when faced with a new problem it is often a good idea to examine one or more simplified versions of the problem, in the hope that this will lead to an understanding of the original problem. In this case, the problem in its “slope” form is somewhat easier to simplify than the original, but equivalent, problem.

Here is a special instance of the problem. Suppose that \(\ds f(t_0)=f(t_1)\text{.}\) Then the two endpoints have the same height and the slope of the line connecting the endpoints is zero. What can we say about the slope between the endpoints? It shouldn't take much experimentation before you are convinced of the truth of this statement: Somewhere between \(\ds t_0\) and \(\ds t_1\) the slope is exactly zero, that is, somewhere between \(\ds t_0\) and \(\ds t_1\) the slope is equal to the slope of the line between the endpoints. This suggests that perhaps the same is true even if the endpoints are at different heights, and again a bit of experimentation will probably convince you that this is so. But we can do better than “experimentation” —we can prove that this is so.

We start with the simplified version:

We know that \(f(x)\) has a maximum and minimum value on \([a,b]\) (because it is continuous), and we also know that the maximum and minimum must occur at an endpoint, at a point at which the derivative is zero, or at a point where the derivative is undefined. Since the derivative is never undefined, that possibility is removed.

If the maximum or minimum occurs at a point \(c\text{,}\) other than an endpoint, where \(f'(c)=0\text{,}\) then we have found the point we seek. Otherwise, the maximum and minimum both occur at an endpoint, and since the endpoints have the same height, the maximum and minimum are the same. This means that \(f(x)=f(a)=f(b)\) at every \(x \in [a,b]\text{,}\) so the function is a horizontal line, and it has derivative zero everywhere in \((a,b)\text{.}\) Then we may choose any \(c\) at all to get \(f'(c)=0\text{.}\)

Rolle's Theorem is illustrated below for a function \(f(x)\) where \(f'(x)=0\) holds for two values of \(x=c_1\) and \(x=c_2\text{:}\)

Perhaps remarkably, this special case is all we need to prove the more general one as well.

Let \(\ds m={f(b)-f(a)\over b-a}\text{,}\) and consider a new function \(g(x)=f(x) - m(x-a)-f(a)\text{.}\) We know that \(g(x)\) has a derivative everywhere, since \(g'(x)=f'(x)-m\text{.}\) We can compute

\begin{equation*} g(a)=f(a)- m(a-a)-f(a) =0 \end{equation*}

and

\begin{equation*} \begin{split} g(b)\amp =f(b)-m(b-a)-f(a) \\ \amp =f(b)-\frac{f(b)-f(a)}{b-a}(b-a)-f(a)\\ \amp =f(b)-(f(b)-f(a))-f(a)=0. \end{split} \end{equation*}

So the height of \(g(x)\) is the same at both endpoints. This means, by Rolle's Theorem, that at some \(c\text{,}\) \(g'(c)=0\text{.}\) But we know that \(g'(c)=f'(c)-m\text{,}\) so

\begin{equation*} 0=f'(c)-m=f'(c)-{f(b)-f(a)\over b-a}\text{,} \end{equation*}

which turns into

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

exactly what we want.

The Mean Value Theorem is illustrated below showing the existence of a point \(x=c\) for a function \(f(x)\) where the tangent line at \(x=c\) (with slope \(f'(c)\)) is parallel to the secant line connecting \(A(a,f(a))\) and \(B(b,f(b))\) (with slope \({f(b)-f(a)\over b-a}\)):

Returning to the original formulation of question (2), we see that if \(f(t)\) gives the position of your car at time \(t\text{,}\) then the Mean Value Theorem says that at some time \(c\text{,}\) \(f'(c)=70\text{,}\) that is, at some time you must have been traveling at exactly your average speed for the trip, and that indeed you exceeded the speed limit.

Now let's return to question (1). Suppose, for example, that two functions are known to have derivative equal to 5 everywhere, \(f'(x)=g'(x)=5\text{.}\) It is easy to find such functions: \(5x\text{,}\) \(5x+47\text{,}\) \(5x-132\text{,}\) etc. Are there other, more complicated, examples? No—the only functions that work are the “obvious” ones, namely, \(5x\) plus some constant. How can we see that this is true?

Although “5” is a very simple derivative, let's look at an even simpler one. Suppose that \(f'(x)=g'(x)=0\text{.}\) Again we can find examples: \(f(x)=0\text{,}\) \(f(x)=47\text{,}\) \(f(x)=-511\) all have \(f'(x)=0\text{.}\) Are there non-constant functions \(f\) with derivative 0? No, and here's why: Suppose that \(f(x)\) is not a constant function. This means that there are two points on the function with different heights, say \(f(a)\not=f(b)\text{.}\) The Mean Value Theorem tells us that at some point \(c\text{,}\) \(f'(c)=(f(b)-f(a))/(b-a)\not=0\text{.}\) So any non-constant function does not have a derivative that is zero everywhere; this is the same as saying that the only functions with zero derivative are the constant functions.

Let's go back to the slightly less easy example: suppose that \(f'(x)=g'(x)=5\text{.}\) Then \((f(x)-g(x))' = f'(x)-g'(x) = 5 -5 =0\text{.}\) So using what we discovered in the previous paragraph, we know that \(f(x)-g(x)=k\text{,}\) for some constant \(k\text{.}\) So any two functions with derivative 5 must differ by a constant; since \(5x\) is known to work, the only other examples must look like \(5x+k\text{.}\)

Now we can extend this to more complicated functions, without any extra work. Suppose that \(f'(x)=g'(x)\text{.}\) Then as before \((f(x)-g(x))' = f'(x)-g'(x) =0\text{,}\) so \(f(x)-g(x)=k\text{.}\) Again this means that if we find just a single function \(g(x)\) with a certain derivative, then every other function with the same derivative must be of the form \(g(x)+k\text{.}\)

Example 5.63. Given Derivative.

Describe all functions that have derivative \(5x-3\text{.}\)

Solution

It's easy to find one:

\begin{equation*} g(x)=(5/2)x^2-3x \implies g'(x)=5x-3\text{.} \end{equation*}

The only other functions with the same derivative are therefore of the form

\begin{equation*} f(x)=(5/2)x^2-3x+k\text{.} \end{equation*}

Alternately, though not obviously, you might have first noticed that

\begin{equation*} g(x)=(5/2)x^2-3x+47\implies g'(x)=5x-3\text{.} \end{equation*}

Then every other function with the same derivative must have the form

\begin{equation*} f(x)=(5/2)x^2-3x+47+k\text{.} \end{equation*}

This looks different, but it really isn't. The functions of the form \(\ds f(x)=(5/2)x^2-3x+k\) are exactly the same as the ones of the form \(\ds f(x)=(5/2)x^2-3x+47+k\text{.}\) For example, \(\ds (5/2)x^2-3x+10\) is the same as \(\ds (5/2)x^2-3x+47+(-37)\text{,}\) and the first is of the first form while the second has the second form.

This is worth calling a theorem:

Example 5.65. Same Derivative.

Describe all functions with derivative \(\ds \sin x + e^x\text{.}\) One such function is \(\ds -\cos x+e^x\text{,}\) so all such functions have the form \(-\cos x+e^x+k\text{.}\)

°Őłó±đ´Ç°ů±đłľĚý5.64 and the above example illustrate what the Mean Value Theorem allows us to say about \(f(x)\) when we have perfect information about \(f^{\prime}(x)\text{.}\) Specifically, \(f(x)\) is determined up to a constant. Our next example illustrates almost the opposite extreme situation, one where we have much less information about \(f^{\prime}(x)\) beyond the fact that \(f^{\prime}(x)\) exists. Specifically, assuming that we know an upper bound on the values of \(f^{\prime}(x)\text{,}\) what can we say about the values of \(f(x)\text{?}\)

Example 5.66. Conclusion Regarding Function Value Based on Derivative Information.

Suppose that \(f\) is a differentiable function such that \(f^{\prime }\left( x\right) \leq 2\) for all \(x\text{.}\) What is the largest possible value of \(f\left( 7\right)\) if \(f\left( 3\right) =5?\)

Solution

We are interested in the values of \(f\left( x\right)\) at \(x=3\) and \(x=7\text{.}\) It makes sense to focus our attention on the interval between 3 and 7. It is given that \(f\left( x\right)\) is differentiable for all \(x\text{.}\) So, \(f\left( x\right)\) is also continuous at all \(x\text{.}\) In particular, \(f\left( x\right)\) is continuous on the interval \(\left[ 3,7\right]\) and differentiable on the interval \(\left( 3,7\right) \text{.}\) By the Mean Value Theorem, we know that there is some \(c\) in \(\left( 3,7\right)\) such that

\begin{equation*} f^{\prime }\left( c\right) =\frac{f\left( 7\right) -f\left( 3\right) }{7-3}\text{.} \end{equation*}

Simplifying and using the given information \(f\left( 3\right) =5\text{,}\) we get

\begin{equation*} f^{\prime }\left( c\right) =\frac{f\left( 7\right) -5}{4}\text{,} \end{equation*}

or, after re-arranging the terms,

\begin{equation*} f\left( 7\right) =4f^{\prime }\left( c\right) +5\text{.} \end{equation*}

We do not know the exact value of \(c\text{,}\) but we do know that \(f^{\prime }\left( x\right) \leq 2\) for all \(x\text{.}\) This implies that \(f^{\prime }\left( c\right) \leq 2\text{.}\) Therefore,

\begin{equation*} f\left( 7\right) \leq 4\cdot 2+5=13\text{.} \end{equation*}

That is, the value of \(f\left( 7\right)\) cannot exceed 13. To convince ourselves that 13 (as opposed to some smaller number) is the largest possible value of \(f\left( 7\right) \text{,}\) we still need to show that it is possible for the value of \(f\left( 7\right)\) to reach 13. If we review our proof, we notice that the inequality will be an equality if \(f^{\prime }\left( c\right) =2\text{.}\) One way to guarantee this without knowing anything about \(c\) is to require \(f^{\prime }\left( x\right) =2\) for all \(x\text{.}\) This means that \(f\left( x\right) =2x+k\) for some constant \(k\text{.}\) From the condition \(f\left( 3\right) =5\text{,}\) we see that \(k=-1\text{.}\) We can easily verify that indeed \(f\left( x\right) =2x-1\) meets all our requirements and \(f\left( 7\right) =13\text{.}\)

Exercises for Section 5.6.

Let \(\ds f(x) = x^2\text{.}\) Find a value \(c\in (-1,2)\) so that \(f'(c)\) equals the slope between the endpoints of \(f(x)\) on \([-1,2]\text{.}\)

Answer

\(c=1/2\)

Verify that \(f(x) = x/(x+2)\) satisfies the hypotheses of the Mean Value Theorem on the interval \([1,4]\) and then find all of the values, \(c\text{,}\) that satisfy the conclusion of the theorem.

Answer

\(\ds c=\sqrt{18}-2\)

Verify that \(f(x) = 3x/(x+7)\) satisfies the hypotheses of the Mean Value Theorem on the interval \([-2 , 6]\) and then find all of the values, \(c\text{,}\) that satisfy the conclusion of the theorem.

Let \(f(x) = \tan x\text{.}\) Show that \(f(\pi ) = f(2\pi)=0\) but there is no number \(c\in (\pi,2\pi)\) such that \(f'(c) =0\text{.}\) Why does this not contradict Rolle's Theorem?

Let \(\ds f(x) = (x-3)^{-2}\text{.}\) Show that there is no value \(c\in (1,4)\) such that

\begin{equation*} f'(c) = (f(4)-f(1))/(4-1)\text{.} \end{equation*}

Why is this not a contradiction of the Mean Value Theorem?

Describe all functions with derivative \(\ds x^2+47x-5\text{.}\)

Solution

\(\ds x^3/3+47x^2/2-5x+k\)

Describe all functions with derivative \(\ds {1\over 1+x^2}\text{.}\)

Solution

\(\arctan x + k\)

Describe all functions with derivative \(\ds x^3-{1\over x}\text{.}\)

Solution

\(\ds x^4/4 -\ln x +k\)

Describe all functions with derivative \(\sin(2x)\text{.}\)

Answer

\(-\cos(2x)/2 +k\)

Find \(f\left( x\right)\) if \(f^{\prime }\left( x\right) =e^{-x}\) and \(f\left( 0\right) =2\text{.}\)

Suppose that \(f\) is a differentiable function such that \(f^{\prime }\left( x\right) \geq -3\) for all \(x\text{.}\) What is the smallest possible value of \(f\left( 4\right)\) if \(f\left( -1\right) =2\text{?}\)

Show that the equation \(\ds 6x^4 -7x+1 =0\) does not have more than two distinct real roots.

Let \(f\) be differentiable on \(\R\text{.}\) Suppose that \(f'(x) \neq 0\) for every \(x\text{.}\) Prove that \(f\) has at most one real root.

Prove that for all real \(x\) and \(y\) \(|\cos x -\cos y | \leq |x-y|\text{.}\) State and prove an analogous result involving sine.

Show that \(\ds \sqrt{1+x} \le 1 +(x/2)\) if \(-1\lt x\lt 1\text{.}\)

Suppose that \(f(a)=g(a)\) and that \(f^{\prime}(x)\leq g^{\prime}(x)\) for all \(x\geq a\text{.}\)

  1. Prove that \(f(x)\leq g(x)\) for all \(x\geq a\text{.}\)

  2. Use part (a) to prove that \(e^x\geq 1+x\) for all \(x\geq 0\text{.}\)

  3. Use parts (a) and (b) to prove that \(e^x\geq 1+x+\dfrac{x^2}{2}\) for all \(x\geq 0\text{.}\)

  4. Can you generalize these results?

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