The Mean Value Theorem

The Mean Value Theorem is one of the better known results about derivatives, and for good reason. It is invoked frequently when one needs to estimate the maximum possible change between the values of a function at two different points. This can be a valuable tool when finding approximations to functions or when it is necessary to know how much variation is exhibited by a particular function. The theorem states that the average rate of change of a function between two points a and b given by ^{f.b/f .a/}_ba is equal to the value of the derivative f⁰.c/ for some c between a and b. This allows you to use information about the derivative to make statements about the change f.b/  f .a/. The theorem is usually proved in two steps by first proving Rolle’s Theorem which is a simpler version of the Mean Value Theorem.

Rolle’s Theorem states that if a< b, and if f is a function continuous on the interval Œa; b, differentiable on the interval .a; b/, and satisfying f .a/ D f .b/, then there is a c2 .a; b/ for which f⁰.c/ D 0.

What tools do you have to prove this result? Your proof needs to conclude that f⁰.c/ D 0. Think through what you know about derivatives, and see if any of the results conclude that the derivative is equal to 0. The only results that come to mind are the result that the derivative of any constant function is 0, and the result that if f reaches a relative extremum at a point where the function is differentiable, then its derivative at that point must be 0. It is unlikely that the first of these two results will be of much help except in the very special case where f is a constant function. So how can you use the result about extreme values to show that there is a place where the function has a derivative of 0? What you do know is that f is continuous on a closed intervalŒa; b, and the Extreme Value Theorem states that such a function obtains its maximum and minimum values on this interval. You also know that these extreme values can only occur at places where the derivative is 0, where the derivative does not exist, or at the endpoints of the interval. OK, there are no places on.a; b/ where the derivative does not exist, but could both the maximum and minimum occur at endpoints of the interval? The hypothesis of Rolle’s Theorem says that f.a/ D f .b/, so the only way that the two endpoints can be both maximum and minimum values of f on the interval is for f to be constant on the interval. In the case of a constant function, the theorem is clearly true. In any other case, it could be that f.a/ and f .b/ are maximum values for f or minimum values for f , but they cannot be both. If f is not constant, its maximum and minimum values must be different. That guarantees that f must have either an absolute maximum or an absolute minimum (possibly both) between a and b. That gives the result (Fig.5.5).

f(x)

a c b

Fig. 5.5 The proof of Rolle’s Theorem finds an extreme point c between a and b for which f⁰.c/ D 0

PROOF (Rolle’s Theorem): For a< b, let f be a function continuous on the interval Œa; b and differentiable on the interval .a; b/ satisfying f.a/ D f.b/. Then there is a c 2 .a; b/ for which f⁰.c/ D 0.

• For a < b, let f be a function continuous on the interval Œa; b and differentiable on the interval.a; b/ satisfying f .a/ D f .b/.

• Because f is continuous on the closed bounded interval Œa; b, it obtains a maximum and a minimum value there.

• If both the maximum and minimum values of f occur at endpoints of the interval, then, since f.a/ D f .b/, the maximum and minimum values of f are equal, and f is constant on the intervalŒa; b.

• In this case, f⁰.c/ D 0 for each c 2 .a; b/, and the conclusion of the theorem holds.

• If the maximum and minimum values of f do not both occur at endpoints of the interval, then there must be a c 2.a; b/ such that f reaches a maximum or a minimum value at c.

• In this case, f⁰.c/ D 0, and the conclusion of the theorem holds.

• In either case, the conclusion of the theorem holds which completes the proof.

Rolle’s Theorem takes care of the case where f.a/ D f .b/. To prove the Mean Value Theorem in the more general case where f.a/ need not equal f .b/, you would want to reduce this general case to the previously proved case where f.a/ and f .b/

are equal. An easy way to do this is to subtract a linear function from f to get a new function h which does satisfy the hypothesis of Rolle’s Theorem. This linear function can be any linear function that takes on a value at b which differs by f.b/  f .a/ from the value it takes on at a. One such function is

f.b/  f .a/

^x_b^a_a because it takes on the value 0 at a and f.b/  f .a/ at b (Fig.5.6).

148 5 Derivatives

Fig. 5.6 Point c between a and b where the tangent line is parallel to the secant line from a to b

a b

f(x)

c

PROOF (Mean Value Theorem): For a< b, let f be a function continuous on the intervalŒa; b and differentiable on the interval .a; b/. Then there is a c2 .a; b/ for which f⁰.c/ D^f.b/f.a/_ba .

• For a < b, let f be a function continuous on the interval Œa; b and differentiable on the interval.a; b/.

• Let h.x/ D f .x/ 

f.b/  f .a/

^x_b^a_a.

• Since both f.x/ and 

f.b/  f .a/

 _b^xa_a are continuous on Œa; b and differentiable .a; b/, h is also continuous on Œa; b and differentiable on .a; b/.

• h.b/ D f .b/ 

f.b/  f .a/

^b_b^a_aD f .b/ 

f.b/  f .a/

D f .a/ D h.a/.

• Thus, h satisfies the hypothesis of Rolle’s Theorem, so there is a c 2.a; b/

such that h⁰.c/ D 0.

• Then0 D h⁰.c/ D f⁰.c/ 

f.b/  f .a/

_ba¹ , so f⁰.c/ D ^{f.b/f .a/}_ba .

• Therefore, there is a c 2 .a; b/ with f⁰.c/ D ^{f.b/f .a/}_ba which completes the proof.

The following are two instructive applications of the Mean Value Theorem. First, if you know that a function f is differentiable on an interval, and its derivative is nonnegative on that interval, then the function must be increasing on the interval. To show that a function is increasing, you need to show that if x and y are in the interval with x< y, then f .x/  f .y/. This would follow from knowing that if y  x  0, then the difference quotient^{f.y/f .x/}_yx  0. What the Mean Value Theorem gives you is that this difference quotient is equal to the derivative of f at some point c between x and y. So, if you know that the derivative on the interval is always nonnegative, then the difference quotient must be nonnegative as needed.

PROOF: Let f be a function whose derivative is nonnegative at every point of an interval. Then f is an increasing function on that interval.

• Let f be a function whose derivative is nonnegative at each point of the interval I.

• Let x and y be in I with x< y.

• Then by the Mean Value Theorem, there is a c between x and y such that

f.y/f .x/

yx D f⁰.c/.

• Since I is an interval and x and y are in I, c is also in I, implying that f⁰.c/  0.

• Thus, ^{f.y/f .x/}_yx  0 so .y  x/  ^{f.y/f .x/}_yx D f .y/  f .x/  0.

• Therefore, f is increasing on I.

Clearly, if f⁰is strictly positive on an interval, then you can prove that f is strictly increasing on the interval. This can be done by altering the above proof by changing the greater than or equal signs to greater than signs where needed. Is the converse of the above theorem true? Well, one cannot conclude that a function is differentiable on an interval by just knowing that the function is increasing there. But what if you are given a differentiable function that is increasing? What can you conclude about the derivative? If a function is increasing, it does mean that every difference quotient ^{f.y/f .x/}_yx will be greater than or equal to 0, and, thus, the derivative which is the limit of such difference quotients will have to be greater than or equal to 0. If f is strictly increasing, can you conclude that its derivative is positive? In this case you cannot. You can conclude that all difference quotients will be positive, but the limit of positive difference quotients can be 0. For example, f.x/ D x³is a function differentiable on the entire real line, and it is strictly increasing, but its derivative is 0 at x D0.

Another important consequence of the Mean Value Theorem is that if a function has a derivative equal to 0 at every point of an interval, then f is constant on that interval. Again, this follows directly from what you can say about any difference quotient.

PROOF: Let f be a function whose derivative is 0 at every point of an interval. Then f is constant on that interval.

• Let f be a function whose derivative is 0 at each point of the interval I.

• Let x and y be in I with x< y.

• Then by the Mean Value Theorem, there is a c between x and y such that

f.y/f .x/

yx D f⁰.c/.

• Since I is an interval and x and y are in I, c is also in I, implying that f⁰.c/ D 0.

• Thus, ^{f.y/f .x/}_yx D 0 so .y  x/ ^{f.y/f .x/}_yx D f .y/  f .x/ D 0.

• Therefore, f.x/ D f .y/ for all x and y in the interval, and, thus, f is constant.

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How important is it that the set where f⁰ is 0 is an interval? The fact that the set is an interval is crucial. For example, the function f.x/ D

0 if x< 0 1 if x> 0



is not defined at 0. The derivative, f⁰, is equal to 0 at each point of the domain of f , but clearly, f is not a constant function, although it is constant on each interval contained in its domain. Looking back to the previous theorem, note that the function f.x/ D ¹_xhas a strictly positive derivative at each point of its domain, but, again, its domain does not include 0. This function is strictly increasing on each interval contained in its domain, but it is not an increasing function because f.1/ > f .1/.

5.7.1 Exercises

Write proofs for each of the following statements.

1. If f is a function whose derivative is negative for all points in an interval, then f is a decreasing function on the interval.

2. If f and g are functions differentiable on an interval with f⁰.x/ D g⁰.x/ for each x in the interval, then there is a constant C such that f.x/ D g.x/ C C for all x in the interval.

3. If f.0/ D g.0/ and f⁰.x/  g⁰.x/ for each x  0, then f .x/  g.x/ for each x > 0.