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

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Functions of Several

Variables

Many variables depend on several other variables. For example, the volume V of a

right circular cone of height h and base radius r depends on both h and r. We say

that V is a function of h and r, and we write V(h, r). In this case, we can describe V

by the rule

Vðh; rÞ5

πr

Similar notation is used for functions of three or more variables. Thus the volume V

of a rectangular box is a function of its height h, width w, and depth d:

Vðh; w; dÞ5 hwd:

Although we will occasionally encounter multivariable functions in the next sev-

eral chapters, we will not study the calculus of multivariable functions until

Chapter 13.

QUICK QUIZ

1. True or false: a is in the domain of a real-valued function f of a real variable if

and only if the vertical line x 5 a intersects the graph of f exactly once.

2. True or false: b is in the range of a real-valued function f of a real variable if and

only if the horizontal line y 5 b intersects the graph of f at least once.

3. True or false: b is in the image of a real-valued function f of a real variable if and

only if the horizontal line y 5 b intersects the graph of f exactly once.

4. What is the domain of the function deﬁned by the expression

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

9 2 x

2 4

5. Suppose that ff

n50

is recursively deﬁned by f

5 1andf

5 n f

n21

for n . 0.

What is f

Answers

1. True 2. False 3. False 4. [23,22) , (2,

3] 5. 720

EXERCISES

Problems for Practice

c In Exercises 18, state the domain of the function deﬁned

by the given expression. b

1. x/(1 1 x)

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ﬃ

1 2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 2 x

5. 1/(x

2 1)

ﬃﬃﬃ

=ðx

1 x 2 6Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4x 1 5

8. 1=

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðx

2 4Þðx 2 1Þ

c In Exercises 914, sketch the graph of the function

deﬁned

by the given expression. b

9. x

1 1

10. x

2 1

11. 1 2 x

12. x

/3 1 3

13. 3 2 x

14. x

1 6x 1 10

c In Exercises 1524, plot several points, and sketch the

graph

of the function deﬁned by the given expression. b

15. x

16.

ﬃﬃﬃﬃﬃ

1.4 Functions and Their Graphs 45

17.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2x 1 4

18.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

x 2 2

19. (x 2 4)

21/2

20. (x 1 1)

21. 1=

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

x 1 1

22. signum(jx

2xj)

23.

if x , 1

2 2 x

if x $ 1



24.

2 4ifx , 2

x 1 2if22 # x , 2

if 2 # x

Further Theory and Practice

25. We say that y is proportional to x if y 5 kx for some

constant k (known as the proportionality constant). Use

the concept of proportionality to determine the arc

length s(α) for the arc of a circle of radius r that subtends

a central angle measuring α degrees. (Deduce the pro-

portionality constant by using the value of s that corre-

sponds to α 5 360



26. Let y 5 f (x) where f (x) 5 mx 1 b for constants m 6¼0 and

b. Show that a change in the value of x from x

to x

1 Δx

results in a change Δ y in the value of f (x) that does not

depend on the initial value x

. In other words, the

increment Δy 5 f (x

1 Δx) 2 f (x

) depends on the

increment Δx but not on the value of x

27. A variable u 5 f (x, y) is said to be jointly proportional to

x and y if f (x, y) 5 kxy for some constant k. The area of a

sector of a circle is jointly proportional to its central angle

and to the square of the radius r of the circle. What is the

area A(r, α) if the degree measure of the central angle of

the sector is α? (Deduce the proportionality constant by

using the value of A that corresponds to α 5 360



28. Calculate the area A(‘, α) of an isosceles triangle having

two sides of length ‘ enclosing an angle α.

29. Let r and s be the roots of x

1 Ax 1 B. Express the

coefﬁcients A and B as functions of r and s .

30. Sketch the graph of the tax function T for 0 , x # 50000

(see Example 3).

31. Figure 10 shows the graph of a function f. Give a formula

for f.

32. The domain of the function f that is graphed in Figure 10

is [0, 5]. For every x A [0, 5], let m(x) be the slope of the

graph of f at (x, f (x)), provided the slope exists. Other-

wise, m(x) is not deﬁned. What is the domain of m?

Sketch the graph of the slope function m. For x A [0, 5],

let A(x) be the area under the graph of f and over the

interval [0, x]. Sketch the graph of the area function A.

33. Let T denote the tax function described in Example 3. In

this exercise, we will restrict the domain of T to

0 , x , 50000. For x in this interval, excluding the points

8025 and 32550, let m(x) denote the slope of the graph of

T at the point (x, T(

x)).

Give a multicase formula for m.

Let A(x) be the area under the graph of m and over the

interval (0, x). Give a multicase formula for A. What is

the relationship of A to T?

c In Exercises 3440, give a recursive deﬁnition of the

sequence. b

34. f

5 2n, n 5 1, 2, 3, . . .

35. f

5 2

, n 5 1, 2, 3, . . .

36. f

5 1 1 2 1 1 n , n 5 1, 2, 3, . . .

37. f

5 n!, n 5 1, 2, 3, . . .

38. f

5 2

((21)n

), n 5 1, 2, 3, . . .

39. f

5 n(n 1 1)/2, n 5 1, 2, 3, . . .

40. f1, 5, 17, 53, 161, . . . g

41. For any real number x, the greatest integer in x is denoted

by bxc and deﬁned to be the unique integer satisfying

bxc# x , b xc1 1. For example, b3:2c5 3 and

b2 3:2c524. Notice that bxc5 x if and only if x is an

integer. The function x / bxc is called the greatest integer

function. (The expression bxc is sometimes read as “the

ﬂoor of x.”) The integer part of x, denoted by Int( x), is

that part of the decimal expansion of x to the left of the

decimal point. Express bxc as a multicase function of

Int(x). Graph x/bxc and x/Int(x)

for 23 # x # 3.

42. A kilowatt-hour is the amount of energy consumed in 1

hour at the constant rate of 1000 watts. At time t 5 0

hour, a three-way lamp is turned on at the 50-watt setting.

An hour later, the lamp is turned up to 100 watts. Forty

minutes after that, the lamp is turned up to 150 watts.

Ninety minutes later, it is turned off. Let E(t) be the

(cumulative) energy consumption in kilowatt-hours

expressed as a function of time measured in hours. Graph

E(t) for 0 # t # 4.

43. A loan of P dollars is paid back by means of monthly

installments of m dollars for n full years. Find a formula

2.5

3.0

0.5

1.0

1.5

2.0

1 2 3 4

m Figure 10

46 Chapter

1 Basics

for the function I(P, m, n) that gives the total amount of

interest paid over the life of the loan.

44. Initially, a solution of salt water has total mass 500 kg and

is 99% water. Over time, the water evaporates. Just for

fun, estimate to within 25 kg the mass of the solution at

the time when water makes up 98% of the solution. Find

a formula for the total mass m( p) of the solution as a

function of p, the percentage contribution of water to the

total mass. Exactly what is m(98)? Graph m(p).

45. Recursively deﬁne two sequences fc

n51

and fC

n51

initializing c

5 C

5 1 and, for each nonnegative integer

n, setting

n11

5 c

1 c

n21

1 1 c

n21

1 c

and

n11

4n 1 2

n 1 2

Calculate c

and C

for 0 # n # 8. (These integer

sequences can be shown to be the same for all non-

negative integers n. The members of these sequences are

called Catalan numbers.)

If P(x) is an assertion about x that is either true or false

depending on the value of x, then we can make P(x) into a

numerical function by using 1 for true and 0 for false. Thus

the function x / (x

, x) has the value 1 for 21 , x , 1 and 0

for all other values of x. Some calculators and computer

programs use this assignment of truth values as a convenient

tool for expressing multicase functions. For instance, we can

describe the function

FðxÞ5

2x 1 3ifx , 0

3if0, x , 2

x 1 1ifx $ 2

with one formula as follows:

FðxÞ52x ðx , 0Þ1 3ð0 , xÞðx , 2Þ1 ðx 1 1Þðx $ 2Þ:

c In each of Exercises 4649, use this method to express the

given

multicase function with one formula. Plot the

function. b

46. HðxÞ5

fx $ 0

0ifx , 0



(H is known as the Heaviside

function).

47. RðxÞ5

1ifx . 1

x if 0 , x # 1

0ifx # 0

ðR is the ramp function:Þ

48. rðxÞ5

0ifx . 1

1if0, x # 1

0ifx # 0

(r is the rectangular bump

function.)

49. bðxÞ5

0ifx # 1

x if 0 , x # 1

2 2 x if 1 , x # 2

0if2, x

(b is

the triangular bump

function.)

Calculator/Computer Exercises

50. In the table, x represents cigarette consumption per adult

(in hundreds) for 1962 for eight countries. The variable y

represents mortality per 100, 000 due to heart disease in

1962.

Australia 32.2 238.1

Belgium 17.0 118.1

Canada 33.5 211.6

West Germany 18.9 150.3

Ireland 27.7 187.3

Netherlands 18.1 124.7

Great Britain 27.9 194.1

United States 39 256.9

Plot the eight points (x, y). Find a function f (x) 5

mx 1 12.5 that could be used to model the relationship

between cigarette consumption and mortality due to

heart disease.

51. Let x A [0, 1] be the incidence of a certain disease in the

general population. For example, if 132 persons out of

100, 000 have the disease, then x 5 0.00132. Suppose that

a screening procedure for the disease results in false

positives 1% of the time and in false negatives 2% of the

time. Then, according to Bayes’s law,

PðxÞ5

0:98x

0:97x 1 0:01

is the probability that a person who tests positive for the

disease actually has the disease. Graph P in an appro-

priate window. What does this graph tell you about the

value of routinely using this screening procedure if the

disease is rare?

52. The astronomical unit (AU) is the mean distance of Earth

to the sun. The table gives the mean distances (D)in

astronomical units between the sun and the six planets

that were known at the time of Johannes Kepler.

The table also provides the planets’ periods of orbit

about the sun (T) in years. Plot the six points (D, T ). In

1.4 Functions and Their Graphs 47

the window [0, 10] 3 [0, 30], graph the function f

(x) 5 x

for each q 5 m/n with 1 # m and n # 3. For what value

of q does the graph of f

ﬁt the data well? Notice that

T , D for D , 1, and T . D for D . 1. Explain why this

indicates that q . 1. Which points show that q , 2?

Mercury 0.24 0.387

Venus 0.615 0.723

Earth 1.00 1.00

Mars 1.88 1.524

Jupiter 11.86 5.203

Saturn 29.547 9.539

53. When the argument of a function f changes by an

increment h from x 2 h to x, the value of the function

changes by the increment F(x, h) 5 f (x) 2 f (x 2 h). If h

is small, then it is to be expected that F(x, h) is also

small. In 1615, Kepler noticed that when f (x

) is a max-

imum value of f, the increment F(x

, h) is even smaller

than one might expect. In this exercise, we use the func-

tion f (x)5 5 2 24x

1 20x

2 3x

to investigate Kepler’s

observation.

a. Graph f in the window [2, 5] 3 [20, 135]. Using the

graph, determine the value x

of x for which f (x)is

maximized.

b. Calculate F(x

, h) and F(3.99, h) for h 5 10

,10

and 10

. You will notice that for each of these values

of h, F(x

, h) is much smaller than F(3.99, h).

c. Plot F(3.99, h) for 10

# h # 10

. You will see that F

(3.99, h) mh for some constant m.

d. Now plot F(x

, h) for 10

# h # 10

. You will see

that F(x

, h) Ah

for some constant A.The reason for

the behavior Kepler observed is that h

is negligible

compared to h when h is small.

54. In this exercise, you will compute the ﬁrst several terms

of a sequence fm

g of rational numbers that approach

ﬃﬃﬃ

. Let I

be the interval [1, 2]. It contains

ﬃﬃﬃ

. Let

5 3/2 be the midpoint of I

so that m

divides I

into

two subintervals. Let I

be the subinterval of I

that

contains

ﬃﬃﬃ

. That is, I

5 [1, m

]if2, m

, and I

5 [m

,2]

otherwise. Let m

be the midpoint of I

. Then m

divides

into two subintervals. Let I

be the subinterval of I

that contains

ﬃﬃﬃ

. Continue this process. Calculate I

and

for 1 # n # 8. What is the length of the interval I

Without calculating

ﬃﬃﬃ

, what is the maximum error that

can result if m

is used to approximate

ﬃﬃﬃ

? (This pro-

cedure for approximating a root is called the bisection

method.)

55. This exercise is based on an ancient Babylonian method

for approximating the square root of a positive number c.

Assume that c lies in the interval (1, 4) for simplicity. Set

5 3/2. For n $ 0, deﬁne

n11





a. Use the deﬁnition of x

n 1 1

to obtain the identity

n11

ﬃﬃﬃ

j5 ðx

ﬃﬃﬃ

=2x

b. Use part (a) to deduce that if x

approximates

ﬃﬃﬃ

to k

decimal places, then x

n 1 1

approximates

ﬃﬃﬃ

to 2k

decimal places.

c. Use this algorithm to compute

ﬃﬃﬃﬃﬃﬃﬃﬃﬃ

3:75

to 8 decimal

places.

56. This exercise is based on an ancient Greek algorithm for

approximating

ﬃﬃﬃ

. Let s

5 d

5 1. Let s

5 s

n21

1 d

n21

and d

5 2s

n21

1 d

n21

for n $ 1. Let r

5 d

a. Calculate s

, d

, and r

for 1 # n # 10.

b. Verify the equation

2 2s

5 ð21Þ

n21

for 0 # n # 10. (It can be proved true for all positive

integers n.)

c. Compare r

with

ﬃﬃﬃ

. Use the equation of part (b) to

explain why r

is a good rational approximation to

ﬃﬃﬃ

for large n.

57. This exercise is based on a formula for π published by

Franc¸ois Vie

te in 1593. Let q

5 1/

ﬃﬃﬃ

. For n $ 2, let

5 q

(1 1 q

n21

)

1/2

. Let Q

5 q

... q

, and p

5 2/Q

Calculate p

for 1 # n # 8.

58. This exercise describes a very fast procedure for calcu-

lating the digits of π (supposing that the digits of

ﬃﬃﬃ

are

already known). Set a

ﬃﬃﬃ

, b

5 0, and p

5 2 1

ﬃﬃﬃ

. For

n $ 0, deﬁne

n11



ﬃﬃﬃ



n11

ﬃﬃﬃﬃﬃ



ð1 1 b

ða

1 b

n11

5 p

 b

n11



ða

n11

1 1Þ

ð1 1 b

n11

Calculate p

for n 5 1, 2, and 3.

48 Chapter 1 Basics

1.5 Combining Functions

Most of the functions that we encounter are built up from simpler functions. This

building process is accomplished using arithmetic and other operations. Although

these two statements may seem perfectly obvious, it is worth noting explicitly how

some of these operations work. The functions that we will be considering from

now until Chapter 10 will have both domain and range that are subsets of the real

number system. We will not always state this fact explicitly.

Arithmetic Operations Let c be a constant. Suppose that f and g are functions with the same domain S.

For sAS,

ðf 1 gÞðsÞ means f ðsÞ1 gðsÞ ;

ðf 2 gÞðsÞ means f ðsÞ2 gðsÞ ;

ðf  gÞðsÞ means f ðsÞgðsÞ;

ðcf ÞðsÞ means c  f ðsÞ; and

ðf =gÞðsÞ means f ðsÞ=gðsÞ; as long as gðsÞ 6¼ 0:

INSIGHT

If f and g are functions, then f 1 g, f 2 g, f g, and f/g are new functions.

The rules deﬁning the new functions are expressed in terms of the original functions f and

g. The next two examples show how these simple ideas are put into practice.

⁄ EXAMPLE 1 Let f (x) 5 2x,andg(x) 5 x

. Calculate f 1 g, f 2 g, f g, and f/g.

Solution We

calculate that

ðf 1 gÞðxÞ5 f ðxÞ1 gðxÞ5 2x 1 x

;

ðf 2 gÞðxÞ5 f ðxÞ2 gðxÞ5 2x 2 x

;

ðf  gÞðxÞ5 f ðxÞgðxÞ5 2x  x

5 2x

; and





ðxÞ5

f ðxÞ

gðxÞ

; as long as x 6¼ 0: ¥

Given a function f,

we let f

denote the function f f. That is, f

(x) 5 f (x)

.We

similarly deﬁne f

for other positive integers k. For instance, if f (x) 5 2x25, then

(x) 5 (2x 2 5)

⁄ EX

AMPLE 2 Let f (x ) 5 x

, and g(x) 5 x 1 2. Compute



f 1 ð2gÞf



ð3Þ:

Solution To

perform this calculation, we apply the rules to write



f 1 ð2gÞf



ð3Þ5

f ð3Þ1 2 gð3 Þf ð3Þ

gð3Þ

9 1 2  5  9

: ¥

1.5 Combining Functions 49

Polynomial Functions A polynomial function p is a funct ion of the form

pðxÞ5 a

1 a

N21

1 1 a

x 1 a

;

where N is a nonnega tive integer, and a

does not equal zero. We say that N is the

degree of the polynomial p(x) and that a

is the leading coefﬁcient. According to

this deﬁnition, if c is a nonzero constant, then the constant function q(x) 5 c is a

degree 0 polynomial. A degree 1 polynomial p has the form p(x) 5 a

x 1 a

and is

sometimes called a linear polynomial because its graph is a line.

If p(r) 5 0, then we say that p vanishes at r and that r is a root,orazero,ofp.If

r is a root of a polynomial p, then there is a polynomial q of degree one less than

that of p such that p(x) 5 (x 2 r) q(x). The polynomial q may be found by division.

The Fundamental Theorem of Algebra implies that every polynomial with real

coefﬁcients can be factored into a product of polynomials with real-valued coefﬁ-

cients that have degree one or two. A degree 2 polynomial ax

1 bx 1 c that cannot

be factored as a product of two degree 1 polynomials with real coefﬁcients is said

to be irreducible. These are the quadratic polynomials with roots

2b 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4ac

and

2b 2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4ac

that have nonzero imaginary parts. Such nonreal roots occur if and only if b

, 4ac.

⁄ EX

AMPLE 3 Factor the polynomial x

1 3x

1 7x 1 10 as a product of

polynomials with real coefﬁcients.

Solution Because

the degree of the given polynomial is greater than two, it can be

factored. A plot of the equation y 5 x

1 3x

1 7x 1 10 reveals that 22 is a root, as

shown in Figure 1. It follows that x 2 (22) 5 x 1 2 is a factor. Division of x 1 2 into

1 3x

1 7x 1 10 yields another factor, x

1 x 1 5:

 x  5

x  2 x

 3x

 7x  10

 2x

 7x

 2x

5x  10

According to the Quadratic Formula, the roots of x

1 x 1 5 are ð21 6

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4  1  5

Þ=ð2  1Þ or 2 1=2 6 i

ﬃﬃﬃﬃﬃ

=2. Because these roots are not real num-

bers, the polynomial x

1 x 1 5 cannot be factored into degree one polynomials

with real-valued coefﬁcients. Thus

1 3x

1 7x 1 10 5 ðx 1 2Þðx

1 x 1 5 Þ

is the complete factorization over the real numbers .

250

150

50

242

m Figure 1

y 5 x

1 3x

1 7x 1 10

50 Chapter 1 Basics

Composition of

Functions

Another way of combining functions is by functional composition. Suppose that f

and g are functions and that the domain of g contains the range of f. This means

that if x is in the domain of f, then g may be app lied to f (x) (see Figure 2). The

result of these two operations, one following the othe r, is called g composed with f,

or the composition of g with f. We write g 3 f for the composition and deﬁne

ðg 3 f ÞðxÞ5 gðf ðx ÞÞ:

INSIGHT

The expression (g 3 f )(x) may look like multiplication of functions, but it is

not. For example, if g(x) 5 x

, and f (x) 5 7x

, then

ðg  f ÞðxÞ5 gðxÞf ðxÞ5 x

 7x

5 7x

;

whereas

ðg 3 f ÞðxÞ5 gðf ð xÞÞ5 gð7x

Þ5 ð7x

5 49x

A good rule when dealing with compositions of functions is to use parentheses to

organize formulas. Also, simplify parenthetical expressions by starting from the inside

and working out.

⁄ EXAMPLE 4 Let f (x ) 5 x

1 1, and g(x) 5 3x 1 5. Calculate g 3 f and f 3 g.

Solution We

calculate that

ðg 3 f ÞðxÞ5 gðf ðxÞÞ5 gðx

1 1Þ: ð1:5:1Þ

We have started to work inside the parentheses: The ﬁrst step is to substitut e the

deﬁnition of f, namely x

1 1, into the equ ation. The deﬁnition of g now says that

the evaluation of g at any argument is obtained by multiplying that argument by 3

and then adding 5. This is the rule no matter what the argum ent of g is. In the

present case, we are applying g to x

1 1. Therefore the right side of equation

(1.5.1) equals 3(x

1 1) 1 5. This simpliﬁes to 3x

1 8. In conclusion, (g 3 f)(x) 5

1 8. We compute f 3 g in a similar manner:

ðf 3 gÞðxÞ5 f ðgðxÞÞ5 ðgðxÞÞ

1 1 5 ð3x 1 5Þ

1 1 5 9 x

1 30x 1 26: ¥

INSIGHT

Example 4 shows that f 3 g can be very different from g 3 f. In general,

composition of functions is not commutative: The order of the operands g and f matters.



f(x)

Domain of g

Domain

of f

g( f(x))

m Figure 2

1.5 Combining Functions 51

⁄ EXAMPLE 5 How can we write the function r(x) 5 (2x 1 7)

as the com-

position of two functions? If u(t) 5 3/(t

1 4), how can we ﬁnd two functions υ and

w for which u 5 υ 3 w?

Solution We

can think of the function r as two operations applied in sequence.

Read the function aloud: First we double and add 7, and then we cube. Thus deﬁne

f (x) 5 2x 1 7 and g(x) 5 x

to get r(x) 5 (g 3 f )(x), or r 5 g 3 f.

Similarly, read aloud the way we evaluate u at w:

First we square t and add 4,

and then we divide 3 by the quantity just obtained. As a result, we deﬁne w(t) 5

1 4 and υ(t ) 5 3/t. It follows that u(t) 5 (υ 3 w)(t), or u 5 υ 3 w. ¥

We can compose three or more functions in a similar way. Thus

ðH 3 G 3 FÞðxÞ5 HðGðFðxÞÞÞ:

⁄ EX

AMPLE 6 Write the function r from Example 5 as the composition of

three functions instead of two.

Solution Read

the function aloud: First we double, then we add 7, and then we cube.

We set F(x) 5 2x, G(x) 5 x 1 7, and H(x) 5 x

.Wegetr(x) 5 (H 3 G 3 F )(x). ¥

Inverse Functions When we work with a function f whose domai n is S and range is T, we generally use

a rule to calculate the value f (s) in the range that corresponds to a given value s in

the domain. Stated in the language of equations, we compute t 5 f (s)inT for a

given value s in S (see Figure 3). There are times when we must “invert” this

process. That is, for a given value t in the range of f, we seek a value s for which

f (s) 5 t. Stated in geometric terms, we seek the abscissa s of a point (s, t)of

intersection of the graph of y 5 f (x) with the horizontal line y 5 t (see Figure 4).

The following deﬁnition describes the functions for which such points of inter-

section always exist.

DEFINITION

A function f : S - T is onto if, for every t in T, there is at least one

s in S for which f (s) 5 t.

y  t

y  f (x)

(s, f (s))

m Figure 4

y  t

y  f (x)

(s, f (s))

m Figure 3

52 Chapter

1 Basics

INSIGHT

In geometric terms, if f is an onto function and t is any value in the range

of f, then the horizontal line y 5 t intersects the graph of f in at least one point.

⁄ EXAMPLE 7 Let S 5 [0, 3] and T 5 [1, 10]. Let f and g be functions with

domain S and range T deﬁned by f (x) 5 3x 1 1 and g(x) 5 2x

24x 1 4. Is f onto?

What about g?

Solution Figure

5 shows the plots of y 5 f (x)andy 5 g(x) for 0 # x # 3. We see

that every horizontal line y 5 t with 1 # t # 10 intersects the graph of f. Analytically,

for every t in the range of f, we can explicitly ﬁnd a value s in the domain for which

f (s) 5 t. We simply set 3s 1 1 5 t and solve s 5 (t 2 1)/3. Notice that 0 # s for 1 # t

and s # 3 for t # 10. Thus s is in the domain of f. We conclude that f is onto. On

the other hand, we see from Figure 5 that there are values t in T for which the

horizontal lines y 5 t do not intersect the graph of g. We can understand the reason

for this as:

gðxÞ5 2ðx

2 2xÞ1 4 5 2ðx

2 2x 1 1Þ1 4 2 2 5 2ðx21Þ

1 2 $ 2;

which shows that the equation g(s) 5 t has no solution s for any t , 2. Figure 5

illustrates this idea with t 5 3/2. We conclude that g is not onto.

Another difﬁculty that can arise when we try to invert a function f is that the

horizontal line y 5 t may intersect the graph of y 5 f (x) in more than one point. The

functions for which this complication doe s not arise are described by the next

deﬁnition.

DEFINITION

A function f : S - T is one-to-one if it takes different elements of

S to different elements of T.

INSIGHT

In geometric terms, no horizontal line intersects the graph of a one-to-one

function in more than one point. Equivalently, a function f is one-to-one if, for any

given t, the equation f (s) 5 t has at most one solution s.

123

y  3/2

f(x)  3x  1

g(x)  2x

 4x  4

m Figure 5

1.5 Combining Functions 53

⁄ EXAMPLE 8 Figure 6 shows the graphs of four funct ions with domain

[0, 2] and range [0, 4]. Determine the functions that are onto. Determine the

functions that are one-to-one.

Solution It

is evident from the graphs that g and k attain every value in the range.

These functions are onto. Because the horizontal line y 5 t

does not intersect the

graph of f, and the horizontal line y 5 t

does not intersect the graph of h, these

functions are not onto.

It is evident that the graphs of f and g do

not have more than one point of

intersection with any horizontal line. Therefore f and g are one-to-one. Because

the horizontal line y 5 t

intersects the graph of h in more than one point and

because the horizontal line y 5 t

intersects the graph of k in more than one point,

these functions are not one-to-one. In summary, f is one-to-one but not onto, g

is one-to-one and onto, h is neither one-to-one nor onto, and k is onto but not

one-to-one.

Suppose that a function f : S - T is both one-to-one and onto. Then, for every t

in the range of f, there is exactly one s in the domain of f such that f (s) 5 t.Ifwe

denote by f

(t) the unique value of s such that f (s) 5 t, then we have f ( f

(t)) 5 t

for every t in the range of f. Notice that f

is a function with domain T and range

S. We say that the function f is invertible, and f

: T - S is the inverse function of f.

If f : S - T is invertible, then the equations f (s) 5 t and s 5 f

(t) yield the

equation f

( f (s)) 5 s for every s in S. This equation tells us that f is the inverse of

its inverse function f

. In symbols, we have ( f

)

5 f. A convenient way to

012

y  t

y  f (x)

012

y  t

y  h(x)

012

y  g(x)

012

y  t

y  k(x)

m Figure 6

54 Chapter

1 Basics