Coburn J.W., Herdlick J.D. College Algebra: Graphs and Models

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5.1 One-to-One and Inverse Functions

Consider the function . If , the equation becomes

and the corresponding value of x can be found using inverse operations. In this section,

we introduce the concept of an inverse function, which can be viewed as a formula for

ﬁnding x-values that correspond to any given value of f(x).

A. Identifying One-to-One Functions

The graphs of and are shown in Figures 5.1 and 5.2. The dashed, verti-

cal lines clearly indicate both are functions, with each x-value corresponding to only

one y. But the points on have one characteristic those from do not—

each y-value also corresponds to only one x (for , 4 corresponds to both and 2).

If each element from the range of a function corresponds to only one element of the do-

main, the function is said to be one-to-one.

Figure 5.1 Figure 5.2

One-to-One Functions

A function f is one-to-one if every element in the range,

corresponds to only one element of the domain.

In symbols, if then , or

if , then .

From this deﬁnition, we note the graph of a one-to-one function must not only pass a

vertical line test (to show each x corresponds to only one y), but also pass a horizontal

line test (to show each y corresponds to only one x).

Horizontal Line Test

If every horizontal line intersects the graph of a function in at most one point,

the function is one-to-one.

Notice the graph of (Figure 5.3) passes the horizontal line test, while the

graph of (Figure 5.4) does not.y  x

y  2x

f 1x

2 f 1x

 x

 x

f 1x

2 f 1x

5

(2, 4)

(0, 0)

(2, 4)

5

(0, 0)

(2, 4)

(2, 4)

2y  x

y  x

y  2x

y  x

y  2x

2x  3  7,f

1x2 7f 1x2 2x  3

480 5–2

College Algebra Graphs & Models—

LEARNING OBJECTIVES

In Section 5.1 you will see

how we can:

A. Identify one-to-one

functions

B. Explore inverse functions

using ordered pairs

C. Find inverse functions

using an algebraic

method

D. Graph a function and its

inverse

E. Solve applications of

inverse functions

55

5

1

(0, 0)

(2, 4)

(2, 4)

542135 4 3 2 1

3

2

1

Figure 5.3

Figure 5.4

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5–3 Section 5.1 One-to-One and Inverse Functions 481

College Algebra Graphs & Models—

If the function is pointwise-deﬁned or given in ordered pair form, we simply check

to see that no given second coordinate is paired with more than one ﬁrst coordinate.

EXAMPLE 1

䊳

Identifying One-to-One Functions

Determine whether each graph or relation shown depicts a function. If so,

determine whether the function is one-to-one.

a. b.

c. d.

e. f.

Solution

䊳

A careful inspection shows all ﬁve graphs depict functions, since each passes the

vertical line test [the relation in (f) is not a function since the input corresponds

to two outputs]. Only (a), (b), and (e) pass the horizontal line test and are one-to-one

functions.

Now try Exercises 7 through 26

䊳

B. Inverse Functions and Ordered Pairs

Consider the function and the solutions shown in Table 5.1. Figure 5.5

shows this function in diagram form (in blue), and illustrates that for each element of

the domain, we multiply by 2, then subtract 3. An inverse function for f is one that

takes the result of these operations (elements of the range), and returns the original

f 1x2 2x  3

3

12, 32, 13, 22, 14, 426

514, 12, 13, 52, 11, 12,

55

5

55

5

55

5

55

5

55

5

A. You’ve just seen how

we can identify one-to-one

functions

cob19545_ch05_479-492.qxd 9/2/10 9:20 PM Page 481

From this illustration we make the following observations regarding an inverse

function, which we actually denote as .

Inverse Functions

If f is a one-to-one function with ordered pairs (a, b),

1. is a one-to-one function with ordered pairs (b, a).

2. The range of f will be the domain of .

3. The domain of f will be the range of .

CAUTION

䊳

The notation is simply a way of denoting an inverse function and has nothing to do

with exponential properties. In particular, does not mean .

EXAMPLE 2

䊳

Finding the Inverse of a Function

Find the inverse of each one-to-one function given:

Solution

䊳

a. When a function is deﬁned as a set of ordered pairs, the inverse function

is found by simply interchanging the x- and y-coordinates:

b. Using diagrams similar to Figures 5.5 and 5.6, we reason that will

subtract 2, then divide the result by . As a test, we ﬁnd

that ( , 8), (0, 2), and (3, ) are solutions to p(x), and note that (8, ),

(2, 0), and ( , 3) are indeed solutions to .

Now try Exercises 27 through 38

䊳

C. Finding Inverse Functions Using an Algebraic Method

The fact that interchanging x- and y-values helps determine an inverse function can be

generalized to develop an algebraic method for ﬁnding inverses. Instead of inter-

changing speciﬁc x- and y-values, we actually interchange the x- and y-variables, then

solve the equation for y. The process is summarized here.

1

1x27

272

3: p

1

1x2

x  2

3

1

1x2

1

1x2 5113, 42, 17, 12, 15, 02, 11, 22, 15, 52, 111, 826

p1x23x  2

1x2 514, 132, 11, 72, 10, 52, 12, 12, 15, 52, 18, 1126

f 1x2

1

1x2

1

1x2

1

1x2

1

1x2

1

1x2

1

1x2

482 CHAPTER 5 Exponential and Logarithmic Functions 5–4

College Algebra Graphs & Models—

domain element. Figures 5.5 and 5.6 show that function F achieves this by “undoing”

the operations in reverse order: add 3, then divide by 2 (in red). A table of values for

F(x) is shown (Table 5.2).

Multiply by 2

Divide by 2

Subtract 3

Add 3

y  2x  3

5 10

Multiply by 2

Divide by 2

Subtract 3

Add 3

Figure 5.5

Figure 5.6

xf(x)

813

3

93

Table 5.1

xF(x)

13 8

3

39

Table 5.2

B. You’ve just seen how

we can explore inverse

functions using ordered pairs

cob19545_ch05_479-492.qxd 9/2/10 9:20 PM Page 482

Finding an Inverse Function

1. Use y instead of f(x). 3. Solve the new equation for y.

2. Interchange x and y. 4. The result gives the inverse function:

substitute for y.

In this process, it might seem like we’re using the same y to represent two different

functions. To see why there is actually no contradiction, see Exercise 97.

EXAMPLE 3

䊳

Finding Inverse Functions Algebraically

State the domain and range of the function given, then use the algebraic method to

ﬁnd the inverse function, and state its domain and range.

a. b.

Solution

䊳

use

instead of

(

)

interchange

and

cube both sides

solve for

the result is

use

instead of

(

)

interchange

and

multiply by and distribute

gather terms with

factor

solve for

Now try Exercises 39 through 46

䊳

In cases where a given function is not one-to-one, we can sometimes restrict the

domain to create a function that is, and then determine an inverse. The restriction

we use is arbitrary, and only requires that the result produce all possible range

values. Most often, we simply choose a limited domain that seems convenient or

reasonable.

EXAMPLE 4

䊳

Restricting the Domain to Create a One-to-One Function

Given , restrict the domain to create a one-to-one function, then

ﬁnd . State the domain and range of both resulting functions.f

1

1x2

1x2 1x  42

1

1x2

2  x

, x  2, y 1

2  x

 y

x  y12  x2

x  2y  xy

y  1 xy  x  2y

x 

y  1

y 

x  1

g1x2

x  1

, x  1, y  2

1

1x2 x

 5, x 僆⺢, y 僆⺢

1

1x2 x

 5  f

1

1x2

 5  y

 y  5

x  1

y  5

y  1

x  5

f 1x2 1

x  5, x 僆⺢, y 僆⺢

g1x2

x  1

f 1x2 1

x  5

1

1x2

5–5 Section 5.1 One-to-One and Inverse Functions 483

College Algebra Graphs & Models—

WORTHY OF NOTE

If a function is not one-to-one,

no inverse function exists since

interchanging the x- and

y-coordinates will result in a

nonfunction. For instance, inter-

changing the coordinates of ( , 4)

and (2, 4) from results in

(4, ) and (4, 2), and we have one x-

value being mapped to two y-values,

in violation of the function deﬁnition.

2

y  x

2

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484 CHAPTER 5 Exponential and Logarithmic Functions 5–6

College Algebra Graphs & Models—

Solution

䊳

The graph of f is a parabola, opening upward with the

vertex at (4, 0). Restricting the domain to

(see ﬁgure) leaves only the “right branch” of the

parabola, creating a one-to-one function without

affecting the range, . For

with restricted domain , we have

given function

use

instead of

(

)

interchange

and

take square roots

solve for

, use since

The result shows , with domain and range

(the domain of f becomes the range of , and the range of f becomes

the domain of ).

Now try Exercises 47 through 52

䊳

We can further illustrate the ideas from Example 4 using a calculator’s ability to draw

inverses. On the screen, set and this function on a

size of for x and for y. Then go to the home screen and access

the DrawInv feature using (DRAW) 8:DrawInv. This will place the

DrawInv feature on the home screen, where we specify that we want the inverse of Y

(Figure 5.7). Pressing returns us to the graph, where we discover that since

is not one-to-one, the calculator has graphed the inverse relation (Figure 5.8).y  x

ENTER

PRGM

2nd

33, 7437.5, 7.54

WINDOW

GRAPH

 X

1

y 僆 34, q2

x 僆 30, q2f

1

1x2 1x  4

x  41x 1x  4  y

1x

 y  4

x  1y  42

y  1x  42

f 1x2 1x  42

x  4

1x2 1x  42

y 僆 30, q2

x  4

2

3

7.5

7.5

Figure 5.7

Figure 5.8

3

7.5

7.5

Figure 5.9

Figure 5.10

Returning to the screen and restricting the domain of Y

to (Figure 5.9), then

repeating the sequence above, produces the graphs shown in Figure 5.10. Note the given

function is now one-to-one, and its inverse is also now a function (and one-to one).

x  0

cob19545_ch05_479-492.qxd 9/2/10 9:21 PM Page 484

5–7 Section 5.1 One-to-One and Inverse Functions 485

College Algebra Graphs & Models—

While we now have the ability to ﬁnd the inverse of a function, we still lack a de-

ﬁnitive method of verifying the inverse is correct. Actually, the diagrams in Figures 5.5

and 5.6 suggest just such a method. If we use the function f itself as an input for ,

or the function as an input for f, the end result should simply be x, as each function

“undoes” the operations of the other. From Section 3.6 this is called a composition of

functions and using the notation for composition we have,

Verifying Inverse Functions

If f is a one-to-one function, then the function exists, where

EXAMPLE 5

䊳

Finding and Verifying an Inverse Function

Use the algebraic method to ﬁnd the inverse function for Then

verify the inverse you found is correct.

Solution

䊳

Since the graph of f is the graph of shifted 2 units left, we know f is one-

to-one with domain and range This is important since the

domain and range values will be interchanged for the inverse function. The domain

of will be and its range

given function;

use

instead of

(

)

interchange

and

solve for

(square both sides)

subtract 2

the result is ;

Verify

䊳

is an input for

adds 2 to inputs, then takes the square root

substitute for

simplify

✓

since the domain of is )

(

) is an input for

squares inputs, then subtracts 2

substitute for

(

)

simplify

✓

result

Now try Exercises 53 through 78

䊳

D. The Graph of a Function and Its Inverse

Graphing a function and its inverse on the same axes reveals an interesting and useful

relationship—the graphs are reﬂections across the line (the identity function).

Consider the function , and its inverse . In

Figure 5.11, the points (1, 5), (0, 3), ( , 0), and ( ) from f (see Table 5.3) are

graphed in blue, with the points (5, 1), (3, 0), (0, ), and ( ) (see Table 5.4) from

graphed in red (note the x- and y-values are reversed). Graphing both lines illus-

trates this symmetry (Figure 5.12).

1

5, 4

4, 5

1

1x2

x  3



x 

f 1x2 2x  3

y  x

 x

 x  2  2

1x  2  32x  24

 2

1

 3f 1x24

 2

1

ⴰ f 21x2 f

1

3f 1x24

x 僆 30, qf

1

1x2  x

 2x

1

1x2x

 2  21x

 22 2

 2f

1

1x2 2

1

1x2 1f ⴰ f

1

21x2 f3 f

1

1x24

y 僆 32, q2x 僆 30, q2,f

1

1x2 f

1

1x2 x

 2

 2  y

 y  2

x  1y  2

y  1x  2

x 2 f 1x2 1x  2

y 僆 32, q2.x 僆 30, q2f

1

y 僆 30, q2.x 僆 32, q2

y  1x

f 1x2 1x  2.

1f ⴰ f

1

21x2 x

and

1

ⴰ f 21x2 x

1

You’ve just seen how we

can ﬁnd inverse functions using

an algebraic method

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486 CHAPTER 5 Exponential and Logarithmic Functions 5–8

College Algebra Graphs & Models—

Figure 5.11Table 5.3

xf(x)

54



Table 5.4

⫺1

(x)

45



y  x

f(x)  2x  3

542135 4 3 2 1

5

4

3

2

1

(x) 

x  3

542135 4 3 2 1

5

4

3

2

1

y  x

f(x)  2x  3

1

(x) 

x  3

Figure 5.12

EXAMPLE 6

䊳

Graphing a Function and Its Inverse

Given the graph shown in Figure 5.13, draw a graph of the inverse function.

Solution

䊳

From the graph, the domain of f appears to be and the range is

This means the domain of will be and the range will be . To

sketch , draw the line interchange the x- and y-coordinates of the

selected points, then plot these points and draw a smooth curve using the domain

and range boundaries as a guide. The result is shown in Figure 5.14.

Now try Exercises 79 through 84

䊳

A summary of important concepts is provided here.

Functions and Inverse Functions

1. If the graph of a function passes the horizontal line test, the function is

one-to-one.

2. If a function f is one-to-one, the function exists.

3. The domain of f is the range of , and the range of f is the domain of .

4. For a function f and its inverse , and

5. The graphs of f and are symmetric to the line .

These ideas can be illustrated and reafﬁrmed using a graphing calculator. To begin,

enter the functions and (which appear to be inverse

functions) on the screen, along with . Press 6:ZStandard, then

5:ZSquare to obtain a screen that is in perspective. The graphs seem to be

reﬂections across the line (Figure 5.15). To verify, we can use the TABLEy  x

ZOOM

 X



 2Y

 21

X  2

y  xf

1

ⴰ f 21x2 x.1f ⴰ f

1

21x2 xf

1

y  x,f

1

y 僆⺢x 僆 10, q2f

1

y 僆 10, q2.x 僆⺢

(2, 4)

(0, 1)

(1, 2)

f(x)

55

5

(4, 2)

(1, 0)

(2, 1)

f(x)

55

5

1

(x)

Figure 5.14Figure 5.13

cob19545_ch05_479-492.qxd 9/2/10 9:21 PM Page 486

feature. As shown in Figure 5.16, the points

and (3, 2) are on the graph of Y

and the points (0, 2), and (2, 3) are on

the graph of Y

. While this seems convincing

(the x- and y-coordinates are interchanged), a

graphing calculator can actually compose the

functions to help verify the inverse relationship.

After entering and

on the screen (deactivat-

ing Y

, Y

, and Y

), the TABLE in Figure 5.17

now shows that one function indeed “undoes”

the other, leaving . See Exercises 85 through 88.Y

 Y

 x

 Y

1X22

 Y

1X22

12, 12,

11, 22, 12, 02

5–9 Section 5.1 One-to-One and Inverse Functions 487

College Algebra Graphs & Models—

D. You’ve just seen how

we can graph a function and

its inverse

10

15

Figure 5.17

Figure 5.15

Figure 5.16

E. Applications of Inverse Functions

Our ﬁnal example illustrates one of the many ways that inverse functions can be applied.

EXAMPLE 7

䊳

Using Volume to Understand Inverse Functions

The volume of an equipoise cylinder (height equal to diameter)

is given by (since ), where v(x)

represents the volume in units cubed and x represents the radius

of the cylinder.

a. Find the volume of such a cylinder if .

b. Find , and discuss what the input and output variables represent.

Solution

䊳

a. given function

substitute 10 for

, exact form

With a radius of 10 ft, the volume of the cylinder would be .

given function

use

instead of

(

)

interchange

and

solve for

result

The inverse function is . In this case, the input x is a given

volume, the output is the radius of an equipoise cylinder that will hold

this volume.

Now try Exercises 91 through 96

䊳

1

1x2

1

1x2

2␲

 y

2␲

 y

x  2␲y

y  2␲x

v1x2 2␲x

2000␲ ft

 1000  2000␲

v1102 2␲1102

v1x2 2␲x

1

1x2

x  10 ft

h  d  2rv1x2 2␲x

E. You’ve just seen how

we can solve applications of

inverse functions

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488 CHAPTER 5 Exponential and Logarithmic Functions 5–10

College Algebra Graphs & Models—

5.1 EXERCISES

2. If every line intersects the graph of a

function in at most point, the function is

one-to-one.

䊳

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. A function is one-to-one if each coordinate

corresponds to exactly ﬁrst coordinate.

3. A certain function is deﬁned by the ordered pairs

, and (4, 19). The inverse

function is .

1⫺2, ⫺112, 10, ⫺52, 12, 12

4. To ﬁnd using the algebraic method, we (1) use

instead of f(x), (2) x and y,

(3) for y and replace y with .f

⫺1

1x2

⫺1

5. State true or false and explain why: To show that g

is the inverse function for f, simply show that

Include an example in your

response.

1f ⴰ g21x2⫽ x.

6. Discuss/Explain why no inverse function exists for

and . How would

the domain of each function have to be restricted to

allow for an inverse function?

g1x2⫽ 24 ⫺ x

f 1x2⫽ 1x ⫹ 32

䊳

DEVELOPING YOUR SKILLS

Determine whether each graph given is the graph of a

one-to-one function. If not, give examples of how the

deﬁnition of one-to-oneness is violated.

7. 8.

9. 10.

11. 12.

54321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

54321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

54321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

54321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

54321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

13. 14.

Determine whether the functions given are one-to-one.

If not, state why.

15.

16.

17.

18.

Determine if the functions given are one-to-one by noting

the function family to which each belongs and mentally

picturing the shape of the graph. If a function is not one-to-

one, discuss how the deﬁnition of one-to-oneness is violated.

19. 20.

21. 22.

23. 24.

25. 26.

For Exercises 27 to 30, ﬁnd the inverse function of the

one-to-one functions given.

27.

28. g1x2⫽ 51⫺2, 302, 1⫺1, 112, 10, 42, 11, 32, 12, 226

f 1x2⫽ 51⫺2, 12, 1⫺1, 42, 10, 52, 12, 92, 15, 1526

y ⫽⫺2xy ⫽ 3

r1t2⫽ 2

t ⫹ 1 ⫺ 2s1t2⫽ 12t ⫺ 1 ⫹ 5

p1t2⫽ 3t

⫹ 5h1x2⫽⫺

冟

x ⫺ 4

冟

⫹ 3

g1x2⫽ 1x ⫹ 22

⫺ 1f 1x2⫽ 3x ⫺ 5

51⫺6, 22, 1⫺3, 72, 18, 02, 112, ⫺12, 12, ⫺32, 11, 326

51⫺6, 12, 14, ⫺92, 10, 112, 1⫺2, 72, 1⫺4, 52, 18, 126

519, 12, 1⫺2, 72, 17, 42, 13, 92, 12, 726

51⫺7, 42, 1⫺1, 92, 10, 52, 1⫺2, 12, 15, ⫺526

54321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

54321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

cob19545_ch05_479-492.qxd 12/1/10 5:51 PM Page 488

5–11 Section 5.1 One-to-One and Inverse Functions 489

College Algebra Graphs & Models—

29. v(x) is deﬁned by the ordered pairs shown.

30. w(x) is deﬁned by the ordered pairs shown.

Find the inverse function using diagrams similar to

those illustrated in Example 2. Check the result using

three test points.

31. 32.

33. 34.

35. 36.

37. 38.

State the domain and range of f(x), then use the

algebraic method to ﬁnd the inverse function and state

its domain and range. Finally, ﬁnd any three ordered

pairs (a, b) that satisfy f, and verify the ordered pairs

(b, a) satisfy the equation for .

39. 40.

41. 42.

43. 44.

45. 46.

The functions given in Exercises 47 through 52 are not

one-to-one. (a) Determine a domain restriction that

preserves all range values and creates a one-to-one

function, then state the new domain and range. (b) State

the domain and range of the inverse function and ﬁnd

its equation.

47. 48.

49. 50.

51. 52. q1x2⫽

1x ⫺ 22

⫹ 1p1x2⫽ 1x ⫹ 42

⫺ 2

V1x2⫽

⫹ 2v1x2⫽

1x ⫺ 32

g1x2⫽ x

⫹ 3f 1x2⫽ 1x ⫹ 52

f 1x2⫽

x ⫹ 2

1 ⫺ x

f 1x2⫽

x ⫹ 1

1x2⫽

x ⫺ 1

f 1x2⫽

x ⫹ 2

1x2⫽ x

⫺ 2f 1x2⫽ x

⫹ 1

1x2⫽ 2

x ⫹ 3f 1x2⫽ 2

x ⫺ 2

ⴚ1

s1x2⫽ 2

x ⫹ 2t1x2⫽ 2

x ⫺ 4

g1x2⫽ 5x ⫺ 2f 1x2⫽ 4x ⫹ 3

r1x2⫽

p1x2⫽⫺

g1x2⫽ x ⫺ 4f

1x2⫽ x ⫹ 5

For each function f(x) given, prove (using a composition)

that g(x) ⫽ f

⫺1

(x).

53.

54.

55.

56.

57.

58.

59.

60.

Find the inverse of each function f(x) given, then prove

(by composition) your inverse function is correct. Note the

domain and range of f in each case is all real numbers.

61. 62.

63. 64.

65. 66.

67. 68.

69. 70.

71. 72.

The functions given in Exercises 73 through 78 are one-

to-one. State the implied domain of each function given,

and use these to state the domain and range of the

inverse function. Then ﬁnd the inverse and prove by

composition that your inverse is correct.

73. 74.

75. 76.

77. 78.

Determine the domain and range for each one-to-one

function whose graph is given, and use this information

to state the domain and range of the inverse function.

Then sketch in the line , estimate the location of

two or more points on the graph, and use this

information to graph (x) on the same grid.

79. 80.

54321

⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

54321⫺5⫺4⫺3⫺2⫺1

⫺1

⫺2

⫺3

⫺4

⫺5

ⴚ1

y ⴝ x

w1x2⫽ x

⫺ 1; x ⱖ 0v1x2⫽ x

⫹ 3; x ⱖ 0

q1x2⫽ 41x ⫹ 1

p1x2⫽ 21x ⫺ 3

g1x2⫽ 12x ⫺ 5f 1x2⫽ 13x ⫹ 2

f 1x2⫽

1x ⫹ 32

⫺27

f 1x2⫽

1x ⫺ 12

1x2⫽ 2

3x ⫺ 2f 1x2⫽ 2

2x ⫹ 1

f 1x2⫽ x

⫺ 4f 1x2⫽ x

⫹ 3

1x2⫽

x ⫹ 1f 1x2⫽

x ⫺ 3

1x2⫽

x ⫹ 4

f 1x2⫽

x ⫺ 5

1x2⫽ 5x ⫹ 4f 1x2⫽ 3x ⫺ 5

1x2⫽ x

⫹ 8; x ⱖ 0, g1x2⫽ 1x ⫺ 8

f 1x2⫽ x

⫺ 3; x ⱖ 0, g1x2⫽ 1x ⫹ 3

f 1x2⫽

x ⫹ 6, g1x2⫽

x ⫺

f 1x2⫽

x ⫺ 6, g1x2⫽

x ⫹ 9

1x2⫽ 2

x ⫺ 4, g1x2⫽ x

⫹ 4

1x2⫽ 2

x ⫹ 5, g1x2⫽ x

⫺ 5

1x2⫽ 3x ⫺ 4, g1x2⫽

x ⫹ 4

1x2⫽⫺2x ⫹ 5, g1x2⫽

x ⫺ 5

⫺2

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