The Mean Value Theorem is one of the most important theorems in Introductory Calculus, and it forms the basis for proofs of many results in subsequent and advanced Mathematics courses. The history of this theorem begins in the 1300's with the Indian Mathematician Parameshvara , and is eventually based on the academic work of Mathematicians Michel Rolle in 1691 and Augustin Louis Cauchy in 1823. The formal statement of this theorem together with an illustration of the theorem will follow. I will also state Rolle's Theorem , which is used in the proof the Mean Value Theorem. Both theorems are given without proof, and all subsequent problems here will be referencing only the Mean Value Theorem. All functions are assumed to be real-valued. ROLLE'S THEOREM: Let $f$ be a continuous function on the closed interval $ [a, b] $ and differentiable on the open interval $ (a, b) $ with $f(a)=f(b)$. Then there is at least one number $c$ ($x$-value) in the interval $(a, b)$ which satifies $$ f'(c)=0 $$ MEAN VALUE THEOREM: Let $f$ be a continuous function on the closed interval $ [a, b] $ and differentiable on the open interval $ (a, b) $. Then there is at least one number $c$ ($x$-value) in the interval $(a, b)$ which satifies $$ f'(c)= \displaystyle{ f(b) - f(a) \over b-a } $$ Numerous proofs for Rolle's Theorem and the Mean Value Theorem can easily be found on the internet. I have attached proofs of both Theorems here , along with other results related to the Mean-Value Theorem. In the list of Mean Value Theorem Problems which follows, most problems are average and a few are somewhat challenging. PROBLEM 1 : Determine if the Mean Value Theorem can be applied to the following function on the the given closed interval. If so, find all possible values of $c$: $ \ f(x)=3+ \sqrt{x} \ $ on $ \ [0, 4] $ Click HERE to see a detailed solution to problem 1.

  • 4.4 The Mean Value Theorem
  • Introduction
  • 1.1 Review of Functions
  • 1.2 Basic Classes of Functions
  • 1.3 Trigonometric Functions
  • 1.4 Inverse Functions
  • 1.5 Exponential and Logarithmic Functions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • 2.1 A Preview of Calculus
  • 2.2 The Limit of a Function
  • 2.3 The Limit Laws
  • 2.4 Continuity
  • 2.5 The Precise Definition of a Limit
  • 3.1 Defining the Derivative
  • 3.2 The Derivative as a Function
  • 3.3 Differentiation Rules
  • 3.4 Derivatives as Rates of Change
  • 3.5 Derivatives of Trigonometric Functions
  • 3.6 The Chain Rule
  • 3.7 Derivatives of Inverse Functions
  • 3.8 Implicit Differentiation
  • 3.9 Derivatives of Exponential and Logarithmic Functions
  • 4.1 Related Rates
  • 4.2 Linear Approximations and Differentials
  • 4.3 Maxima and Minima
  • 4.5 Derivatives and the Shape of a Graph
  • 4.6 Limits at Infinity and Asymptotes
  • 4.7 Applied Optimization Problems
  • 4.8 L’Hôpital’s Rule
  • 4.9 Newton’s Method
  • 4.10 Antiderivatives
  • 5.1 Approximating Areas
  • 5.2 The Definite Integral
  • 5.3 The Fundamental Theorem of Calculus
  • 5.4 Integration Formulas and the Net Change Theorem
  • 5.5 Substitution
  • 5.6 Integrals Involving Exponential and Logarithmic Functions
  • 5.7 Integrals Resulting in Inverse Trigonometric Functions
  • 6.1 Areas between Curves
  • 6.2 Determining Volumes by Slicing
  • 6.3 Volumes of Revolution: Cylindrical Shells
  • 6.4 Arc Length of a Curve and Surface Area
  • 6.5 Physical Applications
  • 6.6 Moments and Centers of Mass
  • 6.7 Integrals, Exponential Functions, and Logarithms
  • 6.8 Exponential Growth and Decay
  • 6.9 Calculus of the Hyperbolic Functions
  • A | Table of Integrals
  • B | Table of Derivatives
  • C | Review of Pre-Calculus

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 .

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 ?

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  • 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

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4.4: Rolle’s Theorem and The Mean Value Theorem

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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.

The figure is divided into three parts labeled a, b, and c. Figure a shows the first quadrant with values a, c, and b marked on the x-axis. A downward-facing parabola is drawn such that its values at a and b are the same. The point c is the global maximum, and it is noted that f’(c) = 0. Figure b shows the first quadrant with values a, c, and b marked on the x-axis. An upward-facing parabola is drawn such that its values at a and b are the same. The point c is the global minimum, and it is noted that f’(c) = 0. Figure c shows the first quadrant with points a, c1, c2, and b marked on the x-axis. One period of a sine wave is drawn such that its values at a and b are equal. The point c1 is the global maximum, and it is noted that f’(c1) = 0. The point c2 is the global minimum, and it is noted that f’(c2) = 0.

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.

The function f(x) = |x| − 1 is graphed. It is shown that f(1) = f(−1), but it is noted that there is no c such that f’(c) = 0.

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.

The function f(x) = x2 +2x is graphed. It is shown that f(0) = f(−2), and a dashed horizontal line is drawn at the absolute minimum at (−1, −1).

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.

The function f(x) = x3 – 4x is graphed. It is obvious that f(2) = f(−2) = f(0). Dashed horizontal lines are drawn at x = ±2/square root of 3, which are the local maximum and minimum.

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 \((b,f(b)).\)

A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a, c1, c2, and b are marked. On the y-axis, f(a) and f(b) are marked. The function f(x) starts at (a, f(a)), decreases to c1, increases to c2, and then decreases to (b, f(b)). A secant line is drawn between (a, f(a)) and (b, f(b)), and it is noted that this line has slope (f(b) – f(a))/(b − a). The tangent lines at c1 and c2 are drawn, and these lines are parallel to the secant line. It is noted that the slopes of these tangent lines are f’(c1) and f’(c2), respectively.

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,

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

Rising but oscillating curve Y equals F of X with secant line Y equals F of B minus F of A all over B minus A plus F of A, distance between the oscillating curve and the line is G of X

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.

The function f(x) = the square root of x is graphed from (0, 0) to (9, 3). There is a secant line drawn from (0, 0) to (9, 3). At point (9/4, 3/2), there is a tangent line that is drawn, and this line is parallel to the secant line.

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.

The function s(t) = −16t2 + 100 is graphed from (0, 100) to (5/2, 0). There is a secant line drawn from (0, 100) to (5/2, 0). At the point corresponding to x = 5/4, there is a tangent line that is drawn, and this line is parallel to the secant line.

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.

A vaguely sinusoidal function f(x) is graphed. It increases from somewhere in the second quadrant to (a, f(a)). In this section it is noted that f’ > 0. Then in decreases from (a, f(a)) to (b, f(b)). In this section it is noted that f’ < 0. Finally, it increases to the right of (b, f(b)) and it is noted in this section that f’ > 0.

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 \(f′(c)=\frac{f(b)−f(a)}{b−a}\)

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Solving Some Problems Using the Mean Value Theorem

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Theorem 1.3. (Flett's theorem) Let f : [a, b] → R be a continuous function on , differentiable on , and such that f (a) = f (b). Then, there is η ∈ (a, b) such that

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Rolle's Theorem Questions and Examples

Rolle's theorem is a special case of the mean value theorem . It is discussed here through examples and questions.

Rolle's Theorem

Solution to question 1, solution to question 2, solution to question 3, more references and links.

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    1 1 Therefore, c = 1 is the only value in (0, 4) that satisfies the conclusion of the theorem. 2. Show that f ( x ) 25 x 2 satisfies the hypothesis of the Mean Value Theorem on [-5, 3], and find all values of c in (-5, 3) that satisfy the conclusion of the theorem. Solution: Note that the domain of f ( x ) 25 x 2 is [-5, 5].

  14. Solving Mean Value Theorem Problems

    In the list of Mean Value Theorem Problems which follows, most problems are average and a few are somewhat challenging. PROBLEM 1 : Determine if the Mean Value Theorem can be applied to the following function on the the given closed interval. If so, find all possible values of c c: f(x) = 3 + x−−√ f ( x) = 3 + x on [0, 4] [ 0, 4]

  15. PDF The Mean Value Theorem and Inequalities

    The mean value theorem tells us that if f and f are continuous on [a, b] then: f(b) − f(a) = f (c) b − a for some value c between a and b. Since f is continuous, f (c) must lie between the minimum and maximum values of f (x) on [a, b]. In other words: f(b) − f(a) min f (x) ≤ = f (c) ≤ max f (x). a≤x≤b b − a a≤x≤b

  16. PDF Calculus I

    Theorem Suppose that f (x) is continuous on the closed interval [a;b]. Then f (x) attains its absolute maximum and minimum values on [a;b] at either: I A critical point I or one of the end points a or b. Note: If f (x) is not continuous on [a;b] then this theorem fails. Example: What is the maximum value of f (x) on the interval [a;b]? Solution ...

  17. 4.4 The Mean Value 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.

  18. 4.4: Rolle's Theorem and The Mean Value Theorem

    Example 4.4.3 4.4. 3: 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) = −16t2 + 100. s ( t) = − 16 t 2 + 100. Determine how long it takes before the rock hits the ground.

  19. Lecture 14: Mean Value Theorem

    Lecture 14: Mean Value Theorem. Topics covered: Mean value theorem; Inequalities. Instructor: Prof. David Jerison. Transcript. Download video. Download transcript. Related Resources. MIT OpenCourseWare is a web based publication of virtually all MIT course content. OCW is open and available to the world and is a permanent MIT activity.

  20. Calculus I

    What the Mean Value Theorem tells us is that these two slopes must be equal or in other words the secant line connecting A A and B B and the tangent line at x =c x = c must be parallel. We can see this in the following sketch. Let's now take a look at a couple of examples using the Mean Value Theorem.

  21. PDF Revisit Mean Value, Cauchy Mean Value and Lagrange Remainder Theorems

    3 Geometric Interpretation of Cauchy Mean Value The-orem We use similar approach mentioned above to demonstrate how geometric interpretations of the Cauchy Mean Value Theorem can be explored with the help of DGS and CAS. The Cauchy Mean Value Theorem can be stated as follows: Theorem 2 Suppose the function f : [a;b] ! R and g : [a;b] !

  22. Solving Some Problems Using the Mean Value Theorem

    Solving Some Problems Using the Mean Value Theorem Truck Driver Theorem 1.3. (Flett's theorem) Let f : [a, b] → R be a continuous function on , differentiable on , and such that f (a) = f (b). Then, there is η ∈ (a, b) such that See Full PDF Download PDF Related Papers MEAN VALUE THEOREMS FOR SOME LINEAR INTEGRAL OPERATORS Cezar Lupu

  23. Rolle's Theorem Questions and Examples

    Rolle's theorem is the result of the mean value theorem where under the conditions: f (x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f ' (c) = [ f (b) - f (a) ] / (b - a).

  24. PDF 14.310x: Data Analysis for Social Scientists Special Distributions, the

    Special Distributions, the Sample Mean, and the Central Limit Theorem . Welcome to your fifth homework assignment! You will have about one week to work through the assignment. We encourage you to get an early start, particularly if you still feel you need more experience using R. We have provided this PDF copy of the assignment so that you can