the mean value theorem homework 5.1

Find what you need to study

5.1 Using the Mean Value Theorem

4 min read • february 15, 2024

Attend a live cram event

Review all units live with expert teachers & students

In the previous unit , we learned all about applying derivatives to different real-world contexts. What else are derivatives useful for? Turns out, we can also use derivatives to determine and analyze the behaviors of functions! 👀

📈  Mean Value Theorem

The Mean Value Theorem states that if a function f is continuous over the interval [ a , b ] [a, b] [ a , b ] and differentiable over the interval ( a , b ) (a, b) ( a , b ) , then there exists a point c c c within that open interval ( a , b ) (a,b) ( a , b ) where the instantaneous rate of change of the function at c c c equals the average rate of change of the function over the interval ( a , b ) (a, b) ( a , b ) .

In other words, if a function f is continuous over the interval [ a , b ] [a, b] [ a , b ] and differentiable over the interval ( a , b ) (a, b) ( a , b ) , there exists some c c c on ( a , b ) (a,b) ( a , b ) such that f ′ ( c ) = f ( b ) − f ( a ) b − a f'(c)=\frac{f(b)-f(a)}{b-a} f ′ ( c ) = b − a f ( b ) − f ( a ) ​ .

Image Courtesy of Sumi Vora and Ethan Bilderbeek

Yet another way to phrase this theorem is that if the stated conditions of continuity and differentiability are satisfied, there is a point where the slope of the tangent line is equivalent to the slope of the secant line between a a a and b b b .

Image Courtesy of Math.net

Remember from Unit 1 , to be continuous over [ a , b ] [a, b] [ a , b ] means that there are no holes, asymptotes, or jump discontinuities between points a and b. Because the interval contains closed brackets, the graph must also be continuous at points a a a and b b b .

Additionally, if we recall from previous guides , to be differentiable over ( a , b ) (a, b) ( a , b ) means that the function is continuous over the interval and that for any point c c c over the interval, lim ⁡ x → c f ( x ) − f ( c ) x − c \lim_{x\to c} \frac{f(x) - f(c)}{x - c} lim x → c ​ x − c f ( x ) − f ( c ) ​ exists.

✏️ Mean Value Theorem: Walkthrough

We can use the Mean Value Theorem to justify conclusions about functions by applying it over an interval. For example:

Let f f f be a differentiable function. The table gives its selected values:

Can we use the Mean Value Theorem to say the equation f ′ ( x ) = 5 f'(x)=5 f ′ ( x ) = 5 has a solution where 3 < x < 9 3< x<9 3 < x < 9 ?

Since it is given that f f f is differentiable, we can apply the Mean Value Theorem (MVT) on the interval ( 3 , 9 ) (3,9) ( 3 , 9 ) . This is what we should write out!

Since f f f is continuous and differentiable on ( 3 , 9 ) (3,9) ( 3 , 9 ) , MVT can be applied. The MVT states that there exists a c c c on ( 3 , 9 ) (3,9) ( 3 , 9 ) such that f ′ ( c ) = f ( 9 ) − f ( 3 ) 9 − 3 = 44 − 20 9 − 3 = 24 6 = 4. f'(c)=\frac{f(9)-f(3)}{9-3}=\frac{44-20}{9-3}=\frac{24}{6}=4. f ′ ( c ) = 9 − 3 f ( 9 ) − f ( 3 ) ​ = 9 − 3 44 − 20 ​ = 6 24 ​ = 4.

4 ≠ 5 4 \neq 5 4  = 5 so MVT cannot be used to say that f ′ ( x ) = 5 f'(x)=5 f ′ ( x ) = 5 has a solution.

📝 Mean Value Theorem: Practice Problems

Now, it’s time for you to do some practice on your own! 🍀

❓ Mean Value Theorem: Practice

Question 1: mean value theorem.

Let h ( x ) = x 3 + 3 x 2 h(x)=x^3+3x^2 h ( x ) = x 3 + 3 x 2 and let c c c be the number that satisfies the Mean Value Theorem for h h h on the interval [ − 3 , 0 ] [-3,0] [ − 3 , 0 ] .

What is c ? c? c ?

Question 2: Mean Value Theorem

Can we use the Mean Value Theorem to say the equation f ′ ( x ) = 2 f'(x)=2 f ′ ( x ) = 2 has a solution where 2 < x < 9 2<x<9 2 < x < 9 ?

✅ Mean Value Theorem: Answers and Solutions

Since h h h is a polynomial, h h h is continuous on [ − 3 , 0 ] [-3,0] [ − 3 , 0 ] and differentiable on ( − 3 , 0 ) (-3,0) ( − 3 , 0 ) . Therefore, the Mean Value Theorem can be applied. By the Mean Value Theorem, there exists a c c c on ( − 3 , 0 ) (-3,0) ( − 3 , 0 ) such that…

To find c c c , we need to differentiate f ( x ) f(x) f ( x ) and find c c c such that f ′ ( c ) = 0. f'(c)=0. f ′ ( c ) = 0.

By the quadratic formula, we have x = − 2 , 0 x=-2,0 x = − 2 , 0 .

Since only x = − 2 x=-2 x = − 2 is in the interval ( − 3 , 0 ) (-3,0) ( − 3 , 0 ) , c = − 2. c=-2. c = − 2. Great work! Make sure you remember to check if the value(s) you get are in the given interval.

Since it is given that f f f is differentiable, we can apply the Mean Value Theorem (MVT) on the interval ( 2 , 9 ) (2,9) ( 2 , 9 ) .

The MVT states that there exists a c c c on ( 2 , 9 ) (2,9) ( 2 , 9 ) such that f ′ ( c ) = f ( 9 ) − f ( 2 ) 9 − 2 = 35 − 14 9 − 2 = 21 7 = 3. f'(c)=\frac{f(9)-f(2)}{9-2}=\frac{35-14}{9-2}=\frac{21}{7}=3. f ′ ( c ) = 9 − 2 f ( 9 ) − f ( 2 ) ​ = 9 − 2 35 − 14 ​ = 7 21 ​ = 3. 3 ≠ 2 3 \neq 2 3  = 2 so MVT cannot be used to say that f ′ ( x ) = 2 f'(x)=2 f ′ ( x ) = 2 has a solution.

The  Mean Value Theorem  states that for any  continuous  function on a  closed interval , there exists a value c in the interval such that the value of the  derivative  of the function at c is equal to the  average rate of change  of the function over the interval. By using this theorem, we can find the mean value of a function on a given interval, which can provide useful information about the behavior of the function. 🧐

Key Terms to Review ( 15 )

Absolute Value Function

Average Rate of Change

Closed Interval

Differentiable

f'(c) = [f(b)-f(a)]/b-a

Instantaneous Rate of Change

lim x->a [f(x) - f(a)]/(x - a)

Mean Value Theorem

Open Interval

Rolle's Theorem

Secant Line

Tangent Line

Stay Connected

© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

4.4 The Mean Value Theorem

Learning objectives.

• 4.4.1 Explain the meaning of Rolle’s theorem.
• 4.4.2 Describe the significance of the Mean Value Theorem.
• 4.4.3 State three important consequences of the Mean Value Theorem.

The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem.

Rolle’s Theorem

Informally, Rolle’s theorem states that if the outputs of a differentiable function f f are equal at the endpoints of an interval, then there must be an interior point c c where f ′ ( c ) = 0 . f ′ ( c ) = 0 . Figure 4.21 illustrates this theorem.

Theorem 4.4

Let f f be a continuous function over the closed interval [ a , b ] [ a , b ] and differentiable over the open interval ( a , b ) ( a , b ) such that f ( a ) = f ( b ) . f ( a ) = f ( b ) . There then exists at least one c ∈ ( a , b ) c ∈ ( a , b ) such that f ′ ( c ) = 0 . f ′ ( c ) = 0 .

Let k = f ( a ) = f ( b ) . k = f ( a ) = f ( b ) . We consider three cases:

• f ( x ) = k f ( x ) = k for all x ∈ ( a , b ) . x ∈ ( a , b ) .
• There exists x ∈ ( a , b ) x ∈ ( a , b ) such that f ( x ) > k . f ( x ) > k .
• There exists x ∈ ( a , b ) x ∈ ( a , b ) such that f ( x ) < k . f ( x ) < k .

Case 1: If f ( x ) = k f ( x ) = k for all x ∈ ( a , b ) , x ∈ ( a , b ) , then f ′ ( x ) = 0 f ′ ( x ) = 0 for all x ∈ ( a , b ) . x ∈ ( a , b ) .

Case 2: Since f f is a continuous function over the closed, bounded interval [ a , b ] , [ a , b ] , by the extreme value theorem, it has an absolute maximum. Also, since there is a point x ∈ ( a , b ) x ∈ ( a , b ) such that f ( x ) > k , f ( x ) > k , the absolute maximum is greater than k . k . Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point c ∈ ( a , b ) . c ∈ ( a , b ) . Because f f has a maximum at an interior point c , c , and f f is differentiable at c , c , by Fermat’s theorem, f ′ ( c ) = 0 . f ′ ( c ) = 0 .

Case 3: The case when there exists a point x ∈ ( a , b ) x ∈ ( a , b ) such that f ( x ) < k f ( x ) < k is analogous to case 2, with maximum replaced by minimum.

An important point about Rolle’s theorem is that the differentiability of the function f f is critical. If f f is not differentiable, even at a single point, the result may not hold. For example, the function f ( x ) = | x | − 1 f ( x ) = | x | − 1 is continuous over [ −1 , 1 ] [ −1 , 1 ] and f ( −1 ) = 0 = f ( 1 ) , f ( −1 ) = 0 = f ( 1 ) , but f ′ ( c ) ≠ 0 f ′ ( c ) ≠ 0 for any c ∈ ( −1 , 1 ) c ∈ ( −1 , 1 ) as shown in the following figure.

Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points c c where f ′ ( c ) = 0 . f ′ ( c ) = 0 .

Example 4.14

Using rolle’s theorem.

For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values c c in the given interval where f ′ ( c ) = 0 . f ′ ( c ) = 0 .

• f ( x ) = x 2 + 2 x f ( x ) = x 2 + 2 x over [ −2 , 0 ] [ −2 , 0 ]
• f ( x ) = x 3 − 4 x f ( x ) = x 3 − 4 x over [ −2 , 2 ] [ −2 , 2 ]

Checkpoint 4.14

Verify that the function f ( x ) = 2 x 2 − 8 x + 6 f ( x ) = 2 x 2 − 8 x + 6 defined over the interval [ 1 , 3 ] [ 1 , 3 ] satisfies the conditions of Rolle’s theorem. Find all points c c guaranteed by Rolle’s theorem.

The Mean Value Theorem and Its Meaning

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions f f defined on a closed interval [ a , b ] [ a , b ] with f ( a ) = f ( b ) f ( a ) = f ( b ) . The Mean Value Theorem generalizes Rolle’s theorem by considering functions that do not necessarily have equal value at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem ( Figure 4.25 ). The Mean Value Theorem states that if f f is continuous over the closed interval [ a , b ] [ a , b ] and differentiable over the open interval ( a , b ) , ( a , b ) , then there exists a point c ∈ ( a , b ) c ∈ ( a , b ) such that the tangent line to the graph of f f at c c is parallel to the secant line connecting ( a , f ( a ) ) ( a , f ( a ) ) and ( b , f ( b ) ) . ( b , f ( b ) ) .

Theorem 4.5

Mean value theorem.

Let f f be continuous over the closed interval [ a , b ] [ a , b ] and differentiable over the open interval ( a , b ) . ( a , b ) . Then, there exists at least one point c ∈ ( a , b ) c ∈ ( a , b ) such that

The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting ( a , f ( a ) ) ( a , f ( a ) ) and ( b , f ( b ) ) . ( b , f ( b ) ) . Since the slope of that line is

and the line passes through the point ( a , f ( a ) ) , ( a , f ( a ) ) , the equation of that line can be written as

Let g ( x ) g ( x ) denote the vertical difference between the point ( x , f ( x ) ) ( x , f ( x ) ) and the point ( x , y ) ( x , y ) on that line. Therefore,

Since the graph of f f intersects the secant line when x = a x = a and x = b , x = b , we see that g ( a ) = 0 = g ( b ) . g ( a ) = 0 = g ( b ) . Since f f is a differentiable function over ( a , b ) , ( a , b ) , g g is also a differentiable function over ( a , b ) . ( a , b ) . Furthermore, since f f is continuous over [ a , b ] , [ a , b ] , g g is also continuous over [ a , b ] . [ a , b ] . Therefore, g g satisfies the criteria of Rolle’s theorem. Consequently, there exists a point c ∈ ( a , b ) c ∈ ( a , b ) such that g ′ ( c ) = 0 . g ′ ( c ) = 0 . Since

we see that

Since g ′ ( c ) = 0 , g ′ ( c ) = 0 , we conclude that

In the next example, we show how the Mean Value Theorem can be applied to the function f ( x ) = x f ( x ) = x over the interval [ 0 , 9 ] . [ 0 , 9 ] . The method is the same for other functions, although sometimes with more interesting consequences.

Example 4.15

Verifying that the mean value theorem applies.

For f ( x ) = x f ( x ) = x over the interval [ 0 , 9 ] , [ 0 , 9 ] , show that f f satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value c ∈ ( 0 , 9 ) c ∈ ( 0 , 9 ) such that f ′ ( c ) f ′ ( c ) is equal to the slope of the line connecting ( 0 , f ( 0 ) ) ( 0 , f ( 0 ) ) and ( 9 , f ( 9 ) ) . ( 9 , f ( 9 ) ) . Find these values c c guaranteed by the Mean Value Theorem.

We know that f ( x ) = x f ( x ) = x is continuous over [ 0 , 9 ] [ 0 , 9 ] and differentiable over ( 0 , 9 ) . ( 0 , 9 ) . Therefore, f f satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value c ∈ ( 0 , 9 ) c ∈ ( 0 , 9 ) such that f ′ ( c ) f ′ ( c ) is equal to the slope of the line connecting ( 0 , f ( 0 ) ) ( 0 , f ( 0 ) ) and ( 9 , f ( 9 ) ) ( 9 , f ( 9 ) ) ( Figure 4.27 ). To determine which value(s) of c c are guaranteed, first calculate the derivative of f . f . The derivative f ′ ( x ) = 1 ( 2 x ) . f ′ ( x ) = 1 ( 2 x ) . The slope of the line connecting ( 0 , f ( 0 ) ) ( 0 , f ( 0 ) ) and ( 9 , f ( 9 ) ) ( 9 , f ( 9 ) ) is given by

We want to find c c such that f ′ ( c ) = 1 3 . f ′ ( c ) = 1 3 . That is, we want to find c c such that

Solving this equation for c , c , we obtain c = 9 4 . c = 9 4 . At this point, the slope of the tangent line equals the slope of the line joining the endpoints.

One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 h down a straight road with an average velocity of 45 mph. Let s ( t ) s ( t ) and v ( t ) v ( t ) denote the position and velocity of the car, respectively, for 0 ≤ t ≤ 1 0 ≤ t ≤ 1 h. Assuming that the position function s ( t ) s ( t ) is differentiable, we can apply the Mean Value Theorem to conclude that, at some time c ∈ ( 0 , 1 ) , c ∈ ( 0 , 1 ) , the speed of the car was exactly

Example 4.16

Mean value theorem and velocity.

If a rock is dropped from a height of 100 ft, its position t t seconds after it is dropped until it hits the ground is given by the function s ( t ) = −16 t 2 + 100 . s ( t ) = −16 t 2 + 100 .

• Determine how long it takes before the rock hits the ground.
• Find the average velocity v avg v avg of the rock for when the rock is released and the rock hits the ground.
• Find the time t t guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is v avg . v avg .
• When the rock hits the ground, its position is s ( t ) = 0 . s ( t ) = 0 . Solving the equation −16 t 2 + 100 = 0 −16 t 2 + 100 = 0 for t , t , we find that t = ± 5 2 sec . t = ± 5 2 sec . Since we are only considering t ≥ 0 , t ≥ 0 , the ball will hit the ground 5 2 5 2 sec after it is dropped.
• The average velocity is given by v avg = s ( 5 / 2 ) − s ( 0 ) 5 / 2 − 0 = 0 − 100 5 / 2 = −40 ft/sec . v avg = s ( 5 / 2 ) − s ( 0 ) 5 / 2 − 0 = 0 − 100 5 / 2 = −40 ft/sec .

Checkpoint 4.15

Suppose a ball is dropped from a height of 200 ft. Its position at time t t is s ( t ) = −16 t 2 + 200 . s ( t ) = −16 t 2 + 200 . Find the time t t when the instantaneous velocity of the ball equals its average velocity.

Corollaries of the Mean Value Theorem

Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.

At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if f ′ ( x ) = 0 f ′ ( x ) = 0 for all x x in some interval I , I , then f ( x ) f ( x ) is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly.

Theorem 4.6

Corollary 1: functions with a derivative of zero.

Let f f be differentiable over an interval I . I . If f ′ ( x ) = 0 f ′ ( x ) = 0 for all x ∈ I , x ∈ I , then f ( x ) = f ( x ) = constant for all x ∈ I . x ∈ I .

Since f f is differentiable over I , I , f f must be continuous over I . I . Suppose f ( x ) f ( x ) is not constant for all x x in I . I . Then there exist a , b ∈ I , a , b ∈ I , where a ≠ b a ≠ b and f ( a ) ≠ f ( b ) . f ( a ) ≠ f ( b ) . Choose the notation so that a < b . a < b . Therefore,

Since f f is a differentiable function, by the Mean Value Theorem, there exists c ∈ ( a , b ) c ∈ ( a , b ) such that

Therefore, there exists c ∈ I c ∈ I such that f ′ ( c ) ≠ 0 , f ′ ( c ) ≠ 0 , which contradicts the assumption that f ′ ( x ) = 0 f ′ ( x ) = 0 for all x ∈ I . x ∈ I .

From Corollary 1: Functions with a Derivative of Zero , it follows that if two functions have the same derivative, they differ by, at most, a constant.

Theorem 4.7

Corollary 2: constant difference theorem.

If f f and g g are differentiable over an interval I I and f ′ ( x ) = g ′ ( x ) f ′ ( x ) = g ′ ( x ) for all x ∈ I , x ∈ I , then f ( x ) = g ( x ) + C f ( x ) = g ( x ) + C for some constant C . C .

Let h ( x ) = f ( x ) − g ( x ) . h ( x ) = f ( x ) − g ( x ) . Then, h ′ ( x ) = f ′ ( x ) − g ′ ( x ) = 0 h ′ ( x ) = f ′ ( x ) − g ′ ( x ) = 0 for all x ∈ I . x ∈ I . By Corollary 1, there is a constant C C such that h ( x ) = C h ( x ) = C for all x ∈ I . x ∈ I . Therefore, f ( x ) = g ( x ) + C f ( x ) = g ( x ) + C for all x ∈ I . x ∈ I .

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function f f is increasing over I I if f ( x 1 ) < f ( x 2 ) f ( x 1 ) < f ( x 2 ) whenever x 1 < x 2 , x 1 < x 2 , whereas f f is decreasing over I I if f ( x ) 1 > f ( x 2 ) f ( x ) 1 > f ( x 2 ) whenever x 1 < x 2 . x 1 < x 2 . Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing ( Figure 4.29 ). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph.

This fact is important because it means that for a given function f , f , if there exists a function F F such that F ′ ( x ) = f ( x ) ; F ′ ( x ) = f ( x ) ; then, the only other functions that have a derivative equal to f f are F ( x ) + C F ( x ) + C for some constant C . C . We discuss this result in more detail later in the chapter.

Theorem 4.8

Corollary 3: increasing and decreasing functions.

Let f f be continuous over the closed interval [ a , b ] [ a , b ] and differentiable over the open interval ( a , b ) . ( a , b ) .

• If f ′ ( x ) > 0 f ′ ( x ) > 0 for all x ∈ ( a , b ) , x ∈ ( a , b ) , then f f is an increasing function over [ a , b ] . [ a , b ] .
• If f ′ ( x ) < 0 f ′ ( x ) < 0 for all x ∈ ( a , b ) , x ∈ ( a , b ) , then f f is a decreasing function over [ a , b ] . [ a , b ] .

We will prove i.; the proof of ii. is similar. Suppose f f is not an increasing function on I . I . Then there exist a a and b b in I I such that a < b , a < b , but f ( a ) > f ( b ) . f ( a ) > f ( b ) . Since f f is a differentiable function over I , I , by the Mean Value Theorem there exists c ∈ ( a , b ) c ∈ ( a , b ) such that

Since f ( a ) > f ( b ) , f ( a ) > f ( b ) , we know that f ( b ) − f ( a ) < 0 . f ( b ) − f ( a ) < 0 . Also, a < b a < b tells us that b − a > 0 . b − a > 0 . We conclude that

However, f ′ ( x ) > 0 f ′ ( x ) > 0 for all x ∈ I . x ∈ I . This is a contradiction, and therefore f f must be an increasing function over I . I .

Section 4.4 Exercises

Why do you need continuity to apply the Mean Value Theorem? Construct a counterexample.

Why do you need differentiability to apply the Mean Value Theorem? Find a counterexample.

When are Rolle’s theorem and the Mean Value Theorem equivalent?

If you have a function with a discontinuity, is it still possible to have f ′ ( c ) ( b − a ) = f ( b ) − f ( a ) ? f ′ ( c ) ( b − a ) = f ( b ) − f ( a ) ? Draw such an example or prove why not.

For the following exercises, determine over what intervals (if any) the Mean Value Theorem applies. Justify your answer.

y = sin ( π x ) y = sin ( π x )

y = 1 x 3 y = 1 x 3

y = 4 − x 2 y = 4 − x 2

y = x 2 − 4 y = x 2 − 4

y = ln ( 3 x − 5 ) y = ln ( 3 x − 5 )

For the following exercises, graph the functions on a calculator and draw the secant line that connects the endpoints. Estimate the number of points c c such that f ′ ( c ) ( b − a ) = f ( b ) − f ( a ) . f ′ ( c ) ( b − a ) = f ( b ) − f ( a ) .

[T] y = 3 x 3 + 2 x + 1 y = 3 x 3 + 2 x + 1 over [ −1 , 1 ] [ −1 , 1 ]

[T] y = tan ( π 4 x ) y = tan ( π 4 x ) over [ − 3 2 , 3 2 ] [ − 3 2 , 3 2 ]

[T] y = x 2 cos ( π x ) y = x 2 cos ( π x ) over [ −2 , 2 ] [ −2 , 2 ]

[T] y = x 6 − 3 4 x 5 − 9 8 x 4 + 15 16 x 3 + 3 32 x 2 + 3 16 x + 1 32 y = x 6 − 3 4 x 5 − 9 8 x 4 + 15 16 x 3 + 3 32 x 2 + 3 16 x + 1 32 over [ −1 , 1 ] [ −1 , 1 ]

For the following exercises, use the Mean Value Theorem and find all points 0 < c < 2 0 < c < 2 such that f ( 2 ) − f ( 0 ) = f ′ ( c ) ( 2 − 0 ) . f ( 2 ) − f ( 0 ) = f ′ ( c ) ( 2 − 0 ) .

f ( x ) = x 3 f ( x ) = x 3

f ( x ) = sin ( π x ) f ( x ) = sin ( π x )

f ( x ) = cos ( 2 π x ) f ( x ) = cos ( 2 π x )

f ( x ) = 1 + x + x 2 f ( x ) = 1 + x + x 2

f ( x ) = ( x − 1 ) 10 f ( x ) = ( x − 1 ) 10

f ( x ) = ( x − 1 ) 9 f ( x ) = ( x − 1 ) 9

For the following exercises, show there is no c c such that f ( 1 ) − f ( −1 ) = f ′ ( c ) ( 2 ) . f ( 1 ) − f ( −1 ) = f ′ ( c ) ( 2 ) . Explain why the Mean Value Theorem does not apply over the interval [ −1 , 1 ] . [ −1 , 1 ] .

f ( x ) = | x − 1 2 | f ( x ) = | x − 1 2 |

f ( x ) = 1 x 2 f ( x ) = 1 x 2

f ( x ) = | x | f ( x ) = | x |

f ( x ) = ⌊ x ⌋ f ( x ) = ⌊ x ⌋ ( Hint : This is called the floor function and it is defined so that f ( x ) f ( x ) is the largest integer less than or equal to x . ) x . )

For the following exercises, determine whether the Mean Value Theorem applies for the functions over the given interval [ a , b ] . [ a , b ] . Justify your answer.

y = e x y = e x over [ 0 , 1 ] [ 0 , 1 ]

y = ln ( 2 x + 3 ) y = ln ( 2 x + 3 ) over [ − 3 2 , 0 ] [ − 3 2 , 0 ]

f ( x ) = tan ( 2 π x ) f ( x ) = tan ( 2 π x ) over [ 0 , 2 ] [ 0 , 2 ]

y = 9 − x 2 y = 9 − x 2 over [ −3 , 3 ] [ −3 , 3 ]

y = 1 | x + 1 | y = 1 | x + 1 | over [ 0 , 3 ] [ 0 , 3 ]

y = x 3 + 2 x + 1 y = x 3 + 2 x + 1 over [ 0 , 6 ] [ 0 , 6 ]

y = x 2 + 3 x + 2 x y = x 2 + 3 x + 2 x over [ −1 , 1 ] [ −1 , 1 ]

y = x sin ( π x ) + 1 y = x sin ( π x ) + 1 over [ 0 , 1 ] [ 0 , 1 ]

y = ln ( x + 1 ) y = ln ( x + 1 ) over [ 0 , e − 1 ] [ 0 , e − 1 ]

y = x sin ( π x ) y = x sin ( π x ) over [ 0 , 2 ] [ 0 , 2 ]

y = 5 + | x | y = 5 + | x | over [ −1 , 1 ] [ −1 , 1 ]

For the following exercises, consider the roots of the equation.

Show that the equation y = x 3 + 4 x + 16 y = x 3 + 4 x + 16 has exactly one real root. What is it?

Find the conditions for exactly one root (double root) for the equation y = x 2 + b x + c y = x 2 + b x + c

Find the conditions for y = e x − b y = e x − b to have one root. Is it possible to have more than one root?

For the following exercises, use a calculator to graph the function over the interval [ a , b ] [ a , b ] and graph the secant line from a a to b . b . Use the calculator to estimate all values of c c as guaranteed by the Mean Value Theorem. Then, find the exact value of c , c , if possible, or write the final equation and use a calculator to estimate to four digits.

[T] y = tan ( π x ) y = tan ( π x ) over [ − 1 4 , 1 4 ] [ − 1 4 , 1 4 ]

[T] y = 1 x + 1 y = 1 x + 1 over [ 0 , 3 ] [ 0 , 3 ]

[T] y = | x 2 + 2 x − 4 | y = | x 2 + 2 x − 4 | over [ −4 , 0 ] [ −4 , 0 ]

[T] y = x + 1 x y = x + 1 x over [ 1 2 , 4 ] [ 1 2 , 4 ]

[T] y = x + 1 + 1 x 2 y = x + 1 + 1 x 2 over [ 3 , 8 ] [ 3 , 8 ]

At 10:17 a.m., you pass a police car at 55 mph that is stopped on the freeway. You pass a second police car at 55 mph at 10:53 a.m., which is located 39 mi from the first police car. If the speed limit is 60 mph, can the police cite you for speeding?

Two cars drive from one stoplight to the next, leaving at the same time and arriving at the same time. Is there ever a time when they are going the same speed? Prove or disprove.

Show that y = sec 2 x y = sec 2 x and y = tan 2 x y = tan 2 x have the same derivative. What can you say about y = sec 2 x − tan 2 x ? y = sec 2 x − tan 2 x ?

Show that y = csc 2 x y = csc 2 x and y = cot 2 x y = cot 2 x have the same derivative. What can you say about y = csc 2 x − cot 2 x ? y = csc 2 x − cot 2 x ?

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax.

Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction

• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 1
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-1/pages/4-4-the-mean-value-theorem

© Feb 5, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

4.5: The Mean Value Theorem

• Last updated
• Save as PDF
• Page ID 143754

• Gilbert Strang & Edwin “Jed” Herman

Learning Objectives

• Explain the meaning of Rolle’s theorem.
• Describe the significance of the Mean Value Theorem.
• State three important consequences of the Mean Value Theorem.

The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem.

Rolle’s Theorem

Informally, Rolle’s theorem states that if the outputs of a differentiable function \(f\) are equal at the endpoints of an interval, then there must be an interior point \(c\) where \(f'(c)=0\). Figure \(\PageIndex{1}\) illustrates this theorem.

Let \(f\) be a continuous function over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\) such that \(f(a)=f(b)\). There then exists at least one \(c∈(a,b)\) such that \(f'(c)=0.\)

Let \(k=f(a)=f(b).\) We consider three cases:

• \(f(x)=k\) for all \(x∈(a,b).\)
• There exists \(x∈(a,b)\) such that \(f(x)>k.\)
• There exists \(x∈(a,b)\) such that \(f(x)<k.\)

Case 1 : If \(f(x)=k\) for all \(x∈(a,b)\), then \(f'(x)=0\) for all \(x∈(a,b).\)

Case 2 : Since \(f\) is a continuous function over the closed, bounded interval \([a,b]\), by the extreme value theorem, it has an absolute maximum. Also, since there is a point \(x∈(a,b)\) such that \(f(x)>k\), the absolute maximum is greater than \(k\). Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point \(c∈(a,b)\). Because \(f\) has a maximum at an interior point \(c\), and \(f\) is differentiable at \(c\), by Fermat’s theorem, \(f'(c)=0.\)

Case 3 : The case when there exists a point \(x∈(a,b)\) such that \(f(x)<k\) is analogous to case 2, with maximum replaced by minimum.

An important point about Rolle’s theorem is that the differentiability of the function \(f\) is critical. If \(f\) is not differentiable, even at a single point, the result may not hold. For example, the function \(f(x)=|x|−1\) is continuous over \([−1,1]\) and \(f(−1)=0=f(1)\), but \(f'(c)≠0\) for any \(c∈(−1,1)\) as shown in the following figure.

Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points \(c\) where \(f'(c)=0.\)

Example \(\PageIndex{1}\): Using Rolle’s Theorem

For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values \(c\) in the given interval where \(f'(c)=0.\)

• \(f(x)=x^2+2x\) over \([−2,0]\)
• \(f(x)=x^3−4x\) over \([−2,2]\)

a. Since \(f\) is a polynomial, it is continuous and differentiable everywhere. In addition, \(f(−2)=0=f(0).\) Therefore, \(f\) satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value \(c∈(−2,0)\) such that \(f'(c)=0\). Since \(f'(x)=2x+2=2(x+1),\) we see that \(f'(c)=2(c+1)=0\) implies \(c=−1\) as shown in the following graph.

b. As in part a. \(f\) is a polynomial and therefore is continuous and differentiable everywhere. Also, \(f(−2)=0=f(2).\) That said, \(f\) satisfies the criteria of Rolle’s theorem. Differentiating, we find that \(f'(x)=3x^2−4.\) Therefore, \(f'(c)=0\) when \(x=±\frac{2}{\sqrt{3}}\). Both points are in the interval \([−2,2]\), and, therefore, both points satisfy the conclusion of Rolle’s theorem as shown in the following graph.

Exercise \(\PageIndex{1}\)

Verify that the function \(f(x)=2x^2−8x+6\) defined over the interval \([1,3]\) satisfies the conditions of Rolle’s theorem. Find all points \(c\) guaranteed by Rolle’s theorem.

Find all values \(c\), where \(f'(c)=0\).

The Mean Value Theorem and Its Meaning

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions \(f\) that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem (Figure \(\PageIndex{5}\)). The Mean Value Theorem states that if \(f\) is continuous over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\), then there exists a point \(c∈(a,b)\) such that the tangent line to the graph of \(f\) at \(c\) is parallel to the secant line connecting \((a,f(a))\) and \((b,f(b)).\)

Mean Value Theorem

Let \(f\) be continuous over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\). Then, there exists at least one point \(c∈(a,b)\) such that

\[f'(c)=\frac{f(b)−f(a)}{b−a} \nonumber \]

The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting \((a,f(a))\) and \((b,f(b)).\) Since the slope of that line is

\[\frac{f(b)−f(a)}{b−a} \nonumber \]

and the line passes through the point \((a,f(a)),\) the equation of that line can be written as

\[y=\frac{f(b)−f(a)}{b−a}(x−a)+f(a). \nonumber \]

Let \(g(x)\) denote the vertical difference between the point \((x,f(x))\) and the point \((x,y)\) on that line. Therefore,

\[g(x)=f(x)−\left[\frac{f(b)−f(a)}{b−a}(x−a)+f(a)\right]. \nonumber \]

Since the graph of \(f\) intersects the secant line when \(x=a\) and \(x=b\), we see that \(g(a)=0=g(b)\). Since \(f\) is a differentiable function over \((a,b)\), \(g\) is also a differentiable function over \((a,b)\). Furthermore, since \(f\) is continuous over \([a,b], \, g\) is also continuous over \([a,b]\). Therefore, \(g\) satisfies the criteria of Rolle’s theorem. Consequently, there exists a point \(c∈(a,b)\) such that \(g'(c)=0.\) Since

\[g'(x)=f'(x)−\frac{f(b)−f(a)}{b−a}, \nonumber \]

we see that

\[g'(c)=f'(c)−\frac{f(b)−f(a)}{b−a}. \nonumber \]

Since \(g'(c)=0,\) we conclude that

\[f'(c)=\frac{f(b)−f(a)}{b−a}. \nonumber \]

In the next example, we show how the Mean Value Theorem can be applied to the function \(f(x)=\sqrt{x}\) over the interval \([0,9]\). The method is the same for other functions, although sometimes with more interesting consequences.

Example \(\PageIndex{2}\): Verifying that the Mean Value Theorem Applies

For \(f(x)=\sqrt{x}\) over the interval \([0,9]\), show that \(f\) satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value \(c∈(0,9)\) such that \(f′(c)\) is equal to the slope of the line connecting \((0,f(0))\) and \((9,f(9))\). Find these values \(c\) guaranteed by the Mean Value Theorem.

We know that \(f(x)=\sqrt{x}\) is continuous over \([0,9]\) and differentiable over \((0,9).\) Therefore, \(f\) satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value \(c∈(0,9)\) such that \(f′(c)\) is equal to the slope of the line connecting \((0,f(0))\) and \((9,f(9))\) (Figure \(\PageIndex{7}\)). To determine which value(s) of \(c\) are guaranteed, first calculate the derivative of \(f\). The derivative \(f′(x)=\frac{1}{(2\sqrt{x})}\). The slope of the line connecting \((0,f(0))\) and \((9,f(9))\) is given by

\[\frac{f(9)−f(0)}{9−0}=\frac{\sqrt{9}−\sqrt{0}}{9−0}=\frac{3}{9}=\frac{1}{3}. \nonumber \]

We want to find \(c\) such that \(f′(c)=\frac{1}{3}\). That is, we want to find \(c\) such that

\[\frac{1}{2\sqrt{c}}=\frac{1}{3}. \nonumber \]

Solving this equation for \(c\), we obtain \(c=\frac{9}{4}\). At this point, the slope of the tangent line equals the slope of the line joining the endpoints.

One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 h down a straight road with an average velocity of 45 mph. Let \(s(t)\) and \(v(t)\) denote the position and velocity of the car, respectively, for \(0≤t≤1\) h. Assuming that the position function \(s(t)\) is differentiable, we can apply the Mean Value Theorem to conclude that, at some time \(c∈(0,1)\), the speed of the car was exactly

\[v(c)=s′(c)=\frac{s(1)−s(0)}{1−0}=45\,\text{mph.} \nonumber \]

Example \(\PageIndex{3}\): Mean Value Theorem and Velocity

If a rock is dropped from a height of 100 ft, its position \(t\) seconds after it is dropped until it hits the ground is given by the function \(s(t)=−16t^2+100.\)

• Determine how long it takes before the rock hits the ground.
• Find the average velocity \(v_{avg}\) of the rock for when the rock is released and the rock hits the ground.
• Find the time \(t\) guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is \(v_{avg}.\)

a. When the rock hits the ground, its position is \(s(t)=0\). Solving the equation \(−16t^2+100=0\) for \(t\), we find that \(t=±\frac{5}{2}sec\). Since we are only considering \(t≥0\), the ball will hit the ground \(\frac{5}{2}\) sec after it is dropped.

b. The average velocity is given by

\[v_{avg}=\frac{s(5/2)−s(0)}{5/2−0}=\frac{0−100}{5/2}=−40\,\text{ft/sec}. \nonumber \]

c. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time \(t\) such that \(v(t)=s′(t)=v_{avg}=−40\) ft/sec. Since \(s(t)\) is continuous over the interval \([0,5/2]\) and differentiable over the interval \((0,5/2),\) by the Mean Value Theorem, there is guaranteed to be a point \(c∈(0,5/2)\) such that

\[s′(c)=\frac{s(5/2)−s(0)}{5/2−0}=−40. \nonumber \]

Taking the derivative of the position function \(s(t)\), we find that \(s′(t)=−32t.\) Therefore, the equation reduces to \(s′(c)=−32c=−40.\) Solving this equation for \(c\), we have \(c=\frac{5}{4}\). Therefore, \(\frac{5}{4}\) sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: \(−40\) ft/sec.

Exercise \(\PageIndex{2}\)

Suppose a ball is dropped from a height of 200 ft. Its position at time \(t\) is \(s(t)=−16t^2+200.\) Find the time \(t\) when the instantaneous velocity of the ball equals its average velocity.

First, determine how long it takes for the ball to hit the ground. Then, find the average velocity of the ball from the time it is dropped until it hits the ground.

\(\frac{5}{2\sqrt{2}}\) sec

Corollaries of the Mean Value Theorem

Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.

At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if \(f′(x)=0\) for all \(x\) in some interval \(I\), then \(f(x)\) is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly.

Corollary 1: Functions with a Derivative of Zero

Let \(f\) be differentiable over an interval \(I\). If \(f′(x)=0\) for all \(x∈I\), then \(f(x)=\) constant for all \(x∈I.\)

Since \(f\) is differentiable over \(I\), \(f\) must be continuous over \(I\). Suppose \(f(x)\) is not constant for all \(x\) in \(I\). Then there exist \(a,b∈I\), where \(a≠b\) and \(f(a)≠f(b).\) Choose the notation so that \(a<b.\) Therefore,

\[\frac{f(b)−f(a)}{b−a}≠0. \nonumber \]

Since \(f\) is a differentiable function, by the Mean Value Theorem, there exists \(c∈(a,b)\) such that

\[f′(c)=\frac{f(b)−f(a)}{b−a}. \nonumber \]

Therefore, there exists \(c∈I\) such that \(f′(c)≠0\), which contradicts the assumption that \(f′(x)=0\) for all \(x∈I\).

From "Corollary 1: Functions with a Derivative of Zero," it follows that if two functions have the same derivative, they differ by, at most, a constant.

Corollary 2: Constant Difference Theorem

If \(f\) and \(g\) are differentiable over an interval \(I\) and \(f′(x)=g′(x)\) for all \(x∈I\), then \(f(x)=g(x)+C\) for some constant \(C\).

Let \(h(x)=f(x)−g(x).\) Then, \(h′(x)=f′(x)−g′(x)=0\) for all \(x∈I.\) By Corollary 1, there is a constant \(C\) such that \(h(x)=C\) for all \(x∈I\). Therefore, \(f(x)=g(x)+C\) for all \(x∈I.\)

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function \(f\) is increasing over \(I\) if \(f(x_1)<f(x_2)\) whenever \(x_1<x_2\), whereas \(f\) is decreasing over \(I\) if \(f(x_1)>f(x_2)\) whenever \(x_1<x_2\). Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure \(\PageIndex{9}\)). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph.

This fact is important because it means that for a given function \(f\), if there exists a function \(F\) such that \(F′(x)=f(x)\); then, the only other functions that have a derivative equal to \(f\) are \(F(x)+C\) for some constant \(C\). We discuss this result in more detail later in the chapter.

Corollary 3: Increasing and Decreasing Functions

Let \(f\) be continuous over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\).

• If \(f′(x)>0\) for all \(x∈(a,b)\), then \(f\) is an increasing function over \([a,b].\)
• If \(f′(x)<0\) for all \(x∈(a,b)\), then \(f\) is a decreasing function over \([a,b].\)

We will prove i.; the proof of ii. is similar. Suppose \(f\) is not an increasing function on \(I\). Then there exist \(a\) and \(b\) in \(I\) such that \(a<b\), but \(f(a)≥f(b)\). Since \(f\) is a differentiable function over \(I\), by the Mean Value Theorem there exists \(c∈(a,b)\) such that

Since \(f(a)≥f(b)\), we know that \(f(b)−f(a)≤0\). Also, \(a<b\) tells us that \(b−a>0.\) We conclude that

\[f′(c)=\frac{f(b)−f(a)}{b−a}≤0. \nonumber \]

However, \(f′(x)>0\) for all \(x∈I\). This is a contradiction, and therefore \(f\) must be an increasing function over \(I\).

Key Concepts

• If \(f\) is continuous over \([a,b]\) and differentiable over \((a,b)\) and \(f(a)=f(b)\), then there exists a point \(c∈(a,b)\) such that \(f′(c)=0.\) This is Rolle’s theorem.
• If \(f\) is continuous over \([a,b]\) and differentiable over \((a,b)\), then there exists a point \(c∈(a,b)\) such that \[f'(c)=\frac{f(b)−f(a)}{b−a}. \nonumber \] This is the Mean Value Theorem.
• If \(f'(x)=0\) over an interval \(I\), then \(f\) is constant over \(I\).
• If two differentiable functions \(f\) and \(g\) satisfy \(f′(x)=g′(x)\) over \(I\), then \(f(x)=g(x)+C\) for some constant \(C\).
• If \(f′(x)>0\) over an interval \(I\), then \(f\) is increasing over \(I\). If \(f′(x)<0\) over \(I\), then \(f\) is decreasing over \(I\).

if \(f\) is continuous over \([a,b]\) and differentiable over \((a,b)\), then there exists \(c∈(a,b)\) such that \(f′(c)=\frac{f(b)−f(a)}{b−a}\)

Return To Contents Go To Problems & Solutions

We have in Section 1.2.2 the definitions of a maximum and a minimum of a function. In Fig. 1.1, the maximum of f is f ( a ) and the minimum of f is f ( x 3 ). The value f ( x 2 ) isn't a maximum of f , but there's a vicinity of x 2 over which f ( x 2 ) is a

maximum. We call f ( x 2 ) a local maximum   of f . The value f ( x 3 ) is the minimum of f ; there's a vicinity of x 3 , which can be as large as [ a , b ], over which f ( x 3 ) is a minimum; so we also call f ( x 3 ) a local minimum   of f . The value f ( a ) is the maximum of f ; there's a vicinity to the right of a , which can be as large as [ a , b ], over which f ( a ) is a maximum; thus   f ( a ) is also a local maximum of f . There's a vicinity to the left of b over which f ( b ) is a minimum; hence f ( b ) is a local minimum of f .

For x 2 , we can say that there exists h > 0 such that f ( x 2 ) is a maximum of f on ( x 2 – h , x 2 + h ). However, for a , we can say that there exists k > 0 such that f ( a ) is a maximum of f on, well, not ( a – k , a + k ), but just [ a , a + k ). Now, [ a , a + k ) is the intersection of ( a – k , a + k ) and dom ( f ). Note that ( x 2 – h , x 2 + h ) is the intersection of itself and dom ( f ).

Definitions 1.1 – Local Maxima And Minima

Relative And Absolute Maximum And Minimum

A local maximum or minimum is also called a relative maximum or minimum. A maximum and minimum are also called   absolute maximum and absolute minimum respectively because they're the maximum and minimum respectively of the function over its entire domain. An absolute maximum (or minimum) is also a local maximum (or minimum). The converse isn't true: a local maximum (or minimum) isn't necessarily an absolute maximum (or minimum).

An extremum is either a maximum or a minimum. Maxima and minima are collectively referred to as extrema .

Go To Problems & Solutions      Return To Top Of Page

Refer to Fig. 2.1. Suppose that f is defined on [ a , b ] and differentiable at every point in ( a , b ) except x 3 , and that f ( x 1 ),   f ( x 2 ), and f ( x 3 ) are local extrema. We observe that the tangent lines to the graph of f at x 1 and x 2 are horizontal, which means that f ' ( x 1 ) = f ' ( x 2 ) = 0. There's no (unique) tangent line at x 3 ; the graph has a sharp point there. It appears that the derivatives of f at all points generating local extrema, if they exist, are 0. Well, they are, as confirmed by the following theorem. Note that since f isn't defined to the left of a or to the right of b , it can't be differentiable at either endpoint a or b .

Theorem 2.1

Consequently, f ' ( x 1 ) = 0. If f ( x 1 ) is a local minimum, a similar argument will lead to the same conclusion that f ' ( x 1 ) = 0.

Let f be continuous on [ a , b ] and differentiable on ( a , b ). See Fig. 3.1. Suppose f ( a ) = f ( b ). We observe that there must be a point on the graph of f where the tangent line to the graph is horizontal, which means that, letting c be the x -coordinate of that point, f ' ( c ) = 0.

Theorem 3.1 - Rolle's Theorem

The following example is an application of Rolle's theorem.

Example 4.1

Prove that the equation x 2 – 3 x + 1 = 0 has a unique solution in [2, 4].

The graph in Fig. 4.1 doesn't constitute “ proof ” ! It simply helps us in doing the proof. The graph of a function is the visualization of the behavior of the function. The graph doesn't explain why the function behaves the way it does.

Suppose f is continuous on [ a , b ] and differentiable on ( a , b ), as shown in Fig. 5.1. Draw the secant line joining the point ( a , f ( a )) to the point ( b , f ( b )). Clearly there's a point c where a < c < b such that the tangent line to the graph of f at x = c is parallel to, thus has the same slope as, the secant line. The slope of the secant line is ( f ( b ) – f ( a ))/( b – a ), and that of the tangent line is f ' ( c ). So ( f ( b ) – f ( a ))/( b – a ) = f ' ( c ). Recall that the quantity ( f ( b ) – f ( a ))/( b – a ) is also the average or mean rate of change of f over [ a , b ] and that the quantity f ' ( c ) is also the instantaneous rate of change of f at c ; see Section 2.3 . Consequently, we see that there's a point c in ( a , b ) such that the average or mean rate of change of f over [ a , b ] equals the instantaneous rate of change of f at c . That's confirmed by the mean-value theorem, below.

Theorem 5.1 – The Mean-Value Theorem

Here we discuss why the function g in the proof of the mean-value theorem is so defined. Refer to Fig. 5.1. The fact that the tangent line is parallel to the secant line reminds us of Rolle's theorem, where the tangent and secant lines are horizontal, ie, parallel to each other and to the x -axis. So we consider the quantity that equals the height (or depth) of the graph of f above (or below) the secant line. This quantity is a function g such that the value g ( x ) at any point x in [ a , b ] equals that height (or depth). Thus g ( a ) = 0 and g ( b ) = 0, yielding g ( a ) = g ( b ). This and the facts that g is also continuous on [ a , b ] and differentiable on ( a , b ) enable us to apply Rolle's theorem to g . Now, g ( x ) is the difference between f ( x ) and s ( x ), where s ( x ) is the equation of the secant line. The secant line passes thru the point ( a , f ( a )) and has slope ( f ( b ) – f ( a ))/( b – a ). Consequently, its equation is:

Proof Of The Mean-Value Theorem Define the function g by:

Remarks 5.1

i.   If f ( a ) = f ( b ), then f ' ( c ) = 0. The mean-value theorem is a generalization of Rolle's theorem. Or, in other words,     Rolle's theorem is a special case of the mean-value theorem.

ii.   The mean-value theorem says that the average rate of   change of f over [ a , b ] is attained at some point in ( a , b ).

iii.   The mean-value theorem assures that there's at least one point c such that ( f ( b ) – f ( a ))/( b – a ) = f ' ( c ). It doesn't      claim that the point c is unique. There may be one or more such points c . For example, in Fig. 5.2, there are 3 points      c 1 , c 2 , and c 3 such that:

     interchangeably.

We saw in Section 3.1 Theorem 2.1 that if a function is constant, then its derivative is 0. We're now going to see that the converse is also true, ie, if the derivative of a function is 0 on an interval, then the function is constant on that interval. This result is an application of the mean-value theorem. Intuitively, this is obvious: since the derivative or rate of change of the function is 0, the function doesn't change, ie, it's constant. Or, geometrically, since the tangent line to the graph of the function at every point of the graph is horizontal, the graph must be a horizontal line or line segment, and thus the function is constant.

Theorem 6.1

Proof Pick an arbitrary point x 1 of I , and let C = f ( x 1 ), so that C is a constant. Let x be any point of I different from x 1 . Applying the mean-value theorem to f on [ x 1 , x ] if x > x 1 or on [ x , x 1 ] if x < x 1 , there exists c between x 1 and x , thus in I , such that ( f ( x ) – f ( x 1 ))/( x – x 1 ) = f ' ( c ). But f ' ( c ) = 0. Hence, f ( x ) = f ( x 1 ) = C . EOP

You may wonder what if we re-start the process and pick a new point x 1 different from the old point x 1 in the proof. Ok, let's see. By the proof, f (new x 1 ) = f (old x 1 ) = C . Well, whatever point x 1 we pick, f ( x 1 ) is always the same value.

The following example is another application of the mean-value theorem.

Example 7.1

Prove that the inequality:

holds true for all x > 0.

Example 8.1

Return To Top Of Page

1.   Prove that the equation 10 – 5 x – x 2 = 0 has a unique solution in the interval [–1, 2].

Let x 1 , x 2 , …, x n be the n distinct points of I where f vanishes. Since f is differentiable on [ x 1 , x 2 ] and f ( x 1 ) = f ( x 2 ), Rolle's theorem assures us that there exists c 1,2 in ( x 1 , x 2 ) such that f ' ( c 1,2 ) = 0. Since f is differentiable on [ x 2 , x 3 ] and   f ( x 2 ) = f ( x 3 ), Rolle's theorem assures us that there exists c 2,3 in ( x 2 , x 3 ) such that f ' ( c 2,3 ) = 0. And so on, until we reach the interval [ x n –1 , x n ]. So, we get at least n – 1 distinct points c 1,2 , c 2,3 , …, c n –1, n of I where f ' vanishes.

3.   For each of the following functions f with the indicated interval of the form [ a , b ] , find the points c in ( a , b ) where the      tangent line to the graph of f is parallel to the secant line joining ( a , f ( a )) and ( b ,   f ( b )).

a.   Let S be the secant line joining ( a , f ( a )) and ( b , f ( b )). The slope of S is:

      Thus c is in ( a , b ) and the tangent line at x = c is parallel to S .

b.   Let S be the secant line joining (1, f (1)) and (2, f (2)). The slope of S is:

4.   Prove that the inequality:

For any x in (–2, 4]   f is differentiable on [–2, x ]. So by the mean-value theorem there exists c in (–2, x ) such that:

The bounding functions of f ( x ) on [–2, 4] are y = – 5 x – 9 and y = 5 x + 11.

6.   Suppose that a function f is twice differentiable on [0, 4] and that f (1) = f (2) = 0 and f (3) = 1. Prove that: a.   f ' ( a ) = 1/2 for some point a in (0, 4). b.   f '' ( b ) > 1/2 for some point b in (0, 4).

a.   Because f is differentiable on [0, 4] it’s also differentiable on [1, 3]. So by the mean-value theorem there exists a in      (1, 3) thus in (0, 4) such that:

Return To Top Of Page      Return To Contents

Helping math teachers bring calculus to life

The Mean Value Theorem  (Topic 5.1)

Unit 5 - day 1, ​learning objectives​.

Justify conclusions about functions by applying the MVT over an interval.

​Success Criteria

I can verify that the conditions of the MVT have been met.

I can use the MVT to justify a conclusion about a function’s average rate of change over an interval and the instantaneous rate of change at a point on that interval.

I can interpret solutions to problems involving the MVT.

Quick Lesson Plan

Activity: can calculus get you fined.

Lesson Handout

Unit 5 focuses on the  analytical applications of differentiation: the MVT, using the First and Second Derivative tests on increasing/decreasing functions, the Candidate Test, concavity, and using derivatives to confirm extrema in context (optimization). These concepts might mean signal drudgery for students in a traditional calculus class, but today’s lesson provides an engaging and interesting launch to the Unit 5 content! Students should be able to work through all questions in the activity given sufficient time. Using the MVT as a justification is likely to be required, so we must insure students can write and use the theorem with fidelity. 

Teaching Tips

Reinforce the connection between  mean  and  average  and  average rate of change  so students understand why this is named the  Mean  Value Theorem. Include a graphical discussion of the relationship between the instantaneous ROC and the average ROC; have students draw functions that do (and do not) obey the MVT so they can visualize the results of this powerful theorem.

Exam Insights

Require students to confirm and state the hypotheses of the theorem in order to earn full points on the Free Response section of the AP Test. Using “MVT” on the AP Test is an acceptable abbreviation when referencing this theorem.

Student Misconceptions

As with the IVT and the EVT, stating that a function has met the hypotheses of the MVT is important and necessary. The MVT is an existence theorem guaranteeing a point on a differentiable function where the slope of the tangent line equals the slope of a secant line. You may discover your students are able to navigate the required calculus and algebra without actually knowing the meaning of their answer! Continuing to require an  interpretation  of results will help students toward better understanding.