The Real Number System

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One of the most famous of all scientific legends has the 16th-century scientist Galileo dropping a wood ball and a lead ball simultaneously from the Leaning Tower of Pisa. In this story, the balls hit the ground at the same time, proving that the acceleration due to gravity is the same for all objects. As we now know, if Galileo had dropped the balls in a vacuum they would have hit the ground at the same time. In fact, Apollo 15 Commander Dave Scott dropped a feather and a hammer in the vacuum of the moon. Just as the theory predicts, the feather and hammer fell at exactly the same rate.

Back in Italy, Galileo had to cope with the reality of air resistance. In fact, one reason that many historians don’t believe that the Leaning Tower experiment ever occurred is that air resistance would cause the heavier lead ball to reach the ground significantly before the lighter wood ball. Whether or not the legend is true, Galileo made several vital contributions to our understanding of how objects move through the air. His most important experiments in this area were precise measurements of balls rolling down inclined planes. The slight slope of his inclined planes meant that the balls rolled slowly, reducing the effect of air resistance. This allowed Galileo to compute time accurately (remember, this was before clocks were invented). In one set of experiments, he set up his inclined plane on a platform a specified height above the floor, rolled a ball down the inclined plane and off the edge of the platform. Having dipped the balls in ink, Galileo could see where the balls landed and then measure the horizontal distance traveled. His data (the units are punti) and a plot of the data are shown below.

400 800 1200 1600

200 400 800 600 1000 y

x Distance (x) Height (y)

800 300

1172 600

1328 800

1340 828

1500 1000

Throughout your study of calculus, you will look at relationships between variables in different forms. Here, we have both a numerical representation and a graphical representation of the same exper- iment. Galileo was also interested in an algebraic representation. That is, he wanted a formula by which he could compute the y-value given any x-value. In this chapter, we will examine a variety of (mostly familiar) functions and their graphs. Knowing these graphs, we can say that Galileo’s data look like points


400 800 1200 1600 200

400 800 600 1000 y


on the graph of a quadratic function. However, it also looks like data from a cubic function or an exponential function. We will consider techniques to help determine which of these functions fits the data best. One simple function that provides a nice fit is y= 0.00044x2+ 26. (See the graph above.) Ultimately, Galileo wanted a theoretical explanation of why such a formula would work. It will take several chapters for us to get there, but the theory of moving objects is one of the great success stories of calculus. Moreover, the study of calculus provides us with numerous explanations of the way things in our world work.


The Real Number System

Although mathematics is far more than just a study of numbers, our journey into calculus begins with the real number system. You might think that this is a fairly mundane starting place. After all, you probably mastered the properties of the real numbers a long time ago.

Nevertheless, we want to give you the opportunity to brush up on those properties that are of particular interest for calculus. A solid foundation here will help you to grasp the many subtleties you will encounter in your study of the calculus.

The most familiar set of numbers is the set of integers. The integers consist of the whole numbers and their additive inverses: 0, ±1, ±2, ±3, . . . . A rational number is any number that can be written in the formp

q, where p and q are integers and q= 0. For example,23, −73 and12527 are all rational numbers. Notice that every integer n is also a rational number, since we can write it as the quotient of two integers: n= n


The irrational numbers are all those real numbers that cannot be written in the formp q, where p and q are integers. The irrationals include the very important numbers

2, π and e. You may recall that rational numbers have a decimal expansion that either terminates or repeats. For instance,12 = 0.5,13 = 0.33333,18 = 0.125 and 16 = 0.166666 are all rational numbers. By contrast, irrational numbers have decimal expansions that do not repeat or terminate. For instance, we have the decimal expansions

√2= 1.4142135623 . . . , π = 3.1415926535 . . . and

e= 2.7182818284 . . . for the irrational numbers√

2, π and e.


0 1 2 3 4 5






2 3 p


FIGURE 0.1 The real line.

We usually picture the real numbers arranged along a number line as displayed in Figure 0.1 (the real line), with center at 0 and extending off forever in both the positive and negative directions. We denote the set of real numbers by the symbolR. The notation x∈ R indicates that x is an element of the set of real numbers. (In other words, x is a real number.)

Recall that for real numbers a and b, where a< b, we define the closed interval [a, b]

to be the set of numbers between a and b, including a and b (the endpoints), that is, [a, b] = {x ∈ R | a ≤ x ≤ b},

as illustrated in Figure 0.2, where we have used the solid circles to indicate that a and b are included in [a, b].

a b

FIGURE 0.2 A closed interval.

a b

FIGURE 0.3 An open interval.

Similarly, we define the open interval (a, b) to be the set of numbers between a and b, but not including the endpoints a and b, that is,

(a, b) = {x ∈ R | a < x < b},

as illustrated in Figure 0.3, where we have used the open circles to indicate that a and b are not included in (a, b).

Equations of Lines

By law, the federal government must conduct a nationwide census every 10 years to deter- mine the population. This is done for a variety of reasons, notably so that the government can track the growth of various segments of the population for the purposes of planning.

Population data for several recent decades are shown in the accompanying table.

Year U.S. population 1960 179,323,175 1970 203,302,031 1980 226,542,203 1990 248,709,873

x y

0 179

10 203

20 227

30 249

One difficulty with analyzing this data is that the numbers are so large. This problem is remedied by transforming the data. We can simplify the year data by defining x to be the number of years since 1960. Then, 1960 corresponds to x = 0, 1970 corresponds to x= 10, and so on. The population data can be simplified by rounding the numbers off to the nearest million. The simplified data are shown in the accompanying table and a scatter plot of these data points is shown in Figure 0.4.


x 50

100 150 200 250

10 20 30

FIGURE 0.4 Population data.

Asked to describe the pattern in Figure 0.4, most people would say that the points appear to form a straight line. (Use a ruler to see if you agree.) To determine whether the points are, in fact, on the same line (such points are called colinear), we might consider the population growth in each of the indicated decades. From 1960 to 1970, the growth was 24 million. (That is, to move from the first point to the second, you increase x by 10 and increase y by 24.) Likewise, from 1970 to 1980, the growth was 24 million. (So, to move


from the second point to the third point, you again increase x by 10 and y by 24.) However, from 1980 to 1990, the growth was only 22 million. (That is, to move from the third point to the fourth point, while x again increases by 10, y increases by only 22.) Since the rate of growth is not constant, this says that the data points do not fall on a line. Notice that to stay on the same line, y would had to have increased by 24 again. (Why is that?) The preceding argument involves the familiar concept of slope.


For x1= x2, the slope of the straight line through the points (x1, y1) and (x2, y2) is the number

m= y2− y1

x2− x1. (1.1)

When x1= x2, the line through (x1, y1) and (x2, y2) is vertical and the slope is undefined.

We often describe slope as “the change in y divided by the change in x,” writteny

x, or more simply asRise

Run (see Figure 0.5a).


x  x2 x1


y  y2 y1

(x1, y1)

(x2, y2)

 Rise y2


x2 x1






y B






FIGURE 0.5a Slope.


Similar triangles and slope.

You should observe that the slope of a straight line is the same no matter which two points on the line you select. Referring to Figure 0.5b (where the line has positive slope), notice that for any four points A, B, D and E on the line, the two right triangles ABC and

DE F are similar. Recall that for similar triangles, the ratios of corresponding sides must be the same. In this case, this says that


x = y


and so, the slope is the same no matter which two points on the line are selected. Furthermore, a line is the only curve with constant slope. Notice that a line is horizontal if and only if its slope is zero.


EXAMPLE 1.1 Finding the Slope of a Line

Find the slope of the line through the points (4, 3) and (2, 5).

Solution From (1.1), we get

m= y2− y1

x2− x1 =5− 3 2− 4 = 2

−2 = −1.

EXAMPLE 1.2 Using the Slope to Determine if Points Are Colinear Use slopes to determine whether or not the points (1, 2), (3, 10), and (4, 14) are colinear.

Solution First, notice that the slope of the line joining (1, 2) and (3, 10) is m1= y2− y1

x2− x1 = 10− 2 3− 1 =8

2 = 4.

Similarly, the slope through the line joining (3, 10) and (4, 14) is m2= y2− y1

x2− x1

= 14− 10 4− 3 = 4.

Since the slopes are the same, the points must be colinear. 

Recall that if you know the slope and a point through which the line must pass, you have enough information to graph the line. The easiest way to graph a line is to plot two points and then draw the line through them. In this case, you only need to find a second point.

EXAMPLE 1.3 Graphing a Line

If a line passes through the point (2, 1) with slope23, find a second point on the line and then graph the line.

Solution Since slope is given by m= y2− y1

x2− x1

, we take m=23, y1= 1 and x1= 2, to obtain


3 = y2− 1 x2− 2.

You are free to choose the x-coordinate of the second point. For instance, to find the point at x2 = 5, substitute this into the equation, and solve. From


3 = y2− 1

5− 2 = y2− 1 3 ,



1 1 2 3 4

1 2 3 4 5


Graph of a straight line.



1 1 2 3 4

1 2 3 4 5

x  3

y  2


Using slope to find a second point.

we get 2= y2− 1 or y2 = 3. A second point is then (5, 3). The graph of the line is shown in Figure 0.6a. An alternative method for finding a second point is to use the slope

m= 2 3 = y


The slope of 23 says that if we move three units to the right, we must move two units up to stay on the line, as illustrated in Figure 0.6b. 

In example 1.3, note that the choice of x = 5 was entirely arbitrary; you can choose any x-value you want to find a second point. Further, since x can be any real number, you can leave x as a variable and write out an equation satisfied by any point (x, y) on the line.


In the general case of the line through the point (x0, y0) with slope m, we have from (1.1) that

m= y− y0

x− x0

. (1.2)

Multiplying both sides of (1.2) by (x− x0), we get y− y0= m(x − x0) or

y= m(x − x0)+ y0. (1.3)

Point-slope form of a line

Equation (1.3) is called the point-slope form of the line. With this, you can find an equation of any line.

EXAMPLE 1.4 Finding the Equation of a Line Given Two Points Find an equation of the line through the points (3, 1) and (4, −1) and graph the line.

Solution From (1.1), the slope is

m= −1 − 1 4− 3 =−2

1 = −2.

Using (1.3) with slope m= −2, x-coordinate x0= 3 and y-coordinate y0= 1, we get the equation of the line:

y= −2(x − 3) + 1. (1.4)



1 1 2 3 4 5 6 7

1 2 3 4


y= −2(x − 3) + 1. To graph the line, plot the points (3, 1) and (4, −1), and you can easily draw the line seen in Figure 0.7. 

In example 1.4, you may be tempted to simplify the expression for y given in (1.4). As it turns out, the point-slope form of the equation is often the most convenient to work with. So, we will typically not ask you to rewrite this expression in other forms. At times, a form of the equation called the slope-intercept form is more convenient. This is an equation of the form

y= mx + b,

where m is the slope and b is the y-intercept (i.e., the place where the graph crosses the y-axis). In example 1.4, you simply multiply out (1.4) to get y= −2x + 6 + 1 or

y= −2x + 7.

As you can see from Figure 0.7, the graph crosses the y-axis at y= 7.

Theorem 1.1 presents a familiar result on parallel and perpendicular lines.


Two (nonvertical) lines are parallel if they have the same slope. Further, any two vertical lines are parallel. Two (nonvertical) lines of slope m1and m2are perpendicular whenever the product of their slopes is−1 (i.e., m1· m2= −1). Also, any vertical line and any horizontal line are perpendicular.

Since we can read the slope from the equation of a line, it’s a simple matter to determine when two lines are parallel or perpendicular. We illustrate this in examples 1.5 and 1.6.


EXAMPLE 1.5 Finding the Equation of a Parallel Line

Find an equation of the line parallel to y= 3x − 2 and through the point (−1, 3).




20 10 20


2 2


FIGURE 0.8 Parallel lines.

Solution It’s easy to read the slope from the equation: m= 3. The equation of the parallel line is then

y= 3[x − (−1)] + 3

or simply y = 3(x + 1) + 3. We show a graph of both lines in Figure 0.8. 

EXAMPLE 1.6 Finding the Equation of a Perpendicular Line

Find an equation of the line perpendicular to y= −2x + 4 and intersecting the line at the point (1, 2).




2 2 4

2 4


FIGURE 0.9 Perpendicular lines.

Solution The slope of y= −2x + 4 is −2. The slope of the perpendicular line is then

−1/(−2) = 12. Since the line must pass through the point (1, 2), the equation of the perpendicular line is now

y= 1

2(x− 1) + 2.

We show a graph of the two lines in Figure 0.9. 

We now return to the section’s introductory example to use the equation of a line to estimate the population in the year 2000.

EXAMPLE 1.7 Using a Line to Estimate Population

Given the population data for the census years 1960, 1970, 1980 and 1990, estimate the population for the year 2000.

Solution First, recall that we began this section by showing that the points in the corresponding table were not colinear. Nonetheless, they were nearly colinear. So, why not use the straight line connecting the last two points (20, 227) and (30, 249) (recall that these correspond to the population in the years 1980 and 1990) to estimate the population in 2000? (This is a simple example of a more general procedure called extrapolation.) The slope of the line joining the two data points is

m=249− 227 30− 20 = 22

10 =11 5 . The equation of the line is then

y= 11

5 (x− 30) + 249.


x 50 40 30 20 10 100 200 300

FIGURE 0.10 Population.

See Figure 0.10 for a graph of the line. If we follow this line to the point corresponding to x = 40 (the year 2000), we have the estimated population


5 (40− 30) + 249 = 271.


That is, the estimated population is 271 million people. The actual census figure for 2000 is 281 million, so that our estimate of 271 million is off by 10 million. This says that the U.S. population has not continued to grow at a linear rate. So, our linear model is not particularly accurate. 

In this section, we have so far drawn connections between the equations and graphs of lines. This process can be extended to a variety of other curves. In fact, most of this introductory chapter is devoted to exploring such connections. Before doing so, we need some general definitions.


For any two subsets A and B of the real line, we make the following familiar definition.


A function f is a rule that assigns exactly one element y in a set B to each element x in a set A. In this case, we write y= f (x).

We call the set A the domain of f . The set of all values f (x) in B is called the range of f . That is, the range of f is{ f (x) | x ∈ A}. Unless explicitly stated otherwise, the domain of a function f is the largest set of real numbers for which the function is defined. We refer to x as the independent variable and to y as the dependent variable.


Functions are often defined by simple formulas, such as

f (x)= 3x + 2, but in general, any correspondence meeting the requirement of matching exactly one y to each x defines a function.

By the graph of a function f , we mean the graph of the equation y= f (x). That is, the graph consists of all points (x, y), where x is in the domain of f and where y = f (x).

Notice that not every curve is the graph of a function, since for a function, there is only one y-value that can correspond to a given value of x. Recall that you can graphically determine whether a curve is the graph of a function by using the vertical line test: if any vertical line intersects the graph in more than one point, the curve is not the graph of a function.

EXAMPLE 1.8 Using the Vertical Line Test Determine which curves correspond to functions.



1 1



0.5 1 2

1 1

FIGURE 0.11a FIGURE 0.11b

Solution Notice that the circle in Figure 0.11a is not the graph of a function, since a vertical line at x= 0.5 intersects the circle twice (see Figure 0.12a). The graph in


Figure 0.11b is the graph of a function, even though it swings up and down repeatedly.

Although horizontal lines intersect the graph repeatedly, vertical lines, such as the one at x= 1.2, intersect only once (see Figure 0.12b).


x 0.5 1




0.5 1 2

1 1

FIGURE 0.12a

Curve fails vertical line test.

FIGURE 0.12b

Curve passes vertical line test.

You are already familiar with a number of different types of functions and we will only briefly review these here and in sections 0.4 and 0.5. The functions that you are probably most familiar with are polynomials. These are the simplest functions to work with because they are defined entirely in terms of arithmetic. 


A polynomial is any function that can be written in the form f (x)= anxn+ an−1xn−1+ · · · + a1x+ a0,

where a0, a1, a2, . . . , anare real numbers (the coefficients of the polynomial) with an = 0 and n ≥ 0 is an integer (the degree of the polynomial).

Note that the domain of every polynomial function is the entire real line. Further, recognize that the graph of the linear (degree 1) polynomial f (x)= ax + b is a straight line.

EXAMPLE 1.9 Sample Polynomials The following are all examples of polynomials:

f (x)= 2 (polynomial of degree 0 or constant), f (x)= 3x + 2 (polynomial of degree 1 or linear polynomial), f (x)= 5x2− 2x + 1 (polynomial of degree 2 or quadratic polynomial),

f (x)= x3− 2x + 1 (polynomial of degree 3 or cubic polynomial), f (x)= −6x4+ 12x2− 3x + 13 (polynomial of degree 4 or quartic polynomial), and

f (x)= 2x5+ 6x4− 8x2+ x − 3 (polynomial of degree 5 or quintic polynomial).

We show graphs of these six functions in Figures 0.13a–0.13f (shown on the next page).



x 1



x 4

2 2




5 5 10 15


x 6 4

2 2



40 80 120

FIGURE 0.13a

f (x)= 2. FIGURE 0.13b

f (x)= 3x + 2. FIGURE 0.13c

f (x)= 5x2− 2x + 1.


x 3 2

1 1




5 5 10


x 2

1 1



10 10

20 y


3 2 1 1

10 10 20

FIGURE 0.13d

f (x)= x3− 2x + 1. FIGURE 0.13e

f (x)= −6x4+ 12x2− 3x + 13. FIGURE 0.13f

f (x)= 2x5+ 6x4− 8x2+ x − 3.


Any function that can be written in the form f (x)= p(x)


where p and q are polynomials, with q(x)= 0, is called a rational function.

Notice that since p(x) and q(x) are polynomials, they are both defined for all x, and so, the rational function f (x)= p(x)

q(x) is defined for all x for which q(x)= 0.

EXAMPLE 1.10 A Sample Rational Function Find the domain of the function

f (x)= x2+ 7x − 11 x2− 4 .


Solution Here, f (x) is a rational function. We show a graph in Figure 0.14. Further, its domain consists of those values of x for which the denominator is nonzero. Notice that

x2− 4 = (x − 2)(x + 2)



1 3




5 5 10


f (x)= x2+ 7x − 11 x2− 4 .

and so, the denominator is zero if and only if x= ±2. This says that the domain of f is {x ∈ R | x = ±2} = (−∞, −2) ∪ (−2, 2) ∪ (2, ∞). 

No doubt, you will recall the following standard definition.


The absolute value of a real number x is|x| =

x, if x ≥ 0

−x, if x < 0 .

Make certain that you read Definition 1.5 correctly. If x is negative, then−x is positive.

This says that|x| ≥ 0 for all real numbers x. For instance, using the definition,

|−4| = −(−4) = 4.

Notice that for any real numbers a and b,

|a · b| = |a| · |b|.


|a + b| = |a| + |b|,

in general. (To verify this, simply take a= 5 and b = −2 and compute both quantities.) However, it is always true that

|a + b| ≤ |a| + |b|. (1.5)

The inequality (1.5) is referred to as the triangle inequality. You will find occasional uses for it as you progress through the calculus.

The square root function is defined in the usual way. When we write y=√ x, we mean that y is that number for which y≥ 0 and y2= x. In particular,

4= 2. Be careful not to write erroneous statements like√

4= ±2. This is a common misconception. While it’s true that 22= 4 and (−2)2= 4, when we write√

4, we are looking for the positive number whose square is 4. In this way,√

x defines a function with domain [0, ∞). If we allowed√

x to take on both positive and negative values, this would not define a function.

In particular, be careful to write

x2= |x|.


x2is asking for the nonnegative number whose square is x2, we are looking for|x|

and not x. We can say

x2 = x, only for x ≥ 0.

EXAMPLE 1.11 Finding the Domain of a Function Involving a Square Root or a Cube Root

Find the domains of f (x)=√

x2− 4 and g(x) =3 x2− 4.


Solution Since even roots are defined only for nonnegative values, f (x) is defined only for x2− 4 ≥ 0. Notice that this is equivalent to having x2≥ 4, which occurs when x ≥ 2 or x≤ −2. The domain of f is then (−∞, −2] ∪ [2, ∞). On the other hand, odd roots are defined for both positive and negative values. Consequently, the domain of g(x) is the entire real line, (−∞, ∞). 

In our study of calculus, we often find it useful to label intercepts and other significant points on a graph. Finding these points often involves solving equations. A solution of the equation f (x)= 0 is called a zero of the function f or a root of the equation f (x) = 0.

Notice that a zero of the function f corresponds to an x-intercept of the graph of y= f (x).

EXAMPLE 1.12 Finding Intercepts of a Graph Find all x- and y-intercepts of y= x2− 4x + 3.


2 4

2 10

2 4 6 8 y


FIGURE 0.15 y= x2− 4x + 3.

Solution To find the y-intercept, recall that we set x= 0 to obtain y= 0 − 0 + 3 = 3.

Notice that in Figure 0.15, the graph crosses the y-axis at y= 3.

Similarly, to find the x-intercepts, we solve the equation f (x)= 0. In this case, we can factor to get

0= x2− 4x + 3 = (x − 1)(x − 3),

so that the x-intercepts are x= 1 and x = 3, as indicated in Figure 0.15.  We briefly discuss several ideas for finding zeros here.

Unfortunately, factoring is not always so easy. Of course, for the quadratic equation ax2+ bx + c = 0

(for a, b and c all real numbers and a= 0), the solution(s) are given by the familiar quadratic formula:

x=−b ±

b2− 4ac

2a .

EXAMPLE 1.13 Finding Zeros Using the Quadratic Formula Find the zeros of f (x)= x2− 5x − 12.

Solution You might try to factor this one, but you probably won’t have much luck.

However, from the quadratic formula, we have x=−(−5) ±

(−5)2− 4 · 1 · (−12)

2· 1 = 5±√

25+ 48

2 = 5±√


2 .

So, the two solutions are given by x= 52+273 ≈ 6.772 and x = 52273 ≈ −1.772.

(No wonder you couldn’t factor the polynomial!) 

Finding zeros of polynomials of degree higher than 2 and other functions is usually quite a bit trickier and is sometimes impossible. At the least, you can always find an approximation of any zeros by using a graph to zoom in closer and closer to any point where the graph crosses the x-axis, as we’ll illustrate shortly. (Keep in mind that the zeros of a function correspond to the x-intercepts of its graph.) A more basic question, though, is to determine how many zeros a given function has. In general, there is no way to answer this


question without the use of calculus. For the case of polynomials, however, the following theorem (a consequence of the Fundamental Theorem of Algebra) provides a clue.


A polynomial of degree n has at most n distinct zeros.

Notice that this theorem does not say how many zeros a given polynomial has, but rather, that the maximum number of distinct (i.e., different) zeros is the same as the degree.

A polynomial of degree n may have anywhere from 0 to n distinct real zeros. [Recall that polynomials may also have complex zeros. For instance, f (x)= x2+ 1 has only the complex zeros x= ±i, where i is the imaginary number defined by i =

−1.] However, polynomials of odd degree must have at least one real zero. (Why is that?) For instance, for the case of a cubic polynomial, we have one of the three possibilities illustrated in Figures 0.16a, 0.16b and 0.16c.


x1 x


x1 x2 x


x1 x2 x3 x

FIGURE 0.16a One zero.

FIGURE 0.16b Two zeros.

FIGURE 0.16c Three zeros.

In these three figures, we show the graphs of cubic polynomials with 1, 2 and 3 distinct, real zeros, respectively. These are the graphs of the functions

f (x)= x3− 2x2+ 3 = (x + 1)(x2− 3x + 3), g(x)= x3− x2− x + 1 = (x + 1)(x − 1)2 and

h(x)= x3− 3x2− x + 3 = (x + 1)(x − 1)(x − 3),

respectively. Note that you can see from the factored form where the zeros are (and how many there are).

The result in Theorem 1.3 provides an important connection between factors and zeros of polynomials.

THEOREM 1.3 (Factor Theorem)

For any polynomial f, f (a) = 0 if and only if (x − a) is a factor of f (x).

EXAMPLE 1.14 Finding the Zeros of a Cubic Polynomial Find the zeros of f (x)= x3− x2− 2x + 2.


Solution You might quickly recognize that one zero of this function is x= 1. [Just calculate f (1) to verify this.] But how many other zeros are there? A graph of the function may be of some value here (see Figure 0.17a). You can see from the graph that the other two zeros of f are near x= −1.5 and near x = 1.5. You can find these zeros more precisely by using your graphing calculator or computer algebra system to zoom in on the locations of these zeros (one at a time, as in Figures 0.17b and 0.17c). From these zoomed graphs, it is clear that the two remaining zeros of f are near x= 1.414 and x= −1.414. Of course, you can make these estimates more precise by zooming in even more closely. Most graphing calculators and computer algebra systems can also find approximate zeros for you, using a built-in “solve” program. In Chapter 3, we present a versatile method (called Newton’s method) for obtaining accurate approximations to zeros. The only way to find the exact solutions is to factor the expression (using either long division or synthetic division). Here, we have

f (x)= x3− x2− 2x + 2 = (x − 1)(x2− 2) = (x − 1)(x −

2)(x+√ 2), from which you can see that the zeros are x= 1, x =

2 and x= −√ 2.



2 3

1 1


2 2 4

1.41 1.39

0.2 0.2

x y

1.40 1.42


0.04 0.02

x y

FIGURE 0.17a

y= x3− x2− 2x + 2.

FIGURE 0.17b

Zoomed in on zero near x= −1.4.

FIGURE 0.17c

Zoomed in on zero near x= 1.4.

Recall that to find the points of intersection of two curves defined by y= f (x) and y = g(x), we set f (x)= g(x) and solve for any x’s satisfying the equation (the x-coordinates of the points of intersection).

EXAMPLE 1.15 Finding the Intersections of a Line and a Parabola Find the points of intersection of the parabola y= x2− x − 5 and the line y = x + 3.



4 6



10 10 20


y= x + 3 and y = x2− x − 5.

Solution First, we draw a sketch of the two curves, as seen in Figure 0.18. Notice from the graph that there are two intersections indicated, one near x = −2 and the other near x= 4. To determine these precisely, we set the two functions equal and solve for x:

x2− x − 5 = x + 3.

Subtracting (x+ 3) from both sides leaves us with

0= x2− 2x − 8 = (x − 4)(x + 2).

This says that the solutions are exactly x= −2 and x = 4. We compute the corresponding y-values from the equation of the line y= x + 3 (or the equation of the parabola). The points of intersection are then (−2, 1) and (4, 7). Notice that these are consistent with the intersections seen in Figure 0.18. 


Unfortunately, you won’t always be able to solve equations exactly, as we did in examples 1.12–1.15. We explore some options for dealing with more difficult equations in section 0.2.



1. If the slope between points A and B equals the slope between points B and C, explain why the points A, B and C are colinear.

2. If a graph fails the vertical line test, it is not the graph of a function. Explain this result in terms of the definition of a function.

3. You should not automatically write the equation of a line in slope-intercept form. Compare the following forms of the same line: y= 2.4(x − 1.8) + 0.4 and y = 2.4x − 3.92. Given x = 1.8, which equation would you rather use to compute y? How about if you are given x= 0? For x = 8, is there any advantage to one equation over the other? Can you quickly read off the slope from either equation? Explain why neither form of the equation is “better.”

4. To understand Definition 1.5, you must believe that|x| = −x for negative x’s. Using x= −3 as an example, explain in words why multiplying x by−1 produces the same result as taking the absolute value of x.

In exercises 1– 4, determine if the points are colinear.

1. (2, 1), (0, 2), (4, 0) 2. (3, 1), (4, 4), (5, 8) 3. (4, 1), (3, 2), (1, 3) 4. (1, 2), (2, 5), (4, 8) In exercises 5–10, find the slope of the line through the given points.

5. (1, 2), (3, 6) 6. (1, 2), (3, 3) 7. (3, −6), (1, −1) 8. (1, −2), (−1, −3) 9. (0.3, −1.4), (−1.1, −0.4) 10. (1.2, 2.1), (3.1, 2.4) In exercises 11–16, find a second point on the line with slope m and point P, graph the line and find an equation of the line.

11. m= 2, P = (1, 3) 12. m= −2, P = (1, 4) 13. m= 0, P = (−1, 1) 14. m= 12, P = (2, 1) 15. m= 1.2, P = (2.3, 1.1) 16. m= −14, P = (−2, 1) In exercises 17–22, determine if the lines are parallel, perpen- dicular, or neither.

17. y= 3(x − 1) + 2 and y = 3(x + 4) − 1 18. y= 2(x − 3) + 1 and y = 4(x − 3) + 1

19. y= −2(x + 1) − 1 and y = 12(x− 2) + 3 20. y= 2x − 1 and y = −2x + 2

21. y= 3x + 1 and y = −13x+ 2 22. x+ 2y = 1 and 2x + 4y = 3

In exercises 23–26, find an equation of a line through the given point and (a) parallel to and (b) perpendicular to the given line.

23. y= 2(x + 1) − 2 at (2, 1) 24. y= 3(x − 2) + 1 at (0, 3) 25. y= 2x + 1 at (3, 1) 26. y= 1 at (0, −1)

In exercises 27–30, find an equation of the line through the given points and compute the y-coordinate of the point on the line cor- responding to x 4.

27. y


2 3 4 5

1 1 2 3 4 5

28. y

x 2

1 3 4 5 6

2 3

1 4 5 6


29. y


1.0 2.0

0.5 1.5

2 3

1 4

30. y


2 1 1

1.0 3.0


In exercises 31–34, use the vertical line test to determine whether or not the curve is the graph of a function.

31. y


2 3




5 5 10

32. y

x 4 2




5 5 10

33. y

x 3 2

1 1



2 4 6

34. y

x 2 1.5 1 0.5 0.5 1

In exercises 35– 40, identify the given function as polynomial, rational, both or neither.

35. f (x)= x3− 4x + 1 36. f (x)= 3 − 2x + x4 37. f (x)= x2+ 2x − 1

x+ 1 38. f (x)= x3+ 4x − 1 x4− 1 39. f (x)=√

x2+ 1 40. f (x)= 2x − x2/3− 6 In exercises 41– 46, find the domain of the function.

41. f (x)=√

x+ 2 42. f (x)=√

2x+ 1 43. f (x)=√3

x− 1 44. f (x)=√

x2− 4 45. f (x)= 4

x2− 1 46. f (x)= 4x

x2+ 2x − 6 In exercises 47–50, find the indicated function values.

47. f (x)= x2− x − 1; f (0), f (2), f (−3), f (1/2) 48. f (x)= x+ 1

x− 1; f (0), f (2), f (−2), f (1/2) 49. f (x)=√

x+ 1; f (0), f (3), f (−1), f (1/2) 50. f (x)= 3

x; f (1), f (10), f (100), f (1/3)

In exercises 51–54, a brief description is given of a physical sit- uation. For the indicated variable, state a reasonable domain.

51. A parking deck is to be built; x= width of deck (in feet).

52. A parking deck is to be built on a 200 -by-200 lot; x= width of deck (in feet).

53. A new candy bar is to be sold; x= number of candy bars sold in first month.


54. A new candy bar is to be sold; x= cost of candy bar (in cents).

In exercises 55–58, discuss whether or not you think y would be a function of x.

55. y= grade you get on an exam, x = number of hours you study

56. y= probability of getting lung cancer, x = number of cigarettes smoked per day

57. y= a person’s weight, x = number of minutes exercising per day

58. y= speed at which an object falls, x = weight of object 59. Figure A shows the speed of a bicyclist as a function of time.

For the portions of this graph that are flat, what is happen- ing to the bicyclist’s speed? What is happening to the bicy- clist’s speed when the graph goes up? down? Identify the por- tions of the graph that correspond to the bicyclist going uphill;




FIGURE A Bicycle speed.

60. Figure B shows the population of a small country as a function of time. During the time period shown, the country experienced two influxes of immigrants, a war and a plague. Identify these important events.



FIGURE B Population.

In exercises 61– 66, find all intercepts of the given graph.

61. y= x2− 2x − 8 62. y= x2+ 4x + 4 63. y= x3− 8 64. y= x3− 3x2+ 3x − 1 65. y= x2− 4

x+ 1 66. y= 2x− 1

x2− 4

In exercises 67–74, factor and/or use the quadratic formula to find all zeros of the given function.

67. f (x)= x2− 4x + 3 68. f (x)= x2+ x − 12 69. f (x)= x2− 4x + 2 70. f (x)= 2x2+ 4x − 1 71. f (x)= x3− 3x2+ 2x 72. f (x)= x3− 2x2− x + 2 73. f (x)= x6+ x3− 2 74. f (x)= x3+ x2− 4x − 4 75. The boiling point of water (in degrees Fahrenheit) at ele-

vation h (in thousands of feet above sea level) is given by B(h)= −1.8h + 212. Find h such that water boils at 98.6. Why would this altitude be dangerous to humans?

76. The spin rate of a golf ball hit with a 9 iron has been measured at 9100 rpm for a 120-compression ball and at 10,000 rpm for a 60-compression ball. Most golfers use 90-compression balls. If the spin rate is a linear function of compression, find the spin rate for a 90-compression ball. Professional golfers often use 100-compression balls. Estimate the spin rate of a 100-compression ball.

77. The chirping rate of a cricket depends on the temperature. A species of tree cricket chirps 160 times per minute at 79F and 100 times per minute at 64F. Find a linear function relating temperature to chirping rate.

78. When describing how to measure temperature by counting cricket chirps, most guides suggest that you count the num- ber of chirps in a 15-second time period. Use exercise 77 to explain why this is a convenient period of time.


1. Suppose you have a machine that will proportionally enlarge a photograph. For example, it could enlarge a 4× 6 photograph to 8× 12 by doubling the width and height. You could make an 8× 10 picture by cropping 1 inch off each side. Explain how you would enlarge a 312× 5 picture to an 8 × 10. A friend returns from Scotland with a 312× 5 picture showing the Loch Ness monster in the outer14 on the right. If you use your proce- dure to make an 8× 10 enlargement, does Nessie make the cut?

2. Solve the equation |x − 2| + |x − 3| = 1. (Hint: It’s an unusual solution, in that it’s more than just a couple of numbers.) Then, solve the equation 

x+ 3 − 4√ x− 1 +

x+ 8 − 6√

x− 1 = 1. (Hint: If you make the correct sub- stitution, you can use your solution to the previous equation.)



The relationship between a function and its graph is one of the central topics in calculus.

Graphing calculators and user-friendly computer software allow you to explore these rela- tionships for a much wider variety of functions than you could with pencil and paper alone.

This section presents a general framework for using technology to explore the graphs of functions.


x 4

4 2 2 20 40 60

FIGURE 0.19a y= 3x2− 1.

As we observed in section 0.1, the graphs of linear functions are straight lines. You probably also remember that the graphs of quadratic polynomials are parabolas. One of the goals of this section is for you to become more familiar with the graphs of other functions.

The best way to become familiar is by experience, working example after example.

EXAMPLE 2.1 Generating a Calculator Graph

Use your calculator or computer to sketch a graph of f (x)= 3x2− 1.


x 2

1 1


4 8

FIGURE 0.19b y= 3x2− 1.

Solution You should get an initial graph that looks something like that in Figure 0.19a.

This is simply a parabola opening upward. A graph is often used to search for important points, such as x-intercepts, y-intercepts, or peaks and troughs. In this case, we could see these points better if we zoom in; that is, display a smaller range of x- and y-values than the technology has initially chosen for us. The graph in Figure 0.19b shows x-values from x= −2 to x = 2 and y-values from y = −2 to y = 10.

You can see more clearly in Figure 0.19b that the parabola bottoms out roughly at the point (0, −1) and crosses the x-axis at approximately x = −0.5 and x = 0.5.

You can make this more precise by doing some algebra. Recall that an x-intercept is a point where y= 0 or f (x) = 0. Solving 3x2− 1 = 0 gives 3x2= 1 or x2= 13, so that x= ±


3 ≈ ±0.57735. 

Notice the interplay in example 2.1 between the graphics and the algebra. The graph suggested approximate values for the two x-intercepts, but you needed the algebra to find the values exactly. (It’s worth noting that since


3 ≈ 0.577, our visual guess of 0.5 was not especially accurate.) Also note that by zooming in, we used the technology to show us a view of the graph that highlighted the features of the graph that we wanted. But, in general, how do you know what the best view is? For example, the graphs in Figures 0.19a and 0.19b indicate a simple parabola, but how do you know that there are no interesting features of the graph lurking just off the screen? In short, you don’t, without some calculus.

Before investigating other graphs, we should say a few words about what a computer- or calculator-generated graph really is. Actually, computers and calculators do not draw graphs. Yes, we usually call them graphs, but what the computer actually does is light up some tiny screen elements called pixels. If the pixels are small enough, the image appears to be a continuous curve or graph.

By graphing window, we mean the rectangle defined by the range of x- and y-values displayed. The graphing window can dramatically affect the look of a graph. For example, suppose the x’s run from x= −2 to x = 2. If the computer or calculator screen is wide enough for 400 columns of pixels from left to right, then points will be displayed for x=

−2, x = −1.99, x = −1.98, . . . . If there is an interesting feature of this function located between x= −1.99 and x = −1.98, you will not see it unless you zoom in some. In this case, zooming in would reduce the difference between adjacent x’s. Similarly, suppose that the y’s run from y= 0 to y = 3 and that there are 600 rows of pixels from top to bottom.


Then, there will be pixels corresponding to y= 0, y = 0.005, y = 0.01, . . . . Now, suppose that f (−2) = 0.0049 and f (−1.99) = 0.0051. Before points are plotted, function values are rounded to the nearest y-value, in this case 0.005. You won’t be able to see any difference in the y-values of these points. If the actual difference is important, you will need to zoom in some to see it.


Most calculators and computer drawing packages follow one of the following two schemes for defining the graphing window for a given function.

r Fixed graphing window: Most calculators follow this method. Graphs are plotted in a preselected range of x- and y-values, unless you specify otherwise. For example, the very popular Texas Instruments graphing calculators will automatically plot points in the rectangle defined by−10 ≤ x ≤ 10 and

−10 ≤ y ≤ 10.

r Automatic graphing window: Most computer drawing packages and some calculators follow this method. Graphs are plotted for a preselected range of x-values and the computer calculates the range of y-values so that all of the calculated points will fit in the screen’s range.

Get to know how your calculator or computer software operates and use it routinely as you progress through this course. Whether you are using a graphing calculator or a computer, you should always be able to reproduce the computer-generated graphs used in this text by adjusting your graphing window appropriately.

Graphs are drawn to provide visual displays of the significant features of a function.

What qualifies as significant will vary from problem to problem, but often the x- and y- intercepts and points known as extrema are of interest. The function value f (M) is called a local maximum of the function f if f (M)≥ f (x) for all x’s “nearby” x = M [more precisely, if there exist numbers a and b with a< M < b such that f (M) ≥ f (x) for all x such that a < x < b]. Similarly, the function value f (m) is a local minimum of the function f if f (m)≤ f (x) for all x’s “nearby” x = m. A local extremum is a function value that is either a local maximum or local minimum. Whenever possible, you should produce graphs that show all intercepts and extrema.

EXAMPLE 2.2 Sketching a Graph

Sketch a graph of f (x)= x3+ 4x2− 5x − 1 showing all intercepts and extrema.

Solution Depending on your calculator or computer software, you may initially get a graph that looks like one of those in Figures 0.20a or 0.20b.


x 4

2 2

4 100 200


x 10

10 5

10 10

FIGURE 0.20a

y= x3+ 4x2− 5x − 1.

FIGURE 0.20b

y= x3+ 4x2− 5x − 1.




2 2

4 20 10 30



y= x3+ 4x2− 5x − 1.

Neither graph is completely satisfactory, although both should give you the idea of a graph that (reading left to right) rises to a local maximum near x= −3, drops to a local minimum near x = 1, and then rises again. To get a better graph, notice the scales on the x- and y-axes. The graphing window for Figure 0.20a is the rectangle defined by

−5 ≤ x ≤ 5 and −6 ≤ y ≤ 203. The graphing window for Figure 0.20b is defined by the rectangle−10 ≤ x ≤ 10 and −10 ≤ y ≤ 10. From either graph, it appears that we need to show y-values larger than 10, but not nearly as large as 203, to see the local maximum.

Since all of the significant features appear to lie between x= −6 and x = 6, one choice for a better window is−6 ≤ x ≤ 6 and −6 ≤ y ≤ 30, as seen in Figure 0.21. There, you can clearly see the three x-intercepts, the local maximum and the local minimum. 

Note that the graph in example 2.2 was produced by a process of trial and error with thoughtful corrections. You are unlikely to get a perfect picture on your first try. But, from this starting place, you can enlarge the graphing window (i.e., zoom out) if you need to see more or shrink the graphing window (i.e., zoom in) if the details are hard to see. You should get comfortable enough with your technology that this revision process is routine (and even fun!).


x y  a1x  a0

FIGURE 0.22a Line, a1< 0.


x y  a1x  a0

FIGURE 0.22b Line, a1> 0.

In the exercises, you will be asked to graph a variety of functions and discuss the shapes of the graphs of polynomials of different degrees. Having some knowledge of the general shapes will help you decide whether or not you have found an acceptable graph. To get you started, we summarize the different shapes of linear, quadratic and cubic polynomials below. Of course, the graphs of linear functions of the form f (x)= a1x+ a0are simply straight lines of slope a1. Two possibilities are shown in Figures 0.22a and 0.22b. Whenever function values get smaller, as you look from left to right, as in Figure 0.22a, we say that the function f is decreasing. If function values get larger, as you look from left to right, as in Figure 0.22b, we say that the function is increasing.

You will also recall that the graphs of quadratic polynomials of the form f (x)= a2x2+ a1x+ a0are parabolas. The parabola opens upward if a2 > 0 and opens downward if a2< 0. We show typical parabolas in Figures 0.23a and 0.23b.


x y  a2x2 a1x  a0


x y  a2x2 a1x  a0

FIGURE 0.23a

Parabola, a2> 0. FIGURE 0.23b

Parabola, a2< 0.

Notice that the parabolas indicated in Figures 0.23a and 0.23b are both increasing and decreasing on different intervals of the x-axis. One of our goals for later in the text is to learn how to precisely determine the intervals on which a given function is increasing and decreasing. The graphs of cubic functions of the form a3x3+ a2x2+ a1x+ a0 are some- what S-shaped. Reading from left to right, the function begins negative and ends positive


if a3> 0 and begins positive and ends negative if a3< 0, as indicated in Figures 0.24a and 0.24b.


x y  a3x3  a2x2  a1x  a0


x y  a3x3  a2x2  a1x  a0

FIGURE 0.24a

Cubic: one max, min, a3> 0.

FIGURE 0.24b

Cubic: one max, min, a3< 0.

Cubics often have one local maximum and one local minimum, as do those in Figures 0.24a and 0.24b. Many cubics, such as those in Figures 0.25a and 0.25b, tem- porarily flatten out without turning around to create a local maximum or minimum. The point where the curve changes its shape (from being bent upward, to being bent downward, or vice versa), is called an inflection point.


x y  a3x3  a2x2  a1x  a0

Inflection point


x y  a3x3  a2x2  a1x  a0

Inflection point

FIGURE 0.25a

Cubic: no max or min, a3> 0.

FIGURE 0.25b

Cubic: no max or min, a3< 0.

We will use the calculus developed over the next several chapters to see how to adjust the graphing window so that all of the significant features of a graph are shown. You can already use your knowledge of the general shapes of certain functions to see how to adjust the graphing window, as in example 2.3.

EXAMPLE 2.3 Sketching the Graph of a Cubic Polynomial Sketch a graph of the cubic polynomial f (x)= x3− 20x2− x + 20.

Solution Your initial graph probably looks like one of those in Figures 0.26a or 0.26b (shown on the next page).

From the preceding discussion on the general shapes of cubics, you should recognize that neither of these graphs looks like a cubic. More than anything else, they look like parabolas. To see the S-shape behavior in the graph, we need to look on a larger range of




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