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Calculus Volume 1

2.4 Continuity

Calculus Volume 12.4 Continuity
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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

  • 2.4.1. Explain the three conditions for continuity at a point.
  • 2.4.2. Describe three kinds of discontinuities.
  • 2.4.3. Define continuity on an interval.
  • 2.4.4. State the theorem for limits of composite functions.
  • 2.4.5. Provide an example of the intermediate value theorem.

Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called continuous. Other functions have points at which a break in the graph occurs, but satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to have a discontinuity at a point where a break occurs.

We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.

Continuity at a Point

Before we look at a formal definition of what it means for a function to be continuous at a point, let’s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.

Our first function of interest is shown in Figure 2.32. We see that the graph of f(x)f(x) has a hole at a. In fact, f(a)f(a) is undefined. At the very least, for f(x)f(x) to be continuous at a, we need the following condition:

i.f(a)is defined.i.f(a)is defined.
A graph of an increasing linear function f(x) which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. The point on the function f(x) above a is an open circle; the function is not defined at a.
Figure 2.32 The function f(x)f(x) is not continuous at a because f(a)f(a) is undefined.

However, as we see in Figure 2.33, this condition alone is insufficient to guarantee continuity at the point a. Although f(a)f(a) is defined, the function has a gap at a. In this example, the gap exists because limxaf(x)limxaf(x) does not exist. We must add another condition for continuity at a—namely,

ii.limxaf(x)exists.ii.limxaf(x)exists.
The graph of a piecewise function f(x) with two parts. The first part is an increasing linear function that crosses from quadrant three to quadrant one at the origin. A point a greater than zero is marked on the x axis. At fa. on this segment, there is a solid circle. The other segment is also an increasing linear function. It exists in quadrant one for values of x greater than a. At x=a, this segment has an open circle.
Figure 2.33 The function f(x)f(x) is not continuous at a because limxaf(x)limxaf(x) does not exist.

However, as we see in Figure 2.34, these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at a. We must add a third condition to our list:

iii.limxaf(x)=f(a).iii.limxaf(x)=f(a).
The graph of a piecewise function with two parts. The first part is an increasing linear function that crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. At this point, there is an open circle on the linear function. The second part is a point at x=a above the line.
Figure 2.34 The function f(x)f(x) is not continuous at a because limxaf(x)f(a).limxaf(x)f(a).

Now we put our list of conditions together and form a definition of continuity at a point.

Definition

A function f(x)f(x) is continuous at a point a if and only if the following three conditions are satisfied:

  1. f(a)f(a) is defined
  2. limxaf(x)limxaf(x) exists
  3. limxaf(x)=f(a)limxaf(x)=f(a)

A function is discontinuous at a point a if it fails to be continuous at a.

The following procedure can be used to analyze the continuity of a function at a point using this definition.

Problem-Solving Strategy: Determining Continuity at a Point
  1. Check to see if f(a)f(a) is defined. If f(a)f(a) is undefined, we need go no further. The function is not continuous at a. If f(a)f(a) is defined, continue to step 2.
  2. Compute limxaf(x).limxaf(x). In some cases, we may need to do this by first computing limxaf(x)limxaf(x) and limxa+f(x).limxa+f(x). If limxaf(x)limxaf(x) does not exist (that is, it is not a real number), then the function is not continuous at a and the problem is solved. If limxaf(x)limxaf(x) exists, then continue to step 3.
  3. Compare f(a)f(a) and limxaf(x).limxaf(x). If limxaf(x)f(a),limxaf(x)f(a), then the function is not continuous at a. If limxaf(x)=f(a),limxaf(x)=f(a), then the function is continuous at a.

The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.

Example 2.26

Determining Continuity at a Point, Condition 1

Using the definition, determine whether the function f(x)=(x24)/(x2)f(x)=(x24)/(x2) is continuous at x=2.x=2. Justify the conclusion.

Solution

Let’s begin by trying to calculate f(2).f(2). We can see that f(2)=0/0,f(2)=0/0, which is undefined. Therefore, f(x)=x24x2f(x)=x24x2 is discontinuous at 2 because f(2)f(2) is undefined. The graph of f(x)f(x) is shown in Figure 2.35.

A graph of the given function. There is a line which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. At a point in quadrant one, there is an open circle where the function is not defined.
Figure 2.35 The function f(x)f(x) is discontinuous at 2 because f(2)f(2) is undefined.

Example 2.27

Determining Continuity at a Point, Condition 2

Using the definition, determine whether the function f(x)={x2+4ifx34x8ifx>3f(x)={x2+4ifx34x8ifx>3 is continuous at x=3.x=3. Justify the conclusion.

Solution

Let’s begin by trying to calculate f(3).f(3).

f(3)=(32)+4=−5.f(3)=(32)+4=−5.

Thus, f(3)f(3) is defined. Next, we calculate limx3f(x).limx3f(x). To do this, we must compute limx3f(x)limx3f(x) and limx3+f(x):limx3+f(x):

limx3f(x)=(32)+4=−5limx3f(x)=(32)+4=−5

and

limx3+f(x)=4(3)8=4.limx3+f(x)=4(3)8=4.

Therefore, limx3f(x)limx3f(x) does not exist. Thus, f(x)f(x) is not continuous at 3. The graph of f(x)f(x) is shown in Figure 2.36.

A graph of the given piecewise function, which has two parts. The first is a downward opening parabola which is symmetric about the y axis. Its vertex is on the y axis, greater than zero. There is a closed circle on the parabola for x=3. The second part is an increasing linear function in the first quadrant, which exists for values of x > 3. There is an open circle at the end of the line where x would be 3.
Figure 2.36 The function f(x)f(x) is not continuous at 3 because limx3f(x)limx3f(x) does not exist.

Example 2.28

Determining Continuity at a Point, Condition 3

Using the definition, determine whether the function f(x)={sinxxifx01ifx=0f(x)={sinxxifx01ifx=0 is continuous at x=0.x=0.

Solution

First, observe that

f(0)=1.f(0)=1.

Next,

limx0f(x)=limx0sinxx=1.limx0f(x)=limx0sinxx=1.

Last, compare f(0)f(0) and limx1f(x).limx1f(x). We see that

f(0)=1=limx0f(x).f(0)=1=limx0f(x).

Since all three of the conditions in the definition of continuity are satisfied, f(x)f(x) is continuous at x=0.x=0.

Checkpoint 2.21

Using the definition, determine whether the function f(x)={2x+1ifx<12ifx=1x+4ifx>1f(x)={2x+1ifx<12ifx=1x+4ifx>1 is continuous at x=1.x=1. If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.

By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.

Theorem 2.8

Continuity of Polynomials and Rational Functions

Polynomials and rational functions are continuous at every point in their domains.

Proof

Previously, we showed that if p(x)p(x) and q(x)q(x) are polynomials, limxap(x)=p(a)limxap(x)=p(a) for every polynomial p(x)p(x) and limxap(x)q(x)=p(a)q(a)limxap(x)q(x)=p(a)q(a) as long as q(a)0.q(a)0. Therefore, polynomials and rational functions are continuous on their domains.

We now apply Continuity of Polynomials and Rational Functions to determine the points at which a given rational function is continuous.

Example 2.29

Continuity of a Rational Function

For what values of x is f(x)=x+1x5f(x)=x+1x5 continuous?

Solution

The rational function f(x)=x+1x5f(x)=x+1x5 is continuous for every value of x except x=5.x=5.

Checkpoint 2.22

For what values of x is f(x)=3x44x2f(x)=3x44x2 continuous?

Types of Discontinuities

As we have seen in Example 2.26 and Example 2.27, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump discontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite discontinuity is a discontinuity located at a vertical asymptote. Figure 2.37 illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.

Three graphs, each showing a different discontinuity. The first is removable discontinuity. Here, the given function is a line with positive slope. At a point x=a, where a>0, there is an open circle on the line and a closed circle a few units above the line. The second is a jump discontinuity. Here, there are two lines with positive slope. The first line exists for x<=a, and the second exists for x>a, where a>0. The first line ends at a solid circle where x=a, and the second begins a few units up with an open circle at x=a. The third discontinuity type is infinite discontinuity. Here, the function has two parts separated by an asymptote x=a. The first segment is a curve stretching along the x axis to 0 as x goes to negative infinity and along the y axis to infinity as x goes to zero. The second segment is a curve stretching along the y axis to negative infinity as x goes to zero and along the x axis to 0 as x goes to infinity.
Figure 2.37 Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.

These three discontinuities are formally defined as follows:

Definition

If f(x)f(x) is discontinuous at a, then

  1. ff has a removable discontinuity at a if limxaf(x)limxaf(x) exists. (Note: When we state that limxaf(x)limxaf(x) exists, we mean that limxaf(x)=L,limxaf(x)=L, where L is a real number.)
  2. ff has a jump discontinuity at a if limxaf(x)limxaf(x) and limxa+f(x)limxa+f(x) both exist, but limxaf(x)limxa+f(x).limxaf(x)limxa+f(x). (Note: When we state that limxaf(x)limxaf(x) and limxa+f(x)limxa+f(x) both exist, we mean that both are real-valued and that neither take on the values ±∞.)
  3. ff has an infinite discontinuity at a if limxaf(x)=±limxaf(x)=± or limxa+f(x)=±.limxa+f(x)=±.

Example 2.30

Classifying a Discontinuity

In Example 2.26, we showed that f(x)=x24x2f(x)=x24x2 is discontinuous at x=2.x=2. Classify this discontinuity as removable, jump, or infinite.

Solution

To classify the discontinuity at 2 we must evaluate limx2f(x):limx2f(x):

limx2f(x)=limx2x24x2=limx2(x2)(x+2)x2=limx2(x+2)=4.limx2f(x)=limx2x24x2=limx2(x2)(x+2)x2=limx2(x+2)=4.

Since f is discontinuous at 2 and limx2f(x)limx2f(x) exists, f has a removable discontinuity at x=2.x=2.

Example 2.31

Classifying a Discontinuity

In Example 2.27, we showed that f(x)={x2+4ifx34x8ifx>3f(x)={x2+4ifx34x8ifx>3 is discontinuous at x=3.x=3. Classify this discontinuity as removable, jump, or infinite.

Solution

Earlier, we showed that f is discontinuous at 3 because limx3f(x)limx3f(x) does not exist. However, since limx3f(x)=−5limx3f(x)=−5 and limx3+f(x)=4limx3+f(x)=4 both exist, we conclude that the function has a jump discontinuity at 3.

Example 2.32

Classifying a Discontinuity

Determine whether f(x)=x+2x+1f(x)=x+2x+1 is continuous at −1. If the function is discontinuous at −1, classify the discontinuity as removable, jump, or infinite.

Solution

The function value f(−1)f(−1) is undefined. Therefore, the function is not continuous at −1. To determine the type of discontinuity, we must determine the limit at −1. We see that limx−1x+2x+1=limx−1x+2x+1= and limx−1+x+2x+1=+.limx−1+x+2x+1=+. Therefore, the function has an infinite discontinuity at −1.

Checkpoint 2.23

For f(x)={x2ifx13ifx=1,f(x)={x2ifx13ifx=1, decide whether f is continuous at 1. If f is not continuous at 1, classify the discontinuity as removable, jump, or infinite.

Continuity over an Interval

Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.

Continuity from the Right and from the Left

A function f(x)f(x) is said to be continuous from the right at a if limxa+f(x)=f(a).limxa+f(x)=f(a).

A function f(x)f(x) is said to be continuous from the left at a if limxaf(x)=f(a).limxaf(x)=f(a).

A function is continuous over an open interval if it is continuous at every point in the interval. A function f(x)f(x) is continuous over a closed interval of the form [a,b][a,b] if it is continuous at every point in (a,b)(a,b) and is continuous from the right at a and is continuous from the left at b. Analogously, a function f(x)f(x) is continuous over an interval of the form (a,b](a,b] if it is continuous over (a,b)(a,b) and is continuous from the left at b. Continuity over other types of intervals are defined in a similar fashion.

Requiring that limxa+f(x)=f(a)limxa+f(x)=f(a) and limxbf(x)=f(b)limxbf(x)=f(b) ensures that we can trace the graph of the function from the point (a,f(a))(a,f(a)) to the point (b,f(b))(b,f(b)) without lifting the pencil. If, for example, limxa+f(x)f(a),limxa+f(x)f(a), we would need to lift our pencil to jump from f(a)f(a) to the graph of the rest of the function over (a,b].(a,b].

Example 2.33

Continuity on an Interval

State the interval(s) over which the function f(x)=x1x2+2xf(x)=x1x2+2x is continuous.

Solution

Since f(x)=x1x2+2xf(x)=x1x2+2x is a rational function, it is continuous at every point in its domain. The domain of f(x)f(x) is the set (,−2)(−2,0)(0,+).(,−2)(−2,0)(0,+). Thus, f(x)f(x) is continuous over each of the intervals (,−2),(−2,0),(,−2),(−2,0), and (0,+).(0,+).

Example 2.34

Continuity over an Interval

State the interval(s) over which the function f(x)=4x2f(x)=4x2 is continuous.

Solution

From the limit laws, we know that limxa4x2=4a2limxa4x2=4a2 for all values of a in (−2,2).(−2,2). We also know that limx−2+4x2=0limx−2+4x2=0 exists and limx24x2=0limx24x2=0 exists. Therefore, f(x)f(x) is continuous over the interval [−2,2].[−2,2].

Checkpoint 2.24

State the interval(s) over which the function f(x)=x+3f(x)=x+3 is continuous.

The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.

Theorem 2.9

Composite Function Theorem

If f(x)f(x) is continuous at L and limxag(x)=L,limxag(x)=L, then

limxaf(g(x))=f(limxag(x))=f(L).limxaf(g(x))=f(limxag(x))=f(L).

Before we move on to Example 2.35, recall that earlier, in the section on limit laws, we showed limx0cosx=1=cos(0).limx0cosx=1=cos(0). Consequently, we know that f(x)=cosxf(x)=cosx is continuous at 0. In Example 2.35 we see how to combine this result with the composite function theorem.

Example 2.35

Limit of a Composite Cosine Function

Evaluate limxπ/2cos(xπ2).limxπ/2cos(xπ2).

Solution

The given function is a composite of cosxcosx and xπ2.xπ2. Since limxπ/2(xπ2)=0limxπ/2(xπ2)=0 and cosxcosx is continuous at 0, we may apply the composite function theorem. Thus,

limxπ/2cos(xπ2)=cos(limxπ/2(xπ2))=cos(0)=1.limxπ/2cos(xπ2)=cos(limxπ/2(xπ2))=cos(0)=1.
Checkpoint 2.25

Evaluate limxπsin(xπ).limxπsin(xπ).

The proof of the next theorem uses the composite function theorem as well as the continuity of f(x)=sinxf(x)=sinx and g(x)=cosxg(x)=cosx at the point 0 to show that trigonometric functions are continuous over their entire domains.

Theorem 2.10

Continuity of Trigonometric Functions

Trigonometric functions are continuous over their entire domains.

Proof

We begin by demonstrating that cosxcosx is continuous at every real number. To do this, we must show that limxacosx=cosalimxacosx=cosa for all values of a.

limxacosx=limxacos((xa)+a)rewritex=xa+a=limxa(cos(xa)cosasin(xa)sina)apply the identity for the cosine of the sum of two angles=cos(limxa(xa))cosasin(limxa(xa))sinalimxa(xa)=0,andsinxandcosxare continuous at 0=cos(0)cosasin(0)sinaevaluate cos(0) and sin(0) and simplify=1·cosa0·sina=cosa.limxacosx=limxacos((xa)+a)rewritex=xa+a=limxa(cos(xa)cosasin(xa)sina)apply the identity for the cosine of the sum of two angles=cos(limxa(xa))cosasin(limxa(xa))sinalimxa(xa)=0,andsinxandcosxare continuous at 0=cos(0)cosasin(0)sinaevaluate cos(0) and sin(0) and simplify=1·cosa0·sina=cosa.

The proof that sinxsinx is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of sinxsinx and cosx,cosx, their continuity follows from the quotient limit law.

As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.

The Intermediate Value Theorem

Functions that are continuous over intervals of the form [a,b],[a,b], where a and b are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the Intermediate Value Theorem.

Theorem 2.11

The Intermediate Value Theorem

Let f be continuous over a closed, bounded interval [a,b].[a,b]. If z is any real number between f(a)f(a) and f(b),f(b), then there is a number c in [a,b][a,b] satisfying f(c)=zf(c)=z in Figure 2.38.

A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.).  A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.
Figure 2.38 There is a number c[a,b]c[a,b] that satisfies f(c)=z.f(c)=z.

Example 2.36

Application of the Intermediate Value Theorem

Show that f(x)=xcosxf(x)=xcosx has at least one zero.

Solution

Since f(x)=xcosxf(x)=xcosx is continuous over (,+),(,+), it is continuous over any closed interval of the form [a,b].[a,b]. If you can find an interval [a,b][a,b] such that f(a)f(a) and f(b)f(b) have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number c in (a,b)(a,b) that satisfies f(c)=0.f(c)=0. Note that

f(0)=0cos(0)=−1<0f(0)=0cos(0)=−1<0

and

f(π2)=π2cosπ2=π2>0.f(π2)=π2cosπ2=π2>0.

Using the Intermediate Value Theorem, we can see that there must be a real number c in [0,π/2][0,π/2] that satisfies f(c)=0.f(c)=0. Therefore, f(x)=xcosxf(x)=xcosx has at least one zero.

Example 2.37

When Can You Apply the Intermediate Value Theorem?

If f(x)f(x) is continuous over [0,2],f(0)>0[0,2],f(0)>0 and f(2)>0,f(2)>0, can we use the Intermediate Value Theorem to conclude that f(x)f(x) has no zeros in the interval [0,2]?[0,2]? Explain.

Solution

No. The Intermediate Value Theorem only allows us to conclude that we can find a value between f(0)f(0) and f(2);f(2); it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function f(x)=(x1)2.f(x)=(x1)2. It satisfies f(0)=1>0,f(2)=1>0,f(0)=1>0,f(2)=1>0, and f(1)=0.f(1)=0.

Example 2.38

When Can You Apply the Intermediate Value Theorem?

For f(x)=1/x,f(−1)=−1<0f(x)=1/x,f(−1)=−1<0 and f(1)=1>0.f(1)=1>0. Can we conclude that f(x)f(x) has a zero in the interval [−1,1]?[−1,1]?

Solution

No. The function is not continuous over [−1,1].[−1,1]. The Intermediate Value Theorem does not apply here.

Checkpoint 2.26

Show that f(x)=x3x23x+1f(x)=x3x23x+1 has a zero over the interval [0,1].[0,1].

Section 2.4 Exercises

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

131.

f(x)=1xf(x)=1x

132.

f(x)=2x2+1f(x)=2x2+1

133.

f(x)=xx2xf(x)=xx2x

134.

g(t)=t−1+1g(t)=t−1+1

135.

f(x)=5ex2f(x)=5ex2

136.

f(x)=|x2|x2f(x)=|x2|x2

137.

H(x)=tan2xH(x)=tan2x

138.

f(t)=t+3t2+5t+6f(t)=t+3t2+5t+6

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

139.

f(x)2x25x+3x1f(x)2x25x+3x1 at x=1x=1

140.

h(θ)=sinθcosθtanθh(θ)=sinθcosθtanθ at θ=πθ=π

141.

g(u)={6u2+u22u1ifu1272ifu=12,g(u)={6u2+u22u1ifu1272ifu=12, at u=12u=12

142.

f(y)=sin(πy)tan(πy),f(y)=sin(πy)tan(πy), at y=1y=1

143.

f(x)={x2exifx<0x1ifx0,f(x)={x2exifx<0x1ifx0, at x=0x=0

144.

f(x)={xsin(x)ifxπxtan(x)ifx>π,f(x)={xsin(x)ifxπxtan(x)ifx>π, at x=πx=π

In the following exercises, find the value(s) of k that makes each function continuous over the given interval.

145.

f(x)={3x+2,x<k2x3,kx8f(x)={3x+2,x<k2x3,kx8

146.

f(θ)={sinθ,0θ<π2cos(θ+k),π2θπf(θ)={sinθ,0θ<π2cos(θ+k),π2θπ

147.

f(x)={x2+3x+2x+2,x2k,x=−2f(x)={x2+3x+2x+2,x2k,x=−2

148.

f(x)={ekx,0x<4x+3,4x8f(x)={ekx,0x<4x+3,4x8

149.

f(x)={kx,0x3x+1,3<x10f(x)={kx,0x3x+1,3<x10

In the following exercises, use the Intermediate Value Theorem (IVT).

150.

Let h(x)={3x24,x25+4x,x>2h(x)={3x24,x25+4x,x>2 Over the interval [0,4],[0,4], there is no value of x such that h(x)=10,h(x)=10, although h(0)<10h(0)<10 and h(4)>10.h(4)>10. Explain why this does not contradict the IVT.

151.

A particle moving along a line has at each time t a position function s(t),s(t), which is continuous. Assume s(2)=5s(2)=5 and s(5)=2.s(5)=2. Another particle moves such that its position is given by h(t)=s(t)t.h(t)=s(t)t. Explain why there must be a value c for 2<c<52<c<5 such that h(c)=0.h(c)=0.

152.

[T] Use the statement “The cosine of t is equal to t cubed.”

  1. Write a mathematical equation of the statement.
  2. Prove that the equation in part a. has at least one real solution.
  3. Use a calculator to find an interval of length 0.01 that contains a solution.
153.

Apply the IVT to determine whether 2x=x32x=x3 has a solution in one of the intervals [1.25,1.375][1.25,1.375] or [1.375,1.5].[1.375,1.5]. Briefly explain your response for each interval.

154.

Consider the graph of the function y=f(x)y=f(x) shown in the following graph.

A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.).  A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.
  1. Find all values for which the function is discontinuous.
  2. For each value in part a., state why the formal definition of continuity does not apply.
  3. Classify each discontinuity as either jump, removable, or infinite.
155.

Let f(x)={3x,x>1x3,x<1.f(x)={3x,x>1x3,x<1.

  1. Sketch the graph of f.
  2. Is it possible to find a value k such that f(1)=k,f(1)=k, which makes f(x)f(x) continuous for all real numbers? Briefly explain.
156.

Let f(x)=x41x21f(x)=x41x21 for x1,1.x1,1.

  1. Sketch the graph of f.
  2. Is it possible to find values k1k1 and k2k2 such that f(−1)=kf(−1)=k and f(1)=k2,f(1)=k2, and that makes f(x)f(x) continuous for all real numbers? Briefly explain.
157.

Sketch the graph of the function y=f(x)y=f(x) with properties i. through vii.

  1. The domain of f is (,+).(,+).
  2. f has an infinite discontinuity at x=−6.x=−6.
  3. f(−6)=3f(−6)=3
  4. limx−3f(x)=limx−3+f(x)=2limx−3f(x)=limx−3+f(x)=2
  5. f(−3)=3f(−3)=3
  6. f is left continuous but not right continuous at x=3.x=3.
  7. limxf(x)=limxf(x)= and limx+f(x)=+limx+f(x)=+
158.

Sketch the graph of the function y=f(x)y=f(x) with properties i. through iv.

  1. The domain of f is [0,5].[0,5].
  2. limx1+f(x)limx1+f(x) and limx1f(x)limx1f(x) exist and are equal.
  3. f(x)f(x) is left continuous but not continuous at x=2,x=2, and right continuous but not continuous at x=3.x=3.
  4. f(x)f(x) has a removable discontinuity at x=1,x=1, a jump discontinuity at x=2,x=2, and the following limits hold: limx3f(x)=limx3f(x)= and limx3+f(x)=2.limx3+f(x)=2.

In the following exercises, suppose y=f(x)y=f(x) is defined for all x. For each description, sketch a graph with the indicated property.

159.

Discontinuous at x=1x=1 with limx−1f(x)=−1limx−1f(x)=−1 and limx2f(x)=4limx2f(x)=4

160.

Discontinuous at x=2x=2 but continuous elsewhere with limx0f(x)=12limx0f(x)=12

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

161.

f(t)=2etetf(t)=2etet is continuous everywhere.

162.

If the left- and right-hand limits of f(x)f(x) as xaxa exist and are equal, then f cannot be discontinuous at x=a.x=a.

163.

If a function is not continuous at a point, then it is not defined at that point.

164.

According to the IVT, cosxsinxx=2cosxsinxx=2 has a solution over the interval [−1,1].[−1,1].

165.

If f(x)f(x) is continuous such that f(a)f(a) and f(b)f(b) have opposite signs, then f(x)=0f(x)=0 has exactly one solution in [a,b].[a,b].

166.

The function f(x)=x24x+3x21f(x)=x24x+3x21 is continuous over the interval [0,3].[0,3].

167.

If f(x)f(x) is continuous everywhere and f(a),f(b)>0,f(a),f(b)>0, then there is no root of f(x)f(x) in the interval [a,b].[a,b].

[T] The following problems consider the scalar form of Coulomb’s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation F(r)=ke|q1q2|r2,F(r)=ke|q1q2|r2, where keke is Coulomb’s constant, qiqi are the magnitudes of the charges of the two particles, and r is the distance between the two particles.

168.

To simplify the calculation of a model with many interacting particles, after some threshold value r=R,r=R, we approximate F as zero.

  1. Explain the physical reasoning behind this assumption.
  2. What is the force equation?
  3. Evaluate the force F using both Coulomb’s law and our approximation, assuming two protons with a charge magnitude of 1.6022×10−19coulombs (C),1.6022×10−19coulombs (C), and the Coulomb constant ke=8.988×109Nm2/C2ke=8.988×109Nm2/C2 are 1 m apart. Also, assume R<1m.R<1m. How much inaccuracy does our approximation generate? Is our approximation reasonable?
  4. Is there any finite value of R for which this system remains continuous at R?
169.

Instead of making the force 0 at R, instead we let the force be 10−20 for rR.rR. Assume two protons, which have a magnitude of charge 1.6022×10−19C,1.6022×10−19C, and the Coulomb constant ke=8.988×109Nm2/C2.ke=8.988×109Nm2/C2. Is there a value R that can make this system continuous? If so, find it.

Recall the discussion on spacecraft from the chapter opener. The following problems consider a rocket launch from Earth’s surface. The force of gravity on the rocket is given by F(d)=mk/d2,F(d)=mk/d2, where m is the mass of the rocket, d is the distance of the rocket from the center of Earth, and k is a constant.

170.

[T] Determine the value and units of k given that the mass of the rocket on Earth is 3 million kg. (Hint: The distance from the center of Earth to its surface is 6378 km.)

171.

[T] After a certain distance D has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by F(d)={mkd2ifd<D10,000ifdD.F(d)={mkd2ifd<D10,000ifdD. Find the necessary condition D such that the force function remains continuous.

172.

As the rocket travels away from Earth’s surface, there is a distance D where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as F(d)={m1kd2ifd<Dm2kd2ifdD.F(d)={m1kd2ifd<Dm2kd2ifdD. Is there a D value such that this function is continuous, assuming m1m2?m1m2?

Prove the following functions are continuous everywhere

173.

f(θ)=sinθf(θ)=sinθ

174.

g(x)=|x|g(x)=|x|

175.

Where is f(x)={0ifxis irrational1ifxis rationalf(x)={0ifxis irrational1ifxis rational continuous?

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