relations and functionsmathsbooks.net/jacplus books/11 advance general/ch03 relations and... · 3a...

3a Relations 3b Functions 3c Inverse functions3

71Chapter 3 relations and functions

Sketching relations in the Cartesian plane • from descriptions, equations or formulas and identifying their key features

Sketching the graph of an inverse function from • the graph of a simple function

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relations and functions

relationsA relation is a set of ordered pairs. The following are examples of relations in listed notation.

A = {(1, 2), (2, 3), (3, 4), (4, 5)} and B = {(1, 1), (1, 2), (2, 1), (3, 2)}

The set of fi rst elements from each ordered pair is the domain. A typical domain element is often represented by x.

The set of second elements from each ordered pair is the range. A typical range element is often represented by y.

For relation A, the domain = {1, 2, 3, 4} and the range = {2, 3, 4, 5}. Alternatively, using set builder notation, the domain = {x: x ∈ Z, 1 ≤ x ≤ 4}, which reads as ‘the set of x-values, such that x is an integer between 1 and 4 inclusive’, and the range = {y: y ∈ Z, 2 ≤ y ≤ 5}, which reads ‘the set of y-values, such that y is an integer between 2 and 5 inclusive’.

For relation B, the domain = {1, 2, 3} and the range = {1, 2}.Some relations have a rule that relates the domain elements with the range elements. For

relation A, the rule is y = x + 1 and using set builder notation A = {(x, y): x ∈ Z, y = x + 1}, which reads as ‘the set of coordinate pairs, (x, y), such that the domain is an element of integer numbers and the rule for y is y = x + 1’. Note that a relation may be uniquely defi ned using the domain and rule.

For relation B, there is no obvious rule, so listing is the only method for representing this relation.

Graphs of relationsWorked exAMPle 1

Graph the following relations and state the range.a {(x, y): x ∈ Z +, y = x + 1} b {(x, y): x ∈ [−1, 2), y = x2}c {(x, y): x2 + y2 = 4}

3A

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Worked example 1

72

Think WriTe

a 1 The graph of y = x + 1 is linear. The domain of the relation is x ∈ Z +, which is all positive integers. Create a table of values, starting with the smallest value of x. A small number of points will suffice.

a The smallest positive integer is 1.

x x + 1 y1 1 + 1 22 2 + 1 33 3 + 1 44 4 + 1 5

2 Determine the range. The range is y ≥ 2, where y ∈ Z +.

3 Draw a pair of axes, showing an appropriate scale and plot the points. Do not join the dots as x ∈ Z +. Place an arrow on the last point to indicate that the pattern continues in that direction.

x

y

y = x + 1

20 4 6

2

4

6

b 1 The graph of y = x2 is parabolic. The domain of the relation is x ∈ [−1, 2), so the graph is continuous over the interval. The square bracket indicates that there should be a closed circle at x = −1 and the round bracket indicates that there should be an open circle at x = 2. Create a table of values to determine the points on the graph.

b x x2 y−1 (−1)2 1

0 (0)2 0

1 (1)2 1

2 (2)2 4

2 Draw a pair of axes, showing an appropriate scale and plot the points. Join the dots.

x

y

y = x2

2 31−10 4 5

21

345

3 Determine the range. The smallest y-value is 0 and the highest is 4, which has an open circle and is therefore not included.

The range is: y ∈ [0, 4).

c 1 x2 + y2 = 4 is a circle with centre (0, 0) and a radius of 2. The domain is not explicitly stated and therefore x ∈ [−2, 2] is implied. Draw a pair of axes, showing an appropriate scale and sketch the circle as described.

c

x0

y

x2 + y2 = 4

−2

2

−2 2

2 Determine the range. The range is: y ∈ [−2, 2].

A relation is a set of ordered pairs.1. The domain of a relation is the set of first elements of a set of ordered pairs.2. The range of a relation is the set of second elements of a set of ordered pairs.3.

reMeMBer

Maths Quest 11 Advanced General Mathematics for the Casio ClassPad


relations 1 For each of the following relations, state the domain and range.

a {(2, 3), (4, 7), (6, 8)} b {(2, 1), (3, 1), (4, 1)}c {(2, 2), (2, 3), (2, 5)} d {(1, 2), (1, 3), (2, 2), (3, 2)}e {(x, y): x ∈ Z +, y = 2x − 1} f {(x, y): x ∈ Z, x ≥ 5, y = −x}g {(x, y): x ∈ Z, 4 ≤ x < 21, y = x − 4} h {(x, y): x ∈[2, 7), y = x + 1}i {(x, y): x ∈[2, 4), y = x2} j {(x, y): x ∈(−1, 1), y = x2}k {(x, y): x ∈(−2, −1), y = x2} l {(x, y): y = x2 + 3}m {(x, y): y = −x2 + 3} n {(x, y): x2 + y2 = 9}o {(x, y): (x − 2)2 + y2 = 4} p {(x, y): x ∈ [0, 3), x2 + y2 = 9}

2 We1 Sketch the graph of each of the following relations. State the domain and range.a {(1, 2), (1, 3), (2, 2), (3, 2)} b {(2, 2), (2, 3), (2, 5)}c {(x, y): x ∈ Z, x ≤ 5, y = −x + 2} d {(x, y): x ∈[−2, 3], y = −x2}e {(x, y): y = (x − 1)2 + 3} f {(x, y): y = −2(x − 1)2 − 2}g {(x, y): x ∈ (−2, 1], y = −2(x + 1)2 + 3} h {(x, y): (x − 2)2 + (y + 1)2 = 9}i {(x, y): (x + 2)2 + (y − 2)2 = 5} j {(x, y): x ∈[0, 2), x2 + y2 = 9}k {(x, y): x ∈[0, 1), (x − 2)2 + (y + 1)2 = 9} l {(x, y): y ∈[0, 3), x2 + y2 = 9}m {(x, y): x ∈[0, 2), y ≥ 0, x2 + y2 = 9} n {(x, y): y ≥ 3, (x + 2)2 + (y − 2)2 = 9}

3 For each of the following graphs, state the domain and range.

a b c

x

y

10

5

−5

−10

−5−10 5 10 x

y

10

5

−5

−10

−5−10 5 10 x

y

10

5

−5

−10

−5−10 5 10

d e f

x

y

10

5

−5

−10

−5−10 5 10 x

y

10

5

−5

−10

−5−10 5 10 x

y

10

5

−5

−10

−5−10 5 10

g h i

x

y

10

5

−5

−10

−5−10 5 10 x

y

10

5

−5

−10

−5−10 5 10 x

y

10

5

−5

−10

−5−10 5 10

exerCise

3A

74

j k l

x

y

10

5

−5

−10

−5−10 5 10 x

y

10

5

−5

−10

−5−10 5 10 x

y

10

5

−5

−10

−5−10 5 10

4 MC The domain and range of ( )( )

xy+ + − =1

34

122

are, respectively:

a [−1, 1] and [−2, 2] b [−2, 0] and [1, 5] c [−2, 0] and [−5, −1]d [0, 2] and [1, 5] e [0, 2] and [−5, 1]

5 MC The domain of ( ) ( )x y− + + =1

42

161

2 2

is:

a R\(−1, 3) b R\(−1, 3]c R\[−1, 3) d R\[−1, 3]e R\{−1, 3}

FunctionsA function is a relation that does not repeat the fi rst element in any of its ordered pairs. That is, for any x-value, there is only one y-value. So, a function is a relation that is either one-to-one (1–1) or many-to-one (many–1).

A one-to-one function has one x-value for a given y-value, whereas a many-to-one function has more than one x-value for a given y-value.

In general terms, a relation can be A–B. To test for A, you can use a horizontal-line test. If the line crosses the graph once, A = 1; if it crosses more than once, A = many. To test for B, you can use a vertical-line test; again, if the line crosses once, B = 1, and if it crosses more than once, B = many. For a relation to be a function, B must always equal 1. Graphs of one-to-one and many-to-one functions are shown below.

y

Vertical linetest (B = 1)

Horizontalline test(A = 1)

x

11

1−1 function

y

Vertical linetest (B = 1)

Horizontalline test(A = 2)

x1

21

Many−1 function

A graph of a relation also represents a function if a vertical line does not cut the graph more than once.

We have seen that listing or set builder notation can be used to state a function, but there is an additional mapping notation for functions only.

f: X → Y, f (x) = .............. For this mapping notation, f is the label for the mapping or function.

X is always clearly expressed as the domain. Y, the co-domain, is a set that is large enough to contain the range, so the range ⊆ co-domain. f (x) or y is the image of x under the mapping. f (x) = ............... or y = f (x) is the rule for f.

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Worked exAMPle 2

Explain why each of the following is an invalid use of mapping notation.

a f R R f xx

: , ( )→ = 1

b f: R+ → R+, f (x) = x − 1

c f: R → R, f (x) = ± x

Think WriTe

a The domain is not specifi ed. a The function not defi ned when x = 0 and this needs to be stated clearly in the mapping notation.

b The range is not a subset of the co-domain. b f12

12

=

−. The range ⊄ co-domain.

c f (x) is not a function. c The graph is not a function since it is one-to-many. Mapping notation cannot be used if the relation is not a function.

Worked exAMPle 3

Find the range for the following functions.a f: R+ → R, f (x) = 4x − 1b f: R → R, f (x) = −x2 − 4x + 5c f: R → R, f (x) = 2x − 1

Think WriTe

a 1 f (x) = 4x − 1 is linear. The domain is x ∈ R+ or x ∈ (0, ∞).

a When x = 0, f (0) = 4(0) − 1 = −1

2 f (0) = −1, but (0, −1) is not included and therefore this lower end of the range must be represented using a round bracket. State the range.

The range: y ∈ (−1, ∞).

b 1 f (x) = −x2 − 4x + 5 is an inverted parabola over the set of real

numbers. Use xba

=−

2 to

determine the x-value of the turning point, as this can be used to indicate the maximum y-value of the graph.

b x =

=

− −

−

−

( )4

22

2 Substitute this x-value into f (x) to determine the maximum y-value.

f (−2) = −4 + 8 + 5 = 9

3 State the range. The range: y ∈ (−∞, 9].

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Worked example 3

76

c 1 c

2 Use the graph and the information described above to state the range.

The range is: y ∈ (−1, ∞).

Worked exAMPle 4

If f: R → R, f (x) = 2x2 − 4x + 1, finda f (x2) b f (2x + 1)

Think WriTe

a To find f (x2), substitute x2 for x and simplify.

a f (x) = 2x2 − 4x + 1f (x2) = 2(x2)2 − 4(x2) + 1

= 2x4 − 4x2 + 1b To find f (2x + 1), substitute

2x + 1 for x and simplify.b f (x) = 2x2 − 4x + 1

f (2x + 1) = 2(2x + 1)2 − 4(2x + 1) + 1= 2[4x2 + 4x + 1] − 8x − 4 + 1= 8x2 + 8x + 2 − 8x − 4 + 1= 8x2 − 1

reMeMBer

A function is a relation that does not repeat the first element in any of its ordered pairs. 1. That is, for any x-value, there is one y-value.A function can be one-to-one or many-to-one.2. The graph of a function cannot be crossed more than once by any vertical line.3.


Use a CAS calculator to draw this graph. f (x) = 2x − 1 is an exponential graph over the set of real numbers. As x → −∞, 2x → −1, y = −1 is an asymptote.

Alternatively, on the Main screen, complete the entry line as:Define f (x) = 2x2 − 4x + 1f (x2)f (2x + 1)Press E after each entry.


Mapping notation for functions: 4. f: X → Y, f (x) = ..............., where X is the domain, Y is the co-domain (range ⊆ Y) and f (x) is the image of x.When required, the range can be determined from the domain and the rule.5.

Functions 1 We2 Explain why the following are an invalid use of mapping notation.

a f R R f x x: , ( )→ = + 5 b f R R f xx x

x: , ( )→ = +2

c f: [1, 5] → R+, f (x) = 4 − x

d f R R f x x: , ( )+ ±→ = e f: R+ → R+, f (x) = x3 − 3x f f: R → R, x = 2

2 We3 Find the range for the following functions.a f: [2, 5) → R, f (x) = 3x − 2 b f: R → R, f (x) = x2 + 2x + 3c f: R → R, f (x) = 3x + 2 d f: (2, 4] → R, f (x) = 3 − 2xe f: R → R, f (x) = −2x2 + 8x + 1 f f: R → R, f (x) = 4 − 2x

g f: (1, 3) → R, f (x) = x2 + 4x + 2 h f: (1, 3) → R, f (x) = x2 − 4x + 2i f: [1, 3) → R, f (x) = −2x + 2 j f: R → R, f (x) = 3−x + 2

k f R f x x: [ , ) , ( )4 5∞ → = + l f R R f xx

: \{ } , ( )12

1→ =

−

m f R f xx

: ( , ) , ( )− ∞ → =−

21

2n f R R f x

x: , ( )→ =

+

−1

2 1

o f R R f xx x

x: \{ } , ( )0

2

→ = +

3 For f (x) = x2 − 5x + 6, find:a the factors of f (x) and the solutions of f (x) = 0 b We4 f (x2 − 1) in its simplest form.

4 The solutions of f (x) = 0 are 2 and 6. Find the solutions of:a f (x2 − x) = 0 b f (2x − 2) = 0.

5 MC Which of the following is a valid use of mapping notation?

a f R R f xx

: , ( )→ =−1

1b f R R f x

x: , ( )+ +→ =

+−1

12

c f R f x x: [ , ) , ( )4 4∞ → = −±

d f: R → R, x = 2

e f R R f xx

: , ( )→ =+

+ 1

12

6 MC The ranges for f R f x x: [ , ) , ( )4 5∞ → = + and f R f xx

: ( , ) , ( )−∞ → =−

21

2 are:

a [3, ∞), (−∞, 0) b [ , ), ( , )5 0∞ ∞− c [3, ∞), (2, ∞)

d [ , ), ( , )5 0∞ ∞ e [ , ), ( , )5 2∞ ∞

inverse functionsUnder a mapping f, X maps onto Y. Under an inverse mapping f −1, Y maps onto X. The following statements result:1. f −1 can only exist if f is a one-to-one function.2. Domain f −1 = range f.3. Range f −1 = domain f.4. If x and y are interchanged, the rule for f −1 is obtained from the rule for f.

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Inverse functions

78

5. If x and y are interchanged, the graph of f −1 can be obtained from the graph of f by reflecting in the line y = x.

x

y = x

f (x)f −1(x)

y

−2

−1−3 1

1

2

3

2 3−2 −1

−3

x

y = xy

1 2 3

(2, −1)

(−1, 2)

−3 −2 −1

−2

−1

2

1

3

−3

It is possible to consider inverse relations. For example, the rule for the inverse of y = x2, a many-to-one function is x = y2, a one-to-many relation. However, if we are to start with a function f and finish with a function f −1, then f has to be a one-to-one function.

Worked exAMPle 5

If f: (−1, 2] → R, f (x) = 2x + 4, find the domain, range and rule of f −1(x), and sketch the graphs of f and f −1 on the same set of axes.

Think WriTe

1 First, determine if the inverse function, f −1(x), exists.

f (x) is linear and a 1−1 function, so f −1(x) exists.

2 Determine the range of f (x). The domain is x ∈ (−1, 2], so substitute the end values of x to determine the range.

f (−1) = −2 + 4 = 2f (2) = 4 + 4 = 8

Therefore, the range of f (x) is y ∈ (2, 8].

3 The domain of f (x) = the range of f −1(x). The range of f (x) = the domain of f −1(x).State the domain and range of f −1(x).

Domain f (x): x ∈ (−1, 2]⇒ range of f −1(x) is y ∈ (−1, 2]Range of f (x): y ∈ (2, 8]⇒ Domain of f −1(x) is (2, 8]

4 To determine the rule of f −1(x), let f (x) = y and interchange x and y. Then make y the subject.

Let y = 2x + 4⇒ x = 2y + 4⇒ −2y = 4 − x

⇒ y x= −12

2

5 Fully define the rule for the inverse function. f R f x x− −→ = −1 12 812

2: ( , ] , ( )

6 Sketch the graphs over the required domains, showing the line y = x.

y

y = x

x−2−4−6−8

2

68

f(x) = 2x + 4

f −1(x) = x − 24

−2−4−6−8 02

4 6 8

12



Worked exAMPle 6

If f: (−∞, −1] → R, f (x) = x2 + 2x + 2, fi nd the domain, range and rule of f −1(x), and sketch the graphs of f and f −1 on the same set of axes.

Think WriTe

1 First, determine if the inverse function, f −1(x), exists. Since f (x) = x2 + 2x + 2 is an upright parabola, it is necessary to locate the

turning point using xba

=−

2.

x = =−

−22

1

Since the turning point occurs at x = −1, and the domain is x ∈ (−∞, −1], f (x) is a 1–1 function and f −1(x) exists.

2 Determine the range of f (x). The domain is x ∈ (−∞, −1], so substitute the end value of x to determine the range (this value is at the turning point).

x = −1⇒ f (−1) = 1 − 2 + 2

= 1The point (−1, 1) is the minimum point on the graph.

3 The domain of f (x) = the range of f −1(x). The range of f (x) = the domain of f −1(x).State the domain and range of f −1(x).

Domain f (x): x ∈ (−∞, −1]⇒ range of f −1(x) is y ∈ (−∞, −1]Range of f (x): y ∈ [1, ∞)⇒ Domain of f −1(x) is [1, ∞)

4 To determine the rule of f −1(x), let f (x) = y and interchange x and y. Then make y the subject.

Let y = x2 + 2x + 2⇒ y = (x + 1)2 + 1Interchange x and y⇒ x = (y + 1)2 + 1⇒ x − 1 = (y + 1)2

⇒ = ± − −y x 1 1

5 Since f (x) = x2 + 2x + 2, and x ∈ (−∞, −1] (left side of the parabola), then f −1 should be

y x= − −−

1 1. Fully defi ne the rule for the inverse function.

f R f x x− − −∞ → = − −1 11 1 1: [ , ) , ( )

6 Sketch the graphs over the required domains, showing the line y = x.

−1

y

x

−2−3−4−5

123

54

−2−3−4−5 10 2 3 4 5

y = x

−1

f −1(x) = −1−

f(x) = x2 + 2x + 2

√x − 1

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Worked example 6

80

7

The inverse function of 1. f, f −1, exists if f is a one-to-one function.For 2. f −1, domain = range of f, range = domain of f and the rule is found by interchanging x and y in the rule for f.To graph 3. f −1 from the graph of f, or f from the graph of f −1, refl ect in the line y = x.

reMeMBer

inverse functions 1 We5 For each of the following functions f, determine the domain, range and rule of f −1.

Sketch the graphs of f and f −1 on the same set of axes.a f: [0, ∞) → R, f (x) = 3x − 2 b f: [1, 3) → R, f (x) = 2x − 4c f: (−1, 2] → R, f (x) = −2x + 4 d f: (−∞, 2] → R, f (x) = 2x + 1e f: (−2, 2] → R, f (x) = −2x − 2 f f: [−4, 2) → R, f (x) = −x − 3

2 We6 For each of the following functions f, determine the domain, range and rule of f −1. Sketch the graphs of f and f −1 on the same set of axes.

a f: (1, ∞) → R, f (x) = x2 − 2x + 2 b f: (−∞, −2]→ R, f (x) = x2 + 4x + 5

c f: (1, ∞) → R, f (x) = −x2 + 2x + 3 d f R f x x x: [ , ) , ( )12

22 2 3∞ → = + +−

e f R f x x: ( , ) , ( )− ∞ → = +2 2 f f R f x x: ( , ) , ( )12 2 1 1∞ → = − +

g f R f x x: ( , ) , ( )− −∞ → = + +2 2 1 h f R f x x: ( , ) , ( )− −

∞ → = − +3 3 2

3 For each of the following functions f, determine if f −1 exists. For those that exist, find the domain, range and rule of f −1.

a f R R f xx

: \{ } , ( )01→ = b f R R f x

x: \{ } , ( )0

12

→ =

c f R R f xx

: \ , ( )12

12 1

{ } → =−

d f R f xx

: ( , ) , ( )( )

21

2 2∞ → =

− 4 MC For f: [1, ∞) → R, f (x) = 2x2 − 4x + 4, the domain and rule of f −1 are:

a [ , ), ( )2 2 12

1∞ = − −−f x x b [ , ), ( )3 1 22

1∞ = + −−f x x c [ , ), ( )2 2 11∞ = − −−f x x

d [ , ), ( )3 1 12

1∞ = − −−f x x e [ , ), ( )2 1 22

1∞ = + −−f x x

exerCise

3C

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To view the graph and its inverse on a CAS calculator, open the Graph & Tab screen. Complete the entry lines as:y1 = x2 + 2x + 2 | x ≤ −1

y2 = −1 − x −1y3 = xTick each of the equations and tap $.


suMMAry

Relations

A relation is a set of ordered pairs.•The domain of a relation is the set of first elements of a set of ordered pairs.•The range of a relation is the set of second elements of a set of ordered pairs.•

Functions

A function is a relation that does not repeat the first element in any of its ordered pairs. That is, for any •x-value, there is one y-value.A function can be one-to-one or many-to-one.•The graph of a function cannot be crossed more than once by any vertical line.•Mapping notation for functions: • f : X → Y, f (x) = ............... where X is the domain, Y is the co-domain (range ⊆ Y) and f(x) is the image of x.When required, the range can be determined from the domain and the rule.•

Inverse functions

The inverse function of • f, f −1, exists if f is a one-to-one function.For • f −1; domain = range of f, range = domain of f and the rule is found by interchanging x and y in the rule for f.To graph • f −1 from the graph of f, or f from the graph of f −1, reflect in the line y = x.

82

ChAPTer revieW

shorT AnsWer

1 For each of the following relations, state the domain and range. Sketch the graph.a {(2, 3), (2, 4), (3, 3)}b {(x, y): x ∈ Z, y = −x}c {(x, y): x ∈ Z +, y = 2x}d {(x, y): (x − 1)2 + (y − 2)2 = 4}

e {( , ):( ) ( )

}x yx y− + − =1

42

161

2 2

f {( , ):( ) ( )

}x yx y− − − =1

42

161

2 2

2 For the graphs of the relations below, state the domain and range.

a

x

y

10

5

−5

−10

−5−10 5 10

b

x

y

10

5

−5

−10

−5−10 5 10

c

x

y

10

5

−5

−10

−5−10 5 10

d

x

y

10

5

−5

−10

−5−10 5 10

e

x

y

10

5

−5

−10

−5−10 5 10

f

x

y

10

5

−5

−10

−5−10 5 10

3 Explain why each of the following are an invalid use of the mapping notation.

a f R R f xx

x: , ( )→ = +2

2

1

b f : R+ → R+, f (x) = x2 − 1

c f R R f x: , ( )→ = ±2 3

4 Find the range for each of the following functions.a f : [3, 4) → R, f (x) = x − 4b f : R → R, f (x) = (x − 4)(x − 2)c f : R → R, f (x) = e x − 4

d f R f x x: [ , ) , ( )4 5∞ → = +

e f R f x x: [ , ] , ( )− → = −2 2 4 2

f f : R → R, f (x) = −2x2 + 4x − 1

g f R R f xx

: \{ } , ( )23

21→ =

−+



h f : (2, ∞) → R, f (x) = log2(x)

i f R R f xx

: , ( )→ =+

1

12

exAM TiP Make sure you state the range taking the domain restriction into account.

5 For f (x) = x2 − 3x + 2:a find the factors of f (x) and the solutions of

f (x) = 0b find f (x2 + 1), the factors of f (x2 + 1) and the

solutions of f (x2 + 1) = 0.

MulTiPle ChoiCe

1 The ranges of ( , ): [ , ],x y xx y∈ + =

−2 04 9

12 2

and

( , ):x yx y2 2

4 91− =

are:

a [−3, 3], [−3, 3] b [−3, 3], R c [−3, 0], Rd [0, 3], R e [−3, 0], [−3, 3]

2 The implied domains for {(x, y): x = 2y} and

{( , ): }x y y x= −25 2 are:a (0, ∞), [−5, 5]b [0, ∞), (−∞, −5] ∪ [5, ∞)c R, [−5, 5]d [0, ∞), [−5, 5]e (0, ∞), (−∞, −5] ∪ [5, ∞)

3 Which of the following is not a proper use of mapping notation?a f : [2, 3] → [3, 4], f (x) = x + 1b f : (2, 3) → [3, 4], f (x) = x + 1

c f f x x: ( , ] ( , ], ( )2 3 7 812

6→ = +

d f : [2, 3] → R−, f (x) = −x + 1e f : [2, 3] → R+, f (x) = −x + 3

4 The ranges of f R R f x x: , ( )+ −→ = + 4 and

f: R → R, f (x) = x2 + 2x + 1 are:a (−∞, 2), [0, ∞) b (−∞, −2), [0, ∞)c (−∞, 0], [0, ∞) d (−∞, 2), R+

e (−∞, −2), R+

5 If under the mapping f : R− → R,

f xx

( ) =−

1

12, and f x( ) = 1

3, then x is:

a − 98

b 124

c −2

d 2 e ±2

6 Which of the following equations represents a function for which an inverse function exists?

a ( , ): [ , ],x y xx y∈ + =

0 24 9

12 2

b ( , ): [ , ],x y yx y∈ + =

0 34 9

12 2

c ( , ): [ , ),x y xx y∈ ∞ − =

24 9

12 2

d ( , ): [ , ),x y yx y∈ ∞ − =

04 9

12 2

e ( , ): [ , ), ( , ],x y x yx y∈ ∞ ∈ ∞ − =

−2 0

4 91

2 2

7 For f : (−∞, 2] → R, f (x) = x2 − 4x + 1, the domain

and rule of f −1 are:

a ( , ], ( )− −∞ = − +2 2 31f x x

b [ , ], ( )− −∞ = − +3 2 31f x x

c [ , ], ( )− −∞ = + +3 2 31f x x

d ( , ), ( )3 2 31∞ = − −−f x x

e ( , ), ( )3 2 31∞ = + −−f x x

exTended resPonse

1 For each of the following functions f, find the domain, range and rule of f −1. Sketch the graphs of f and f −1 on the same set of axes.a f : [0, ∞) → R, f (x) = 2x + 1b f : (−2, 4] → R, f (x) = −2x + 1c f : [1, ∞) → R, f (x) = x2 − 2x + 3d f : (−∞, 0) → R, f (x) = x2 − 2x − 1

e f R f x x: [ , ) , ( )2 2∞ → = −f f R f x x: ( , ] , ( )− −

∞ → = −1 1

g f R f x x: [ , ] , ( )− → = −3 0 2 19

2

h f R f x x: [ , ) , ( )+ ∞ → = −3 29

12

exAM TiP Make sure you write the equation of the asymptotes.

84

2 The perimeter for a new seal enclosure is to have a maximum side length of 8 m.The width is to be twice the length (x).a Draw a diagram of the enclosure and label the sides.b Defi ne a rule that gives the perimeter, P, of the new enclosure.c What is the largest value that x can be?d State the domain and range.e Write in function notation the rule for the perimeter.f Defi ne a function for the area of the enclosure, A(x).g If the maximum area allowed is 18 m2, fi nd the dimension of the enclosure.

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Digital docTest YourselfChapter 3



eBookpluseBookplus ACTiviTies

chapter openerDigital doc

10 Quick Questions: Warm up with ten quick •questions on relations and functions. (page 71)

3a RelationsTutorial

We1 • int-1036: Watch how to graph relations and state their range. (page 71)

Digital doc

WorkSHEET 3.1: Use set notation to identify •the domain and range of relations and functions. (page 74)

3b FunctionsTutorial

We3 • int-1037: Watch how to determine the determine the range for functions given in function notation. (page 75)

3c Inverse functionsInteractivity

Inverse functions • int-0970: Consolidate your understanding of graphs of inverse functions. (page 77)

Tutorial

We6 • int-1038: Watch how to fi nd the domain and range of an inverse function. (page 79)

Digital doc

WorkSHEET 3.2: Identify types of relations and •their domain and range. (page 80)

chapter reviewDigital doc

Test Yourself: Take the end-of-chapter test to test •your progress. (page 84)

To access eBookPLUS activities, log on to

www.jacplus.com.au

86

exAM PrACTiCe 1 ChAPTers 1 To 3

shorT AnsWer 20 minutes

1 Simplify 3 72 4 12 300+ − . 2 marks

2 Sketch the graph of f : [−3, 3] → R, where f (x) = [2x + 1]. 3 marks

3 Let f : R → R, where f (x) = x – 2 and let

g: R\{2} → R, where g xf x

( )( )

= 1.

Sketch the graphs of f (x) and g(x) on the same set of axes. Label all key features. 4 marks

4 Show that if z1 = a + bi and z2 = c − di, then z

z1

2

equals ac bd bc ad i

c d

− + ++( )

2 2. 3 marks

MulTiPle ChoiCe 10 minutes

Each question is worth 1 mark.

1 If z1 = 4 − 3i and z2 = 1 + 2i, then z1z2 equals:a 5 − ib 4 − 6ic 2 − 5id 10 + 5ie −2 + 5i

2 Point A (x, y) is translated by a

b

and then reflected

in the y-axis to the image B. The coordinates of point B are:a (−x + a, y + b)b (−x − a, y + b)c (x + a, −y + b)d (x + a, −y − b)e (−x − a, −y − b)

3 The points z1 and z2 are shown on the Argand diagram below.

Re (z)

Im (z)

z2z1

M

L

K

HJ

Which point represents z1 − z2?a H b J c Kd L e M

4 The graph of {(x, y): y = | x − 2 |} is transformed by a dilation of factor 2 from the x-axis. The image of this transformation can be represented by the rule:a {(x, y): y = | 2x − 2 |}

b ( , ):x y y x= −

12

2| |

c ( , ):x y y x= −

12

2

d {(x, y): y = | x − 4 |}e {(x, y): y = 2 | x − 2 |}

5 The graph of f is shown below.

x

y

Which one of the following represents f −1?

a

x

y b

x

y

c

x

y d

x

y

e

x

y


87exam practice 1

exTended resPonse 30 minutes

1 Graphic designer Rhonda has been contracted by Wacky World to design a new logo for their T-shirts. Her first design is shown below.

To allow Rhonda to have the fl exibility to change her designs, she determines a function that models her design. For her design, 1 cm represents 1 unit on the Cartesian plane.

x

y

y1y2

B

C

1 cm

O A

Graph 1

She fi nds that part of her design follows the rule y = | 2x − 3 |.a For the rule y1 = | 2x − 3 |, determine the coordinates of: i A ii B iii C 1 + 1 + 2 = 4 marksThe second part of the design is defi ned by y2, which is determined by transforming y1 onto the image of y2 as shown in Graph 1 above. b Describe a transformation that maps the graph of y1 to y2. 1 mark c Use set notation to determine the rule for y2. 3 marks d In the context of this problem, determine the feasible domain and range. Write your answer using

function notation. 1 + 1 = 2 marks

2 Rhonda inserts another ‘W’ in the design. The second ‘W’ is smaller than the first. Rhonda uses the method of finding a rule to describe part of the ‘W’ and then applies transformations to complete the design. Using Graph 1, her new design for ‘W’ will include points AB′C′. a The coordinates of B′ are (0, 1). Describe a transformation that maps B to the image B′. 2 marksThe second ‘W’ includes the original point A.b Determine a rule that describes the line that passes through points A and B′. 3 marks

c The point C′ lies on the line found in part c. If C′ has coordinates x,43

, fi nd the exact value of x.

2 marksd Rhonda is going to place the second smaller ‘W’ above the fi rst ‘W’. The position of B′ will become

(−1, 3). Write down in matrix form, the combination of transformations that map B′ to its new position. 2 marks

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