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CHAPTER
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Functions SPM Topical Analysis
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ONCEPT MAP Domain Domain
RELATIONS HUBUNGAN Types of Relations Jenis Hubungan • One-to-one • Many-to-one • One-to-many • Many-to-many
Codomain Kodomain FUNCTIONS FUNGSI
Objects Objek Images Imej Range Julat
Absolute Value Functions Fungsi Nilai Mutlak f : x→|g(x)|
Graphs of Absolute Value Functions Graf Fungsi Nilai Mutlak The graph of a linear absolute value function has a V shape COMPANION WEBSITE
Learning Objectives
Composite Functions Fungsi Gubahan The function of f followed by g is gf.
Inverse Functions Fungsi Songsang Given that y = f(x), then f – 1(y) = x.
Problems that involve Composite Functions and Inverse Functions Masalah melibatkan Fungsi Gubahan dan Fungsi Songsang
1.1 Relations 1 1.1a Representation of a Relation
1 CHAPTER
4
1
F O R M
CHAPTER
1 A relation from set A to set B is the linking (or pairing) of the elements of set A to the elements of set B. 2 A relation between two sets can be represented by (a) an arrow diagram, (b) ordered pairs, (c) a graph. F O R M
14
A relation from set A = {12, 14, 23, 25, 43} to set B = {3, 5, 7} is defined by ‘sum of digits of’. Represent the relation by (a) an arrow diagram, (b) ordered pairs, (c) a graph. Solution (a) Arrow diagram A
‘sum of digits of’ B
12 14
3
23
5
25
7
43
(b) Ordered pairs {(12, 3), (14, 5), (23, 5), (25, 7), (43, 7)} (c) Graph 7 Set B 5 3 12 14 23 25 43 Set A Try: Question 1, Self Assess 1.1.
1.1b Domain, Codomain, Objects, Images and Range In a relation between set (A) and another set (B), • the first set (A) is known as domain, • the second set (B) is known as codomain, • the elements in the domain are known as objects, • the elements in the codomain that are linked to the objects are known as images, • the set of images is known as range. Functions
2
Facts A set is a well-defined collection of objects. For example: Universal set, = {x : 10 x 30, x is an integer} Set A = {Factors of 36} Set B = {Prime numbers} Set C = {Numbers where the sum of the digits is 3} The list of elements of each of the sets , A, B and C is as follows: = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} A = {12, 18} B = {11, 13, 17, 19, 23, 29} C = {12, 21, 30}
3 2 A relation from set P = {16, 36, 49, 64} to set Q = {4, 6, 7, 8, 11} is defined by ‘factor(s) of’. (a) Represent this relation using an arrow diagram. (b) State (i) the domain, (ii) the codomain, (iii) the images of 36, (iv) the images of 64, (v) the object of 7, (vi) the objects of 4, (vii) the range of this relation. Solution (a) The arrow diagram that represents the given relation is as shown in the next column.
P
‘factor(s) of’
Q
16
4
36
6
49
7
64
8
Range
11 Domain
Non-image
Codomain
1
1
CHAPTER
CHAPTER
(b) (i) The domain is {16, 36, 49, 64}. F 8, 11}. (ii) The codomain is {4, 6, 7, O (iii) The images of 36 are 4 and 6. R (iv) The images of 64 are 4 and 8. M (v) The object of 7 is 49. (vi) The objects of 4 are 16, 436 and 64. (vii) The range is {4, 6, 7, 8}.
Try: Question 2, Self Assess 1.1.
1.1c Types of Relations
3 State the type of relation shown by each of the following arrow diagrams. (a) (b)
Types of Relations 1 One-to-one relation Each object in the domain has only one image in the codomain. 2
Many-to-one relation There are more than one object in the domain that have the same image in the codomain.
3
One-to-many relation There is at least one object in the domain that has more than one image in the codomain.
4
Many-to-many relation There is at least one object in the domain that has more than one image in the codomain and there is at least one element in the codomain that is linked to more than one object in the domain.
A
‘reciprocal of’
1 3 1 4 1 5
3 Domain
Codomain
4
5
Domain
P
B
‘factors of’
Q 2
10
3
21
5 7
Solution (a) One-to-one relation (b) One-to-many relation
Codomain
Try: Question 3, Self Assess 1.1.
4
Domain
State the type of relation shown by (a) the ordered pairs: {(4, 8), (4, 10), (4, 12), (6, 10), (7, 12)} (b) the graph:
Codomain
Set Y M P
Solution Domain
5
7
9
12 15
Set X
Codomain
Draw arrow diagrams to represent the relations because it is easier to interpret arrow diagrams compared to ordered pairs or graphs.
3
Functions
F O R M 4
(a)
4
8
6
10
7
12
(b)
∴ The above arrow diagram shows many-to-many relation. Try: Question 4, Self Assess 1.1.
P M
SPM Clone
’07
F O The following ordered pairs represent a relation R from set A = {3, 5, M 7} to set B = {9, 15, 21}.
1
1
SPM Clone
’06
A
CHAPTER
4
CHAPTER
F O R M
Y
∴ The above arrow diagram shows many-to-one relation.
1 1
X 5 7 9 12 15
{(3,4 9), (5, 15), (7, 21)}
(a) State the type of the above relation. (b) State the range. (c) Using function notation, write down a relation between set A and set B. Solution
B
3
9
5
15
7
21
Based on the arrow diagram, (a) the relation is a one-to-one relation. (b) the range is {9, 15, 21}. (c) the function notation is f(x) = 3x. Each image is three times its objects.
It is better to draw an arrow diagram to represent the above relation because it is easier to interpret an arrow diagram compared to ordered pairs. Try: Question 5, Self Assess 1.1.
1.1 (iv) the objects of 12, (v) the range of the relation. 3 State the type of relation shown by each of the following arrow diagrams. (a) M N ‘capital of’
1 Represent each of the following relations by (a) an arrow diagram, (b) ordered pairs, (c) a graph. (i) The relation ‘remainder when divided by 5’ from set A = {6, 12, 18, 24} to set B = {1, 2, 3, 4}. (ii) The relation ‘factors of’ from set P to set Q where P = Q = {2, 3, 4, 8}. (iii) The relation ‘difference of digits of’ from set X = {14, 21, 23, 34, 48} to set Y = {1, 3, 4}. 2 A relation from set P = {26, 34, 45, 62} to set Q = {12, 20, 24, 48} is defined by ‘product of digits of’. (a) Represent the relation by an arrow diagram. (b) State (i) the domain, (ii) the codomain, (iii) the image of 45, Functions
Sarawak
Kuching
Perak
Ipoh
Kedah
(b)
P methane water carbon dioxide
Alor Setar
‘elements of’
Q carbon hydrogen oxygen
4 State the type of relation of (a) the ordered pairs: {(4, 0), (4, 1), (5, 2), (6, 7), (6, 8)} 4
(b) the graph:
Set B
{(2, 1), (6, 3), (10, 5)}
11 10 8 3 4 6 7 9
(a) State the type of the above relation. (b) State the range. (c) Using function notation, write down a relation between set P and set Q.
Set A
5 The following ordered pairs represent a relation from set P = {2, 6, 10} to set Q = {1, 3, 5}.
1.2 Functions
a
x
b
y
c
z
a
w
b
x
c
y
d
z
1.2a To Recognise Functions 5 Explain whether each of the following relations is a function. (a) (b) (c) ‘has’
B
January
28 days
June
30 days
August
31 days
M ‘three times of’ N
P ‘consists of ’ Q Bronze Brass
1 3 5 7
cuprum stanum zinc
1 CHAPTER
4
Facts
3 The main properties for a relation to be considered a function are: (a) Each and every object must have one and only one image. [If there are objects that have two or more images, the relation is not a function.] (b) Two or more objects are allowed to have the same image. [Many-to-one relation is considered a function.] (c) It is not necessary that all elements in the codomain are images. [Non-image elements are allowed in a function.] 4 If a function links x to x 2 – 3x + 5, by using function notation, it is written as f : x → x2 – 3x + 5 or f(x) = x2 – 3x + 5. It is read as ‘function f that maps x onto x2 – 3x + 5’.
A
mapping.
1
5
F O Facts R M known as Function is also
CHAPTER
1 A function is a special type of relation where each and every object in the domain has only one image. 2 The types of relations that are considered as functions are: (a) one-to-one relation (b) many-to-one relation
3 9 15
Solution (a) It is a function because • each and every object has only one image, • many-to-one relation is allowed, • non-image element, i.e. ‘28 days’ is allowed. (b) It is not a function because there are objects that have more than one image. (c) It is not a function because there is an element in the domain that does not have an image, i.e. ‘7’.
The types of relations that are not considered as functions are: (a) one-to-many relation x
a b Facts
y z
[Not a function because element ‘a’ has more than one image.] (b) many-to-many relation a
x
b
y
c
z
[Not a function because element ‘b’ has more than one image.] (c) relation where there are elements in the domain that do not have an image a b c d
x y z
[Not a function because element ‘d’ does not have an image.]
Try: Question 1, Self Assess 1.2.
5
Functions
F O R M 4
1.2b To Solve Problems involving Functions 6 (b)
When the function f has an image 4, it means that f(x) = 4.
When f(x) = 4, x 2 – 4x – 1 = 4 x 2 – 4x – 5 = 0 (x + 1)(x – 5) = 0 x = –1 or 5 Hence, the objects that have the image 4 are –1 or 5.
Object
1
F When x = 2, O R f(2) = 22 – 4(2) –1 M = 4 – 8 – 1
CHAPTER
4
CHAPTER
F O R M
1
Given the function f : x → x2 – 4x – 1, find (a) the image of 2, (b) the objects that have the image 4. Solution (a) f : x → x 2 – 4x – 1 f(x) = x 2 – 4x – 1
= –5 74 of 2 is –5. Hence, the image Try: Questions 2–3, Self Assess 1.2.
7 The arrow diagram below shows the function 15 f : x → ———— . ax + b x
f
–3
f(–4) = –3 15 Substitute x = –4. –––––––– = –3 –4a + b 12a – 3b = 15 4a – b = 5 ........... 2 1 – 2 : –a = –2 a = 2 From 1 , 3(2) – b = 3 b = 3 (b)
15 ax + b –3
–4 –5
(a) Find the value of a and of b. (b) State the value of x such that the function f is undefined.
A function f is undefined when its denominator is zero.
Solution 15 (a) f : x → —— —— ax + b 15 f(x) = —— —— ax + b f(–3) = –5 15 ––––––– = –5 Substitute x = –3. –3a + b 15a – 5b = 15 3a – b = 3 ........... 1
15 f(x) = — — — —— 2x + 3 When the denominator 0, 2x + 3 0 x – –3– 2 x –1–1– 2 Hence, the value of x such that the function f is undefined is –1–1– . 2
Try: Questions 4–8, Self Assess 1.2.
Functions
6
8
9
Press 0.8
SHIFT
4
1
R M
(b) g(x) = 1.6 2 cos x = 1.6 10 1.6 cos x = ––– 2 cos x = 0.8 x = 36.87°
CHAPTER
Solution 3x + k , (a) For f : x → ––––––– 2x – 4 8 Denominator ≠ 0 2x – 4 ≠ 0 x≠2 It is given that x ≠ p ∴p=2 (b)
1
(a) Given the function f(x) = 3 sin x – tan x, find the image of 30°. (b) Given the function g(x) = 2 cos x, find the value of x such that the function g has the image 1.6, for the domain 0° x 90°. Solution (a) f(x) = 3 sin x – tan x f (30°) = 3 sin 30° – tan 30° = 3(0.5) – 0.5774 F = 0.9226 O Hence, the image of 30° is 0.9226. CHAPTER
3x + k for all A function f is defined by f : x → –––––– 2x – 4 values of x, except x = p and k is a constant. (a) State the value of p. (b) Given that the value 5 is mapped onto itself under f, find (i) the value of k, (ii) another value of x that is mapped onto itself.
cos
=
Answer display: 36.86989765
Self-mapping is given by f(x) = x, where both the object and the image have the same value.
The value of x such that the function g has the image 1.6 is 36.87°.
(i) 5 is mapped onto itself, thus f(5) = 5 3(5) +— k = 5 —— —— 2(5) – 4 15— +— k = 5 — — 6 15 + k = 30 k = 15 (ii) For self-mapping, f(x) = x 3x + —–——15 — = x 2x – 4 3x + 15 = 2x 2 – 4x 2 2x – 7x – 15 = 0 (2x + 3)(x – 5) = 0 x = – –3– or 5 2
Try: Question 11, Self Assess 1.2.
1.2c Domain, Range, Objects and Images of a Function
10 The arrow diagram represents the function f : x → 2x 2 – 5. State f x (a) the domain, (b) the range, 2 (c) the image of 0, 1 (d) the objects of 0 (i) 3, –1 –2 (ii) –3. Solution (a) Domain = {–2, –1, 0, 1, 2} (b) Range = {–5, –3, 3} (c) The image of 0 is –5. (d) (i) The objects of 3 are 2 and –2. (ii) The objects of –3 are 1 and –1.
Hence, another value of x that is mapped onto itself, apart from 5, is –1–1– . 2
2x2 – 5 3
–3 –5
Try: Question 12, Self Assess 1.2.
Try: Questions 9–10, Self Assess 1.2.
7
Functions
F O R M 4
1.2 (a) the image of 10, (b) the object that has the image – –1– , 2 (c) the value of x such that the function f is undefined. 6 Given that f : x → f(3) = 4, find (a) the value of p and of q, (b) the values of x such that f(x) = 4–– x. 3 7 Given that g : x → a + bx, g(1) = –3 and g(–2) = 3, find (a) the value of a and of b, (b) the values of n if g(n2 + 1) = 5n – 6.
1 State whether each of the following relations is a function. (a) A
‘type of number’
B
6 14
multiple of 3
15
multiple of 7
24
(b)
P
Q
‘consist of letters’
i v
1
1
F O Consonants R Vowels M
(c)
CHAPTER
4
CHAPTER
F O R M
w u
4
M ‘product of digits of’ N
21
2
32
6
53
15
8 A function f is defined by f : x → 5x – 2. Find (a) the object that has the image 5, (b) the object that is mapped onto itself. a— . The 9 A function f is defined by f : x → —— b–x values 3 and 5 are mapped onto themselves under f. (a) Find the value of a and of b. (b) State the value of x such that the function f is undefined.
64
(d)
X
‘has’
Y
PERAK
2 vowels
KEDAH
3 vowels
SELANGOR
4 vowels
18 9 2 Given the function f : x → — , x ≠ –– , find 2x – 9 2 (a) the image of (i) 0, (ii) 12, (b) the object that has the image (i) 2, (ii) 6 . 3 Given that f : x → –––a––– , f(3) = –5 and x–b f(–5) = –1, find (a) the value of a and of b, (b) the value of x such that the function f is undefined. f
10 The arrow diagram x shows the function –1 f : x → px + qx 2. –2 Find (a) the value of p and of q, (b) the values of x that are mapped onto themselves.
px + qx2
–5 –16
11 (a) Given the function f: x → sin x + cos x, find the image of (i) 45°, (ii) 60°. (b) Given the function g: x → tan x, find the value of x such that the image is 1.527, if x is an acute angle. 12 The arrow diagram f x2 + 2 x represents the function 6 f : x → x 2 + 2. State (a) the domain, 3 (b) the range, 2 2 1 (c) the image of 0, 0 (d) the objects of –1 (i) 3, –2 (ii) 6.
b
ax + – x 4 The arrow diagram x shows the function 7 f : x → ax + –b– . x 2 Find –1 (a) the value of a and of b, –5 (b) the value of x such that the function f is undefined, (c) the object that has the image 7, apart from x = 2.
2x 5 The function g is defined by g : x → — — — —. x+m If g(5) = 3g(2), find the value of m. Hence, find
Functions
f
8
1.3
Absolute Value Functions
1 The absolute value of x is written as _x_ and it is read as ‘the modulus of x ’. 2 The definition of _x_ is
Facts _x_ is the numerical value of x, i.e. positive numbers are still positive but negative numbers will be changed to positive. For example, _5_ = 5 and _–5_ = 5.
x, if x 0 _x_ = –x, if x 0 3 The absolute value function is defined by
f(x), if f(x) 0 _ f(x)_ = –f(x), if f(x) 0
1 CHAPTER
4
SPM Clone
1
The graph of O a linear absolute value functionRhas a V shape. M CHAPTER
2 2
Facts F
’07
f(x) = 3 _4x – 5_ = 3 4x – 5 = ± 3
Given the function f : x → _4x – 5_, find (a) the image of (i) –1, (ii) 4, (b) the objects that have the image 3. Solution (a) f : x → _4x – 5_ f(x) = _4x – 5_ (i) f(–1) = _4(–1) – 5_ = _–9_ = 9 Hence, the image of –1 is 9. (ii) f(4) = _4(4) – 5_ = _11_ = 11 Hence, the image of 4 is 11. (b)
The first equation is 4x – 5 = 3 4x = 8 x = 2 The second equation is 4x – 5 = –3 4x = 2 x = –1– 2 Hence, the objects that have the image 3 are 2 or –1–. 2
For any _f(x)_ = k, there are two equations that can be formed, i.e. f(x) = k or f(x) = –k. Try: Questions 1–2, Self Assess 1.3.
3
SPM Clone
’08
Sketch the graph of each of the following absolute value functions. (a) f(x) = _x + 3_ for the domain – 4 x 1 (b) f(x) = _4x – 7_ for the domain 0 x 4 State the corresponding range of values of f(x).
The sketch of the graph of f(x) = _x + 3_ is as shown below. y 4 3
Solution (a) Prepare a table as shown below.
x
Range
2 1
–4 –3 –2 –1 0 1
–4 –3 –2 –1 O
1
x
Domain
f (x) 1 0 1 2 3 4 9
Functions
F O R M 4
3 Hence, the graph touches the x-axis at 1 –– , 0 . 4 Prepare a table as shown below.
The corresponding range of values of f(x) means the range from the smallest value of y to the largest value of y, based on the given domain.
x
0 1 2 3 4
f (x) 7 3 1 5 9 ∴ The corresponding range of values of f(x) is 0 f(x) 4. (b)
The sketch of graph of f(x) = _4x – 7_ is as shown below. y 8
1
F First of all, determine the point where the graph O touches the x-axis.
CHAPTER
4
CHAPTER
F O R M
1
10 6
Range
4
R M
2
At the x-axis, y = 0 Thus, _4x – 7_ = 0 4 4x – 7 = 0 4x = 7 x = –7 – = 1–3– 4 4
3 1– 4
O
1
2
3
4
x
Domain
∴ The corresponding range of values of f(x) is 0 f(x) 9.
Try: Question 3, Self Assess 1.3.
Facts
1.3 3 Sketch the graph of each of the following absolute value functions. (a) f(x) = _ x + 2_ for the domain –3 x 3 (b) f(x) = _ 2x – 5_ for the domain 0 x 7 (c) f(x) = _ 3 – 2x_ for the domain –3 x 4 (d) f(x) = _ 3x – 5_ for the domain –2 x 4 State the corresponding range of values of f(x).
1 Given the function f : x → _ x 2 – 4x – 3_ , find the image of (a) –3, (b) 0, (c) 2. 2 Given the function f(x) = _ 2 – 5x_ , find (a) the image of (i) 2, (ii) –2, (b) the objects that have the image 7.
1.4
Composite Functions
1.4a To Find Composite Functions 1 If f is a function which maps set A onto set B and g is a function which maps set B onto set C, then gf is a composite function of f followed by g which maps set A onto set C. gf
x A Functions
f
f(x) B
g
g[f(x)] = gf(x) C
10
Facts • fg(x) means f[g(x)]. • In general, fg ≠ gf. • f 2 = ff, f 3 = fff or ff 2 and so on.
11 11 The functions f and g are defined by f : x → 2x + 1 and g : x → x 2 – 2 respectively. Find (a) the value of fg(3) and of gf(–2), (b) the composite functions (i) fg (iii) f2 (ii) gf (iv) g 2, (c) the values of x if gf(x) = 23. Solution (a) fg(3) = f[g(3)] = f(32 – 2) = f(7) = 2(7) + 1 = 15 gf(–2) = g[2(–2) + 1] = g(–3) = (–3)2 – 2 =7 (b) f : x → 2x + 1 g : x → x2 – 2 f(x) = 2x + 1 g(x) = x 2 – 2 (i) fg(x) = f[g(x)] = f(x 2 – 2) Substitute the x in
(iii) f 2(x) = ff(x) = f(2x + 1)
↓
↓
= 2(x – 2) + 1 = 2x 2 – 3 ∴ fg : x → 2x 2 – 3 (ii) gf(x) = g[ f(x)] = g(2x + 1)
1
= (x – 2) – 2 = x 4 – 4x 2 + 4 – 2 = x 4 – 4x 2 + 2 ∴ g 2 : x → x 4 – 4x 2 + 2
CHAPTER
2
F O R Substitute the x in M g(x) = x2 – 2
1
2
with (2x + 1).
CHAPTER
= 2(2x + 1) + 1 = 4x + 3 ∴ f 2: x → 4x + 3 (iv) g2(x) = gg(x) = g(x 2 – 2)
f(x) = 2x + 1 2
Substitute the x in f(x) = 2x + 1
↓ 4 with (x2 – 2).
(c) gf(x) = 23 4x 2 + 4x – 1 = 23 4x 2 + 4x –24 = 0 x 2 + x – 6 = 0 (x – 2)(x + 3) = 0 x = 2 or –3
with (x2 – 2).
Substitute the x in g(x) = x2 – 2
↓
with (2x + 1).
= (2x + 1)2 – 2 = 4x 2 + 4x + 1 – 2 = 4x 2 + 4x – 1 ∴ gf : x → 4x 2 + 4x – 1
Try: Questions 1–5, Self Assess 1.4.
12 Given the function f : x →
x— +— 1 +1 = — x – 1 = ————— — x + 1 = ——— – 1 x – 1 x +— 1— +— x— ––— 1 —— x – 1 = ——————–— x+ 1— –— (x— ––— 1) — —— x – 1 2x = –– 2 = x
expression for each of the following functions. (a) f 2 (b) f 8 (c) f 9 Solution f:x→— x— +— 1 (a) x–1 x— +— 1 f(x) = — x–1 f 2(x) = ff (x) x— +— 1 = f — x – 1
11
Functions
F O R M 4
(b) f 8(x) = f 2f 2f 2f 2(x) = f 2f 2f 2(x)
(c) f 9(x) = ff 8(x)
f 8(x) = x f 2(x) = x
= f(x) x— +–––1 , x ≠ 1 = — x – 1
= f 2f 2(x) = f 2(x) = x
In the SPM marking scheme, it is not compulsory for students to write down x ≠ 1 in the answer.
1 CHAPTER
4
CHAPTER
F O R M
1
Try: Questions 6–7, Self Assess 1.4.
F O R M
1.4b To Find the Related Function Given the 4 Composite Function and One of the Functions 4
SPM Clone
’06
(b)
(a) The function f is defined by f : x → 2x + 4. Another function g is such that fg : x → 3x – 8. Find the function g. (b) The function f is defined by f : x → x – 2. Another function g is such that gf : x → 1 , x ≠ 11 . Find the function g. 11 – 3x 34
It is given that f : x → x – 2 and 1 gf : x → —–—— —. 11 – 3x g is outside. 1 gf(x) = —–—— — 11 – 3x Change each x to 1— g[f(x)] = —–— — the in u. 11 – 3x 1— g(x – 2) = —–— — 11 – 3x Let x – 2 = u, then x = u + 2 ∴ g(u) = —–——1——— — 11 – 3(u + 2) 1—— = —–——— — 11 – 3u – 6 1— = — — — — 5 – 3u ∴ g : x → 1— , x ≠ –5– 5 – 3x 3
Solution (a) Case where the function g that has to be determined is situated ‘inside’.
It is given that f : x → 2x + 4 and fg : x → 3x – 8. g is inside.
fg(x) = 3x – 8 f [g(x)] = 3x – 8 Substitute the x in f(x) = 2x + 4
↓
with g(x).
2g(x) + 4 = 3x – 8 2g(x) = 3x – 12 3x—— – –12 g(x) = — –— 2 3x – –12 ∴ g : x → ——— –— 2 Try: Questions 8–10, Self Assess 1.4.
Functions
Case where the function g that has to be determined is situated ‘outside’.
12
1.4c Further Examples 5 on Composite Functions SPM Clone
’07
It is given that f : x → hx + k and g : x → (x + 1)2 + 2. Thus, fg(x) = f[(x + 1)2 + 2] = h[(x + 1)2 + 2] + k = h(x + 1)2 + 2h + k
Given the functions f : x → hx + k, g : x → (x + 1)2 + 2 and fg : x → 2(x + 1)2 + 1, find (a) the value of g2(2), (b) the value of h and of k. Solution (a) g2(2) = gg(2) = g[(2 + 1)2 + 2] = g(11) = (11 + 1)2 + 2 = 146 (b)
1 CHAPTER
4
1
But it is given that fg(x) = 2(x F+ 1)2 + 1. Hence, by comparison, O R h = 2 and 2h + k = 1 M 2(2) + k = 1 k = –3 CHAPTER
5
13
The following function problem can be solved by making a comparison. Try: Questions 11–12, Self Assess 1.4.
13 Given the functions f(x) = 3x + 7 and fg(x) = 22 – 3x, find gf(x). Solution Find g(x) first. fg(x) = 22 – 3x f(g(x)) = 22 – 3x 3g(x) + 7 = 22 – 3x
3g(x) = 22 – 3x – 7 3g(x) = 15 – 3x 15 –—3x g(x) = — — — — 3 g(x) = 5 – x Hence, gf(x) = g(3x + 7) = 5 – (3x + 7) Substitute the x in g(x) = 5 – x
Substitute the x in f(x) = 3x + 7
↓
↓
with (3x – 7).
with g(x).
= 5 – 3x – 7 = –3x – 2
Try: Question 8, SPM Exam Practice 1 – Paper 2.
1.4 (c) f : x → x – 3 (a) f : x → x 2 – 1 g : x → 3x + 1 g : x →_ x_ (b) f : x → (x + 1)2 (d) f: x → 2 – x g g : x → 1 – 3x 1— : x → —— — x2 + 2 3 The functions f and g are defined by f : x → 4x – 3 and g : x → x + 1 respectively.
1 Given the functions f : x → _ 4 – 5x_ and pooooooo g : x → x – 2 , x 2, find the value of (a) fg(6) (c) f 2(0) (b) gf(2) (d) g2(27) 2 Find the composite functions fg and gf for each of the following pairs of functions f and g.
13
Functions
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10 Find the expressions for the functions g and h in each of the following. (a) f : x → 2x, fg : x → 4x – 12, 2x + 1 hf : x → ———— 2 2x –—, 1 (b) f : x → 2x – 2, gf : x → — —— 3 Facts fh : x → 2x 2 (c) f : x → x 2 – 2, fg : x → x 2 + 6x + 7, hf : x → 2x 2 – 7
6 Given the functions f : x → 5x – 7 and g : x → 2, x ≠ 0, find the expression for x each of the following functions. (a) fg (c) g6 2 (b) g (d) g7
11 The functions f and g are defined by f : x → 1 – x and g : x → px 2 + q respectively. If the composite function gf is given by gf : x → 3x 2 – 6x + 5, find (a) the value of p and of q, (b) the value of g 2(0). Facts 12 The functions f and f 2 are such that f : x → hx + k and f 2 : x → 9x + 16. (a) Find the values of h and the corresponding values of k. (b) Considering h 0, find the values of x such that f(x 2) = 8x.
x–1 , x ≠ –1, find the x+1 expression for each of the following functions. (a) f 2 (c) f 16 4 (b) f (d) f 17 8 The function f and the composite function fg are defined as follows. Find the function g. (a) f : x → x + 2, fg : x → 3x – 2 (b) f : x → 3x + 2, 7 If f is defined by f : x →
1.5
fg : x →
9 The function f and the composite function gf are defined as follows. Find the function g. (a) f : x → x + 1, gf : x → 3 , x ≠ 2 x–2 1 5 , x ≠ –1–– (b) f : x → ––, gf : x → x 10 10x – 1 (c) f : x → 3x + 2, gf : x → 9x 2 + 9x + 2
1
1
CHAPTER
4
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2x + 5 ,x≠2 x–2 (c) f : x → x 2 – 1, fg : x → x 2 + 4x + 3
Find the expressions for f 2 and g2. Hence, find the value of x such that (a) f = g, (b) f 2 = g2. 4 The functions f and g are defined by f : x → mx + 2 and g : x → kx – 3 respectively. If fg = gf, find the relation between m and k. If m = 5, find the value of x that satisfies each of the following equations. (a) f 2 = f (b) g2 = g 5 The functions f and g are defined by f : x → (x + 1)2 and g : x → x – 2 respectively. F Find O (a) the composite functions fg and gf, R M of x if (b) the values (i) fg = 9, (ii) gf =414, (c) the value of x if fg = gf.
Inverse Functions
Facts –1
1.5a To Find Inverse Functions 1 If f : x → y is a function that maps x onto y, its inverse function is denoted by f –1. Inverse function is a function that maps y back to x.
Facts
f
x
y
f –1
2 The main step to find the inverse function is ‘Let y = f(x), then x = f –1 (y)’ or ‘Let f –1(x) = y, then f(y) = x’. Functions
• (fg) = g–1f –1 and (gf )–1 = f –1g –1. • (f 2)–1 = (f –1)2. • ff –1(x) = f –1f(x) = x.
14
The conditions for the existence of inverse functions are (a) the function must be one-toone, (b) that every element in the codomain must be linked to an object.
6
SPM Clone
SPM Clone
SPM Clone
SPM Clone
’03
’05
’08
’10
Try: Question 1, Self Assess 1.5.
F O R M
14 Find the inverse function of f : x → Solution Method 1 1 f 2x : x → ——––— x+2 2x—––— 1 Let y = — x + 2 y(x + 2) = 2x – 1 yx + 2y = 2x – 1 yx – 2x = –1 – 2y x(y – 2) = –1 – 2y –1 – 2y x = —–––—–— y – 2 –(1 + 2y) x = —–––—–—––— –(2 – y) 1 + 2y x = ——–— 2 – y
2x – 1 , x ≠ –2. x+2
1 + 2y f –1(y) = ——–— ∴ 2 – y 1 + 2x f –1(x) = ——–— 2 – x
4
1
Solution Let y = f –1(3) f(y) = 3
CHAPTER
y +— 8 = 3 —— y – 6 y + 8 = 3(y – 6) y + 8 = 3y – 18 2y = 26 y = 13 ∴ f –1(3) = 13
1
Given the function f –1 : x → find the value of f –1(3).
CHAPTER
14
\ f –1 : Method 2 Let f –1(x) = y Thus, f(y) = x 2y 1 = x ——––— y+2 2y – 1 = x(y + 2) 2y – 1 = xy + 2x 2y – xy = 2x + 1 y(2 – x) = 2x + 1 2x—+–— 1 y = — 2 – x 2x—+–— 1 f –1(x) = — 2 – x
Rearrange, making x the subject. Cross multiplication. Expand the left-hand side. Group the in x together. Factorise the left-hand side. Making x the subject
In the SPM marking scheme, it is not compulsory for students to write x ≠ 2. Rearrange the formula, making y the subject.
\ f –1 : x → Try: Question 2, Self Assess 1.5.
15 Given the functions f(x) = 6 – 2x and g(x) = –x1– , x ≠ 0, find f –1g–1.
6 ––––— x ∴ f –1(x) = — 2 g(x) = –x1– Let g –1(x) = y Thus, g(y) = x –y1– = x y = –x1– ∴ g–1(x) = –x1–
Solution First of all, find the inverse functions f –1 and g–1. f(x) = 6 – 2x Let f –1(x) = y Thus, f(y) = x 6 – 2y = x 6 – x = 2y 6 ––––— x y = — 2 15
Functions
F O R M 4
6 – –x1– = —–——–— 2 6x ––—1 , x ≠ 0 = — — –— 2x
Hence, f –1g–1 (x)
= f –1 –x1–
Apply the composite function of g –1 followed by f –1.
Try: Questions 3–7, Self Assess 1.5.
SPM Clone
’04
F hx – 5 ———, x ≠ 2 and its Given the function O f : x → — x–2 R 2x + k inverse function M f –1: x → ————, x ≠ 3, find x–3 (a) the value of h4and of k, (b) the values of t such that f(t) = –4– t. 3
1
2x +— k. But it is given that f –1(x) = — —— x–3 Hence, by comparison, k = –5, h = 3.
CHAPTER
4
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1
7 7
f(t) = –4– t 3 3t – 5 4 — — — — = –– t t–2 3 3(3t – 5) = 4t (t – 2) 9t – 15 = 4t 2 – 8t 0 = 4t 2 – 17t + 15 0 = (4t – 5)(t – 3) t = ––5 or 3 4 (b)
Solution (a) Let f –1(x) = y Thus, f(y) = x hy – 5 — ——— = x y–2 hy – 5 = x(y – 2) hy – 5 = xy – 2x 2x – 5 = xy – hy 2x – 5 = y(x – h) 2x –— 5 y = — —— x – h 2x –— 5 ∴ f –1(x) = — —— x – h Try: Questions 8–9, Self Assess 1.5.
8 x
SPM Clone
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y
z g
f 1
Solution 5–— (a) f : y → py + q g : y → —–— 3y – q 5–— f(y) = py + q g(y) = —–— 3y – q f –3– = 1 g –3– = 10 2 2 5 3 –– p + q = 1 —–————— = 10 2 3 –3– – q 2 1 3p + 2q = 2 ... 5— = 10 —–— 9 –– – q 2
10
3 – 2
The above diagram shows the representation of the mapping of y onto x by the function f : y → py + q and the mapping of y onto z by q 5 the function g : y → ——–— , y ≠ –– . Find 3y – q 3 (a) the value of p and of q, (b) the function that maps x onto y, (c) the function that maps x onto z.
Functions
16
5—–— = 10 —–— 9 – 2q —–——— 2 10 —–——— = 10 9 – 2q 1 —–——— = 1 9 – 2q 1 = 9 – 2q 2q = 8 q = 4 From 1 : When q = 4, 3p + 2(4) = 2 p = –2
(c) Based on the diagram, the function that maps x onto z is gf –1(x). f
–1
g
x
gf
(b) The function that maps x onto y is f (x). It has been found that f(y) = –2y + 4. Let f –1(x) = w Thus, f(w) = x –2w + 4 = x 4 – x = 2w 4–––— x w = — 2 4–x ∴ f –1(x) = — — — — 2
F O R M 4
1
CHAPTER
1
gf (x)
4 –––— x = g — 2 5 ——–— —–——— = 4 – x –4 3 —––— 2 5 –—— = —–——— 12 – 3x –— 8 ———–— —— 2 = —–10 —— 4 – 3x 10 4 ∴ gf –1 : x → ,x≠ 4 – 3x 3
–1
–1
–1
CHAPTER
z
y
Try: Question 10, Self Assess 1.5.
1.5b To Find a Function Given its Inverse Function
1.5c Further Examples on Inverse Functions
The inverse of an inverse function, f –1(x), will give 9 f(x). us back the original function,
9 Given that g–1(x) =
16 The function f is defined by f (x) = 2x + 1 , x ≠ 1. x–1 If f –1(k) = –4– k, find the values of k. 3
SPM Clone
’11
2x + 1 1 , x ≠ , find g(x). 2x – 1 2
Solution 2x +— 1 Let y = — —— 2x – 1 y(2x – 1) = 2x + 1 2xy – y = 2x + 1 2xy – 2x = y + 1 x(2y – 2) = y + 1 y + 1 x = ——— — 2y – 2 y + 1 g(y) = ——— — 2y – 2 x+1 ∴ g(x) = ,x≠1 2x – 2
Solution 4 f –1(k) = –– k 3 If f –1 (x) = y, k = f –4– k then x = f (y). 3 4k 2 –– + 1 3 k = —————— 4k – 1 –– 3 8k +— 3 — —— 3 k = — — — — — — 4k –— 3 — —— 3 8k +— 3 ——— 3— k = ——— 3 4k – 3 8k +— 3 k = — —— 4k – 3
Rearrange, making x the subject.
Try: Question 11, Self Assess 1.5.
17
Functions
F O R M 4
4k 2 – 3k = 8k + 3 4k 2 – 11k – 3 = 0 (4k + 1)(k – 3) = 0 k = – –1– or 3 4
18 Given the function f : x → 5x – 2, find ( f 2)–1 in the same form. Solution Find f 2(x) first. f 2(x) = ff(x) = f(5x – 2) = 5(5x – 2) – 2 = 25x – 10 – 2 = 25x – 12
Try: Question 30, SPM Exam Practice 1 – Paper 1.
Next, find the inverse of f 2(x). Let ( f 2)–1 (x) = y Thus, f 2(y) = x If the question requires 25y – 12 = x us to state the answer 25y = x + 12 in the ‘same form,’ we x— +––12 have to express the y = — –— 25 answer in the form (f 2)–1 : x → ————— x + 12 x + 12 ( f 2)–1(x) = — —–––— 25 25 and not x— +––12 (f 2)–1 (x) = —————. x + 12 ∴ ( f 2)–1 : x → — –— 25 25
1
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F O R M
Given that g–1 4 2 – kx (x) 17= ———— and f(x) = 3x, find 4 (a) g(x) in of k, (b) the values of k such that fg(1) = – ––3 k. 2 Solution
The inverse of g–1 will give us back g.
2— –— kx — (a) g–1(x) = — 4 2— –— kx Let y = — — 4 4y = 2 – kx kx = 2 – 4y 2 – 4y x = ———— k 2 – 4y g(y) = ———— k 2 – 4x g(x) = ——— — k
Try: Question 21, SPM Exam Practice 1 – Paper 2.
Rearrange, making x the subject.
19 1 Given that f –1(x) = x + m , x ≠ –m and
g(x) = 3 – x, find 1 –— – –3x (a) the value of m if f(x) = — x— , (b) the values of k if ff –1(k 2 – 7) = g[(k + 2)2]. Solution (a) Let f –1(x) = y f(y) = x 1–— –––3y — — = x y 1 – 3y = xy 1 = xy + 3y 1 = y(x + 3) 1— y = —–— x + 3 1 ∴ f –1(x) = —–— — x + 3 1–––— . But it is given that f –1(x) = —–— x+m Hence, by comparison, m = 3.
fg(1) = – ––3 k 2 2 – 4(1) f ————–— = – ––3 k k 2 –2 f — — = – ––3 k k 2 –2 3 — — = – ––3 k k 2 –6 –3k = k 2 –3k 2 = –12 k 2 = 4 k = ± 2 (b)
Try: Question 20, SPM Exam Practice 1 – Paper 2.
Functions
18
(b) ff –1(k 2 – 7) = g[(k + 2)2] k 2 – 7 = 3 – (k + 2)2
k 2 – 7 = 3 – (k 2 + 4k + 4) k 2 – 7 = 3 – k 2 – 4k – 4 2k 2 + 4k – 6 = 0 k 2 + 2k – 3 = 0 (k + 3)(k – 1) = 0 k = –3 or 1
ff –1(x) = x is always true and it has to be memorised. Try: Question 22, SPM Exam Practice 1 – Paper 2.
1.5 (a) f –1 (b) g–1 (c) g–1f –1 Is ( fg)–1 = g–1f –1?
1 CHAPTER
O R M
1
(d) fg F –1 (e) ( fg) CHAPTER
1 Evaluate f –1(4) for each of the following functions. (a) f : x → 5 – 4x 5 (b) f : x → 6 – —, x x≠0 (c) f : x → 3x + 2 , x ≠ – 3 2x + 3 2 2 Find the inverse function of each of the following functions. (a) f : x → 7x – 4 3x –— 4 (b) g : x → — —–— 2 3 x≠0 (c) h : x → 9 – —, x (d) m : x → 2x + 2 , x ≠ 3 5x – 3 5 pooooooo (e) n : x → 2 – x – x, x 2
7 The function f is defined by f : x4→ 2x – 1. (a) Find the expressions for f 2 dan f –1. (b) Show that ( f –1)2 = ( f 2)–1. 8 Given the function f : x → 4x + h and its inverse kx +—, 5 find function f –1 : x → — –—— 4 (a) the value of h and of k, (b) the expression for f –1 f. x+p , x ≠ 5 and 9 Given the function f : x → x–5 qx + 6 its inverse function f –1 : x → —–––——, x–1 x ≠ 1, find (a) the value of p and of q, (b) the values of x for which f –1(x) = 8x.
4 3 Given the functions f : x → —, x ≠ 0 and x g : x → 2x + 3, find each of the following functions. (a) fg (d) g2 (g) f –1g–1 –1 (b) gf (e) f (h) g–1f –1 2 –1 (c) f (f) g
10
x
y
z
2
3x –—, 1 4 The function f is defined by f : x → — —–— x–2 x ≠ k, find (a) the value of k, (b) f 2, (c) f –1.
–1 –2
The diagram shows the representation of the mapping of y onto x by the function f : y → my + n and the mapping of y onto z by n m the function g : y → — y ≠ — . Find n – 2y 2 (a) the value of m and of n, (b) the function that maps x onto y, (c) the function that maps x onto z.
5 Given the functions f : x → 2 – 4x and g : x → 3 , x ≠ 1, find x–1 –1 (a) f (c) f –1g–1 (e) (gf )–1 –1 (b) g (d) gf Is (gf )–1 = f –1g–1? 6 Given the functions f : x → 1 – 2x and
3x 10 11 (a) Given that f –1(x) = ——+ —— —, find f(x). 4
x–— +— 2 x ≠ 2, find g : x →— , x–2
2x +—, 3 find g(x). (b) Given that g–1(x) = — —— x+4
19
Functions
F O R M 4
1 Paper 1 Short Questions Multiple-choice Questions 1.1 P
1
1
15 24
CHAPTER
4
1
CHAPTER
F O R M
Based on the above information, the relation from P to Q is defined by the ordered pairs {(2, 1), (2, 3), (6, 5), (6, 7)}. State (a) the images of 2, (b) the object of 7. [2 marks]
Relations
35 45
Q
F O R M
6 7 8
The arrow diagram shows the relation between set P and set Q. Based on the arrow diagram, state (a) the domain, (b) the range, (c) the codomain. 2 The arrow diagram shows the relation between set A and set B. A
B
1
6
e
8
’03
Functions
n
20
y z
Set P
30
m
40
k
Set Q
h
State the cant rangeFigure of the relation, 1.1 (a) Signifi
(b) the type of the relation. [2 marks]
6 In the diagram below, set Q shows the images of the elements of set P.
’06
3 –2
16
7 The following arrow diagram SPM Clone represents a linear function f.
’07
4
6
8
Set X
State (a) the relation in the form of ordered pairs, (b) the type of the relation, (c) the domain of the relation. [3 marks]
1.2
Functions
1.3
Absolute Value Functions
Set Q
(a) State the type of relation between set P and set Q. (b) Using function notation, write down a relation between set P and set Q. [2 marks]
Set P
2
81
2
Set P
P = {2, 6, 8} Q = {1, 3, 5, 7, 9}
Set Y
10
x
–3
The above graph represents the relation from set P = {2, 6, 8} to set Q = {5, 7, 9, 11}. State (a) the images of 6, (b) the objects of 7, (c) the range.
4
’10
Set Q 11 9 7 5 2 6 8
SPM Clone
SPM Clone
’04
SPM Clone
(a) Represent the given relation using ordered pairs. (b) State the type of the given relation. 3
5 The diagram below shows the relation between set P and set Q.
SPM Clone
a
5
8 The diagram shows the relation between set X and set Y in the form ’09 of a graph. SPM Clone
9
4
(a) State the value of k. (b) Using the function notation, express f (x) in of x. [2 marks]
x
f(x)
2 5 7 k
1 4 6 8
20
9
x
f
a bx – 2 10
4 1
1
The arrow diagram represents a , x ≠ k, the function f : x → bx – 2 where a, b and k are constants. Find (a) the value of a and of b, (b) the value of k.
1.1
Sig
1.1
Sig
m
x
19 Given the functions f : x → 3x – 7 and fg : x → x 2 + 1, find the function g. 20 The function f is defined by f : x → x – 2. Another function g is such that gf : x → x2 + 1. Find the function g. 21 Given the functions f(x) = 2x – 4 and fg(x) = 8x + 2, find gf(x).
x–h h 7
2
Find the value of h. [2 marks] 13 Given the function f : x → _x – 4_, SPM Clone find the values of x such that ’07 f (x) = 9. [2 marks] 14 The given diagram shows the graph SPM Clone of the function f(x) = |2x – 3| for the ’08 domain 0 ≤ x ≤ 4. y
22 Given the functions h(x) = 7x – 1 and the composite function ’04 hg(x) = 35x + 13, find (a) g(x), (b) the value of x when gh(x) = 4. [4 marks] SPM Clone
23 Given the function g: x → px + q and its composite function ’07 g2: x → 49x – 32, find the value of p and of q such that p > 0. [3 marks]
x ≠ –4, find g–1.
2 , x –3 x ≠ 3 and g(x) = 4x – 1, find 1 f –1g –– . 2 28 Given the functions f(x) =
29 Given the function f : x → 3x + h and its inverse Ffunction 2O f –1: x → kx – ––R, find the value of h 3 and of k. M 4
1
18 Given that f : x → 2x + m and f 2 : x → px + 6, find the value of m and of p.
4x + 1 , x+4
CHAPTER
12 The diagram shows the function x–h ’06 m : x → h , where h is a constant. SPM Clone
17 Given the function f : x → 4 – 5x, find f 2.
27 Given the function g : x →
1
11 Given the function g(x) = 2x – 9, find the possible values of x if g(x) = 4.
16 Given the functions g : x → 4 + x 2 and h : x → 2x – 4, find gh.
CHAPTER
10 (a) Sketch the graph of the absolute value function f (x) = 3 – x for the domain 0 x 4. (b) Hence, state the corresponding range of values of f(x).
24 30 Given the function f(x) = , px + q f(1) = 8, find (a) the value of p and of q, (b) the values of k if f –1 (k) = k. 31 Given the inverse function f –1(x) = 2x – 5, find (a) f(x), (b) the values of k if f –1 f (k) = k 2 – 12.
SPM Clone
32 x
f –1
4x + p
–2
24 Given the function f (x) = x + 4 and g(x) = tx – 6, find ’08 (a) f (6), SPM Clone (b) the value of t such that ’10 gf (6) = 24. [3 marks] SPM Clone
3
O
k
4
x
State (a) the value of k, (b) the range of values of f (x) corresponding to the given domain.[3 marks]
1.4
25 Given the function g: x → 4x – 1 and h: x → 8x, find ’09 (a) hg (x), (b) the value of x if hg(x) = 2g(x). [4 marks] SPM Clone
1.5 Composite Functions
15 The functions f and g are defined by f : x → 3x – 2 and g : x → x + 2 respectively. Calculate fg(5).
–6
The arrow diagram represents f –1(x) = 4x + p, where p is a constant. Find (a) the value of p, (b) f(x).
33 Given the functions f(x) =
2 , 3x + 1
Inverse Functions
4x + 3 , x+5 x ≠ –5, calculate the value of
26 Given the function f(x) = f –1(2).
21
(a) g–1 f(x), (b) the values of x which are mapped onto themselves under the function f. Functions
F O R M 4
x
f
g
y
4
34 Given that g : x → 4x – 1 and h : x → x 2 – 3x + 5, find ’03 (a) g –1(7), SPM Clone (b) hg(x). ’08 [4 marks]
1 CHAPTER
4
CHAPTER
F O R M
1
SPM Clone
m 35 Given the functions h : x → — – – 3, x SPM Clone 10 ’04 x ≠ 0 and h –1 : x → —— — , x ≠ –k, x+k where m and k are constants, find F the value of m and O of k. [3 marks] R M
36 In the following arrow diagram, the function f maps x4onto y and the ’05 function g maps y onto z. SPM Clone
(a) h–1(6), (b) w–1(x). [4 marks]
z
38 Given the functions m: x → 4x – 6 3 , x ≠ 0, find nm–1. ’05 and n: x → — x [3 marks] SPM Clone
1
–2
Write down the value of (a) g –1(4), (b) gf(1). [2 marks] 37 The functions h and w are defined 2 by h(x) = 3x + 5 and w(x) = , ’05 1 – 4x SPM Clone
39 Given the function g : x → 6 – 2x, find ’09 (a) g(–4), (b) the value of p such that g–1(p) = 8. [3 marks] SPM Clone
40 Given the functions g : x → 2x + 3 and h : x → 5x – 8, find ’10 (a) g –1(x), (b) hg–1(11). [3 marks] SPM Clone
Paper 2 Long Questions 1.4
Composite Functions
1 The following diagram shows the functions f and g that are defined 10 by f : x → ax + b and g : x → , c–x x ≠ c respectively. 17 f g
4
m –2
–3
g f
–4
Find (a) the value of a, of b and of c, (b) the value of m, (c) the expression for gf.
2 The functions f and g are defined by f : x → hx – 5 and g : x → x 2 + 3x + 5 respectively. If the composite function fg is defined by fg : x → 2x 2 + kx + 5, find the value of h and of k. Functions
3 The functions f and g are defined ax by f : x → ——— , x ≠ 1 and 1–x g : x → bx – 1 respectively. If f(4) = –4 and g(2) = 3, find (a) the value of a and of b, (b) the value of x for which fg = gf.
4 Given the functions f : x → 3 – x 1 and g : x → ——— , x ≠ 1, find the 1–x expression for each of the following. (a) fg (b) f 2 (c) f 13 g2 1.1 (d) Signifi cant Figure (e) g3 19 (f) g
5 The function f is defined by f : x → 2x + 3. Another function g is such that gf : x → 4x2 + 12x + 15. Find (a) the function g, (b) the values of c if g(c) = 7c, (c) the values of x if fg = gf.
22
6
x
y 8
z g
9
f 5
The above arrow diagram shows the representation of the mapping of x onto y by the function f : x → 4x – a and the mapping of y onto z b by the function g : y → ——––— , 12 – y y ≠ 12. Find (a) the value of a and of b, (b) the expression for the function that maps x onto z, (c) the element x that does not change when it is mapped onto z.
7 The functions f and g are defined by f : x → 2x + 3 and g : x → x 2 + bx + c respectively. If the composite function fg is given by fg : x → 2x 2 + 4x – 3, find (a) the value of b and of c, (b) the value of g2(1).
10 The function g is defined by g : x → x + 2. Another function f is such that fg : x → x 2 + 5x + 7. Find (a) the function f(x), (b) the values of c if f(2c) = 7c.
1.5
Inverse Functions
11 The function f is defined by f : x → mx + k, where m and k are constants. The function g is 12 defined by g : x → ——— , x ≠ –1. x+1 (a) Find the expression for g–1. (b) Find the expression for fg in of m and k. (c) If f(3) = g–1(3) and fg(–2) = –2, find the value of m and of k.
12 Given the functions f : x →
x+1 , x–2
kx + 3 x ≠ 2 and g : x → — —— —, x ≠ 0, x find the value of k if gf –1(0) = 3.
5
f
g–1 3
–1
15 Given the function f : x → 5x + h and its inverse function 2 f –1 : x → kx + —, find 5 (a) the value of h and of k, (b) (i) f(3), (ii) f –1f(3).
mx – n 16 Given the function f : x → ——— ––—, x–2 x ≠ 2 and its inverse function 5 – 2x f –1 : x → ———— , x ≠ 2, find 2–x (a) the value of m and of n, (b) the value of k for which f(k) = k + 2.
17 The following diagram represents the mapping of y onto x by the function f : y → hy + k and the mapping of y onto z by the 6 k function g : y → ,y≠ 2y – k 2 . x
1.1
y
z
19 Given that f : x → p – qx, find (a) f –1(x) in of p and q, (b) the value Fof p and of q if O and f(1) = –2. f –1(8) = –1 R M 4
1
The diagram shows the mapping of the functions f and g–1, such that f : x → 2x + a and g : x → bx + c. Given that g maps 1 onto itself, find (a) the value of a, (b) the value of b and of c.
7
18 Given the functions f(x) = 2x – 5 3x and g(x) = ——— , x ≠ –3, find x+3 (a) f –1g(x), (b) the value of x for which gf(–x) = f 2(2).
CHAPTER
9 The functions f and f 2 are such that f : x → ax + b and f 2 : x → 9x + 8. (a) Find the values of a and the corresponding values of b. (b) Considering a 0, find the values of k for which f(2k + 1) = k(k + 2).
C
B
A
1
14
CHAPTER
8 Given the functions f : x → 3x – 2 and fg : x → 4 – 15x, find (a) the expression for gf, (b) the values of n if gf(n2 – 1) = 6n + 6.
3 – kx 20 Given that g–1(x) = ———— and 2 f(x) = 2x 2 – 3, find (a) g(x) in of k, (b) the value of k for which g(x 2) = 2f(–x).
21 Given the functions f : x → hx + k, h 0 and f 2 : x → 25x – 18, find (a) the value of h and of k, (b) ( f –1)2 in the same form.
1 ,x≠k k–x and g(x) = 2 + x, find (a) f(x) in of k, (b) the value k if ff –1(k 2 + 2) = g[(5 + k)2].
22 Given that f –1(x) =
Signifi cant Figure 3 2
–2
x+p , x+q x ≠ –q and its inverse function 2 – 3x f –1 : x → , x ≠ 1, find 1–x (a) the value of p and of q, 1 (b) the values of k if f 2(k) = —k. 3
13 Given the function f : x →
Find (a) the value of h and of k, (b) the function that maps x onto y, (c) the function that maps x onto z.
23
23 Given that f : x → 2x – 3 and x g : x → + 2, find ’06 2 SPM Clone
(a) f –1 g,[3 marks] (b) the function h such that hg : x → 2x + 4. [3 marks] Functions
F O R M 4
Paper 1
Short Questions 1 The relation between set X = {6, 12, 15, 21} and set ’11 Y = {3, 5, 7} is ‘factor of’. (a) Find the image of 12. (b) Express the relation in the form of ordered pairs. F O [3 marks]
1 CHAPTER
2 Given the functions g(x) = 5x – 11 and h(x) = 3x, find the value of ’11 gh(2). [2 marks]
3 The inverse function h–1 is defined 3 by h–1 : x → , x ≠ 4. Find ’11 4–x (a) h(x), (b) the value of x such that h(x) = –14. [4 marks]
State (a) the object of 40, (b) the type of the relation. [2 marks]
3 Given the function f (x) = px + q and f 2 (x) = 4x + 9, where p and q are constants, find the value of p and of q if p < 0. [3 marks]
2 The relation between two variables is represented by the following set of ordered pairs:
4 Given the function f : x → x + 4 and the composite function gf : x → x 2 + 6x + 2, find (a) g(x), (b) fg(4). [4 marks]
SPM Clone
R M
Paper 1
1 The graph below shows the relation between set A and B.
40 30
{(–4, 16), (–3, 9), (–2, 4), (2, 4), (3, 9), (4, 16)}
20 10
1
2
3 Set A
4
(a) State the type of the above relation. (b) Represent the above relation using a function notation. [2 marks]
2x + p and 5 5x + 3 , its inverse function f –1(x) = q find the value of p and of q. [4 marks]
5 Given the function f (x) =
(a) Find the value of h and of k. [4 marks] (b) If gf –1(x) = –5x, find the values of x. [3 marks]
Paper 2
Long Questions 1 The function f and its inverse function f –1 are hx kx , x ≠ 3, f –1(x) = ,x≠2 defined by f(x) = x–3 x–2 respectively, where h and k are constants. Another 1 function g is defined by g(x) = , x ≠ 0. x
Functions
SPM Clone
4
Short Questions
Set B
4
CHAPTER
F O R M
1
SPM Clone
2 Given the functions f : x →
x – 2 and g : x → 3x + k, 2
where k is a constant, find (a) f –1(3), [2 marks] (b) the value of k if f –1g: x → 6x – 4, [2 marks] (c) h(x) such that hf : x → 9x – 3. [2 marks]
24
COMPANION WEBSITE
Online Tests
FORM
CHAPTER
1
4
Functions SPM Topical Analysis
Year
2004
2005
2006
2007
2008
2009
2010
2011
Paper
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
Number of questions
3
0
3
0
2
1
3
0
3
0
3
0
3
0
3
0
ONCEPT MAP Domain Domain
RELATIONS HUBUNGAN Types of Relations Jenis Hubungan • One-to-one • Many-to-one • One-to-many • Many-to-many
Codomain Kodomain FUNCTIONS FUNGSI
Objects Objek Images Imej Range Julat
Absolute Value Functions Fungsi Nilai Mutlak f : x→|g(x)|
Graphs of Absolute Value Functions Graf Fungsi Nilai Mutlak The graph of a linear absolute value function has a V shape COMPANION WEBSITE
Learning Objectives
Composite Functions Fungsi Gubahan The function of f followed by g is gf.
Inverse Functions Fungsi Songsang Given that y = f(x), then f – 1(y) = x.
Problems that involve Composite Functions and Inverse Functions Masalah melibatkan Fungsi Gubahan dan Fungsi Songsang
1.1 Relations 1 1.1a Representation of a Relation
1 CHAPTER
4
1
F O R M
CHAPTER
1 A relation from set A to set B is the linking (or pairing) of the elements of set A to the elements of set B. 2 A relation between two sets can be represented by (a) an arrow diagram, (b) ordered pairs, (c) a graph. F O R M
14
A relation from set A = {12, 14, 23, 25, 43} to set B = {3, 5, 7} is defined by ‘sum of digits of’. Represent the relation by (a) an arrow diagram, (b) ordered pairs, (c) a graph. Solution (a) Arrow diagram A
‘sum of digits of’ B
12 14
3
23
5
25
7
43
(b) Ordered pairs {(12, 3), (14, 5), (23, 5), (25, 7), (43, 7)} (c) Graph 7 Set B 5 3 12 14 23 25 43 Set A Try: Question 1, Self Assess 1.1.
1.1b Domain, Codomain, Objects, Images and Range In a relation between set (A) and another set (B), • the first set (A) is known as domain, • the second set (B) is known as codomain, • the elements in the domain are known as objects, • the elements in the codomain that are linked to the objects are known as images, • the set of images is known as range. Functions
2
Facts A set is a well-defined collection of objects. For example: Universal set, = {x : 10 x 30, x is an integer} Set A = {Factors of 36} Set B = {Prime numbers} Set C = {Numbers where the sum of the digits is 3} The list of elements of each of the sets , A, B and C is as follows: = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} A = {12, 18} B = {11, 13, 17, 19, 23, 29} C = {12, 21, 30}
3 2 A relation from set P = {16, 36, 49, 64} to set Q = {4, 6, 7, 8, 11} is defined by ‘factor(s) of’. (a) Represent this relation using an arrow diagram. (b) State (i) the domain, (ii) the codomain, (iii) the images of 36, (iv) the images of 64, (v) the object of 7, (vi) the objects of 4, (vii) the range of this relation. Solution (a) The arrow diagram that represents the given relation is as shown in the next column.
P
‘factor(s) of’
Q
16
4
36
6
49
7
64
8
Range
11 Domain
Non-image
Codomain
1
1
CHAPTER
CHAPTER
(b) (i) The domain is {16, 36, 49, 64}. F 8, 11}. (ii) The codomain is {4, 6, 7, O (iii) The images of 36 are 4 and 6. R (iv) The images of 64 are 4 and 8. M (v) The object of 7 is 49. (vi) The objects of 4 are 16, 436 and 64. (vii) The range is {4, 6, 7, 8}.
Try: Question 2, Self Assess 1.1.
1.1c Types of Relations
3 State the type of relation shown by each of the following arrow diagrams. (a) (b)
Types of Relations 1 One-to-one relation Each object in the domain has only one image in the codomain. 2
Many-to-one relation There are more than one object in the domain that have the same image in the codomain.
3
One-to-many relation There is at least one object in the domain that has more than one image in the codomain.
4
Many-to-many relation There is at least one object in the domain that has more than one image in the codomain and there is at least one element in the codomain that is linked to more than one object in the domain.
A
‘reciprocal of’
1 3 1 4 1 5
3 Domain
Codomain
4
5
Domain
P
B
‘factors of’
Q 2
10
3
21
5 7
Solution (a) One-to-one relation (b) One-to-many relation
Codomain
Try: Question 3, Self Assess 1.1.
4
Domain
State the type of relation shown by (a) the ordered pairs: {(4, 8), (4, 10), (4, 12), (6, 10), (7, 12)} (b) the graph:
Codomain
Set Y M P
Solution Domain
5
7
9
12 15
Set X
Codomain
Draw arrow diagrams to represent the relations because it is easier to interpret arrow diagrams compared to ordered pairs or graphs.
3
Functions
F O R M 4
(a)
4
8
6
10
7
12
(b)
∴ The above arrow diagram shows many-to-many relation. Try: Question 4, Self Assess 1.1.
P M
SPM Clone
’07
F O The following ordered pairs represent a relation R from set A = {3, 5, M 7} to set B = {9, 15, 21}.
1
1
SPM Clone
’06
A
CHAPTER
4
CHAPTER
F O R M
Y
∴ The above arrow diagram shows many-to-one relation.
1 1
X 5 7 9 12 15
{(3,4 9), (5, 15), (7, 21)}
(a) State the type of the above relation. (b) State the range. (c) Using function notation, write down a relation between set A and set B. Solution
B
3
9
5
15
7
21
Based on the arrow diagram, (a) the relation is a one-to-one relation. (b) the range is {9, 15, 21}. (c) the function notation is f(x) = 3x. Each image is three times its objects.
It is better to draw an arrow diagram to represent the above relation because it is easier to interpret an arrow diagram compared to ordered pairs. Try: Question 5, Self Assess 1.1.
1.1 (iv) the objects of 12, (v) the range of the relation. 3 State the type of relation shown by each of the following arrow diagrams. (a) M N ‘capital of’
1 Represent each of the following relations by (a) an arrow diagram, (b) ordered pairs, (c) a graph. (i) The relation ‘remainder when divided by 5’ from set A = {6, 12, 18, 24} to set B = {1, 2, 3, 4}. (ii) The relation ‘factors of’ from set P to set Q where P = Q = {2, 3, 4, 8}. (iii) The relation ‘difference of digits of’ from set X = {14, 21, 23, 34, 48} to set Y = {1, 3, 4}. 2 A relation from set P = {26, 34, 45, 62} to set Q = {12, 20, 24, 48} is defined by ‘product of digits of’. (a) Represent the relation by an arrow diagram. (b) State (i) the domain, (ii) the codomain, (iii) the image of 45, Functions
Sarawak
Kuching
Perak
Ipoh
Kedah
(b)
P methane water carbon dioxide
Alor Setar
‘elements of’
Q carbon hydrogen oxygen
4 State the type of relation of (a) the ordered pairs: {(4, 0), (4, 1), (5, 2), (6, 7), (6, 8)} 4
(b) the graph:
Set B
{(2, 1), (6, 3), (10, 5)}
11 10 8 3 4 6 7 9
(a) State the type of the above relation. (b) State the range. (c) Using function notation, write down a relation between set P and set Q.
Set A
5 The following ordered pairs represent a relation from set P = {2, 6, 10} to set Q = {1, 3, 5}.
1.2 Functions
a
x
b
y
c
z
a
w
b
x
c
y
d
z
1.2a To Recognise Functions 5 Explain whether each of the following relations is a function. (a) (b) (c) ‘has’
B
January
28 days
June
30 days
August
31 days
M ‘three times of’ N
P ‘consists of ’ Q Bronze Brass
1 3 5 7
cuprum stanum zinc
1 CHAPTER
4
Facts
3 The main properties for a relation to be considered a function are: (a) Each and every object must have one and only one image. [If there are objects that have two or more images, the relation is not a function.] (b) Two or more objects are allowed to have the same image. [Many-to-one relation is considered a function.] (c) It is not necessary that all elements in the codomain are images. [Non-image elements are allowed in a function.] 4 If a function links x to x 2 – 3x + 5, by using function notation, it is written as f : x → x2 – 3x + 5 or f(x) = x2 – 3x + 5. It is read as ‘function f that maps x onto x2 – 3x + 5’.
A
mapping.
1
5
F O Facts R M known as Function is also
CHAPTER
1 A function is a special type of relation where each and every object in the domain has only one image. 2 The types of relations that are considered as functions are: (a) one-to-one relation (b) many-to-one relation
3 9 15
Solution (a) It is a function because • each and every object has only one image, • many-to-one relation is allowed, • non-image element, i.e. ‘28 days’ is allowed. (b) It is not a function because there are objects that have more than one image. (c) It is not a function because there is an element in the domain that does not have an image, i.e. ‘7’.
The types of relations that are not considered as functions are: (a) one-to-many relation x
a b Facts
y z
[Not a function because element ‘a’ has more than one image.] (b) many-to-many relation a
x
b
y
c
z
[Not a function because element ‘b’ has more than one image.] (c) relation where there are elements in the domain that do not have an image a b c d
x y z
[Not a function because element ‘d’ does not have an image.]
Try: Question 1, Self Assess 1.2.
5
Functions
F O R M 4
1.2b To Solve Problems involving Functions 6 (b)
When the function f has an image 4, it means that f(x) = 4.
When f(x) = 4, x 2 – 4x – 1 = 4 x 2 – 4x – 5 = 0 (x + 1)(x – 5) = 0 x = –1 or 5 Hence, the objects that have the image 4 are –1 or 5.
Object
1
F When x = 2, O R f(2) = 22 – 4(2) –1 M = 4 – 8 – 1
CHAPTER
4
CHAPTER
F O R M
1
Given the function f : x → x2 – 4x – 1, find (a) the image of 2, (b) the objects that have the image 4. Solution (a) f : x → x 2 – 4x – 1 f(x) = x 2 – 4x – 1
= –5 74 of 2 is –5. Hence, the image Try: Questions 2–3, Self Assess 1.2.
7 The arrow diagram below shows the function 15 f : x → ———— . ax + b x
f
–3
f(–4) = –3 15 Substitute x = –4. –––––––– = –3 –4a + b 12a – 3b = 15 4a – b = 5 ........... 2 1 – 2 : –a = –2 a = 2 From 1 , 3(2) – b = 3 b = 3 (b)
15 ax + b –3
–4 –5
(a) Find the value of a and of b. (b) State the value of x such that the function f is undefined.
A function f is undefined when its denominator is zero.
Solution 15 (a) f : x → —— —— ax + b 15 f(x) = —— —— ax + b f(–3) = –5 15 ––––––– = –5 Substitute x = –3. –3a + b 15a – 5b = 15 3a – b = 3 ........... 1
15 f(x) = — — — —— 2x + 3 When the denominator 0, 2x + 3 0 x – –3– 2 x –1–1– 2 Hence, the value of x such that the function f is undefined is –1–1– . 2
Try: Questions 4–8, Self Assess 1.2.
Functions
6
8
9
Press 0.8
SHIFT
4
1
R M
(b) g(x) = 1.6 2 cos x = 1.6 10 1.6 cos x = ––– 2 cos x = 0.8 x = 36.87°
CHAPTER
Solution 3x + k , (a) For f : x → ––––––– 2x – 4 8 Denominator ≠ 0 2x – 4 ≠ 0 x≠2 It is given that x ≠ p ∴p=2 (b)
1
(a) Given the function f(x) = 3 sin x – tan x, find the image of 30°. (b) Given the function g(x) = 2 cos x, find the value of x such that the function g has the image 1.6, for the domain 0° x 90°. Solution (a) f(x) = 3 sin x – tan x f (30°) = 3 sin 30° – tan 30° = 3(0.5) – 0.5774 F = 0.9226 O Hence, the image of 30° is 0.9226. CHAPTER
3x + k for all A function f is defined by f : x → –––––– 2x – 4 values of x, except x = p and k is a constant. (a) State the value of p. (b) Given that the value 5 is mapped onto itself under f, find (i) the value of k, (ii) another value of x that is mapped onto itself.
cos
=
Answer display: 36.86989765
Self-mapping is given by f(x) = x, where both the object and the image have the same value.
The value of x such that the function g has the image 1.6 is 36.87°.
(i) 5 is mapped onto itself, thus f(5) = 5 3(5) +— k = 5 —— —— 2(5) – 4 15— +— k = 5 — — 6 15 + k = 30 k = 15 (ii) For self-mapping, f(x) = x 3x + —–——15 — = x 2x – 4 3x + 15 = 2x 2 – 4x 2 2x – 7x – 15 = 0 (2x + 3)(x – 5) = 0 x = – –3– or 5 2
Try: Question 11, Self Assess 1.2.
1.2c Domain, Range, Objects and Images of a Function
10 The arrow diagram represents the function f : x → 2x 2 – 5. State f x (a) the domain, (b) the range, 2 (c) the image of 0, 1 (d) the objects of 0 (i) 3, –1 –2 (ii) –3. Solution (a) Domain = {–2, –1, 0, 1, 2} (b) Range = {–5, –3, 3} (c) The image of 0 is –5. (d) (i) The objects of 3 are 2 and –2. (ii) The objects of –3 are 1 and –1.
Hence, another value of x that is mapped onto itself, apart from 5, is –1–1– . 2
2x2 – 5 3
–3 –5
Try: Question 12, Self Assess 1.2.
Try: Questions 9–10, Self Assess 1.2.
7
Functions
F O R M 4
1.2 (a) the image of 10, (b) the object that has the image – –1– , 2 (c) the value of x such that the function f is undefined. 6 Given that f : x → f(3) = 4, find (a) the value of p and of q, (b) the values of x such that f(x) = 4–– x. 3 7 Given that g : x → a + bx, g(1) = –3 and g(–2) = 3, find (a) the value of a and of b, (b) the values of n if g(n2 + 1) = 5n – 6.
1 State whether each of the following relations is a function. (a) A
‘type of number’
B
6 14
multiple of 3
15
multiple of 7
24
(b)
P
Q
‘consist of letters’
i v
1
1
F O Consonants R Vowels M
(c)
CHAPTER
4
CHAPTER
F O R M
w u
4
M ‘product of digits of’ N
21
2
32
6
53
15
8 A function f is defined by f : x → 5x – 2. Find (a) the object that has the image 5, (b) the object that is mapped onto itself. a— . The 9 A function f is defined by f : x → —— b–x values 3 and 5 are mapped onto themselves under f. (a) Find the value of a and of b. (b) State the value of x such that the function f is undefined.
64
(d)
X
‘has’
Y
PERAK
2 vowels
KEDAH
3 vowels
SELANGOR
4 vowels
18 9 2 Given the function f : x → — , x ≠ –– , find 2x – 9 2 (a) the image of (i) 0, (ii) 12, (b) the object that has the image (i) 2, (ii) 6 . 3 Given that f : x → –––a––– , f(3) = –5 and x–b f(–5) = –1, find (a) the value of a and of b, (b) the value of x such that the function f is undefined. f
10 The arrow diagram x shows the function –1 f : x → px + qx 2. –2 Find (a) the value of p and of q, (b) the values of x that are mapped onto themselves.
px + qx2
–5 –16
11 (a) Given the function f: x → sin x + cos x, find the image of (i) 45°, (ii) 60°. (b) Given the function g: x → tan x, find the value of x such that the image is 1.527, if x is an acute angle. 12 The arrow diagram f x2 + 2 x represents the function 6 f : x → x 2 + 2. State (a) the domain, 3 (b) the range, 2 2 1 (c) the image of 0, 0 (d) the objects of –1 (i) 3, –2 (ii) 6.
b
ax + – x 4 The arrow diagram x shows the function 7 f : x → ax + –b– . x 2 Find –1 (a) the value of a and of b, –5 (b) the value of x such that the function f is undefined, (c) the object that has the image 7, apart from x = 2.
2x 5 The function g is defined by g : x → — — — —. x+m If g(5) = 3g(2), find the value of m. Hence, find
Functions
f
8
1.3
Absolute Value Functions
1 The absolute value of x is written as _x_ and it is read as ‘the modulus of x ’. 2 The definition of _x_ is
Facts _x_ is the numerical value of x, i.e. positive numbers are still positive but negative numbers will be changed to positive. For example, _5_ = 5 and _–5_ = 5.
x, if x 0 _x_ = –x, if x 0 3 The absolute value function is defined by
f(x), if f(x) 0 _ f(x)_ = –f(x), if f(x) 0
1 CHAPTER
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The graph of O a linear absolute value functionRhas a V shape. M CHAPTER
2 2
Facts F
’07
f(x) = 3 _4x – 5_ = 3 4x – 5 = ± 3
Given the function f : x → _4x – 5_, find (a) the image of (i) –1, (ii) 4, (b) the objects that have the image 3. Solution (a) f : x → _4x – 5_ f(x) = _4x – 5_ (i) f(–1) = _4(–1) – 5_ = _–9_ = 9 Hence, the image of –1 is 9. (ii) f(4) = _4(4) – 5_ = _11_ = 11 Hence, the image of 4 is 11. (b)
The first equation is 4x – 5 = 3 4x = 8 x = 2 The second equation is 4x – 5 = –3 4x = 2 x = –1– 2 Hence, the objects that have the image 3 are 2 or –1–. 2
For any _f(x)_ = k, there are two equations that can be formed, i.e. f(x) = k or f(x) = –k. Try: Questions 1–2, Self Assess 1.3.
3
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Sketch the graph of each of the following absolute value functions. (a) f(x) = _x + 3_ for the domain – 4 x 1 (b) f(x) = _4x – 7_ for the domain 0 x 4 State the corresponding range of values of f(x).
The sketch of the graph of f(x) = _x + 3_ is as shown below. y 4 3
Solution (a) Prepare a table as shown below.
x
Range
2 1
–4 –3 –2 –1 0 1
–4 –3 –2 –1 O
1
x
Domain
f (x) 1 0 1 2 3 4 9
Functions
F O R M 4
3 Hence, the graph touches the x-axis at 1 –– , 0 . 4 Prepare a table as shown below.
The corresponding range of values of f(x) means the range from the smallest value of y to the largest value of y, based on the given domain.
x
0 1 2 3 4
f (x) 7 3 1 5 9 ∴ The corresponding range of values of f(x) is 0 f(x) 4. (b)
The sketch of graph of f(x) = _4x – 7_ is as shown below. y 8
1
F First of all, determine the point where the graph O touches the x-axis.
CHAPTER
4
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1
10 6
Range
4
R M
2
At the x-axis, y = 0 Thus, _4x – 7_ = 0 4 4x – 7 = 0 4x = 7 x = –7 – = 1–3– 4 4
3 1– 4
O
1
2
3
4
x
Domain
∴ The corresponding range of values of f(x) is 0 f(x) 9.
Try: Question 3, Self Assess 1.3.
Facts
1.3 3 Sketch the graph of each of the following absolute value functions. (a) f(x) = _ x + 2_ for the domain –3 x 3 (b) f(x) = _ 2x – 5_ for the domain 0 x 7 (c) f(x) = _ 3 – 2x_ for the domain –3 x 4 (d) f(x) = _ 3x – 5_ for the domain –2 x 4 State the corresponding range of values of f(x).
1 Given the function f : x → _ x 2 – 4x – 3_ , find the image of (a) –3, (b) 0, (c) 2. 2 Given the function f(x) = _ 2 – 5x_ , find (a) the image of (i) 2, (ii) –2, (b) the objects that have the image 7.
1.4
Composite Functions
1.4a To Find Composite Functions 1 If f is a function which maps set A onto set B and g is a function which maps set B onto set C, then gf is a composite function of f followed by g which maps set A onto set C. gf
x A Functions
f
f(x) B
g
g[f(x)] = gf(x) C
10
Facts • fg(x) means f[g(x)]. • In general, fg ≠ gf. • f 2 = ff, f 3 = fff or ff 2 and so on.
11 11 The functions f and g are defined by f : x → 2x + 1 and g : x → x 2 – 2 respectively. Find (a) the value of fg(3) and of gf(–2), (b) the composite functions (i) fg (iii) f2 (ii) gf (iv) g 2, (c) the values of x if gf(x) = 23. Solution (a) fg(3) = f[g(3)] = f(32 – 2) = f(7) = 2(7) + 1 = 15 gf(–2) = g[2(–2) + 1] = g(–3) = (–3)2 – 2 =7 (b) f : x → 2x + 1 g : x → x2 – 2 f(x) = 2x + 1 g(x) = x 2 – 2 (i) fg(x) = f[g(x)] = f(x 2 – 2) Substitute the x in
(iii) f 2(x) = ff(x) = f(2x + 1)
↓
↓
= 2(x – 2) + 1 = 2x 2 – 3 ∴ fg : x → 2x 2 – 3 (ii) gf(x) = g[ f(x)] = g(2x + 1)
1
= (x – 2) – 2 = x 4 – 4x 2 + 4 – 2 = x 4 – 4x 2 + 2 ∴ g 2 : x → x 4 – 4x 2 + 2
CHAPTER
2
F O R Substitute the x in M g(x) = x2 – 2
1
2
with (2x + 1).
CHAPTER
= 2(2x + 1) + 1 = 4x + 3 ∴ f 2: x → 4x + 3 (iv) g2(x) = gg(x) = g(x 2 – 2)
f(x) = 2x + 1 2
Substitute the x in f(x) = 2x + 1
↓ 4 with (x2 – 2).
(c) gf(x) = 23 4x 2 + 4x – 1 = 23 4x 2 + 4x –24 = 0 x 2 + x – 6 = 0 (x – 2)(x + 3) = 0 x = 2 or –3
with (x2 – 2).
Substitute the x in g(x) = x2 – 2
↓
with (2x + 1).
= (2x + 1)2 – 2 = 4x 2 + 4x + 1 – 2 = 4x 2 + 4x – 1 ∴ gf : x → 4x 2 + 4x – 1
Try: Questions 1–5, Self Assess 1.4.
12 Given the function f : x →
x— +— 1 +1 = — x – 1 = ————— — x + 1 = ——— – 1 x – 1 x +— 1— +— x— ––— 1 —— x – 1 = ——————–— x+ 1— –— (x— ––— 1) — —— x – 1 2x = –– 2 = x
expression for each of the following functions. (a) f 2 (b) f 8 (c) f 9 Solution f:x→— x— +— 1 (a) x–1 x— +— 1 f(x) = — x–1 f 2(x) = ff (x) x— +— 1 = f — x – 1
11
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(b) f 8(x) = f 2f 2f 2f 2(x) = f 2f 2f 2(x)
(c) f 9(x) = ff 8(x)
f 8(x) = x f 2(x) = x
= f(x) x— +–––1 , x ≠ 1 = — x – 1
= f 2f 2(x) = f 2(x) = x
In the SPM marking scheme, it is not compulsory for students to write down x ≠ 1 in the answer.
1 CHAPTER
4
CHAPTER
F O R M
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Try: Questions 6–7, Self Assess 1.4.
F O R M
1.4b To Find the Related Function Given the 4 Composite Function and One of the Functions 4
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(b)
(a) The function f is defined by f : x → 2x + 4. Another function g is such that fg : x → 3x – 8. Find the function g. (b) The function f is defined by f : x → x – 2. Another function g is such that gf : x → 1 , x ≠ 11 . Find the function g. 11 – 3x 34
It is given that f : x → x – 2 and 1 gf : x → —–—— —. 11 – 3x g is outside. 1 gf(x) = —–—— — 11 – 3x Change each x to 1— g[f(x)] = —–— — the in u. 11 – 3x 1— g(x – 2) = —–— — 11 – 3x Let x – 2 = u, then x = u + 2 ∴ g(u) = —–——1——— — 11 – 3(u + 2) 1—— = —–——— — 11 – 3u – 6 1— = — — — — 5 – 3u ∴ g : x → 1— , x ≠ –5– 5 – 3x 3
Solution (a) Case where the function g that has to be determined is situated ‘inside’.
It is given that f : x → 2x + 4 and fg : x → 3x – 8. g is inside.
fg(x) = 3x – 8 f [g(x)] = 3x – 8 Substitute the x in f(x) = 2x + 4
↓
with g(x).
2g(x) + 4 = 3x – 8 2g(x) = 3x – 12 3x—— – –12 g(x) = — –— 2 3x – –12 ∴ g : x → ——— –— 2 Try: Questions 8–10, Self Assess 1.4.
Functions
Case where the function g that has to be determined is situated ‘outside’.
12
1.4c Further Examples 5 on Composite Functions SPM Clone
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It is given that f : x → hx + k and g : x → (x + 1)2 + 2. Thus, fg(x) = f[(x + 1)2 + 2] = h[(x + 1)2 + 2] + k = h(x + 1)2 + 2h + k
Given the functions f : x → hx + k, g : x → (x + 1)2 + 2 and fg : x → 2(x + 1)2 + 1, find (a) the value of g2(2), (b) the value of h and of k. Solution (a) g2(2) = gg(2) = g[(2 + 1)2 + 2] = g(11) = (11 + 1)2 + 2 = 146 (b)
1 CHAPTER
4
1
But it is given that fg(x) = 2(x F+ 1)2 + 1. Hence, by comparison, O R h = 2 and 2h + k = 1 M 2(2) + k = 1 k = –3 CHAPTER
5
13
The following function problem can be solved by making a comparison. Try: Questions 11–12, Self Assess 1.4.
13 Given the functions f(x) = 3x + 7 and fg(x) = 22 – 3x, find gf(x). Solution Find g(x) first. fg(x) = 22 – 3x f(g(x)) = 22 – 3x 3g(x) + 7 = 22 – 3x
3g(x) = 22 – 3x – 7 3g(x) = 15 – 3x 15 –—3x g(x) = — — — — 3 g(x) = 5 – x Hence, gf(x) = g(3x + 7) = 5 – (3x + 7) Substitute the x in g(x) = 5 – x
Substitute the x in f(x) = 3x + 7
↓
↓
with (3x – 7).
with g(x).
= 5 – 3x – 7 = –3x – 2
Try: Question 8, SPM Exam Practice 1 – Paper 2.
1.4 (c) f : x → x – 3 (a) f : x → x 2 – 1 g : x → 3x + 1 g : x →_ x_ (b) f : x → (x + 1)2 (d) f: x → 2 – x g g : x → 1 – 3x 1— : x → —— — x2 + 2 3 The functions f and g are defined by f : x → 4x – 3 and g : x → x + 1 respectively.
1 Given the functions f : x → _ 4 – 5x_ and pooooooo g : x → x – 2 , x 2, find the value of (a) fg(6) (c) f 2(0) (b) gf(2) (d) g2(27) 2 Find the composite functions fg and gf for each of the following pairs of functions f and g.
13
Functions
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10 Find the expressions for the functions g and h in each of the following. (a) f : x → 2x, fg : x → 4x – 12, 2x + 1 hf : x → ———— 2 2x –—, 1 (b) f : x → 2x – 2, gf : x → — —— 3 Facts fh : x → 2x 2 (c) f : x → x 2 – 2, fg : x → x 2 + 6x + 7, hf : x → 2x 2 – 7
6 Given the functions f : x → 5x – 7 and g : x → 2, x ≠ 0, find the expression for x each of the following functions. (a) fg (c) g6 2 (b) g (d) g7
11 The functions f and g are defined by f : x → 1 – x and g : x → px 2 + q respectively. If the composite function gf is given by gf : x → 3x 2 – 6x + 5, find (a) the value of p and of q, (b) the value of g 2(0). Facts 12 The functions f and f 2 are such that f : x → hx + k and f 2 : x → 9x + 16. (a) Find the values of h and the corresponding values of k. (b) Considering h 0, find the values of x such that f(x 2) = 8x.
x–1 , x ≠ –1, find the x+1 expression for each of the following functions. (a) f 2 (c) f 16 4 (b) f (d) f 17 8 The function f and the composite function fg are defined as follows. Find the function g. (a) f : x → x + 2, fg : x → 3x – 2 (b) f : x → 3x + 2, 7 If f is defined by f : x →
1.5
fg : x →
9 The function f and the composite function gf are defined as follows. Find the function g. (a) f : x → x + 1, gf : x → 3 , x ≠ 2 x–2 1 5 , x ≠ –1–– (b) f : x → ––, gf : x → x 10 10x – 1 (c) f : x → 3x + 2, gf : x → 9x 2 + 9x + 2
1
1
CHAPTER
4
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2x + 5 ,x≠2 x–2 (c) f : x → x 2 – 1, fg : x → x 2 + 4x + 3
Find the expressions for f 2 and g2. Hence, find the value of x such that (a) f = g, (b) f 2 = g2. 4 The functions f and g are defined by f : x → mx + 2 and g : x → kx – 3 respectively. If fg = gf, find the relation between m and k. If m = 5, find the value of x that satisfies each of the following equations. (a) f 2 = f (b) g2 = g 5 The functions f and g are defined by f : x → (x + 1)2 and g : x → x – 2 respectively. F Find O (a) the composite functions fg and gf, R M of x if (b) the values (i) fg = 9, (ii) gf =414, (c) the value of x if fg = gf.
Inverse Functions
Facts –1
1.5a To Find Inverse Functions 1 If f : x → y is a function that maps x onto y, its inverse function is denoted by f –1. Inverse function is a function that maps y back to x.
Facts
f
x
y
f –1
2 The main step to find the inverse function is ‘Let y = f(x), then x = f –1 (y)’ or ‘Let f –1(x) = y, then f(y) = x’. Functions
• (fg) = g–1f –1 and (gf )–1 = f –1g –1. • (f 2)–1 = (f –1)2. • ff –1(x) = f –1f(x) = x.
14
The conditions for the existence of inverse functions are (a) the function must be one-toone, (b) that every element in the codomain must be linked to an object.
6
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’05
’08
’10
Try: Question 1, Self Assess 1.5.
F O R M
14 Find the inverse function of f : x → Solution Method 1 1 f 2x : x → ——––— x+2 2x—––— 1 Let y = — x + 2 y(x + 2) = 2x – 1 yx + 2y = 2x – 1 yx – 2x = –1 – 2y x(y – 2) = –1 – 2y –1 – 2y x = —–––—–— y – 2 –(1 + 2y) x = —–––—–—––— –(2 – y) 1 + 2y x = ——–— 2 – y
2x – 1 , x ≠ –2. x+2
1 + 2y f –1(y) = ——–— ∴ 2 – y 1 + 2x f –1(x) = ——–— 2 – x
4
1
Solution Let y = f –1(3) f(y) = 3
CHAPTER
y +— 8 = 3 —— y – 6 y + 8 = 3(y – 6) y + 8 = 3y – 18 2y = 26 y = 13 ∴ f –1(3) = 13
1
Given the function f –1 : x → find the value of f –1(3).
CHAPTER
14
\ f –1 : Method 2 Let f –1(x) = y Thus, f(y) = x 2y 1 = x ——––— y+2 2y – 1 = x(y + 2) 2y – 1 = xy + 2x 2y – xy = 2x + 1 y(2 – x) = 2x + 1 2x—+–— 1 y = — 2 – x 2x—+–— 1 f –1(x) = — 2 – x
Rearrange, making x the subject. Cross multiplication. Expand the left-hand side. Group the in x together. Factorise the left-hand side. Making x the subject
In the SPM marking scheme, it is not compulsory for students to write x ≠ 2. Rearrange the formula, making y the subject.
\ f –1 : x → Try: Question 2, Self Assess 1.5.
15 Given the functions f(x) = 6 – 2x and g(x) = –x1– , x ≠ 0, find f –1g–1.
6 ––––— x ∴ f –1(x) = — 2 g(x) = –x1– Let g –1(x) = y Thus, g(y) = x –y1– = x y = –x1– ∴ g–1(x) = –x1–
Solution First of all, find the inverse functions f –1 and g–1. f(x) = 6 – 2x Let f –1(x) = y Thus, f(y) = x 6 – 2y = x 6 – x = 2y 6 ––––— x y = — 2 15
Functions
F O R M 4
6 – –x1– = —–——–— 2 6x ––—1 , x ≠ 0 = — — –— 2x
Hence, f –1g–1 (x)
= f –1 –x1–
Apply the composite function of g –1 followed by f –1.
Try: Questions 3–7, Self Assess 1.5.
SPM Clone
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F hx – 5 ———, x ≠ 2 and its Given the function O f : x → — x–2 R 2x + k inverse function M f –1: x → ————, x ≠ 3, find x–3 (a) the value of h4and of k, (b) the values of t such that f(t) = –4– t. 3
1
2x +— k. But it is given that f –1(x) = — —— x–3 Hence, by comparison, k = –5, h = 3.
CHAPTER
4
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1
7 7
f(t) = –4– t 3 3t – 5 4 — — — — = –– t t–2 3 3(3t – 5) = 4t (t – 2) 9t – 15 = 4t 2 – 8t 0 = 4t 2 – 17t + 15 0 = (4t – 5)(t – 3) t = ––5 or 3 4 (b)
Solution (a) Let f –1(x) = y Thus, f(y) = x hy – 5 — ——— = x y–2 hy – 5 = x(y – 2) hy – 5 = xy – 2x 2x – 5 = xy – hy 2x – 5 = y(x – h) 2x –— 5 y = — —— x – h 2x –— 5 ∴ f –1(x) = — —— x – h Try: Questions 8–9, Self Assess 1.5.
8 x
SPM Clone
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y
z g
f 1
Solution 5–— (a) f : y → py + q g : y → —–— 3y – q 5–— f(y) = py + q g(y) = —–— 3y – q f –3– = 1 g –3– = 10 2 2 5 3 –– p + q = 1 —–————— = 10 2 3 –3– – q 2 1 3p + 2q = 2 ... 5— = 10 —–— 9 –– – q 2
10
3 – 2
The above diagram shows the representation of the mapping of y onto x by the function f : y → py + q and the mapping of y onto z by q 5 the function g : y → ——–— , y ≠ –– . Find 3y – q 3 (a) the value of p and of q, (b) the function that maps x onto y, (c) the function that maps x onto z.
Functions
16
5—–— = 10 —–— 9 – 2q —–——— 2 10 —–——— = 10 9 – 2q 1 —–——— = 1 9 – 2q 1 = 9 – 2q 2q = 8 q = 4 From 1 : When q = 4, 3p + 2(4) = 2 p = –2
(c) Based on the diagram, the function that maps x onto z is gf –1(x). f
–1
g
x
gf
(b) The function that maps x onto y is f (x). It has been found that f(y) = –2y + 4. Let f –1(x) = w Thus, f(w) = x –2w + 4 = x 4 – x = 2w 4–––— x w = — 2 4–x ∴ f –1(x) = — — — — 2
F O R M 4
1
CHAPTER
1
gf (x)
4 –––— x = g — 2 5 ——–— —–——— = 4 – x –4 3 —––— 2 5 –—— = —–——— 12 – 3x –— 8 ———–— —— 2 = —–10 —— 4 – 3x 10 4 ∴ gf –1 : x → ,x≠ 4 – 3x 3
–1
–1
–1
CHAPTER
z
y
Try: Question 10, Self Assess 1.5.
1.5b To Find a Function Given its Inverse Function
1.5c Further Examples on Inverse Functions
The inverse of an inverse function, f –1(x), will give 9 f(x). us back the original function,
9 Given that g–1(x) =
16 The function f is defined by f (x) = 2x + 1 , x ≠ 1. x–1 If f –1(k) = –4– k, find the values of k. 3
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2x + 1 1 , x ≠ , find g(x). 2x – 1 2
Solution 2x +— 1 Let y = — —— 2x – 1 y(2x – 1) = 2x + 1 2xy – y = 2x + 1 2xy – 2x = y + 1 x(2y – 2) = y + 1 y + 1 x = ——— — 2y – 2 y + 1 g(y) = ——— — 2y – 2 x+1 ∴ g(x) = ,x≠1 2x – 2
Solution 4 f –1(k) = –– k 3 If f –1 (x) = y, k = f –4– k then x = f (y). 3 4k 2 –– + 1 3 k = —————— 4k – 1 –– 3 8k +— 3 — —— 3 k = — — — — — — 4k –— 3 — —— 3 8k +— 3 ——— 3— k = ——— 3 4k – 3 8k +— 3 k = — —— 4k – 3
Rearrange, making x the subject.
Try: Question 11, Self Assess 1.5.
17
Functions
F O R M 4
4k 2 – 3k = 8k + 3 4k 2 – 11k – 3 = 0 (4k + 1)(k – 3) = 0 k = – –1– or 3 4
18 Given the function f : x → 5x – 2, find ( f 2)–1 in the same form. Solution Find f 2(x) first. f 2(x) = ff(x) = f(5x – 2) = 5(5x – 2) – 2 = 25x – 10 – 2 = 25x – 12
Try: Question 30, SPM Exam Practice 1 – Paper 1.
Next, find the inverse of f 2(x). Let ( f 2)–1 (x) = y Thus, f 2(y) = x If the question requires 25y – 12 = x us to state the answer 25y = x + 12 in the ‘same form,’ we x— +––12 have to express the y = — –— 25 answer in the form (f 2)–1 : x → ————— x + 12 x + 12 ( f 2)–1(x) = — —–––— 25 25 and not x— +––12 (f 2)–1 (x) = —————. x + 12 ∴ ( f 2)–1 : x → — –— 25 25
1
1
F O R 17 M
CHAPTER
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F O R M
Given that g–1 4 2 – kx (x) 17= ———— and f(x) = 3x, find 4 (a) g(x) in of k, (b) the values of k such that fg(1) = – ––3 k. 2 Solution
The inverse of g–1 will give us back g.
2— –— kx — (a) g–1(x) = — 4 2— –— kx Let y = — — 4 4y = 2 – kx kx = 2 – 4y 2 – 4y x = ———— k 2 – 4y g(y) = ———— k 2 – 4x g(x) = ——— — k
Try: Question 21, SPM Exam Practice 1 – Paper 2.
Rearrange, making x the subject.
19 1 Given that f –1(x) = x + m , x ≠ –m and
g(x) = 3 – x, find 1 –— – –3x (a) the value of m if f(x) = — x— , (b) the values of k if ff –1(k 2 – 7) = g[(k + 2)2]. Solution (a) Let f –1(x) = y f(y) = x 1–— –––3y — — = x y 1 – 3y = xy 1 = xy + 3y 1 = y(x + 3) 1— y = —–— x + 3 1 ∴ f –1(x) = —–— — x + 3 1–––— . But it is given that f –1(x) = —–— x+m Hence, by comparison, m = 3.
fg(1) = – ––3 k 2 2 – 4(1) f ————–— = – ––3 k k 2 –2 f — — = – ––3 k k 2 –2 3 — — = – ––3 k k 2 –6 –3k = k 2 –3k 2 = –12 k 2 = 4 k = ± 2 (b)
Try: Question 20, SPM Exam Practice 1 – Paper 2.
Functions
18
(b) ff –1(k 2 – 7) = g[(k + 2)2] k 2 – 7 = 3 – (k + 2)2
k 2 – 7 = 3 – (k 2 + 4k + 4) k 2 – 7 = 3 – k 2 – 4k – 4 2k 2 + 4k – 6 = 0 k 2 + 2k – 3 = 0 (k + 3)(k – 1) = 0 k = –3 or 1
ff –1(x) = x is always true and it has to be memorised. Try: Question 22, SPM Exam Practice 1 – Paper 2.
1.5 (a) f –1 (b) g–1 (c) g–1f –1 Is ( fg)–1 = g–1f –1?
1 CHAPTER
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1
(d) fg F –1 (e) ( fg) CHAPTER
1 Evaluate f –1(4) for each of the following functions. (a) f : x → 5 – 4x 5 (b) f : x → 6 – —, x x≠0 (c) f : x → 3x + 2 , x ≠ – 3 2x + 3 2 2 Find the inverse function of each of the following functions. (a) f : x → 7x – 4 3x –— 4 (b) g : x → — —–— 2 3 x≠0 (c) h : x → 9 – —, x (d) m : x → 2x + 2 , x ≠ 3 5x – 3 5 pooooooo (e) n : x → 2 – x – x, x 2
7 The function f is defined by f : x4→ 2x – 1. (a) Find the expressions for f 2 dan f –1. (b) Show that ( f –1)2 = ( f 2)–1. 8 Given the function f : x → 4x + h and its inverse kx +—, 5 find function f –1 : x → — –—— 4 (a) the value of h and of k, (b) the expression for f –1 f. x+p , x ≠ 5 and 9 Given the function f : x → x–5 qx + 6 its inverse function f –1 : x → —–––——, x–1 x ≠ 1, find (a) the value of p and of q, (b) the values of x for which f –1(x) = 8x.
4 3 Given the functions f : x → —, x ≠ 0 and x g : x → 2x + 3, find each of the following functions. (a) fg (d) g2 (g) f –1g–1 –1 (b) gf (e) f (h) g–1f –1 2 –1 (c) f (f) g
10
x
y
z
2
3x –—, 1 4 The function f is defined by f : x → — —–— x–2 x ≠ k, find (a) the value of k, (b) f 2, (c) f –1.
–1 –2
The diagram shows the representation of the mapping of y onto x by the function f : y → my + n and the mapping of y onto z by n m the function g : y → — y ≠ — . Find n – 2y 2 (a) the value of m and of n, (b) the function that maps x onto y, (c) the function that maps x onto z.
5 Given the functions f : x → 2 – 4x and g : x → 3 , x ≠ 1, find x–1 –1 (a) f (c) f –1g–1 (e) (gf )–1 –1 (b) g (d) gf Is (gf )–1 = f –1g–1? 6 Given the functions f : x → 1 – 2x and
3x 10 11 (a) Given that f –1(x) = ——+ —— —, find f(x). 4
x–— +— 2 x ≠ 2, find g : x →— , x–2
2x +—, 3 find g(x). (b) Given that g–1(x) = — —— x+4
19
Functions
F O R M 4
1 Paper 1 Short Questions Multiple-choice Questions 1.1 P
1
1
15 24
CHAPTER
4
1
CHAPTER
F O R M
Based on the above information, the relation from P to Q is defined by the ordered pairs {(2, 1), (2, 3), (6, 5), (6, 7)}. State (a) the images of 2, (b) the object of 7. [2 marks]
Relations
35 45
Q
F O R M
6 7 8
The arrow diagram shows the relation between set P and set Q. Based on the arrow diagram, state (a) the domain, (b) the range, (c) the codomain. 2 The arrow diagram shows the relation between set A and set B. A
B
1
6
e
8
’03
Functions
n
20
y z
Set P
30
m
40
k
Set Q
h
State the cant rangeFigure of the relation, 1.1 (a) Signifi
(b) the type of the relation. [2 marks]
6 In the diagram below, set Q shows the images of the elements of set P.
’06
3 –2
16
7 The following arrow diagram SPM Clone represents a linear function f.
’07
4
6
8
Set X
State (a) the relation in the form of ordered pairs, (b) the type of the relation, (c) the domain of the relation. [3 marks]
1.2
Functions
1.3
Absolute Value Functions
Set Q
(a) State the type of relation between set P and set Q. (b) Using function notation, write down a relation between set P and set Q. [2 marks]
Set P
2
81
2
Set P
P = {2, 6, 8} Q = {1, 3, 5, 7, 9}
Set Y
10
x
–3
The above graph represents the relation from set P = {2, 6, 8} to set Q = {5, 7, 9, 11}. State (a) the images of 6, (b) the objects of 7, (c) the range.
4
’10
Set Q 11 9 7 5 2 6 8
SPM Clone
SPM Clone
’04
SPM Clone
(a) Represent the given relation using ordered pairs. (b) State the type of the given relation. 3
5 The diagram below shows the relation between set P and set Q.
SPM Clone
a
5
8 The diagram shows the relation between set X and set Y in the form ’09 of a graph. SPM Clone
9
4
(a) State the value of k. (b) Using the function notation, express f (x) in of x. [2 marks]
x
f(x)
2 5 7 k
1 4 6 8
20
9
x
f
a bx – 2 10
4 1
1
The arrow diagram represents a , x ≠ k, the function f : x → bx – 2 where a, b and k are constants. Find (a) the value of a and of b, (b) the value of k.
1.1
Sig
1.1
Sig
m
x
19 Given the functions f : x → 3x – 7 and fg : x → x 2 + 1, find the function g. 20 The function f is defined by f : x → x – 2. Another function g is such that gf : x → x2 + 1. Find the function g. 21 Given the functions f(x) = 2x – 4 and fg(x) = 8x + 2, find gf(x).
x–h h 7
2
Find the value of h. [2 marks] 13 Given the function f : x → _x – 4_, SPM Clone find the values of x such that ’07 f (x) = 9. [2 marks] 14 The given diagram shows the graph SPM Clone of the function f(x) = |2x – 3| for the ’08 domain 0 ≤ x ≤ 4. y
22 Given the functions h(x) = 7x – 1 and the composite function ’04 hg(x) = 35x + 13, find (a) g(x), (b) the value of x when gh(x) = 4. [4 marks] SPM Clone
23 Given the function g: x → px + q and its composite function ’07 g2: x → 49x – 32, find the value of p and of q such that p > 0. [3 marks]
x ≠ –4, find g–1.
2 , x –3 x ≠ 3 and g(x) = 4x – 1, find 1 f –1g –– . 2 28 Given the functions f(x) =
29 Given the function f : x → 3x + h and its inverse Ffunction 2O f –1: x → kx – ––R, find the value of h 3 and of k. M 4
1
18 Given that f : x → 2x + m and f 2 : x → px + 6, find the value of m and of p.
4x + 1 , x+4
CHAPTER
12 The diagram shows the function x–h ’06 m : x → h , where h is a constant. SPM Clone
17 Given the function f : x → 4 – 5x, find f 2.
27 Given the function g : x →
1
11 Given the function g(x) = 2x – 9, find the possible values of x if g(x) = 4.
16 Given the functions g : x → 4 + x 2 and h : x → 2x – 4, find gh.
CHAPTER
10 (a) Sketch the graph of the absolute value function f (x) = 3 – x for the domain 0 x 4. (b) Hence, state the corresponding range of values of f(x).
24 30 Given the function f(x) = , px + q f(1) = 8, find (a) the value of p and of q, (b) the values of k if f –1 (k) = k. 31 Given the inverse function f –1(x) = 2x – 5, find (a) f(x), (b) the values of k if f –1 f (k) = k 2 – 12.
SPM Clone
32 x
f –1
4x + p
–2
24 Given the function f (x) = x + 4 and g(x) = tx – 6, find ’08 (a) f (6), SPM Clone (b) the value of t such that ’10 gf (6) = 24. [3 marks] SPM Clone
3
O
k
4
x
State (a) the value of k, (b) the range of values of f (x) corresponding to the given domain.[3 marks]
1.4
25 Given the function g: x → 4x – 1 and h: x → 8x, find ’09 (a) hg (x), (b) the value of x if hg(x) = 2g(x). [4 marks] SPM Clone
1.5 Composite Functions
15 The functions f and g are defined by f : x → 3x – 2 and g : x → x + 2 respectively. Calculate fg(5).
–6
The arrow diagram represents f –1(x) = 4x + p, where p is a constant. Find (a) the value of p, (b) f(x).
33 Given the functions f(x) =
2 , 3x + 1
Inverse Functions
4x + 3 , x+5 x ≠ –5, calculate the value of
26 Given the function f(x) = f –1(2).
21
(a) g–1 f(x), (b) the values of x which are mapped onto themselves under the function f. Functions
F O R M 4
x
f
g
y
4
34 Given that g : x → 4x – 1 and h : x → x 2 – 3x + 5, find ’03 (a) g –1(7), SPM Clone (b) hg(x). ’08 [4 marks]
1 CHAPTER
4
CHAPTER
F O R M
1
SPM Clone
m 35 Given the functions h : x → — – – 3, x SPM Clone 10 ’04 x ≠ 0 and h –1 : x → —— — , x ≠ –k, x+k where m and k are constants, find F the value of m and O of k. [3 marks] R M
36 In the following arrow diagram, the function f maps x4onto y and the ’05 function g maps y onto z. SPM Clone
(a) h–1(6), (b) w–1(x). [4 marks]
z
38 Given the functions m: x → 4x – 6 3 , x ≠ 0, find nm–1. ’05 and n: x → — x [3 marks] SPM Clone
1
–2
Write down the value of (a) g –1(4), (b) gf(1). [2 marks] 37 The functions h and w are defined 2 by h(x) = 3x + 5 and w(x) = , ’05 1 – 4x SPM Clone
39 Given the function g : x → 6 – 2x, find ’09 (a) g(–4), (b) the value of p such that g–1(p) = 8. [3 marks] SPM Clone
40 Given the functions g : x → 2x + 3 and h : x → 5x – 8, find ’10 (a) g –1(x), (b) hg–1(11). [3 marks] SPM Clone
Paper 2 Long Questions 1.4
Composite Functions
1 The following diagram shows the functions f and g that are defined 10 by f : x → ax + b and g : x → , c–x x ≠ c respectively. 17 f g
4
m –2
–3
g f
–4
Find (a) the value of a, of b and of c, (b) the value of m, (c) the expression for gf.
2 The functions f and g are defined by f : x → hx – 5 and g : x → x 2 + 3x + 5 respectively. If the composite function fg is defined by fg : x → 2x 2 + kx + 5, find the value of h and of k. Functions
3 The functions f and g are defined ax by f : x → ——— , x ≠ 1 and 1–x g : x → bx – 1 respectively. If f(4) = –4 and g(2) = 3, find (a) the value of a and of b, (b) the value of x for which fg = gf.
4 Given the functions f : x → 3 – x 1 and g : x → ——— , x ≠ 1, find the 1–x expression for each of the following. (a) fg (b) f 2 (c) f 13 g2 1.1 (d) Signifi cant Figure (e) g3 19 (f) g
5 The function f is defined by f : x → 2x + 3. Another function g is such that gf : x → 4x2 + 12x + 15. Find (a) the function g, (b) the values of c if g(c) = 7c, (c) the values of x if fg = gf.
22
6
x
y 8
z g
9
f 5
The above arrow diagram shows the representation of the mapping of x onto y by the function f : x → 4x – a and the mapping of y onto z b by the function g : y → ——––— , 12 – y y ≠ 12. Find (a) the value of a and of b, (b) the expression for the function that maps x onto z, (c) the element x that does not change when it is mapped onto z.
7 The functions f and g are defined by f : x → 2x + 3 and g : x → x 2 + bx + c respectively. If the composite function fg is given by fg : x → 2x 2 + 4x – 3, find (a) the value of b and of c, (b) the value of g2(1).
10 The function g is defined by g : x → x + 2. Another function f is such that fg : x → x 2 + 5x + 7. Find (a) the function f(x), (b) the values of c if f(2c) = 7c.
1.5
Inverse Functions
11 The function f is defined by f : x → mx + k, where m and k are constants. The function g is 12 defined by g : x → ——— , x ≠ –1. x+1 (a) Find the expression for g–1. (b) Find the expression for fg in of m and k. (c) If f(3) = g–1(3) and fg(–2) = –2, find the value of m and of k.
12 Given the functions f : x →
x+1 , x–2
kx + 3 x ≠ 2 and g : x → — —— —, x ≠ 0, x find the value of k if gf –1(0) = 3.
5
f
g–1 3
–1
15 Given the function f : x → 5x + h and its inverse function 2 f –1 : x → kx + —, find 5 (a) the value of h and of k, (b) (i) f(3), (ii) f –1f(3).
mx – n 16 Given the function f : x → ——— ––—, x–2 x ≠ 2 and its inverse function 5 – 2x f –1 : x → ———— , x ≠ 2, find 2–x (a) the value of m and of n, (b) the value of k for which f(k) = k + 2.
17 The following diagram represents the mapping of y onto x by the function f : y → hy + k and the mapping of y onto z by the 6 k function g : y → ,y≠ 2y – k 2 . x
1.1
y
z
19 Given that f : x → p – qx, find (a) f –1(x) in of p and q, (b) the value Fof p and of q if O and f(1) = –2. f –1(8) = –1 R M 4
1
The diagram shows the mapping of the functions f and g–1, such that f : x → 2x + a and g : x → bx + c. Given that g maps 1 onto itself, find (a) the value of a, (b) the value of b and of c.
7
18 Given the functions f(x) = 2x – 5 3x and g(x) = ——— , x ≠ –3, find x+3 (a) f –1g(x), (b) the value of x for which gf(–x) = f 2(2).
CHAPTER
9 The functions f and f 2 are such that f : x → ax + b and f 2 : x → 9x + 8. (a) Find the values of a and the corresponding values of b. (b) Considering a 0, find the values of k for which f(2k + 1) = k(k + 2).
C
B
A
1
14
CHAPTER
8 Given the functions f : x → 3x – 2 and fg : x → 4 – 15x, find (a) the expression for gf, (b) the values of n if gf(n2 – 1) = 6n + 6.
3 – kx 20 Given that g–1(x) = ———— and 2 f(x) = 2x 2 – 3, find (a) g(x) in of k, (b) the value of k for which g(x 2) = 2f(–x).
21 Given the functions f : x → hx + k, h 0 and f 2 : x → 25x – 18, find (a) the value of h and of k, (b) ( f –1)2 in the same form.
1 ,x≠k k–x and g(x) = 2 + x, find (a) f(x) in of k, (b) the value k if ff –1(k 2 + 2) = g[(5 + k)2].
22 Given that f –1(x) =
Signifi cant Figure 3 2
–2
x+p , x+q x ≠ –q and its inverse function 2 – 3x f –1 : x → , x ≠ 1, find 1–x (a) the value of p and of q, 1 (b) the values of k if f 2(k) = —k. 3
13 Given the function f : x →
Find (a) the value of h and of k, (b) the function that maps x onto y, (c) the function that maps x onto z.
23
23 Given that f : x → 2x – 3 and x g : x → + 2, find ’06 2 SPM Clone
(a) f –1 g,[3 marks] (b) the function h such that hg : x → 2x + 4. [3 marks] Functions
F O R M 4
Paper 1
Short Questions 1 The relation between set X = {6, 12, 15, 21} and set ’11 Y = {3, 5, 7} is ‘factor of’. (a) Find the image of 12. (b) Express the relation in the form of ordered pairs. F O [3 marks]
1 CHAPTER
2 Given the functions g(x) = 5x – 11 and h(x) = 3x, find the value of ’11 gh(2). [2 marks]
3 The inverse function h–1 is defined 3 by h–1 : x → , x ≠ 4. Find ’11 4–x (a) h(x), (b) the value of x such that h(x) = –14. [4 marks]
State (a) the object of 40, (b) the type of the relation. [2 marks]
3 Given the function f (x) = px + q and f 2 (x) = 4x + 9, where p and q are constants, find the value of p and of q if p < 0. [3 marks]
2 The relation between two variables is represented by the following set of ordered pairs:
4 Given the function f : x → x + 4 and the composite function gf : x → x 2 + 6x + 2, find (a) g(x), (b) fg(4). [4 marks]
SPM Clone
R M
Paper 1
1 The graph below shows the relation between set A and B.
40 30
{(–4, 16), (–3, 9), (–2, 4), (2, 4), (3, 9), (4, 16)}
20 10
1
2
3 Set A
4
(a) State the type of the above relation. (b) Represent the above relation using a function notation. [2 marks]
2x + p and 5 5x + 3 , its inverse function f –1(x) = q find the value of p and of q. [4 marks]
5 Given the function f (x) =
(a) Find the value of h and of k. [4 marks] (b) If gf –1(x) = –5x, find the values of x. [3 marks]
Paper 2
Long Questions 1 The function f and its inverse function f –1 are hx kx , x ≠ 3, f –1(x) = ,x≠2 defined by f(x) = x–3 x–2 respectively, where h and k are constants. Another 1 function g is defined by g(x) = , x ≠ 0. x
Functions
SPM Clone
4
Short Questions
Set B
4
CHAPTER
F O R M
1
SPM Clone
2 Given the functions f : x →
x – 2 and g : x → 3x + k, 2
where k is a constant, find (a) f –1(3), [2 marks] (b) the value of k if f –1g: x → 6x – 4, [2 marks] (c) h(x) such that hf : x → 9x – 3. [2 marks]
24
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