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Calculus Volume 1

4.4 The Mean Value Theorem

Calculus Volume 14.4 The Mean Value Theorem
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

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 ff are equal at the endpoints of an interval, then there must be an interior point cc where f(c)=0.f(c)=0. Figure 4.21 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.
Figure 4.21 If a differentiable function f satisfies f(a)=f(b),f(a)=f(b), then its derivative must be zero at some point(s) between aa and b.b.
Theorem 4.4

Rolle’s Theorem

Let ff 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.

Proof

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

  1. f(x)=kf(x)=k for all x(a,b).x(a,b).
  2. There exists x(a,b)x(a,b) such that f(x)>k.f(x)>k.
  3. There exists x(a,b)x(a,b) such that f(x)<k.f(x)<k.

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

Case 2: Since ff 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 ff has a maximum at an interior point c,c, and ff 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)<kf(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 ff is critical. If ff is not differentiable, even at a single point, the result may not hold. For example, the function f(x)=|x|1f(x)=|x|1 is continuous over [−1,1][−1,1] and f(−1)=0=f(1),f(−1)=0=f(1), but f(c)0f(c)0 for any c(−1,1)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.
Figure 4.22 Since f(x)=|x|1f(x)=|x|1 is not differentiable at x=0,x=0, the conditions of Rolle’s theorem are not satisfied. In fact, the conclusion does not hold here; there is no c(−1,1)c(−1,1) such that f(c)=0.f(c)=0.

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

  1. f(x)=x2+2xf(x)=x2+2x over [−2,0][−2,0]
  2. f(x)=x34xf(x)=x34x over [−2,2][−2,2]

Solution

  1. Since ff is a polynomial, it is continuous and differentiable everywhere. In addition, f(−2)=0=f(0).f(−2)=0=f(0). Therefore, ff satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value c(−2,0)c(−2,0) such that f(c)=0.f(c)=0. Since f(x)=2x+2=2(x+1),f(x)=2x+2=2(x+1), we see that f(c)=2(c+1)=0f(c)=2(c+1)=0 implies c=−1c=−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).
    Figure 4.23 This function is continuous and differentiable over [−2,0],[−2,0], f(c)=0f(c)=0 when c=−1.c=−1.
  2. As in part a. ff is a polynomial and therefore is continuous and differentiable everywhere. Also, f(−2)=0=f(2).f(−2)=0=f(2). That said, ff satisfies the criteria of Rolle’s theorem. Differentiating, we find that f(x)=3x24.f(x)=3x24. Therefore, f(c)=0f(c)=0 when x=±23.x=±23. Both points are in the interval [−2,2],[−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.
    Figure 4.24 For this polynomial over [−2,2],[−2,2], f(c)=0f(c)=0 at x=±2/3.x=±2/3.

Checkpoint 4.14

Verify that the function f(x)=2x28x+6f(x)=2x28x+6 defined over the interval [1,3][1,3] satisfies the conditions of Rolle’s theorem. Find all points cc 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 ff 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 ff 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 ff at cc is parallel to the secant line connecting (a,f(a))(a,f(a)) and (b,f(b)).(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.
Figure 4.25 The Mean Value Theorem says that for a function that meets its conditions, at some point the tangent line has the same slope as the secant line between the ends. For this function, there are two values c1c1 and c2c2 such that the tangent line to ff at c1c1 and c2c2 has the same slope as the secant line.
Theorem 4.5

Mean Value Theorem

Let ff 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

f(c)=f(b)f(a)ba.f(c)=f(b)f(a)ba.

Proof

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

f(b)f(a)baf(b)f(a)ba

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

y=f(b)f(a)ba(xa)+f(a).y=f(b)f(a)ba(xa)+f(a).

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,

g(x)=f(x)[f(b)f(a)ba(xa)+f(a)].g(x)=f(x)[f(b)f(a)ba(xa)+f(a)].
A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a 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, then increases, 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 equation y = ((f(b) – f(a))/(b − a)) (x − a) + f(x). A line is drawn between the maximum of f(x) and the secant line and it is marked g(x).
Figure 4.26 The value g(x)g(x) is the vertical difference between the point (x,f(x))(x,f(x)) and the point (x,y)(x,y) on the secant line connecting (a,f(a))(a,f(a)) and (b,f(b)).(b,f(b)).

Since the graph of ff intersects the secant line when x=ax=a and x=b,x=b, we see that g(a)=0=g(b).g(a)=0=g(b). Since ff is a differentiable function over (a,b),(a,b), gg is also a differentiable function over (a,b).(a,b). Furthermore, since ff is continuous over [a,b],[a,b], gg is also continuous over [a,b].[a,b]. Therefore, gg 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

g(x)=f(x)f(b)f(a)ba,g(x)=f(x)f(b)f(a)ba,

we see that

g(c)=f(c)f(b)f(a)ba.g(c)=f(c)f(b)f(a)ba.

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

f(c)=f(b)f(a)ba.f(c)=f(b)f(a)ba.

In the next example, we show how the Mean Value Theorem can be applied to the function f(x)=xf(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)=xf(x)=x over the interval [0,9],[0,9], show that ff 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 cc guaranteed by the Mean Value Theorem.

Solution

We know that f(x)=xf(x)=x is continuous over [0,9][0,9] and differentiable over (0,9).(0,9). Therefore, ff 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 cc are guaranteed, first calculate the derivative of f.f. The derivative f(x)=1(2x).f(x)=1(2x). The slope of the line connecting (0,f(0))(0,f(0)) and (9,f(9))(9,f(9)) is given by

f(9)f(0)90=9090=39=13.f(9)f(0)90=9090=39=13.

We want to find cc such that f(c)=13.f(c)=13. That is, we want to find cc such that

12c=13.12c=13.

Solving this equation for c,c, we obtain c=94.c=94. 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.
Figure 4.27 The slope of the tangent line at c=9/4c=9/4 is the same as the slope of the line segment connecting (0,0)(0,0) and (9,3).(9,3).

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 0t10t1 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

v(c)=s(c)=s(1)s(0)10=45mph.v(c)=s(c)=s(1)s(0)10=45mph.

Example 4.16

Mean Value Theorem and Velocity

If a rock is dropped from a height of 100 ft, its position tt seconds after it is dropped until it hits the ground is given by the function s(t)=−16t2+100.s(t)=−16t2+100.

  1. Determine how long it takes before the rock hits the ground.
  2. Find the average velocity vavgvavg of the rock for when the rock is released and the rock hits the ground.
  3. Find the time tt guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is vavg.vavg.

Solution

  1. When the rock hits the ground, its position is s(t)=0.s(t)=0. Solving the equation −16t2+100=0−16t2+100=0 for t,t, we find that t=±52sec.t=±52sec. Since we are only considering t0,t0, the ball will hit the ground 5252 sec after it is dropped.
  2. The average velocity is given by
    vavg=s(5/2)s(0)5/20=11005/2=−40ft/sec.vavg=s(5/2)s(0)5/20=11005/2=−40ft/sec.
  3. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time tt such that v(t)=s(t)=vavg=−40ft/sec.v(t)=s(t)=vavg=−40ft/sec. Since s(t)s(t) is continuous over the interval [0,5/2][0,5/2] and differentiable over the interval (0,5/2),(0,5/2), by the Mean Value Theorem, there is guaranteed to be a point c(0,5/2)c(0,5/2) such that
    s(c)=s(5/2)s(0)5/20=−40.s(c)=s(5/2)s(0)5/20=−40.

    Taking the derivative of the position function s(t),s(t), we find that s(t)=−32t.s(t)=−32t. Therefore, the equation reduces to s(c)=−32c=−40.s(c)=−32c=−40. Solving this equation for c,c, we have c=54.c=54. Therefore, 5454 sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: −40−40 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.
    Figure 4.28 At time t=5/4t=5/4 sec, the velocity of the rock is equal to its average velocity from the time it is dropped until it hits the ground.

Checkpoint 4.15

Suppose a ball is dropped from a height of 200 ft. Its position at time tt is s(t)=−16t2+200.s(t)=−16t2+200. Find the time tt 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)=0f(x)=0 for all xx 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 ff be differentiable over an interval I.I. If f(x)=0f(x)=0 for all xI,xI, then f(x)=f(x)= constant for all xI.xI.

Proof

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

f(b)f(a)ba0.f(b)f(a)ba0.

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

f(c)=f(b)f(a)ba.f(c)=f(b)f(a)ba.

Therefore, there exists cIcI such that f(c)0,f(c)0, which contradicts the assumption that f(x)=0f(x)=0 for all xI.xI.

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 ff and gg are differentiable over an interval II and f(x)=g(x)f(x)=g(x) for all xI,xI, then f(x)=g(x)+Cf(x)=g(x)+C for some constant C.C.

Proof

Let h(x)=f(x)g(x).h(x)=f(x)g(x). Then, h(x)=f(x)g(x)=0h(x)=f(x)g(x)=0 for all xI.xI. By Corollary 1, there is a constant CC such that h(x)=Ch(x)=C for all xI.xI. Therefore, f(x)=g(x)+Cf(x)=g(x)+C for all xI.xI.

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function ff is increasing over II if f(x1)<f(x2)f(x1)<f(x2) whenever x1<x2,x1<x2, whereas ff is decreasing over II if f(x)1>f(x2)f(x)1>f(x2) whenever x1<x2.x1<x2. 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 FF such that F(x)=f(x);F(x)=f(x); then, the only other functions that have a derivative equal to ff are F(x)+CF(x)+C for some constant C.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.
Figure 4.29 If a function has a positive derivative over some interval I,I, then the function increases over that interval I;I; if the derivative is negative over some interval I,I, then the function decreases over that interval I.I.

Theorem 4.8

Corollary 3: Increasing and Decreasing Functions

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

  1. If f(x)>0f(x)>0 for all x(a,b),x(a,b), then ff is an increasing function over [a,b].[a,b].
  2. If f(x)<0f(x)<0 for all x(a,b),x(a,b), then ff is a decreasing function over [a,b].[a,b].

Proof

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

f(c)=f(b)f(a)ba.f(c)=f(b)f(a)ba.

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

f(c)=f(b)f(a)ba0.f(c)=f(b)f(a)ba0.

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

Section 4.4 Exercises

148.

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

149.

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

150.

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

151.

If you have a function with a discontinuity, is it still possible to have f(c)(ba)=f(b)f(a)?f(c)(ba)=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.

152.

y=sin(πx)y=sin(πx)

153.

y=1x3y=1x3

154.

y=4x2y=4x2

155.

y=x24y=x24

156.

y=ln(3x5)y=ln(3x5)

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

157.

[T] y=3x3+2x+1y=3x3+2x+1 over [−1,1][−1,1]

158.

[T] y=tan(π4x)y=tan(π4x) over [32,32][32,32]

159.

[T] y=x2cos(πx)y=x2cos(πx) over [−2,2][−2,2]

160.

[T] y=x634x598x4+1516x3+332x2+316x+132y=x634x598x4+1516x3+332x2+316x+132 over [−1,1][−1,1]

For the following exercises, use the Mean Value Theorem and find all points 0<c<20<c<2 such that f(2)f(0)=f(c)(20).f(2)f(0)=f(c)(20).

161.

f(x)=x3f(x)=x3

162.

f(x)=sin(πx)f(x)=sin(πx)

163.

f(x)=cos(2πx)f(x)=cos(2πx)

164.

f(x)=1+x+x2f(x)=1+x+x2

165.

f(x)=(x1)10f(x)=(x1)10

166.

f(x)=(x1)9f(x)=(x1)9

For the following exercises, show there is no cc 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].

167.

f(x)=|x12|f(x)=|x12|

168.

f(x)=1x2f(x)=1x2

169.

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

170.

f(x)=xf(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.

171.

y=exy=ex over [0,1][0,1]

172.

y=ln(2x+3)y=ln(2x+3) over [32,0][32,0]

173.

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

174.

y=9x2y=9x2 over [−3,3][−3,3]

175.

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

176.

y=x3+2x+1y=x3+2x+1 over [0,6][0,6]

177.

y=x2+3x+2xy=x2+3x+2x over [−1,1][−1,1]

178.

y=xsin(πx)+1y=xsin(πx)+1 over [0,1][0,1]

179.

y=ln(x+1)y=ln(x+1) over [0,e1][0,e1]

180.

y=xsin(πx)y=xsin(πx) over [0,2][0,2]

181.

y=5+|x|y=5+|x| over [−1,1][−1,1]

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

182.

Show that the equation y=x3+3x2+16y=x3+3x2+16 has exactly one real root. What is it?

183.

Find the conditions for exactly one root (double root) for the equation y=x2+bx+cy=x2+bx+c

184.

Find the conditions for y=exby=exb 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 aa to b.b. Use the calculator to estimate all values of cc 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.

185.

[T] y=tan(πx)y=tan(πx) over [14,14][14,14]

186.

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

187.

[T] y=|x2+2x4|y=|x2+2x4| over [−4,0][−4,0]

188.

[T] y=x+1xy=x+1x over [12,4][12,4]

189.

[T] y=x+1+1x2y=x+1+1x2 over [3,8][3,8]

190.

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?

191.

Two cars drive from one spotlight 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.

192.

Show that y=sec2xy=sec2x and y=tan2xy=tan2x have the same derivative. What can you say about y=sec2xtan2x?y=sec2xtan2x?

193.

Show that y=csc2xy=csc2x and y=cot2xy=cot2x have the same derivative. What can you say about y=csc2xcot2x?y=csc2xcot2x?

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