Blank B.E., Krantz S.G. Calculus: Single Variable

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½a; bÞ5 fx : a # x , bg

ða; b5 fx : a , x # bg

Intervals of the form (a, b) are called open, whereas those of the form [a, b] are

called closed. The other two types are called either half open or half closed (these

two terms are interchangeable) . If I is any one of these four types of bounded

intervals, then we say that a is the left endpoint of I, and b is the right endp oint of I.

The interval [a, a] is simply the set fag, which means it has only the number a in it.

Notice that an interval such as (a, a) does not contain any point. When a set has no

elements, it is called the empty set and is denoted by the symbol ;.

If a an d b are the left and right endpoints, respectively, of an interval I , then the

midpoint (or center) c of I is given by

c 5

a 1 b

It is easy to verify that c is equally distant from each of the endpoints:

jc 2 aj5





a 1 b



2 a



b 2 a



b 2



a 1 b





5 jb 2 cj:

⁄ EX

AMPLE 4 Write the interval (3, 11) in the form fx : jx 2 cj, rg .

Solution Becau

se I 5 fx : jx 2 cj, rg is the set of points with a distance from

c that is less than r, we can deduce that c is the midpoint of I. The midpoint of

the interval (3, 11) is (3 1 11)/2, or 7. Therefore c 5 7. The distance r between

c and the endpoints 3 and 11 is 4. We can therefore write the interval (3, 11) as

fx : jx 2 7j, 4g.

When working with inequalities, we often want to multiply or divide each

side by a nonzero number c. Inequalities are preserved under multiplication or

division by positive numbers. For example, if x # y and 0 , c, then cx # cy.

Inequalities are reversed under multiplication or division by negative numbers.

Thus if x # y and c , 0, then cx $ cy.

⁄ EX

AMPLE 5 Write the set W 5 fx : j3x 2 18j # 6g as a closed interval [a, b].

Solution We

ﬁrst multiply each side of j3x218j# 6 by 1/3 to obtain

j3x 2 18j #

 6; or



ð3x 2 18Þ



# 2; or

jx 2 6j # 2:

The inequality jx 2 6j# 2 is shorthand for the two inequalities 22 # x 2 6 # 2.

By adding 6 to each side, we obtai n 4 # x # 8. Thus W is the closed interval

[4, 8].

We can use the symbols Nand 2Ntogether with interval notation to describe

some intervals with only one endpoint or with no endpoints (see Figure 8). These

intervals are

(a, )

[a, )

(, b)

(, b]

(, )

m Figure 8

1.1 Number Systems 5

ða; NÞ5 f x : a , xg

½a; NÞ5 fx : a # xg

ð2N; bÞ5 fx : x , bg

ð2N; b5 fx : x # bg

ð2N; NÞ5 fx : 2N , x , Ng

Each of these intervals is an unbounded interval, as opposed to the bounded

intervals described earlier.

INSIGHT

The symbols N and 2N do not represent real numbers. They are simply

a convenient notational device. We never write [2N, a) or any other expression

suggesting that 1N or 2N is an element of a set of real numbers.

⁄ EXAMPLE 6 Solve the inequality

3x 2 2

jx 1 1j

. 1:

Solution Observe

that x cannot equal 21. For other values of x, we clear the

denominator by multiplying both sides of the equation by jx 1 1j. This quantity

is never negative, so the multiplication preserves the inequality. The result is

3x 2 2 . jx 1 1j. We divide our analysis of this last inequality into two cases,

according to the sign of the expression inside the absolute value.

If x .21,

then x 1 1 . 0, and jx 1 1j5 x 1 1. The inequality 3x 2 2 . jx 1 1j

becomes 3x 2 2 . (x 1 1), which simpliﬁes to x . 3/2. The points that satisfy both

x .21 and x . 3/2 make up the interval (3/2,N)

In the second case, when x ,21, the expression x 1 1 is negative, and, as a

consequence, jx 1 1j52(x 1 1). The inequality 3x 2 2 . jx 1 1j becomes

3x 2 2 .2(x 1 1), which simpliﬁes to x . 1/4. There are, however, no points that

satisfy both x ,21 and x . 1/4.

In summary, we see that the set of points satisfying the given inequality is

simply the interval (3/2, N) that we found in the ﬁrst case.

If A is a set of real numb ers, and x is a number in A,thenwewritex A A. Given

two sets A and B of real numbers, we can form the union A,B 5 fx: x 2 A or x 2 Bg

and the intersection A-B 5 fx : x 2 A and x 2 Bg. In words, the union A , B consists

of every real number that is in at least one of the two sets A and B.Theintersection

A - B consists of every real number that is in both sets A and B.

⁄ EX

AMPLE 7 Let A 5 fx : jxj# 5g and B 5 fx : jxj. 3g. Describe the

intersection A - B as a union of intervals.

Solution Using

set equation (1.1.1) with r 5 5 and c 5 0, we have A 5 fx : 25 #

x # 5g5 [25, 5]. To determine if a real number x is in the set B, we consider the

cases x . 0, x 5 0, and x , 0 separately. If x . 0, then jxj5 x. Therefore a positive

number x is in B if and only if x . 3. If x 5 0, then jxj5 0, and x is not in B.Ifx , 0,

then 2x 5 j x j. Therefore a negative number x is in B if and only if 2 x . 3, or

x ,23. Thus B 5 (2N, 23) , (3,N). Figure 9 shows that A - B 5 [ 25, 23) ,

(3,

5].

0510 10

A  B

0510 10

m Figure 9

6 Chapter

1 Basics

The Triangle Inequality It is not difﬁcult to see that ja 1 bj5 jaj1 jbj when a and b have the same sign (see

Exercise 53). If a and b have opposite signs, however, then cancellation occurs in

the sum a 1 b but not in the sum jaj1 jbj. In this case, ja 1 bj, jaj1 jbj. In sum-

mary, all real numbers a and b satisfy

ja 1 bj # jaj1 jbj:

This relationship, called the Triangle Inequality, is a useful tool for estimating the

magnitude of a sum in terms of the magnitudes of its summands.

⁄ EX

AMPLE 8 Suppose the distance of a to b is less than 3, and the distance

of b to 2 is less than 6. Estimate the distance of a to 2.

Solution We

have ja 2 2j5 j(a 2 b) 1 (b 2 2)j. By the Triangle Inequality,

ja 2 2j # ja 2 bj1 jb 2 2j, 3 1 6 5 9:

Therefor

e the distance of a to the number 2 is not greater than 9.

The Triangle Inequality can be extended to three or more summands. For

example,

ja 1 b 1 cj # j a j1 jbj1 jcj:

Approximation Calculus deals with many “inﬁnite processes.” One inﬁnite process that we have

already considered is the decimal expansion of an irrational number. Although we

can think abstractly of irrational numbers such as π and

ﬃﬃﬃ

, we cannot write all the

digits of thei r decimal expansions. If a calculator returns the value 3.14159265359

for π, then it is really displaying an approximation of π. A small error has resulted.

In general, the type of error that results when an inﬁnite process is terminated is

known as a truncation error.

In numerical work, we always have the equation

True value 5 approximate value 1 error:

If a is an approximation of the number x, then jx 2 aj is called the absolute error,

and jx 2 aj/ jxj is called the relative error. Relative error is often of greater concern

than absolute error. For example, an ab solute error of $1000 in the national budget

is not serious because it is a very small relative error. The same absolute error in

your personal budget, however, would be far more signiﬁcant. When we calculate

an approximation, we must ensure that the error term is acceptably small. Often

the size of an allowa ble error is stated in terms of decimal places.

DEFINITION

If the absolute error jx 2 aj is less than or equal to 5 3 10

2(q 1 1)

then a approximates (or agrees with) x to q decimal places.

⁄ EX

AMPLE 9 Does 22/7 approxi mate π to as many decimal places as does

3.14? Does 1.418 approxi mate

ﬃﬃﬃ

to two decimal places?

1.1 Number Systems 7

Solution Because jπ 2 22/7j5 1.26 . . . 3 10

, we have

5 3 10

2ð311Þ

5 5 3 10



π 2



, 5 3 10

5 5 3 10

2ð211Þ

Therefore 22/7 approximates π to two, but not three, decimal places. Similarly,

π 2 3.14 5 1.59 . . . 3 10

and so

5 3 10

2ð311Þ

5 5 3 10

, jπ 2 3:14j, 5 3 10

5 5 3 10

2ð211Þ

Thus the number 3.14, like 22/7, approximates π to two, but not three, decimal

places.

Because j

ﬃﬃﬃ

2 1:418j5 3:78 :::3 10

, 5 3 10

2ð211Þ

, we conclude that 1.418

does approximate

ﬃﬃﬃ

to two decimal places.

INSIGHT

Even if two numbers a and b agree to q decimal places, we cannot be

certain that the results of rounding a and b to q decimal places will be the same. In

Example 9, we see that

ﬃﬃﬃ

and 1.418 agree to two decimal places. Rounding these

numbers to two decimal places, however, results in 1.41 and 1.42, respectively.

When working with a calc ulator or computer software, there are several

sources of error. The errors that we now consider are caused by the way real

numbers are stored in calculators or computers.

The ﬂoating point decimal representation of a nonzero real number x is

x 560:a

:::3 10

where p is an integer, and a

, a

, . . . are natural numbers from 0 to 9 with a

6¼0.

The decimal point is thought of as “ﬂoating” to the left of the ﬁrst nonzero digit. A

number with a ﬂoating point decimal representation of 6 0. a

:::a

3 10

said to have k signiﬁcant digits.

Because the memory of a calculator or computer is ﬁnite, it must limit the

number of signiﬁcant digits with which it works. This can lead to a type of error

known as roundoff error. It is important to understand that a computation can

reduce the number of signiﬁcant digits. Such a loss of signiﬁcant digits, or, more

simply, a loss of signiﬁcance, can occur when two very nearly equal numbers are

subtracted. For example, 0.124 and 0.123 have three signiﬁcant digits, whereas their

difference, 0.1 3 10

, has only one signiﬁcant digit. The next example illustrates

why loss of signiﬁcance is sometimes called catastrophic cancellation.

⁄ EX

AMPLE 10 Suppose that x 5 0.412 3 0.300 2 0.617 3 0.200. A calcu-

lator that displays three signiﬁcant digits is used to evaluate the products before

subtraction. What relative error results?

Solution When

rounded to three signiﬁcant digits, the products 0.412 3 0.300 5 0.1236

and 0.617 3 0.200 5 0.1234 become 0.124 and 0.123. The difference of these

numbers is x 5 0.124 2 0.123 5 0.001 5 0.1 3 10

. Of course, x really equals

0.1236 2 0.1234, or 0.2 3 10

. The relative error is therefore

jx 2 aj

jxj

j0:2 3 10

2 0:1 3 10

j0:2 3 10

0:8 3 10

0:2 3 10

5 4;

or 400%.

8 Chapter

1 Basics

Computer Algebra

Systems

From time to time, this book refers to software packages known as computer

algebra systems. A computer algebra system not only performs arithmetic opera-

tions with speciﬁc numbers but also carries out those operations symbolically.

When the “factor” command of a computer algebra system is applied to x

, for

example, the result is

ðx 2 yÞðx 1 yÞðx

1 xy 1 y

Þðx

2 xy 1 y

Þ:

Several calculators now offer some of the capabilities of computer algebra systems.

QUICK QUIZ

1. What interval does the set fx : jx 2 2j, 9g represent?

2. Write the interval (25, 1) in the form fx : jx 2 cj, rg.

3. Write fx : jx 1 8j. 7g as the union of two intervals.

4. For which numbers b is j25 1 bj less than 5 1 jbj?

5. What is the smallest number that approximates 0.997 to two decimal places?

The largest?

Answers

1. (27,

11) 2. fx : jx2(22)j, 3g 3. (2N, 215) , (21,N)

4. b . 0 5. 0.992, 1.002

EXERCISES

Problems for Practice

c In Exercises 16, convert the decimal to a rational frac-

tion. (Ellipses are included in some exercises to indicate

repetition.) b

1. 2.13

2. 0.00034

3. 0.232323

. . .

4. 2.222 . . .

5. 5.001001001 . . .

6. 15.7231231231 . . .

c In Exercises 712, use long division to convert the rational

fraction

to a (possibly nonterminating) decimal with a

repeating block. Identify the repeating block. b

7. 1/40

8. 25/8

9. 5/3

10. 2/7

11. 18/25

12. 31/14

c In Exercises 1318, write the set using interval notation.

Use

the symbol , where appropriate. b

13. fx :1# x # 3g

14. fx : jx 2 2j, 5g

15. ft : t . 1g

16. fu : ju 2 4j$ 6g

17. fy : jy 1 4j# 10 g

18. fs : js 2 2j. 8g

c In Exercises 1924, sketch the set on a real number

line. b

19. fx :2x 2 5 , x 1 4g

20. fs : js 2 2j, 1g

21. ft :(t 2 5)

, 9/4 g

22. fy :7y 1 4 $ 2y 1 1g

23. fx : j3x 1 9j# 15g

24. fw : j2w 2 12j$ 1g

c In Exercises 2528, write the interval in the form

fx : jx 2 cj, rg or fx : jx 2 cj# rg. b

25. [21,

26. ½3; 4

ﬃﬃﬃ



27. (2π, π 1 2)

28. ðπ 2

ﬃﬃﬃ

; πÞ

Further Theory and Practice

29. Explain why the sum and product of two rational num-

bers are always rational.

30. Give an example of two irrational numbers with a

rational product; give an example of two irrational

numbers with a rational sum.

31. Two commonly used approximations of π are 22/7 and

3.14. How can you tell at a glance that these approx-

imations cannot be exact?

1.1 Number Systems 9

32. Find the smallest interval of the form fx : jx 2 π j# rg that

contains both 22/7 and 3.14. Is r rational or irrational?

33. A chemistry experiment, if followed exactly, results in

the production of 34.5 g of a compound. To allow for a

small amount of experimental error, the lab instructor

will accept, without penalty, any reported measurement

within 1% of the correct mass. In what interval must a

measurement lie if a penalty is to be avoided? Describe

this interval as a set of the form fx : jx 2 cj# ε g.

34. A door designed to be 885 mm wide and 1475 mm high is

to ﬁt in a 895 mm 3 1485 mm rectangular frame. The

clearance between the door and the frame can be no

greater than 7 mm on any side for an acceptable seal to

result. Under anticipated temperature ranges, the door

will expand by at most 0.2% in height and width. The

frame does not expand signiﬁcantly under these tem-

perature ranges. At all times, even when the door has

expanded, there must be a 1 mm clearance between the

door and the frame on each of the four sides. The door

does not have to be manufactured precisely to speciﬁca-

tions, but there are limitations. In what interval must the

door width lie? The height?

c In Exercises 3543, sketch the set on a real number

line. b

35. fx : x .22

and x

, 9g

36. fs : js 1 3j, j2s 1 7jg

37. fy: y 2

ﬃﬃﬃ

, 3y 1 4 # 4y 1

ﬃﬃﬃ

38. ft : jt

1 6tj# 10g

39. fx : jx

2 5j$ 4g

40. fs : js 1 5j, 4 and js 2 2j# 8g

41. fx : x 1 1 $ 2x 1 5 . 3x 1 8g

42. ft :(t 2 4)

, (t 2 2)

or jt 1 1j# 4g

43. fx : jx

1 xj. x

2 xg

c Describe each set in Exercises 4447 using interval nota-

tion

and the notation fx : P(x)g. Use the symbol , where

appropriate. b

44. The

set of points with a distance from 2 that does not

exceed 4

45. The set of numbers that are equidistant from 3 and 29

46. The set of numbers with a square that lies strictly

between 2 and 10

47. The set of all numbers with a distance less than 2 from 3

and with a square that does not exceed 8

c In Exercises 4851, if the set is given with absolute value

signs,

then write it without absolute value signs. If it is given

without absolute value signs, then write it using absolute

value signs. b

48. fx : x 1 5 , jx 1 1jg

49. fs : js 2 4j. j2s 1 9jg

50. ft : t

2 3t , 2t

2 5tg

51. fw : w/(w 1 1) , 0g

52. Suppose p(x) 5 x

1 Bx 1 C has two real roots, r

and r

with r

, r

. Write the set fx : p(x) # 0g as an interval.

53. Show that equality holds in the Triangle Inequality when

both summands are nonpositive.

54. Use the Triangle Inequality to prove that

jaj2 jbj # ja 1 bj

for all a; b 2 R.

55. Prove that there is no smallest positive real number.

56. At ﬁrst glance, it may appear that the number

y 5 0:

9 5 0:999 ::: is just a little smaller than 1. In fact,

y 5 1. Let x 5 1 2 y. Explain why it is true that x , 10

for every positive n. Deduce that 0 # x , b for every

positive b. Explain why this implies that y 5 1.

57. About 2400 years ago, the followers of Pythagoras dis-

covered that if x is a positive number such that x

5 2,

then x is irrational. Complete the following outline to

obtain a proof of this fact.

a. Suppose that x 5 a/b where a and b are integers with

no common factors.

b. Conclude that 2 5 x

5 a

c. Conclude that 2b

5 a

d. Conclude that 2 divides a evenly, with no remainder.

Therefore a 5 2α for some integer α .

e. Conclude that b

5 2 α

f. Conclude that 2 divides b evenly, with no remainder.

g. Notice that parts d and f contradict part a.

58. Explain the error in the following reasoning: Let x 5

(π 1 3)/2. Then 2x 5 π 1 3, and 2x(π 2 3) 5 π

2 9. It

follows that x

1 2πx 2 6x 5 x

1 π

2 9, and x

2 6x 1

9 5 x

2 2πx 1 π

. Each side is a perfect square:

(x 2 3)

5 (x 2 π )

. Therefore x 2 3 5 x 2 π , and 3 5 π .

Calculator/Computer Exercises

c In Exercises 5962, determine the interval that y must lie

in to agree with x to q decimal places. b

59. x 5 0.449, q 5 3

60. x 5 24, q 5 2

61. x 5 0.999 3 10

, q 5 3

62. x 5 0.213462 3 10

, q 5 5

63. Write a number with four signiﬁcant digits that agrees

with x 5 3.996 to two decimal places but that differs from

x in each digit.

64. Let x 5 0.4449. Round this number directly to three, to two,

and to one signiﬁcant digit(s). Now successively round x to

three, to two, and to one signiﬁcant digit(s). The mathe-

matical phrasing of this phenomenon is that rounding is not

transitive. Describe this idea in your own words.

65.

α 5 1728148040 2 140634693

ﬃﬃﬃﬃﬃﬃﬃﬃ

151

and

β 5 1728148040 1 140634693

ﬃﬃﬃﬃﬃﬃﬃﬃ

151

10 Chapter 1 Basics

A direct computation (with integers) shows that α β 5 1.

Calculate decimal representations of α and β. Use them

to compute the product α β. Note: To twenty signiﬁcant

digits, 140634693

ﬃﬃﬃﬃﬃﬃﬃﬃ

151

equals

1728148039:9999999997:

Calculators that do not record enough signiﬁcant digits

will give 0 for the value of α.

66. Find two numbers 0.a

and 0.b

with

5 b

that agree when rounded to four signiﬁcant digits

but that do not agree when rounded directly to one sig-

niﬁcant digit.

c The term a

is said to be the leading term of the polynomial

1 a

n21

1 1 a

x 1 a

;

where a

6¼0. The othe r terms are referred to as lower-order

terms. In calculus, for large values of jxj, neglecting all the lower-

order terms of a polynomial introduces only a small relative

error; this relative error tends to zero as jxj tends to inﬁnity. In

Exercises 6770, present two tables, one for x . 0

and one for

x , 0, to support this assertion. For example, for p(x) 5

1 100x

21000, the table below supports the assertion that

the relative error jp(x)23x

j/ jp(x)j tends to zero as x tends to

positive inﬁnity. b

x Relative

Error

3.2258 3 10

3.3223 3 10

3.3322 3 10

3.3332 3 10

3.3333 3 10

67. x 1 10

68. x

1 12345x

69. 23x

1 10

70. jxj

1 1000x

1.2 Planar Coordinates and Graphing in the Plane

A set such as fa, bg does not have an implied ordering of its elements. We do not,

for example, interpret a as the ﬁrst element of fa, bg and b as the second. When an

ordering of a and b is required, we use the ordered pair notation (a, b). We say that

a is the ﬁrst coordinate and b the second coordinate of the ordered pair (a, b).

Although fa , bg and fb, a g are exactly the same set, (a, b)and(b, a) are different

ordered pairs (if a 6¼b). The set of all ordered pairs of real numbers is called the

Cartesian plane and is denoted by R

5 fða; bÞ : a; b 2 Rg:

Just as we can describe a point on a line with a real number, so too can we

describe a point in a plane with an ordered pair of real numbers. Begi n with

intersecting horizontal and vertical lines. Put scales of real numbers on each line so

that the zero of each scale is at the point of intersection. The point of intersection of

the two lines is called the origin and is represented by the symbol 0. The vertical

and horizontal lines are called axes. It is convenient to name each axis by using a

variable. Commonly, the horizontal axis is named the x-axis and the vertical axis is

named the y-axis. The Cartesian plane is then called the xy-plane. The two axes

divide the plane into four regions that are known as quadrants, which are ordered

as in Figure 1.

Each point P is determined by its displacement from the axes. Here the word

“displacement” is interpreted in the signed sense: Positive displacement is to the

right or upward, and negative displacement is to the left or downward. The dis-

placement a of the point P from the y-axis is said to be the x-coordinate,orabscissa,

of P. The displacement b of the point P from the x-axis is said to be the

y-coordinate,orordinate,ofP. We represent the point P by the ordered pair (a, b),

Quadrant I

Quadrant II

Quadrant IVQuadrant III

m Figure 1 The xy-plane

1.2 Planar Coordinates and Graphing in the Plane 11

as shown in Figure 2. The signs of the x- and y-coordinates in each of the four

quadrants are also illustrated in Figure 2. Figure 3 exhibits several points and their

coordinates.

It is easy to plot single points in the xy-plane. Often our job, however, is to plot

a set of points with coordinates that satisfy some given equation in the variables x

and y. This set is often referred to as the locus of the equation, or the curve deﬁned

by the equation. The equation is called the Cartesian equation of the locus of

points. For example, the graph of the Cartesian equation x 5 3 is the vertical line

consisting of every point (3, y). Here y may be any real number because the

Cartesian equation x 5 3 imposes no restrictions on y. The plot of x 5 3 is shown in

Figure 4. In a similar way, we see that the graph of y 522 is the horizontal line that

consists of every point (x, 22). This line is also shown in Figure 4. In the next

example we consider a Cartesian equation that involves both x and y.

⁄ EX

AMPLE 1 Sketch or graph all points (x, y) in the plane that satisfy the

equation y 5 2x.

Solution For

now, we perform this task by plotting some points and connecting

them. Notice that the points (0, 0), (25, 210), (3, 6), (22, 24), and (9/2, 9) all satisfy

the equation y 5 2x. Plotting these points suggests that the collection of all points

satisfying y 5 2x is a straight line (see Figure 5).

INSIGHT

When we plot a few points and connect them with a smooth curve, we are

assuming that the curve is well behaved between the points we have actually plotted.

Later, we will learn how calculus provides ways to know when this assumption is justiﬁed.

The Distance Formula

and Circles

We can learn more from Cartesian coordinates if we use some geometric insight.

First we review the distance formula. Let

denote the line segment between

points P

5 (a

, b

) and P

5 (a

, b

) in the plane. The shortest distance between P

and P

is the length of line segment P

, which we denote by jP

j. We can

calculate this distance by realizing line segment

as the hypotenuse of a right

triangle (see Figure 6). By the Pythagorean Theorem, the length d of the hypo-

tenuse is given by

(2, 4)

(5, 10)

(3, 6)



4 242

4

8

m Figure 5

, b

)

, b

)

a

 a



b

 b



m Figure 6

d 5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ða

2 a

1 ðb

2 b

⫺3 ⫺2 ⫺132

(2, 0)

(0, ⫺2.7)

⫺1

⫺2

⫺3

, 0

⫺



⫺2.5,

␲



⫺



⫺2,

⫺







m Figure 3

(a, b)

(, )(, )

(, )(, )

m Figure 2

y  2

x  3

5

m Figure 4

12 Chapter

1 Basics

5 ja

2 a

1 jb

2 b

;

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ða

2 a

1 ðb

2 b

Because the quantities under the square root sign are squared, we may omit the

absolute value signs.

⁄ EX

AMPLE 2 Calculate the distance between (2, 6) and (24, 8).

Solution The

requested distance is

d 5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð2 2 ð24ÞÞ

1 ð6 2 8Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 ð22Þ

ﬃﬃﬃﬃﬃ

5 2

ﬃﬃﬃﬃﬃ

: ¥

⁄ EX

AMPLE 3 Graph the set of points satisfying (x 2 4)

1 ( y 2 2)

5 9.

Solution We

rewrite the equation as

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðx2 4Þ

1 ðy2 2Þ

5 3:

According to this equation, 3 is the distance between the point (4, 2) and any point

(x, y) that satisﬁes the equation. The collection of such points is a circle of radius 3

and center (4, 2). The locus is exhibited in Figure 7.

The Equation of a

Circle

Example 3 illustrates a general principle: An equation of the form

ðx2 hÞ

1 ðy2 kÞ

5 r

ð1:2:1Þ

with r . 0 has a graph that is a circle of radius r and center (h, k). Conversely, every

circle is described by an equation of this form. We call this equation the standard

form for a circle. The circle centered at the origin with radius 1 is called the unit

circle. It is the locus of the equation

1 y

5 1:

⁄ EX

AMPLE 4 Write the equation of the circle with center (8, 23/5) and

radius 6.

Solution We

apply formula (1.2.1) to obtain ðx 2 8Þ



y 2ð23=5Þ



5 6

,or

ðx 2 8Þ

1 ðy 1 3=5Þ

5 36: ¥

The Method of

Completing the Square

To work efﬁciently with expressions of the form Ax

1 Bx 1 C, we must use an

algebraic procedure known as the method of completing the square. It is a good idea

to review this procedure because it will be needed from time to time. (Althou gh

you should not memorize the following formulas, you should certainly remember

the steps.)

7543

1

 4)

 (y

 2)

 9

(4,

m Figure 7

1.2 Planar Coordinates and Graphing in the Plane 13

Completing the Square

Step 1: Starting from Ax

1 Bx 1 C, group together the terms that involve x, and from this group factor out the

coefﬁcient A of x

1 Bx 1 C 5 A





1 C:

Step 2: To the group of terms that involve x, add the square of half the coefﬁcient of x. From the constant term

subtract an equal amount. In doing so, do not forget to account for the coefﬁcient A:

1 Bx 1 C 5 A



x 1





1 C 2 A





Step 3: Identify the square:

1 Bx 1 C 5 A



x 1



1 C 2 A





5 A



x 1



4AC 2 B

⁄ EX

AMPLE 5 Apply the method of completing the square to 3x

1 18x 1 16

and to 5 2 4x 2 4x

Solution We

have

1 18x 1 16 5 3ðx

1 6xÞ1 16 ðStep 1Þ

5 3



1 6x 1





1 16 2 3 





ðStep 2Þ

5 3



x 1



2 11 ðStep 3Þ

and

5 2 4x 2 4x

524ðx

1 xÞ1 5 ðStep 1Þ

524



1 x 1





1 5 1 4





ðStep 2Þ

524



x 1



1 6: ðStep 3Þ ¥

⁄ EXAMPLE 6 Suppose that A 6¼0.

Use the method of completing the

square to show that the roots of the quadratic equation Ax

1 Bx 1 C 5 0 satisfy

the Quadratic Formula

x 5

2 B 2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4AC

or x 5

2B 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4AC

Solution From

Step 3, we see that a root x of Ax

1 Bx 1 C 5 0 satisﬁes

14 Chapter 1 Basics