Modul 1 Functions EDISI GURU

1.1 FUNCTIONS Learning Outcome:
a) State the domain and range of a function, and find composite functions.

➢ DEFINITION
A function ia a special type of relation where each and every element in the domain has only
one image in the range.

➢ NOTES
The types of relations that are connected as a functions are:

a) One-to-one relation b) many-to-one relation

a. .d a.
b. .e b. .d
c. .f c. .e

Domain Range Domain Range

A function f from a set X to a set Y is defined as a rule that associates exactly one element of Y
with each element of X. We say, f maps X into Y and we write f ∶ →

If ∈ and ∈ such that y is assigned to x under f, we say f maps x to y and write
f ∶ → y or y = f(x), where y is the image of x.

In general, a function f is defined for certain values of x only. This set of values of x for which
f is defined is called the domain of f. The set of values of f(x) for a given domain is called the
range of f.

For more informations, scan this QR code.

Composite Functions
If f is a function which maps set X to set Y and g is a fuction which maps set Y to set Z, then
g ∘ f is a composite function which maps set X directly to set Z, as illustrated in figure.

g∘f

x f y=f(x) g z
=g(y)

=gff(x)
]

XY Z

2

For g∘f to be defined, the range of f must be a subset of the domain of g,
⊆ .

For more informations, scan this QR code.

EXAMPLE 1
The domain of a function h, where h: x → 2 , is ℝ
a) Find the images of -1 and 6.

b) Find the element in the domain with images 16.

c) Find the range of the function h.

Solution :

(a) For x = -1,

h(−1) = 2−1

= 1
2

For x = 6,
h(6) = 26

= 64

(b) h( ) = 16
2 = 16
2 = 24

= 4

(c) The range is ℝ+.

EXAMPLE 2
State the set of value of x for which each of the following function is defined.
(a) f: → √ − 1
(b) h: → 1

−1

(c) g ∶ → lg( − 2)

Solution :
(a) For f defined, − 1 ≥ 0

≥ 1
Hence, the domain of the function f is { : ≥ 1}

(b) For g defined, − 2 > 0
> 2

Hence, the domain of the function g is { : > 2}

(c) For g defined, − 1 ≠ 0
≠ 1

Hence, the domain of the function g is { : ≠ 1}

3

EXAMPLE 3
Given the function f( ) = √ and g( ) = 2 + 9
(a) Determine the domain and the range of f and g.
(b) Explain why f∘g exist.
(c) Find f∘g and state its domain and range.

Solution :

(a) For f defined, ≥ 0
Then, the domain of f is { : ≥ 0}
The domain of g is { : ∈ ℝ}.

(b) The range of g is { : ≥ 9}
The domain of f is { : ≥ 0}
Since ⊆ , f ∘ g exist.

(c) f ∘ g(x) =f[g( )]
= f( 2 + 9)
= √ 2 + 9

The domain of f ∘ g is the same as the domain of g, that is { : ∈ ℝ}.
The range of f ∘ g is { : ≥ 3}.

EXAMPLE 4
If f( ) = cos and g( ) = 1 + , ∈ ℝ, find
(a) f ∘ g( )
(b) f ∘ f( )
(c) g ∘ f( )

Solution :

(a) f ∘ g( ) = f[g( )]
= f(1 + )
= cos (1 + )

(b)f ∘ f( ) = f[f( )]
= f( cos )
= cos (cos )

(c) g ∘ f( ) = g[f( )]
= g(cos )
= 1 + cos

4

EXAMPLE 5

If f( ) = 2 − 3 and f ∘ g( ) = 2 + 1 , find g(x)

Solution :

If f ∘ g( ) = 2 + 1
f[g( )] = 2 + 1
2g( ) − 3 = 2 + 1
2g( ) = 2 + 4
g( ) = + 2

EXAMPLE 6

Given that the function f: → 2 + 1, find the function g if the composite function
g ∘ f ∶ → 1 , ≠ 1.

−1

Solution:

If g ∘ f( ) =1

−1

g[f( )] = 1

−1

g(2 + 1) = 1

−1

Let = 2 + 1

Then = 1 ( − 1)
2
1
Hence, g( ) = 12( −1)−1

=2
−3

Then, g( ) = 2 , ≠3

−3

5

WORKSHEET 1.1 (a)

Q1 Given that g: → 2 + + 1 and ∈ X, where = {−2, −1, 0, 1, 2}..
a) Sketch the diagram to represent g.
b) State the domain of g.
c) State the range Y of g.
d) Is g a one-to-one functions?

Solution :

1.(a) -2 . .1
-1 . .3
0. .7
1
2

X Y
(b) Domain of g is {-2, -X1, 0, 1, 2}
(c) Range of g is {1, 3, 7}
(d) No. g is many-to-one functions

Q2 Function f is defined by f: → 3 − 5 , ∈ ℝ. Find
a) f(2)
b) The value of x whose image is 13
c) f( 2)
d) f(2x-1)
e) the value of x if f( 2) = f(2 − 1)

Solution:

2(a) f(2) = 3 − 5(2)
= -7

f( ) = 3 − 5

(b) 3 − 5 = 13
−5 = 10

= −2

6

(c) f( 2) = 3 − 5 2

(d) f(2 − 1) = 3 − 5(2 − 1)
= 8 − 10

f( 2) = f(2 − 1)
3 − 5 2 = 8 − 10
(e) 5 2 − 10 + 5 = 0
2 − 2 + 1 = 0
( − 1)2 = 0
= 1

Q3 Determine the domain of f such that f is a function:

a) f ∶ → ln ( + 2) + 1
√1−
1
b) f ∶ → 2+2 −8 + √

Solution :

For f defined :
+ 2 > 0 and 1 − x > 0
3(a) > −2 and < 1
Then, domain of f is { : − 2 < < 1}

For f defined,

(b) 2 + 2 − 8 > 0 and ≥ 0
≠ −4 or 2 and ≥ 0

Then, domain of f is { : 0 ≤ < 2 > 2}

7

Q4 Given that g: → a + b , ∈ ℝ, g(1)= -3 and g(-1) =1, find
a) the values of a and b,
b) The values of n if g( 2 + 1) = 5 − 6

Solution :

g(1) = −3
a + b(1) = −3
a + b = −3 …… ①

g(−1) =1
a + b(−1) = 1

a−b = 1
a = 1 + b …..②

4.(a)

① → (1 + b) + b = −3
1 + 2b = −3
b = −2

a − 2 = −3
a= -1

Then, a = -1 and b= -2
g( 2 + 1) = 5 − 6

−1 − 2( 2 + 1) = 5 − 6
(b) 2 2 + 5 − 3 = 0

( + 3)(2 − 1) = 0
= −3 = 1

2

8

Q5 The functions f and g are defined by f( ) = 2 − 4 and g( ) = √ − 2 respectively.
a) Determine the domain and the range of
i) f ii) g
b) Explain why f ∘ g exist.
c) Find f ∘g and determine its domain.

Solution :
i) Domain of f is { : ∈ ℝ}
Range of f is { ∶ ≥ −4}

5.(a)
ii) Domain of g is { : ≥ 2}
Range of f is { ∶ ≥ 0}

(b) Because ⊆

fg( ) = f(√ − 2)
= (√ − 2)2 − 4

(c)
= − 6

Domain of f ∘ g is { : ≥ 2}

Q6 If f( ) = √ , ≥ 0 and g( ) = 1 − 2, ∈ ℝ, find
a) f ∘ g( )
b) g ∘ f( )
c) g ∘ g( )

Solution :

fg( ) = f(1 − 2)
= 1− x2 , x  −1 and x  1

6.(a)

gf( ) = g(√ )
(b) = 1 − (√ )2

= 1 −

gg( ) = g(1 − 2)
= 1 − (1 − 2)

(c)
= 2

9
EXERCISE 1.1 (a)

Q1 Given that f( ) = √ + 1, ≥ −1 ,

a) Find the value of

i) f(8) ii) f(0)

b) Find the value of x such that

i) f( ) = 5 ii) f( ) = 10

Q2 Determine the domain of f such that f is a function:

(a) f: → √16 − 2 − ln ( 2 − − 2)

(b) f: → 4 − ln (1 − )
√ +4

Q3 The functions f and g are defined by f: → and g: → +4 respectively.

+2

(a) State the domain of f and g.
(b) 5 + 8 , ≠ 0Find the composite function g ∘ f and state its domain.



,

Q4 The functions g and h are defined by f( ) = 25 , 1 < < 9, ∈ ℝ and
3 −2
g(x) = 2, 1 < < 3, ∈ ℝ .

a) Determine the range of

i) f ii) g

b) Explain why f ∘ g exists. Hence, find f ∘g, starting its domain.

Q5 The functions f and g are defined by f: → 2 − 1, ∈ ℝ and g: x → 2 + 3, ∈ ℝ respectively.

Find

a) (g ∘ f)(3)

b) (f ∘ g) (1)

2

10 Functions - Module 1

ANSWERS 1(a) i) f(8) = √8 + 1
Section =3
1.1
ii) f(0) = √0 + 1
Exercise =1
1.1(a)
(b)
i) f( ) = 5
√ + 1 = 5
+ 1 = 25
x = 24

ii) f( ) =10
√ + 1 =10
+ 1 =100
x = 99

2(a) For f defined:
16 − 2 ≥ 0 and 2 − − 2 > 0
(4 − )(4 + ) ≥ 0 and ( + 1)( − 2) > 0
−4 ≤ ≤ 4 and < −1 or > 2
Then, domain of f is { : − 4 < < −1 2 < ≤ 4}

(b) For f defined:

+ 4 > 0 and 1 − > 0
> −4 and < 1
> −4 and < 0
Then, domain of f is { : − 4 < < 0}

11 Domain of f is {x: x ∈ ℝ, x ≠ −2}
3(a) Domain of g is {x: x ∈ ℝ, x ≠ 0}
Section
1.1 gf( ) = g ( )
(b) +2
Exercise
1.1(a) = + 2+4

+2

= +4( +2)



= 5 +8



= 5 + 8 , ≠ 0 ……. ①



The domain of g ∘f follows the domain of f, that is { : ∈ ℝ, ≠
−2}. ………②

Combining ①and ②, the domain of g ∘f is { : ∈ ℝ, ≠ −2, ≠
0}

4(a) i) Range of f is { : 1 ≤ ≤ 25}
ii) Range of g is { : 1 ≤ ≤ 9}

Because ⊆ .
(b) fg( ) = f( 2)

= 25 , ≠ ±√2
3 2−2
3

Domain of f ∘g is { : 1 ≤ ≤ 3}

5(a) Method 1
gf(3) = g(32 − 1)
= g(8)
= 2(8) + 3
= 19

12 Method 2
gf( ) = g[f( )]
Section = g( 2 − 1)
1.1 = 2( 2 − 1) + 3
= 2 2 + 1
Exercise gf(3) = 2(3)2 + 1
1.1(a) = 19

(b) Method 1

fg (1) = f(2 (1) + 3)

2 2

= f(4)

= 42 − 1

= 15

Method 2

fg( ) = f(2 + 3)

= (2 + 3)2 − 1

= 4 2 + 12 + 8

fg (1) = 4 (1)2 + 12 (1) + 8
22 2

= 15

13 Learning Outcome:
1.1(b) b) Determine whether a function is one-to-one, and find the inverse of a one-to-one
ONE-TO-ONE function
FUNCTIONS

➢ DEFINITION

One-to-one function is a function that maps distinct elements of its domain to distinct elements
of its codomain.

➢ NOTES

1. Every element of the function’s codomain is the image of at most one element of its domain.
A function f : D → S is a one-to-one function if and only if a,bD, f (a) = f (b) implies
that a = b .
2. An inverse function ( or anti-function) is a function that “reverses” another function . If the
function f applied to an input x gives a result of y , then applying its inverse function g to
y gives the result x , that is g( y) = x if and only if f (x) = y . The inverse function of f is
also denoted as f −1 .

➢ CONCEPT
1. To determine whether or not a function is a one to one function , we can use the horizontal
line test.

2. If there is only one intersection point, the function is a one to one function.

3. If f is a one-to-one function :

• the domain of f −1is the range of f
• the range of f −1is the domain of f
4. The graph of f and f −1are symmetrical about the straight line y=x

5. Inverse functions ONLY exist for one-to-one functions.

6. A function which is NOT one-to-one can have an inverse function if we RESTRICT the
domain of the function so that it is one-to-one. ****USE COMPLETING THE SQUARE****

REMEMBER THIS !

: → 2, ∈ ℝ ( not one-to-one )
: → 2, ∈ ℝ, ≥ 0 ( one-to-one
)

EXTRA KNOWLEDGE !

EXAMPLE 1
Determine whether each of the following functions is one-to-one.

(a) f (x) = x3

(b) g(x) = x

14

Solution :
(a) Method 1 (using definition)

Let, x1 = x2
Then, x13 = x23

f (x1) = f (x2)
 f (x) = x3 , is one-to-one function
Method 2 (graphical method)

100 y

0 5
-5 0

-100

Since no horizontal lines intersect the graph of f (x) = x3 more than once, by the Horizontal
Line Test the function f is one-to-one.

(b) Method 1 (using definition)
Let x1 = x2

Then, x1 = x2

g(x1) = g(x2 )

g(x) = x is one-to-one function

Method 2 (graphical method)

y

2.5
2

1.5
1

0.5
0
012345

Since no horizontal lines intersect the graph of g(x) = x more than once, by the Horizontal
Line Test the function g is one-to-one.

15

EXAMPLE 2
A function is defined by :
f :x → 3−x
a) find f −1
b) determine the domain and the range of

i) f
ii) f −1
c) sketch the graph of y = f (x) and y = f −1(x) on the same axes.

Solution : y = f −1(x)
a) Let
f (y) = x

3− y = x
3− y = x2

y = 3− x2
 f −1 : x → 3 − x2

b) i) domain = x x  3

range = y y  0

ii) domain = x x  0

range = y y  3

c)

16

EXAMPLE 3
The function g is defined by :

g : x → 2
x2 +1

a) Sketch the graph of y = g(x)

b) State the domain of g so that g −1exists. Determine the corresponding range of g .

c) Find g −1, hence determine the domain and range of g −1.

d) Sketch the graph of y = g(x) and y = g −1(x) based on the domains and ranges determined in (b)
and (c).

Solution :
a)

y

3 y
2 5
1
0
-5 0

b) In order for g −1to exist, g must be one-to-one. Hence, the domain of g must be restricted to only

x x  0 and range of g is y 0  y  2 .

c) Let y = g −1(x) d)

g(y) = x

2 = x
y2 +1

2 = xy2 + x

y2 = 2− x
x

y = 2−x
x

g −1(x) = 2 − x
x

17

WORKSHEET 1.1(b)

Q1 Determine whether each of the following functions is one-to-one.

a) f (x) = 2x − 5

b) h(x) = 5 − x2

c) p(x) = 1
x

Solution:

(a) Method 1 (using definition)

Let 1 = 2 ℎ ,
2 1 = 2 2
2 1 − 5 = 2 2 − 5
( 1) = ( 2)

Therefore, f (x) = 2x − 5 is one-to-one
Method 2 (graphical method)

10 y

5 y
5 10
0
0

-5

-10

Since no horizontal lines intersect the graph of f (x) = 2x − 5 more than once, by the

horizontal line test, the function is one-to-one.

(b)

Method 1 (using definition)

Let 1 = −2 2 = 2
ℎ(−2) = 5 − (−2)2 = 1
ℎ(2) = 5 − (2)2 = 1
, −2 ≠ 2 ℎ(−2) = ℎ(2) = 1
ℎ , ℎ ℎ( ) = 5 − 2

18

Method 2 (graphical method)

y y
5
6
4
2
0
-5 -2 0
-4
-6

As seen in the above graph, the horizontal line intersects the graph of h(x) = 5 − x2 more
than once. By the horizontal line test, the function is not one to one.

(c) Method 1 (using definition)

Let 1 = 2, ℎ
11
1 = 2
( 1) = ( 2)

Method 2 (graphical method

y y

1.5 5
1

0.5
0

-5 -0.5 0
-1

-1.5

Since no horizontal lines intersect the graph of p(x) = 1 more than once, by the
x

horizontal line test the function is one to one

19

Q2 Given the function f, f : x → x2 . Show that the function is not one-to-one.

Solution:
Since -1,1 ϵ ʀ (domain) and f(-1)=1=f(1) we can see that the function f is not one to one.

Q3 Given that :

f (x) = 3x + 1, x0
2x −1, x 0

Show that the above function is one to one .

Solution:
Suppose a,b ϵ ʀ (domain), there are 4 cases to consider :
Case 1 : a.b ≥ 0

( ) = ( )
∴ 3 + 1 = 3 + 1
∴ =

Case 2 : a ≥ 0, b < 0

( ) = ( )

∴ 3 + 1 = 2 − 1
2

∴ = 3 ( − 1)
∴ < 0

This contradicts the choice a ≥0. Therefore, when a ≥0, b <0, f(a) ≠f(b).
Case 3 : a < 0, b ≥ 0
Argue as in case 2, we can see that in this case, f(a) ≠ f(b).

Case 4 : a <0, b <0

f (a) = f (b)
2a −1 = 2b −1

a = b

We see that in any case, f(a)=f(b) always implies a=b, so f is one to one.

20

Q4 Determine if :
f (x) = sin x − sin x , 0  x  2

is one to one and give reason to your answer.
Solution:

If 0  x   ,sin x is positive or zero, so sin x = sin x, and f (x) = 0 .
If   x  2 ,sin x is negative or zero, so sin x = −sin x, and f (x) = 2sin x .

From the graph, the function is not one-to-one since f ( ) = f (2 ) = 0

Q5 Given that the following functions are not one- to- one . Determine the maximum domain of

each function so that it can have an inverse.

a) ( ) = 3 + 4 2, ∈ ℝ

b) ( ) = 2 − 4 + 1, ∈ ℝ

c) ( ) = 3 + 2 − 2, ∈ ℝ

d) ( ) = 2( + 1) 2, ∈ ℝ

e) ( ) = 1 , ∈ ℝ
2+1

Solution:
(a) x ϵ ʀ, x ≥ 0 or x ≤ 0
(b) x ϵ ʀ, x ≤ 2
(c) x ϵ ʀ, x ≤ 1
(d) x ϵ ʀ, x ≥ 0
(e) x ϵ ʀ, x ≥ 0 or x ≤ 0

21
EXERCISE 1.1(b)

Q1 Given the function f (x) = 4 − x .

a) Find the domain and range of the function.
b) Determine whether f −1 exists. If it exists, find its expression, domain and range.

Q2 Given the function f (x) = x +1 , x  2 is one to one. Find f −1(x) .
x−2

Q3 A function f is defined by f : x → 2 + 2x −1, x  1.

a) Find f −1(x) .
b) Determine its domain and range.

Q4 The function f is defined by f (x) = ex − e−x , x  . Find the inverse of f.
ex + e−x

Q5 The function f is defined by f (x) = x2 − 2x + 2,0  x 1. Sketch the graphs of
y = f (x) and y = f −1(x) on the same axes.

22

ANSWER Module 1

Section 1. (a) 4−x0 x4
1.1  Df = (−, 4]

Exercise 4 − x  0  f (x)  0
1.1(b) Rf = [0, )

1.(b) Let x1, x2  Df , where x1 = x2 , then

4 − 1 = 4 − 2
Thus √, t(4h e1−)fu = n1c =t i(o √n 24i)s−on e2to one. Hence, f −1 exists.
Let :

= √4 −
2 = 4 −
= 4 − 2
−1( ) = 4 − 2
∴ −1( ) = 4 − 2

Domain of f −1 = Range of f = [0, ) Range of f −1 = Domain of f = (−, 4]

2. Let

+ 1
= − 2
( − 2) = + 1

− 2 = + 1

− = 2 + 1

( − 1) = 2 + 1

2 + 1
= − 1

−1 ( ) = 2 + 1
− 1

∴ −1( ) = 2 −1 , ≠ 1

−1

3. Let
= −1( )
( ) =

2 + √ − 1 =

√ − 1 = − 2
− 1 = ( − 2)2
− 1 = 2 − 4 + 4
= 2 − 4 + 5
∴ −1( ) = 2 − 4 + 5

23 The domain of f −1 is the same as the range of f that is { | ≥ 2, ∈ ℝ}.
The range of f −1 is the same as the domain of f that is { | ≥ 1, ∈ ℝ}.
Section
1.1 4. Let

Exercise
1.1(b)

−1( ) = ⇒ ( ) =

− −
+ − =
2 − 1
2 + 1 =
2 − 1 = 2 +

(1 − ) 2 = 1 +

2 = 1 +
1 −
1 1 +
= 2 (1 − )
1 1 +
∴ −1 ( ) = 2 (1 − )

∈ ℝ, −1

5.

24 Learning Outcome:
1.1(c) a) sketch the graphs of simple functions, including piecewise-defined

GRAPH OF functions.

FUNCTION

GRAPH OF FUNCTION
➢ DEFINITION

The graph of a function f is the set of all points in the plane of the form (x, f(x)). We could also
define the graph of f to be the graph of the equation y = f(x).

➢ CONCEPT

A. Basic shape of the graph > 0 < 0

Type of function
Linear function

( ) = + , ≠ 0

Domain = ℝ Domain = ℝ
Range = ℝ Range = ℝ

Quadratic function
( ) = 2 + + , ≠ 0

Cubic function Domain = ℝ Domain = ℝ
( ) = 3 + 2 + + , Range = [ , ∞) Range = (−∞ , ]

≠ 0 Domain = ℝ Domain = ℝ
Range = ℝ Range = ℝ

Reciprocal function

( ) = 1 , ≠ 0
−

Domain = (−∞, ) ∪ ( , ∞) Domain = (−∞, ) ∪ ( , ∞)


Range = (−∞, 0) ∪ (0, ∞)
Range = (−∞, 0) ∪ (0, ∞)
2
( ) = , ≠ 0

Domain = (−∞, 0) ∪ (0, ∞) Domain = (−∞, 0) ∪ (0, ∞)
Range = (0, ∞) Range = (−∞, 0)

25

Square root function
( ) = ±√ −

Domain = [ , ∞) Domain = [ , ∞)
Range = [0, ∞) Range = (−∞, 0]

( ) = ±√ −

Domain = (−∞, ] Domain = (−∞, ]
Range = [0, ∞) Range = (−∞, 0]

B. Graphs of modulus functions

A modulus function is a function which gives the absolute value of a number or variable. The
outcome of this function is always positive, no matter what input has been given to the function.

The graphs of = + and = | + | are as below:

For x < − , the graph = | + | is the reflection of the graph



= + about x-axis

C. Graphs of piecewise functions
A piecewise-defined function is a function defined by multiple sub-functions, where each sub-function
applies to a different interval in the domain.

26 Quick Notes 1.1(c)

➢ NOTES https://youtu.be/-
P5VUP_NQVM
Step:
To sketch the graph of the function, we need to perform the
following:
1. Determine type of function and corresponding shape.
2. Find y-intercept (point where (0).
3. Find x-intercepts (points where ( ) = 0.
4. Find what asymptotes does function have, if any.
5. Find stationary points (minimum, maximum and inflection point).
6. Use test to classify stationary points.
7. Add "control" points (some arbitrary points), if needed.
8. Draw "important" and "control" points and connect them by lines

taking into account found behaviour of the function.
9. If function is even, odd or periodic then perform corresponding

reflection.
10. If function is obtained by transforming simpler function, perform

corresponding shift, compressing/stretching.

LLET
DANCE
TOGETHE

R!

27

EXAMPLE 1:

Sketch on separate diagram, the graphs of

(a) = 1

−2
|1|
(b) =
−2
1
(c) 2 = −2

Solution:

(a) = 1
−2

a>0 shape
The asymptotes are x=2 , y = 0

(b) = | 1 | = 2
= 2
−2

The asymptotes are x = 2 , y = 0

Reflection y negative become y positive

(c) 2 = 1

−2

The asymptotes are x = 2 , y = 0

Reflection x < 2 become x > 2

= 2

28

EXAMPLE 2

Sketch the graph of = 2 + − 2, noting the points of intersection of the graph with x-axis and y-
axis.
On a separate diagram, sketch the graph of = |2 + − 2|

Solution:

= 2 + − 2
• a = -1 < 0 shape
• When y = 0 2 + − 2 = 0
(1 + )(2 − ) = 0
= −1 2
• When x = 0, y = 2

Graph of = 2 + − 2 shown in diagram (a)

Diagram (a)

The graph of = |2 + − 2|
can be drawn by reflecting the section
of graph that is below the x-axis
as shown in diagram (b).

Diagram (b)

29

EXAMPLE 3: Quick Notes
Example 3
Sketch the graphs of
Note :
( ) = {−2 2++32, < 1 ● indicates this point
, ≥ 1 is included
o indicates this point
Solution: is excluded

• For the region x<1,
✓ we have a straight line with slope 2 and y-intercept 3.

✓ As x approaches 1, the value of the function approaches 5
(but does not reach it because of the “<” sign)

• For the region ≥ 1
✓ When x =1,

the function has value (1) = −(1)2 + 2 = 1
✓ As we go further to the right,

the function takes value based on ( ) = − 2 + 2
✓ It is a parabola.

30 2. ( ) = √ − 2

WORKSHEET 1.1(c) Solution:
Sketch the graph of the following: • shape
1. ( ) = 3 2 − 2 − 1 • − 2 ≥ 0
≥ 2
Solution: • When y = 0, = 2
• a = 3 > 0 shape • When ≥ 0, ≥ 2
• When y = 0 3 2 − 2 − 1 = 0
(3 + 1)( − 1) = 0
= − 1 = 2

3

• When x = 0, = −1

3(a). ( ) = | + 1| 3(b) With the aid of graph of ( ) in 3(a),
sketch the graph of 2 ( ).
Solution:
Solution:
( ) = | + 1| = {− ( ++11,), > −1 y value of graph 3(a) multiple by 2
≤ −1

3(c) With the aid of graph of ( ) in 3(a), 3(d) With the aid of graph of ( ) in 3(a),
sketch the graph of ( + 2). sketch the graph of ( ) − 2.

Solution: Solution:
The graph of 3(a) transform 2 step to left The graph of 3(a) transform 2 step downward

31

4(a) = 4(b) 2 =

−2 −2

Solution: Solution:

shape The asymptotes are = 2 , = ±1

The asymptotes are x = 2 , y = 1 Since 2 ≥ 0 , hence ≥ 0

lim = +∞ −2

→2+ ≤ 0 , ≥ 2

lim = −∞ The curve is symmetrical about the x-axis.

→2−

2 , < 0
5. ( ) = { , 0 ≤ < 2

−2 , ≥ 2

Solution:
• For the region < 0
✓ When → 0 , → 0
✓ As we go further to the left,
the function takes value based on ( ) = 2
✓ It is a parabola.

• For the region 0 ≤ < 2,
✓ we have a straight line with slope 1.
✓ As → 2 , → 2

• For the region ≥ 2
✓ It is a horizontal line where ( ) = −2

32

EXERCISE 1.1(c)

1. Sketch the graph of (b) = −3 + 4
(a) = 2 − 3

2. Sketch the graph of

(a) = 2 − 4 (c) 2 = 2 − 4

(b) = 1
2−4

3. Given ( ) = − 1 3 + 3 + 1. Sketch the following graphs separately
2 2

(a) = ( ) (c) = −3 ( )

(b) = ( + 1) (d) = (2 )

4. Sketch in separate diagrams, the graphs of

(a) = +1 (c) 2 = +1

−2 −2
(b) = | +1|

−2

5. Sketch the graph of (c) = −| + 2| + 3

(a) = | |
(b) = 2| − 1| − 4

6. The function f is defined by ( ) = {−32 +−28, , < −2
Sketch the graph of f . > −2

7. The function f is defined by ( ) = {5 −− 1 2,, ≤2
Sketch the graph of f . >2

8. The function f is defined on the domain [-3,3] as follows:

( ) = {−11 , −3 ≤ < 0
, 0 ≤ <3

Sketch the graph of f .

9. The function f is defined by sin , < −2
Sketch the graph of f

( ) = { 2 − 2 , −2 ≤ < 2
2 − 8 + 10 , ≥ 2

33

ANSWER Module 1

Section 1. (a)
1.1 = −

Exercise • Linear graph
1.1(c) • Gradient = 2
• y-intercept = -3

(b)
= −3 + 4

• Linear graph
• Gradient = 2
• y-intercept = -3

2.(a) (0,4)
= 2 − 4 (2,4)

• a = 1 > 0 shape
• When y = 0 2 − 4 = 0

( − 4) = 0
= 0 4

• When x = 0, y = 0
• = 2 − 4

= ( − 2)2 − 4
The graph has minimum point
at (2,4)

34

Section 2.(b)
1.1
= = X=4
Exercise − ( − )
1.1(c)
• Since ( − ) ≠ ,
The asymptotes are x = 0, x = 4 and y = 0

2(c) (4,0)
2 = 2 − 4
Since 2 ≥ 0
Therefore 2 − 4 ≥ 0

( − 4) ≥ 0

≤ 0, ≥ 4

The curve is symmetrical about the x-axis

since the multiples of y are even.

3(a) (1,2)
(-1,0)
( ) = − 1 3 + 3 + 1

22

• a = − 1 < 0 shape

2

• When f (x) = 0 − 1 3 + 3 + 1 = 0

22

= −1 2

• When x = 0, y = 1

• ′( ) = − 3 2 + 3
22
3 3
When ′( ) = 0 , − 2 2 + 2 =0

= −1 1

When = −1 , ( ) = 0

When = 1 , ( ) = 2

Minimum point at (-1,0)

Maximum point at (1,2)

35

Section 3.(b)
1.1
= ( + )
Exercise Graph 3(a) move 1 step to left
1.1(c)
(0,2)
(-2,0)

3(c)
= −3 ( )
Stretch Graph 3(a) with factor scale 3
then reflect on x-axis

(-1,0)

3(d) (1,-6)
= (2 )
Compress Graph 3(a) with scale factor 2 (1/2,2)
In direction parallel with x-axis, (1/2,0)

36

Section 4.(a)
1.1
= +
Exercise
1.1(c) −

a>0 shape y=1

The asymptotes are x=2 , y = 1

When y = 0 = −
When x = 0, = −



4(b) x=2 y=1
+ 1
(- y=1
= | − 2| 1,0) y= -1
= | +1| ≥ 0 for all values of x
x=2
−2
(-
4(c) 1,0)
2 = +1

−2

Since 2 ≥ 0 , hence +1 ≥ 0

−2

≤ −1 , ≥ 2

x=2

37

Section 5(a) = | |
1.1
• Sketch line y = x
Exercise • For x < − ,
1.1(c)


the graph is the reflection
about x-axis

5(b) (1,-4)
= 2| − 1| − 4 (-2,3)

• Sketch line = 2( − 1)
• for x < 1 ,

the graph is the reflection
about x-axis
• the graph move down 4 steps

5(c)
= −| + 2| + 3

• Sketch line = −( + 2)
• for x < −1 ,

the graph is the reflection
about x-axis
• the graph move up 3 steps

38

Section 6.
1.1
( ) = {− +− , , < −
> −

Exercise • For the region x < -2, use ( ) = − −
1.1(c) ✓ we have a straight line with slope -2.
✓ As x approaches -2, the value of the function approaches -4 (but does
not reach it because of the “<” sign)

• For the region > − , use ( ) = +
✓ we have a straight line with slope 3.
✓ As x approaches -2, the value of the function approaches -4 (but does
not reach it because of the “>” sign)

7.

( ) = {5 −− 1 2,, ≤2
>2

• For the region ≤ 2

✓ When x =2, the function has value (2) = 5 − (2)2 = 1
✓ As we go further to the left, the function takes value based on

( ) = 5 − 2
✓ It is a parabola with maximum point (0,5).

• For the region x>2,
✓ we have a straight line with slope 1.

✓ As x approaches 2, the value of the function approaches 1 (but does
not reach it because of the “>” sign)

39

Section 8.
1.1
( ) = {− , − ≤ <
Exercise , ≤ <
1.1(c)
• For the region − ≤ <
✓ It is a horizontal line y = -1 including x = -3 and excluding x = 0

• For the region ≤ <

✓ It is a horizontal line y = 1 including x = 0 and excluding x = 3

9.

sin , < −2

( ) = { 2 − , −2 ≤ < 2
2

2 − 8 + 10 , ≥ 2

• For the region x< -2,
✓ we have a sin x curve with maximum point(−1.5 , 1), … and

minimum points (− , −1) (−2 , −1),…
✓ As x approaches -2, the value of the function approaches -0.9093 (but

does not reach it because of the “<” sign)

• For the region −2 ≤ < 2,

✓ we have a straight line with slope − 1 and y-intercept 2.
2
(−2)
✓ when x = -2, the function has value (−2) = 2 − 2 = 3

✓ As x approaches 2, the value of the function approaches 1 (but does

not reach it because of the “<” sign)

• For the region ≥ 2

✓ When x =2, the function has value (2) = 22 − 8(2) + 10 = −2
✓ As we go further to the right, the function takes value based on

( ) = 2 − 8 + 10
✓ It is a parabola.

40
CLONE STPM 1.1

Q1 The function f and g are defined by f( ) = 2 ln , > 0 and g( ) = √ − 1, >1

(a)Sketch the graphs of f and g.
(b)Find f −1 and g−1, stating the domains and ranges of f −1 and g−1.
(c)Determine whether g ∘ f exist.

Q2 Functions f and g are defined by f(x) = ex+2 and (g o f)(x) = x , for all x ≥ 0.

(a) Find the function g, and state its domain.
(b) Determine the value of (f o g)(e3).

Q3 (a) The function is defined as : h( ) = | + 3| − | − 3|, ∈ .
(i) Sketch the graph of h.
(ii) State the range of h.

(b) The functions f and g are defined as:

f( ) = +| | , ∈
2
− 1, < 0,
g( ) = { 2, > 0.

Sketch the graphs of the function f and g.

From the graphs, determine the continuity of the composite function fog and hence discuss

whether the composite function fog is defined.

Q4 The function is defined by ( ) → 2 − 3 , ∈ ℝ.
a) Sketch the graph of g(x) . Hence, show that g(x) is not a one-to-one function.

b) Determine the set of values of domain g, Dg , for which g(x) is a one-to-one function.

Q5 The function is defined by : → √3 + 1, ∈ ℝ, ≥ −31. Find f −1 and state its domain and
range.

Q6 The function is defined as follows :
: → 4 + ( − 1)2, ∈ ℝ

a) Sketch the graph of f .

b) State the range of f .

c) Determine if f −1exist.

Q7 The functions f is defined by

( ) = 4 − 2 , − 3 ≤ < −1
|2 − 1|, > −1

a) Find lim ( ).

→ −1

b) Determine whether ( ) is continous at = −1. Give your reason.

c) Sketch the graph of ( ).

Q8 a) A functions f is defined by : → √ + 1, ∈ [−1, ∞)
(i) Define −1 in the similar form and state its domain and range.

(ii) Sketch the graph of f and −1 on the same axes.

41

ANSWER Module 1

Section i) ii) y=g(x)
1.1 1(a) y x

Clone y
STPM
y=f(x
)

01 x 01

i) Let = f −1( )
1(b) f( ) =

2 ln =



f −1( ) = 2
Df−1 = { : ∈ ℜ}
Rf−1 = { : ∈ ℜ, > 1}

ii) Let = −1( )
g( ) =
√ − 1 =
g−1( ) = 2 + 1

Dg−1 = { : ∈ ℜ, > 0} ;
−1 = { : ∈ ℜ, > 1}

1(c) Rf = { : ∈ ℛ} ; D = { : > 1}

1(d) Since ⊈ , ° does not exist

42

Section 2(a) Let f (x) = u
1.1 ex+2 = u
x + 2 = ln u
Clone x = ln u − 2
STPM g[f( x)] = x
g(u) = ln u - 2
g(x) = ln x - 2 , x  e2
g : x → ln x − 2

fg(e3 ) = f ( ln e3 − 2)
2(b)

= f( 3−2)
= f (1)
= e1+2
= e3

(i) ℎ( ) = | + 3| − | − 3|, ∈

3(a) | + 3| = {− +−33, , ≥<−−33

| − 3| = {− −+33, , ≥<33

For < −3,
ℎ( ) = (− − 3) − (− + 3)

= −6

For −3 ≤ <3,
ℎ( ) = ( + 3) − (− + 3)

= 2

For ≥ 3,
ℎ( ) = ( + 3) − ( − 3)

=6

−6, < −3
Thus ℎ( ) = { 2 , − 3 ≤ < 3

6, ≥ 3

43
Sketch graph of function h is

(ii) Range of h is { |−6 ≤ ≤ 6}

f( ) = +| | , ∈

2

3(b) For < 0, ( ) = +(− )=0

2

For > 0, f( ) = +( ) =

2

Therefore, f( ) = { 0 ,, < 0
≥ 0

g( ) = { − 1, < 0
2, > 0

Sketch of the graph of function f:

44
Sketch of the graph of function g:

Since there is a break in the graph of g at x = 0 , then fog is discontinuous
(@not continuous)
The range of g is the subset of f , then the composite function fog is defined.

y

15

10

4(a) 5 y ( ) = 2 − 3
= ( − 3)
0
5 10
-5 0
-5

Since g(0)=0 and g(3)=0, the function g is not one to one.

( ) = 2 − 3
32 9

4(b) = ( − 2) − 4

 Domain :   3  3   
Dg : − Dg 2  or Dg : 2 Dg

45 Let = −1, ( ) =
5.
Section √3 + 1 =
1.1

Clone
STPM

3 + 1 = 2

2 − 1
= 3
2 −
∴ −1( ) = 3 1

Since f (x) = 3x +1 0 , we can see that f −1(x)  [0, ) .
Domain f −1(x) = [0, ) and range of f −1(x) = [− 1 , )

3

y

40

30

6(a) 20 y

10

0 5
-5 0

6(b) From the graph, f(1)=4 is the minimum then range of f is [4, ) .

Inverse of f does not exist since f is not a one to one function as can be seen
6(c) from the graph.

7(a) ( )

→−1−
= (4 − 2)
→1−

=3

( )

→−1+

= |2 − 1|

→1−

=3

∴ ( )

→−1−

=3

7(b) lim ( ) exist but (−1) is not defined, thus ( ) is not continuous at x = -1

→− −

46

Section 7(c)
1.1

Clone
STPM

8(a) (i) : → √ + 1, ∈ [−1, ∞)

Let, = −1( ), ↔ = ( )
= √ + 1
2 = + 1
= 2 − 1

−1( ) = 2 − 1
∴ −1( ) = 2 − 1
Domain of −1 is [0, ∞) or { : ≥ 0} and range is [−1, ∞)or { : ≥ 1}

8(b)

y=x

f
1

−1

-1 1

-1

47 Learning Outcome:
1.2(a) Polynomial
a) use the factor theorem and the remainder theorem
and rational b) solve polynomial and rational equations.
functions

POLYNOMIAL FUNCTION

1. A function of x in the form

( ) ≡ + 1 −1 + ⋯ + − + ⋯ + −1 +
where ∈ ℝ and ≠ 0, ∈ ℤ+ is called a polynomial function of degree n.

2. Polynomials of degrees 1, 2, 3 and 4 are also known as linear, quadratic, cubic and quartic

functions respectively.

3. The value of a polynomial, ( ), when = is written as ( ).

For example,
If ( ) = 3 2 + 4 − 1
Then when = 1, (1) = 3(1)2 + 4(1) − 1 = 6
And when = 2, (2) = 3(2)2 + 4(2) − 1 = 19

4. A number ∈ ℝ is the zeros of ( )

Example: If ( ) = ( − 2)( + 1)(2 − 1), then zeros of ( ) are 2, −1, 1 .
2

5. When polynomial ( ) is divided by ( − ), then ( ) = ( − ) ( ) + , where

( ) = , =

THE REMAINDER THEOREM

When the polynomial ( ) with degree n is divided by (x-a), the quotient, ( ), is a polynomial of
degree (n-1), and the remainder R is a constant,

( ) = ( ) +
− −

( ) ≡ ( )( − ) +

By substituting = , we see that ( ) =
When a polynomial ( ) is divided by ( − ), the remainder is ( )

48

EXAMPLE 1
Find the remainder when a polynomial ( ) = 3 5 − 2 3 + 7 is divided by (x-2).

(2) = 3(2)5 − 2(2)3 + 7
= 87

EXAMPLE 2
Find the value of a and b when a polynomial ( ) = 3 − 6 2 + is divided by
( − 1)( + 2) , its remainder is 8x-16.

3 − 6 2 + ≡ ( − 1)( + 2) ( ) + (8 − 16)
When = 1, − 6 + = 8 − 16

+ = −8 -----(1)

When = −2, − 8 − 24 − 2 = 8(−2) − 16
8 + 2 = 8
4 + = 4 -----(2)

(2) − (1), 3 = 6
= 2

∴ = −4

THE FACTOR THEOREM
From the remainder theorem, we have shown that when a polynomial ( ) is divided by (x-a),
its remainder, R is P(a). On the other hand, if (x-a) is a factor of ( ), then its remainder is zero,
R = P(a) = 0.

For a polynomial P(x), (x-a) is a factor of ( ) if and only if ( ) = 0

For a polynomial ( ), ( − ) is a factor of ( ) if and only if ( ) = 0



EXAMPLE 3
Given that( − 1) and(3 + 1) are factors of polynomial ( ) = 3 3 − 2 + + 2 , find the value of

a and b.

Given that( − 1) is a factor of ( ),

By the factor theorem,
(1) = 0
3 − + + 2 = 0
− + = −5 ------- (1)

49

(1) − (2), 1
(− 3) = 0

13 12 1
3 (− 3) − (− 3) + (− 3) + 2 = 0

− 1 − − + 2 = 0

993

− − 3 = −17 ------- (2)

4 = 12
=
∴ =

EXAMPLE 4
Given that( − 3) is a factor of ( ) = 3 + 2 − − . When the polynomial is divided

by( − 1), the remainder is 24. Find the value of p and q.

Given that( − 3) is a factor of ( ),

By the factor theorem,
(3) = 0

27 + 9 − 3 − = 0
9 − = −24 ------- (1)

Given that when ( ) is divided by ( − 1) the remainder is 24.
By the remainder theorem,

(1) = 24
1 + − 1 − = 24

− = 24 ------(2)

(1) − (2), 8 = −48
= −
∴ = −

50

WORKSHEET 1.2(a)
Answer the following question.

Q1 When the polynomial ( ) is divided by Q2 . Find the value of a if the remainder 18
( + 1), its remainder is 3, and the when a polynomial
remainder is 1 when its divided by ( − 1). ( ) = 2 3 + 2 − 5 + 6 is divided by
( − 4).
Find the remainder of the polynomial when
its divided by 2 − 1. Solution:
( ) = 2 3 + 2 − 5 + 6
Solution:
( ) = ( 2 − 1) ( ) + + (4) = 18
2(4)3 + (4)2 − 5(4) + 6 = 18
( ) = ( − 1)( + 1) ( ) + +
16 = −96
when = −1, (−1) = 3 = −
− + = 3 -----(1)

when = 1, (1) = 1
+ = 1 -----(2)

(1) + (2), 2 = 4

= 2

= −1

Hence, the remainder is − +

Q3 Given the function f(x) = (2x - l)(x - 2)(x + 3). Q4 Given that (x + 2) is a factor of the polynomial

Find the value of constant h such that (x + 2) is a ( ) = 2 3 − 2 + − 6. Determine the

factor of f(x) + hx. value of a. Factorise ( ) completely.

Solution : ( ) = 2 3 − 2 + − 6
Given ( ) = (2 − 1)( − 2)( + 3) (−2) = 0
Let ( ) = ( ) + ℎ
2(−2)3 − (−2)2 + (−2) − 6 = 0
= (2 − 1)( − 2)( + 3) + ℎ 2 = −26
= −
Given ( + 2) is a factor of ( ),
∴ (−2) = 0 ( ) = 2 3 − 2 − 13 − 6
= ( + 2)(2 2 − 5 − 3)
(2(−2) − 1)(−2 − 2)(−2 + 3) + ℎ(−2) = 0 = ( + )( + )( − )
=