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National Junior College Mathematics Department 2010

2010 / SH1 / H2 Maths / Functions of 43

National Junior College

2010 H2 Mathematics (Senior High 1)

Functions (Lecture Notes)

Topic 6: Functions

Objectives:

At the end of this topic, students should be able to

� understand that a function is a rule or relationship where for every input there is only

one output, and that two functions having the same rule are different if they have

different domains.

� use the function notations.

� use the vertical line test on a graph to determine whether it represents a function.

� determine the range of a function from its graph.

� understand the definition of a one-one function and identify it using a horizontal line

test.

� explain the condition for a function to have an inverse function.

� obtain the inverse of a one-one function and state its domain and range.

� illustrate in graphical terms the relation between a one-one function and its inverse.

� restrict the domain of a function so that the inverse exists.

� understand that a composite function gf is a composition of two functions, f followed

by g.

� determine if the composition of two functions exists and obtain the composite

function if it exists.

� determine the domain and range of a composite function.

� recognise through examples that composition of functions is not commutative.

§1 Set Notations

The symbol ‘∈’ represents ‘belongs to’ or ‘is an element of’. For example, ‘ x ∈� ’ means x

belongs to the set of real numbers.

�+ represents the set of positive real numbers. It does not include zero.

�− represents the set of negative real numbers. It does not include zero.

{ }0+ ∪� represents the set of non-negative real numbers.



§2 Representation of Range of Values

{ }( , ) |a b x a x b= ∈ < <�

{ }[ , ] |a b x a x b= ∈ ≤ ≤�

{ }[ , ) |a b x a x b= ∈ ≤ <�

{ }[ , ) |a x x a∞ = ∈ ≥�

In short, when we are expressing a range of values, we use round brackets if we want to

exclude the extreme values of the range and square brackets if we want to include them.

§3 Relations

An association between the elements of two sets is called a relation.

Definition. A relation R from a set X to Y (written :R X Y→ ) is a rule to determine how the

members in set X are related to the members in set Y.

Figure 1.1 illustrates a relation between the elements of two sets A = {1,2,3,4} and

B = {3,5,7,9}. It is obvious that the relation maps 1 to 5, 2 to 5, 3 to 7, and 4 to 9.

Different Ways to Represent Mappings

Let R be the relation as shown in Figure 1.1. Then the following are ways to represent the

mapping from 1 to 5.

(a) :1 5R → or

(b) (1) 5R = or

(c) 1R5.

*Note: (c) is seldom encountered in ‘A’ levels. They are used mainly in pure algebra texts.

B A 1

2

3

4

3

5

7

9

Figure 1.1



§4 Functions

A function is a special type of relation.

Definition. A relation f : X Y→ is a function if and only if for each element x X∈ , there

exists only one element y Y∈ such that ( )f x y= .

The diagram below illustrates how the elements in X can be mapped onto elements in Y under

the function f.

§5 Differences between Relations and Functions

Three relations are illustrated by figures 1.1, 1.2, and 1.3 below. Observe and see why 1.1 is a

function and 1.2, 1.3 are not functions.

A relation may map each element in A to one or more elements in B.

A function must map each element in A to at most one element in B.

X Y

f 1

2

3

4

a

b c

d

e

f

B A

Figure 1.2

1

2

3

4

3

5

7

9

� a function � NOT a function

B A 1

2

3

4

3

5

7

9

Figure 1.1

� NOT a function

B A

Figure 1.3

1

2

3

4

3

5

7

9



Exercise. Determine which of the following (is a function)/(are functions).

§6 Components of Functions

Domains, Codomains, Ranges, and Images

Let f : A → B , x f ( )x� be a function, then f(x) is called the image of x under f, or

more commonly, the value of f at x. .

The set A is called the domain of the function f, usually denoted by fD .

It is not necessary for all the elements of B to be the image of some x ∈A. The range of the

function is the subset of B which contains all the possible images under f of all the elements

of A. It is usually denoted by fR .

Set B is the codomain of f. It is the general set that contains the range of f.

E.g. The set of real numbers and the set of positive integers are some of the more commonly

encountered codomains in ‘A’ levels.

B

x

f

fDA = , domain

fR , range

f ( )y x=

Not a function.

Last element in the

domain has no image

Not a function.

3rd element in the domain

has two images

A function.

Every element in the domain

has one and only one image



Example

Let f be a function from set X to set Y.

In this case, domain of f is { }fD 1,2,3,4= , codomain of f is { }1,2,8,24,27Y = , and range of f

is { }fR 1,8,27= . In addition, 8 is the image of 2, or 2 is the pre-image of 8.

Note:

• Range of f is a subset of the codomain of f,

• At ‘A’ level, most functions are considered relations between real numbers. Therefore

the codomain is usually not given. We take it that codomain = domain = �.

Rules of Functions

A function is defined by both the rule and the domain. For example, if a function f has the

rule 2x x� and the domain is � , then we write it as follows, 2f : ,x x x ∈� � or 2f( ) ,x x x= ∈� .

Two functions with the same rule but different domains are considered different functions.

So, 2f : ,x x x ∈� � and 2g : , 1 1x x x− < ≤� are different functions. This has implications

on how the graphs look.

x x

y y

– 1 1

2

1

8

27

24

1

2

3

X Y

f



Exercise. State the domain and find the range of each of the following functions.

(a) 2f : ,x x x ∈� �

(b) [ ]2g : , 1,1x x x ∈ −�

(c) 2h : , [ 1,2)x x x ∈ −�

Solution

(a) The domain of f ( , )= −∞ ∞

The range of f [0, )= ∞

(b) The domain of g [ 1, 1]= −

The range of g [0, 1]=

(c) The domain of h [ 1, 2)= −

The range of h [0,4)=

§7 Graphs of Relations and Functions

When sketching graphs of functions, we need to take note of the rule and domain of the

function. The rule gives us the shape of the graph and the domain gives us the horizontal

expanse of the graph.

Example

Take note that the figure 1.4 illustrates a piece-wise function. A piece-wise function is a

function that has a different rule for each domain component.

y

x

x

y

2x y=

1, 0

, 0

x xy

x x

− + <=

≥

x y=

f( )x

x

f( )x

x x

f( )x

x

Figure 1.4

y

x 1− 1

1



§8 Vertical Line Test (Testing for Functions)

A relation f is a function if and only if every vertical line f, Dx a a= ∈ cuts the graph

of f exactly once.

Revisiting the previous examples, the graph on the left does not represent a function, because

you can find a line, say 1x = , that cuts the graph at two points.

The graph on the right represents a function, because regardless of the vertical line you draw,

it will only cut the graph at exactly one point.

QUESTION. What about the point at 0x = ? Does the vertical line cuts the graph once?

Exercise. Determine if the following relations defined on � are functions.

(i) 3x y= (ii) | |y x x= (iii)

2

, 0

, 0

x xy

x x

− ≤=

≥

x a=

x a=

x

x

y y

1, 0

, 0

x xy

x x

− + <=

≥

2x y=



§9 Restriction Functions

Given any function f with domain X, we can define a function g on a smaller domain, say

W X⊆ which has the same rule as f. The function g is known as the restriction of f on W.

These restriction functions are determined based on the restrictions we place on the function.

Example. Find the maximal domain of the restriction function of 2g( ) 4x x= − such that

the range of the function is [ ]1, 2 .

When 1y = , 2 24 1 3 3x x x− = ⇒ = ⇒ = ± .

From graph, maximal domain is 3, 3 − .

§10 One-One Functions

DEFINITION. A function f is said to be one-to-one (1-1) or injective if no two distinct

elements in its domain have the same image under f. Mathematically, this is defined

as follows.

Given a function f : X Y→ , f is 1-1 if for all 1 2,x x X∈ ,

if 1 2x x≠ , then ( ) ( )1 2f fx x≠ .

We can easily observe from the graph if the function is 1-1.

Not a one-one function A one-one function

x

y

2

1

3− 3 O



Exercise. Using the definition of 1-1, can you identify which of the following are graphs of

1-1 functions?

(Ans: No) (Ans: No) (Ans: Yes)

§11 Horizontal Line Test (Testing for 1-1 Functions)

A function f is one-to-one if and only if every horizontal line f, Ry a a= ∈ cuts the

graph of f exactly once.

Example. Determine if the following are one-one functions, giving your reasons. If it is not

one-one, find a restriction of the function such that it is one-one and has the same range as the

original function.

(i) 2f : ,x x x ∈� �

(ii) 2g : 4 1, , 1x x x x x− + ∈ ≥� �

(iii) h : | |,x x x x ∈� �

Solution

(i) Since f( 1) 1 f(1)− = = , f is not 1-1.

Alternatively, if we use a horizontal line test,

we see that there is at least a horizontal line,

say 1y = that cuts the curve at two points.

Now [ )fR 0,= ∞ . From the graph, if we

restrict the domain to be [ )0,∞ , then every

horizontal line will cut the graph at most

once. Furthermore, the range is the same as fR .

Therefore the required restriction function is 2

1f : , 0x x x ≥� .

QUESTION. Is this the only answer?

x

y

x

y

x

y

x

y



(ii) ( )22g( ) 4 1 2 3, 1x x x x x= − + = − − ≥

From the graph, the horizontal line 2y = −

cuts the curve at 2 points. Hence the function

is not 1-1.

[ )gR 3,= − ∞ . From the graph, a restriction

function is 2

1g : 4 1, 2x x x x− + ≥� .

QUESTION. Is this the only answer? (Ans: Yes)

(iii) h : | |,x x x x ∈� �

From the graph, any horizontal line cuts the graph

at most once. Therefore by the horizontal line test,

h is 1-1 function.

§12 Inverse Functions

It is very natural to ask if a process is reversible. Similarly for functions, it is also very natural

for us to ask if there is a function that reverses what another function does. This leads to the

notion of inverse functions.

Let f : X Y→ be a function such that ,f : x Xx y ∈� there is always an inverse relation 1f −

such that for all fRy ∈ 1f( ) f ( )y x x y

−= ⇔ =

In this case x is called the pre-image(s) of y under f. The relation 1f − may or may not be a

function.

Note. The notation 1f

− means inverse relation/function of f. Do not in this case see “– 1” as a

power and treat it as 1f

. This can be seen in inverse trigonometric functions, e.g. 1cos x

− is the

arc cosine of x and is not equals to sec x .

x

f

y

1f

−

x

x

y

x

y



In the diagram below, the inverse relation is not a function. Why? What is the cause of the

‘problem’?

(Ans: In the second diagram, the 3

rd element in the domain has two distinct images under

1f − )

In the diagram below, is 1f − is a function? What do you observe about the domain and range

of 1f − ?

Domain of 1f − is the range of f and range of 1f − is the domain of f., i.e.

1 ffD R− = and 1 ff

R D− =

f 1f −

f 1f −



12.1 Graphs of a Function and its Inverse

For a function, say f, defined on real values, we can obtain the graph of the inverse relation, 1f − from the graph of f.

Consider the graph f( )y x= where 2f( ) ,x x x= ∈� .

Recall that the domain of 1f − is the range of f.

Therefore in order to obtain the graph of 1f − , we

simply replace the y-axis by the x-axis and vice

versa. At the same time, the graph becomes

f( )x y= which is 1f ( )y x

−= .

Note also that when sketching graphs, the domain

is always part of the x-axis. Hence the next step is

to make the graph look ‘normal’ again, and this is

achieved by reflecting the whole diagram about the

line y x= .

Observe now that the graph of 1f − is obtained from

the graph of f by a reflection about the line y x= .

Therefore, we see that

Graph of 1f − is the reflection of graph of f about

the line y x= .

Note: f-1

in this case is a relation but not a function. For f-1

to be a function, f has to be a 1-1

function.

Together with the earlier observation on the domain and range of 1f − we have the following,

A function f : X Y→ has an inverse function 1f − if and only if f is one-to-one. In

which case, 1 ffD R− = and 1 ff

R D− =

x

y

a

1f ( )a

− 1f ( )a

−

x

y

a

1f ( )a

−

1f ( )a

−

y

x

a

1f ( )a

− 1f ( )a

−

f( )y x=

f( )x y=

1f ( )y x−=

y x=



12.2 Finding Inverse of a Function

Let f( )y x= . Then 1f ( )x y−= . Therefore if we express x in terms if y, we will get the

expression for 1f − in terms of y. Since y is a dummy variable, we simply replace y by x to get

the expression of 1f − in x. It sounds abstract, so let us look at a simple example.

Example. Given f : 2 2,x x x− ∈� � , find 1f − , if it exists.

Solution

The graph of f is a straight line, as such any horizontal line cuts the graph at most once.

Therefore by the horizontal line test, f is 1-1 function and 1f − exists.

Let ( )2 2 f( )y x x= − = . Therefore ( )12f ( )

2

yx y

−+= = .

Since -1 ffD R= = � , we have 1 2

f ( ) ,2

xx x

− += ∈� .

Example. Given a function f where 2f : 2 3,x x x x S+ + ∈� where S is a subset of � . Find

the largest possible domain S consisting of negative real numbers in a single interval such

that f is one-one. Hence, define 1f − and sketch the graph of f and 1f − on the same diagram.

Solution

( )22f( ) 2 3 1 2x x x x= + + = + +

From the graph of f, f is one-one if the domain is

( , 1]−∞ − or [ )1,− ∞ , Since we are looking for a

single interval of negative numbers, ( , 1]S = −∞ − .

Note that f fD ( , 1], R [2, )= −∞ − = ∞ .

Let ( )2

f( ) 1 2, 1y x x x= = + + ≤ − ,

( )2

2 1 1 2y x x y− = + ⇒ = − ± − .

Now since 1x ≤ − , therefore 1 2x y= − − − . Therefore 1f ( ) 1 2, 2x x x− = − − − ≥ .

Recall that the inverse function 1f − is defined to “reverse” the effects of f. In other words, if

we apply 1f − after f, we will get back the original element. In mathematics, we say that this is

a composite function of ‘f followed by 1f − ’, written 1f f− . (In this case 1f f− is actually an

identity function.) We shall discuss composite functions in general in the following section.

Note: Students are expected to know how to sketch inverse graphs without actually

finding the rule of the inverse function, and to label the respective axial intercepts,

range, and domains of the inverse graphs accurately.

y

f( )y x=

x

1f ( )y x

−=

-1

2

y x=



§13 Composite Functions

If there are two or more functions, say f and g, it is natural (again) to ask if we can apply one

function after another.

If it is possible, then this composite function denoted by g f� or just gf is defined as

( ) fgf( ) g f( ) , Dx x x= ∈

OBSERVATIONS.

(A) For the composite function gf to exist, f gR D⊆ .

In the figures below, { } { }f gR 1,8,24 D 1,8,27= ⊄ = , hence gf is undefined because gf(3) is

not defined.

x

f

f( )x

g

( )g f( )x

gf

1

8

27

24

1

2

3

X Y

f 3

6

7

45

89

Z

Wg

1

2

3

X

3

6

7

45

89

Z gf

Undefined!



In the figures below, { } { }u vR 1,8 D 1,8,27= ⊂ = , hence vu is defined.

(B) gf fD D=

E.g. In the figures below , { }gf fD 1,2,3 D= = .

1

8

27

1

2

3

X Y

f 3

6

7

45

89

Z

Wg

1

2

3

X

3

6

7

45

89

Z gf

1

8

27

1

2

X Y

u 3

6

7

45

89

Z

Wv

1

2

X

3

6

7

45

89

Z vu

Defined!



(C) gf gR R⊆

E.g. In the figures below, { } { }gf gR 3,89 R 3,45,89= ⊆ = .

(D) When f gR D= , then g f gR R= .

E.g. In the figures below, { }gf gR 3,45,89 R= = because { }g fD 1,8,27 R= = .

1

8

27

1

2

3

X Y

f 3

6

7

45

89

Z

Wg

1

2

3

X

3

6

7

45

89

Z gf

1

8

27

1

3

X Y

f 3

6

7

45

89

Z

Wg

1

3

X

3

6

7

45

89

Z gf



(E) When g f gR R= , f gR may not equal D . (i.e. converse of observation D is not true)

E.g. In the figures below, { }gf gR 3,45,89 R= = but g fD R≠ .

In conclusion, two cases can happen when we want to compose two functions, say f followed

by g.

(a) f gR D⊆

Therefore gf exists. gf fD D= and g fR is the set of images of fR under g.

(b) f gR D⊄

We need to find a restriction of f such that gf exists. Then it will be as in (a), except

that we are working with the restriction function of f, instead of f itself.

We can also sketch the graph of gf using GC and determine the range from the graph.

• In general, the functions gf and fg are not equivalent. i.e. composition of functions are

not commutative. i.e.

( ) ( )

( ) ( )

2

2 2

f 2 , g .

fg 2 , gf 4 ,fg gf.

x x x x

x x x x

= =

= = ≠

Example. For the following functions

1

2

3

X

3

6

7

45

89

gf

1

8

27

24

1

2

3

X Y

f 3

6

7

45

89

Z

Wg

Z



( )

2

f : e ,

g : ,

h : ln 1 , 1

xx x

x x x

x x x

∈

− ∈

− <

� �

� �

�

determine whether the composite functions fg, gh, hg, hf exist. If the composite function

exists, define the function and state its range. If the composite function does not exist, find

the maximal domain such that it exists.

Solution

( )f fD , R 0,= = ∞� ( ]g gD ,R ,0= = −∞� h hD ( ,1), R= −∞ = �

Since g fR D⊆ , fg exists.

Since h gR D⊆ , gh exists.

Since g hR D⊆ , hg exists.

Since f hR D⊄ , hf does not exist.

( )2

2f g( ) f e ,xx x x−= − = ∈�

For the range, ( ] ( ]g f,0 0,1→ −∞ →�

( )( ) ( )( )2

g h( ) g ln 1 ln 1 , ( ,1)x x x x= − = − − ∈ −∞

For the range, ( ) ( ]h g,1 ,0−∞ → → −∞�

( ) ( )2 2h g( ) h ln 1 ,x x x x= − = + ∈�

For the range, ( ] [ )g h,0 0,→ −∞ → ∞�

f hR D (0,1)∩ = . Therefore, we need to find a subset of fD which gives ( )0,1 under f. From

the graph of f, we see that this subset is ( ,0)−∞ .

Hence define f : e , 0xx x <� , then hf exists and ( ) ( )h f( ) h e ln 1 e , 0x x

x x= = − < .

For the range, ( ) ( )f h( ,0) 0,1 ,0−∞ → → −∞

Example. Let f be a one-one function defined by f : 2 3, 2 2x x x+ − ≤ ≤� . Find

x

y

x

y

x

y

f( ) ex

x = 2g( )x x= − ( )h( ) ln 1x x= −

1

1



(i) 1f − (ii) 1f f− (iii) 1f f −

Solution

fD [ 2,2]= − , fR [ 1,7]= −

(i) 13 32 3 f ( ) , [ 1,7]

2 2

y xy x x x x

−− −= + ⇒ = ⇒ = ∈ −

(ii) ( )1 1

ff f( ) f f( ) , D [ 2,2]x x x x− −= = ∈ = −

(iii) ( ) 1

1 1

ff f ( ) f f ( ) , D [ 1,7]x x x x −

− −= = ∈ = −

Although 1 1f f( ) f f ( )x x x− −= = , they are in general two different functions because the

domains might be different. Therefore care must be taken when you sketch these graphs.

Note:

1 1

1

1

f f f

1

ff f

f f ( ) , D D

f f ( ) , D D

x x

x x

− −

−

−

−

= =

= =

Try using ( ) ( )2 -1f , fx x x x= = to see for yourself the above results.

i.e. what is the difference between ( )2

x and ( )2x ?

Example.

The functions f and g are defined as 1f : e , , 0xx x x

− ∈ >� � and

( )g : ln 1 , , 1x x x x− ∈ >� � .

(i) By sketching the graphs of f and g, or otherwise, state the range of f and g.

(ii) State why the composite function fg does not exist.

(iii) By restricting the domain of g to ( ),a ∞ , where a is a real number, find the least value

of a such that the composite function fg exists. Define fg in similar form. State the

range of fg.

Solution

(i) From graphs, ( )fR 0,e= and gR = � . Alternatively,

10 1 1 e exx x −> ⇒ − < ⇒ < . However, exponential functions

are greater than 0, therefore, ( )fR 0,e= .

(ii) Since ( )g fR D 0,= ⊄ = ∞� , therefore fg does not exist.



(iii) For the composite function to exist, we must find a subset of the domain of g such that

range of the restriction function will be a subset of fD , i.e. ( )g fR D 0,∩ = ∞ .

From the graph of g, we see that if we restrict the domain to ( )2,∞ , then range of the

restriction is ( ) f0, D∞ ⊆ .

( ) ( )( ) ( )1 ln 1 efg( ) f g( ) f ln 1 e

1

xx x x

x

− −= = − = =

−.

efg : , 2

1x x

x>

−�

( )f gR 0,e=

Note: When defining a function, the rule and domain of the function must be stated

2

2



Appendix: Use of Graphing Calculators in Functions

A1 Sketching Functions

Before we begin, always check the mode by pressing MODE key on your calculator. Since

we are sketching graphs of the form f( )y x= , set the fourth line of the mode screen to FUNC.

Let us use an example to illustrate the process.

EXAMPLE. Sketch the graph of 2f : , 2x x x ≥ −� .

Key Press Screen Shot Steps/Notes/Descriptions

(first of the five buttons

immediately below the screen)

This brings up the input window

for keying in your function.

You can plot a maximum number

of 10 functions at any one time.

The first button is the “variable”

button. The four symbols

corresponds to the four different

types of graphs you can plot.

Since we are in FUNC mode, X

will appear as you press the

button.

However, if the function has a

domain that is not � , it is

advisable to take that into

consideration before plotting it.

The inequality signs can be

found by pressing TEST, i.e. 2nd

MATH.

This syntax sets the domain of

the function*.

If the functions consists of more

than one term, e.g. 2

4x + , then

you need to put the function in

parentheses, i.e.

( )( )24 2x x+ ≥ −

If the domain is 2 2x− ≤ < , then

enter “(X<2 and X 2≥ − )”. The

“and” can be found in TEST >

LOGIC tab.



After plotting the graph,

sometimes you need to adjust

WINDOW in order to see some

features that are not inside the

current window. Alternatively

you can use ZOOM.

* The syntax “(X>0)” or the like can be seen to take the value “1” when the condition inside

the parentheses is true and the value “0” if it is false. Therefore “(X+4)(X>0)” means that the

function will be ( )( )4 1 4x x+ = + when 0x > and ( )( )4 0 0x + = otherwise. That is why you

will see the graph 4y x= + when 0x > . If you understand how this works, you will be able

to use it to plot piece-wise defined functions, such as

2

, 0f( )

, 0

x xx

x x

− ≤=

≥.

A2 Finding Range of Functions

EXAMPLE. Find the range of 2 1

f : , , 11

xx x x

x

+∈ ≠

−� � using GC.


Key in the function and plot the

graph. Since the domain consists

of all real values except 1x = ,

we can safely ignore the point.

Anyway it will not show on the

GC. However when you sketch it

on paper, do remember to use a

circle to indicate that the point is

not included.

Observe that there are two

turning points. We need to find

the y-values of the two turning

points.



The calculator enables you to

find out more about the graph.

Press 2nd

TRACE to access the

CALCULATE menu.

We need the MINIMUM and

MAXIMUM functions in this

menu.

Select MINIMUM will allow

you to find the coordinates of the

minimum point. However, it

requires three inputs from you:

Left Bound, Right Bound,

Guess.

If you want, you can zoom into

the right half of the graph before

finding the MINIMUM point.

Use the direction key to move

the cursor to a point on the left

of the minimum point. Then

press ENTER. Alternative you

can also enter a value for the x-

ordinate of the Left Bound.

The chosen point and the

minimum point should be

connected via the graph. In this

case, we should not choose any

point on the “downward” curve

for the Left Bound.

Upon enter it will ask for a Right

Bound. Shift the cursor to a

point on the right of the

minimum point. Press ENTER.

Now choose move the cursor to

a point near the minimum point.

Then press ENTER.

This screen shows the

coordinates of the minmum

point.

(2.41, 4.83)



Repeat the same steps to find the

coordinates of the MAXIMUM

point.

The maximum point is

(-0.414, -0.828).

Since the range consists of all

possible y-values that the graph

takes, we see that the range is

( ] [ ), 0.828 4.83,−∞ − ∪ ∞ .

The most important thing when finding range is to adjust the window or zoom to ensure that

you did not leave out any features of the graph. For example, there might be a turning point

for large values of x. You might need to analyse the function to get a rough picture of the

graph. A good start is to check the signs of the function when | |x is large, i.e. when x tends

to positive or negative infinity.

EXERCISE. Find the range of 0.05f : e xx x −� .

Ans: ( ],39.9−∞

A3 Sketching Inverse Functions

There are a few ways to sketch the inverse function. The method introduced here is an

‘artificial’ method because it sort of draws on top of the graph. As you are not really plotting

it, you will not be able to perform calculations on the inverse graph.

EXAMPLE. Sketch the function 2f : 4 1, 2x x x x+ + ≥ −� and its inverse on the same

diagram.


Key in the function, take note of

the use of parentheses to group

the terms of the function.

Try it without the parentheses and

see if you can figure out why it

appears the way it does.

Adjust WINDOW to get a better

picture of the graph.

(59.9, 39.9)



Access the DRAW menu.

Browse through the objects that

you can draw. You will find

8: DrawInv

Press ENTER and you will be

prompted to enter a function

Access the VARS > Y-VARS.

Since we are in FUNC mode,

select Function.

Choose Y1 since the function is

entered into Y1. Press ENTER.

Note that the graph that was

newly drawn does not look like a

reflection of the original graph

about the line y x= . The reason

for this is because the scale is

different for both axes. To rectify

this problem, we use ZSquare

under the ZOOM menu.

Now check the markings on both

axes. You should see that they are

equally spaced for both axes.

However, the graph for the

inverse function is gone. This is

exactly what is meant by

“artificial”. The inverse is not

plotted as a function but rather is

drawn as a picture. Hence you

should repeat DrawInv Y1.

Since you have already entered

the command previously, you can

easily recall it by pressing 2nd

ENTER which gives ENTRY.

You can repeated press ENTRY

to recall the pass commands you

have entered.



Now you should get the correct

graphs. Therefore if you want to

draw the inverse function, do

remember to use ZSquare so that

the scale for both axes are the

same.

A4 Sketching Composite Functions

EXAMPLE. Sketch gf given 2f : ,x x x ∈� � and g : e ,xx x ∈� � .


Key in the two functions into Y1

and Y2. It does not matter which

one you key as f and g.

Since you do not want to show

these graphs later, hide them by

moving the cursor to the “=”

signs and press ENTER. The

“equal” signs will no longer be

highlighted.

Enter into Y3 the composite

function. Since the functions f

and g are assigned to Y2 and Y1

respectively, the composite

function is entered as Y2(Y1(X))

Graph it and adjust the display

window accordingly.

There are just too many functions to be covered. For some of these functions, you might need

to, as suggested, do some preliminary analysis to have a rough sense of how they will look

like. It would be a good idea too if you can invest some time to “explore” the calculator.



National Junior College

2010 H2 Mathematics (Senior High 1)

Functions (Tutorial)

Basic Mastery Questions

1. Complete the following diagrams to illustrate your understanding of the essential

concepts on functions.

2. Determine whether the following functions are one-one. Where the function is one-one,

find the inverse function, stating the domain and range.

(a) 2 4 4, , 2x x x x x− + ∈ ≥� �

(b) 2 ,x x x− ∈� �

(c) { }2 3

, \ 11

xx x

x

+∈

−� �

3. Given that the functions gf and f are such that 2gf : 3 5,x x +� and 2f : 1x x +� ,

x ∈� , find the function g.

Function Not a function

1-1 function Not a 1-1 function



Tutorial Questions

1. The function f is given by

2f : 6 , forx x x xλ− ∈� � ,

where λ is a positive constant. Find in terms of λ ,

(i) ff ( )λ

(ii) the range of f.

Give a reason why f does not have an inverse.

The function f has an inverse if its domain is restricted to x k≥ and also has an inverse

if its domain is restricted to x k≤ . Find k in terms of λ , and find an expression for 1f ( )x

− corresponding to each of these domains for f.

2. Given the following functions,

f : e , 0xx x

− >�

2g : 3 5, ( , )x x x− ∈ −∞ ∞�

( 5)

h : , 53

xx x

− +< −�

determine whether the composite functions fg, gf, gh and hg exist. If it does, give its

rule and range. If it does not, analyze if it is possible to restrict its domain such that the

composition exist. Show your working clearly.

3. The functions f and g are defined by

f : ( 2)( 8), x x x x− − ∈� � and 2g : 2 1, , 0x x x x− ∈ >� �

(i) Explain why the composite function gf does not exist.

(ii) Explain briefly why 1f − does not exist. If h is a restriction of f such that the

inverse of h exists, find the maximal domain of h in the form ( , ],a−∞ where

a ∈� is to be determined. Hence, find h and its range.

(iii) Find the solution of 1 1h h ( ) h h( )x x− −= .

4. The functions g is defined by 2g : 4 4 , x x x x− ∈� �

(i) By means of a graphical argument, or otherwise, show that 1g− does not exist.

A restriction function h of g is defined by 2h : 4 4 , , x x x x x k− ∈ <� � .

(ii) State the maximum value of k such that 1h− exists

(iii) Sketch the graph of 1h− , showing its relation to the graph of h. Hence indicate

a point P on the curve of 1h ( )y x

−= that satisfies hh( )x x= .




3

f : 2 , , 04

xx x x− ∈ ≥� �

g : e , , 0x ax x x

+ ∈ <� � where a is a positive constant.

(i) Sketch the graph of f and find its range. Hence state the set of values of x for

which 1 1ff ( ) f f ( )x x− −= .

(ii) Sketch the graph of g and define the inverse function 1g

− .

(iii) Show that gf does not exist. Find the maximal domain of f for which gf exists

and state the range.

6. The functions f, g and h are defined as follows:

2 12

f : 6 ,

g : e 2,

h : ln(ln ), 1

x

x x x x

x x

x x x

− − ≤ −

− ∈

>

�

� �

�

(i) Find the inverse function of f in similar form.

(ii) Find the range of g.

(iii) State why the composite function hg does not exist.

(iv) By restricting the domain of g to ( , )α ∞ , where α ∈� , find the smallest value

of α in exact form such that the composite function hg exists. Define hg.

(v) Solve the equation hg( ) ln 2x = . (RJC 2006)

7. The function f is defined as follows

1

f : , , 11

xx x x

x

+→ ∈ ≠

−�

(i) Sketch the graph of f and state the range of f.

(ii) Find the subset S, of the form { },x a x b∈ ≤ <� where a and b are constants to

be determined, such that the function 1

1f : ,

1

xx x S

x

+→ ∈

− is one-one and has

the same range as f.

(iii) Find1

1f ( )x−

.

8. The function f is defined by

f : , for , ,ax a

x x xbx a b

→ ∈ ≠−

�

where a and b are non-zero constants.

(i) Find1f ( )x−

. Hence or otherwise find 2f ( )x and state the range of

2f .

(ii) The function g is defined by 1

g : xx

→ for all real non-zero x. State whether

the composite function fg exists, justifying your answer.

(iii) Solve the equation 1f ( )x x− = .

[N2009\P2\Q3]



9. It is given that

( )27 for 0 2,

f2 1 for 2 4,

x xx

x x

− < ≤=

− < ≤

and that ( ) ( )f f 4x x= + for all real values of x.

where a and b are non-zero constants.

(i) Evaluate f(27) + f(45).

(ii) Sketch the graph of ( )fy x= for 7 10x− ≤ ≤ .

[N2009\P1\Q4]

Challenging Questions

1. Find the range of2 1

f : , , 11

xx x x

x

+∈ ≠

−� � . Leave your answer in exact values.

2. Let g :+ →� � be a function defined by

1g : x x

x−� . Using your GC, sketch g and

show that g has an inverse function 1g

− . Define 1g

− . Sketch 1g

− on the same diagram

and identify a relationship between the graphs of g and 1g− .

Assignment Questions

1. The function f is defined by

f : ( 1)(3 ) , , 2x x x x x− − ∈ <� �

(i) Define f −1

in similar form. [3]

(ii) Sketch the graphs of f and f −1

on the same diagram. [2]


2

1f : , for , 3

3

g : , for .

x x xx

x x x

∈ ≠−

∈

� �

� �

(i) Only one of the composite functions fg and gf exists. Give a definition

(including the domain) of the composite that exists, and explain why the other

composite does not exist. [3]

(ii) Show that 1f − exists. [2]

(iii) Sketch the graph of 1ff − . [1]




3

f : ex

x−

� , x ∈�

2g : ( 1)x x a− +� , x ∈� and a is a positive constant.

Sketch the graph of y = f(x), and show that f does not have an inverse. [2]

(i) The function f has an inverse if its domain is restricted to x b≥ . State the

smallest possible value of b and define, in similar form, the inverse function 1f − corresponding to this domain for f. [4]

(ii) Using the value of b in (i), find the smallest possible value of a such that the

composite function 1f g

− exists. State the range of 1f g

− for this value of a. [3]

--- End of Assignment ---

Numerical Answers to Functions Tutorial


1. Anything reasonable and correct.

2. (a) 1-1. 1f ( ) 2 , 0x x x− = + ≥ , [ )1f

R 2,− = ∞ . 3. g : 3 2x x +�

(b) Not 1-1

(c) 1-1. { }1 3h ( ) , \ 2

2

xx x

x

− += ∈

−�

{ }1hR \ 1− = �

Tutorial Questions

1. (i) 4 325 30λ λ+ (ii) ( )2

fR 9 ,λ= − ∞ ,

3k λ= , 1 2 2f ( ) 3 9 , 9x x xλ λ λ− = + + ≥ −

2. fg does not exist. 15

3| |x >

gf exists. ( )2 2gf( ) 3 f( ) 5 3e 5, 0x

x x x−= − = − >

( )gfR 5, 2= − −

gh exists. gh( ) 10, 5x x x= − − < −

( )ghR 5,= − ∞

hg does not exist. Not possible.



3. (ii) h : ( 2)( 8), 5x x x x− − ≤� (iii) [ ]9,5−

[ )hR 9,= − ∞

4. (ii) 12

k = (iii)

5. (i) fR ( , 2]= −∞ , Set of values of x = [ 0 , 2 ]

(ii) 1g : ln , 0 eax x a x

− − < <�

(iii) Maximal domain is ( )2,∞ , ( )gfR 0,ea=

6. (i) 1 25 2514 2 4

f : , ,x x x x− − − − ∈ ≤� �

(ii) ( )gR 2,= − ∞

(iv) ln 3α =

( )( )hg( ) ln ln e 2 , ln 3x

x x= − >

(v) ( )2ln e 2x = +

7. (i) [ )fR 0,= ∞

(ii) [ )1,1S = −

(iii) 1

1

2f ( ) 1 , , 0

1x x x

x

− = + ∈ ≥− −

�

8. (i) ( )-1f ,ax a

x xbx a b

= ≠−

( )2f ,a

x x xb

= ≠ 2fR \

a

b

=

�

(ii) Does not exist.

(iii) 2

0,a

x xb

= =

9. (i) 11


1. ( )fR , 2 2 2 2 2 2, = −∞ − ∪ + ∞

Point P is the Origin.



2010 H2 Maths Functions Tutorial (Suggested Solutions)


1. Anything reasonable and correct.

2.(a) ( )22f( ) 4 4 2 , 2x x x x x= − + = − ≥

Since every horizontal line y k= , 0k ≥ cuts the graph at most once, hence f is 1-1.

( )2

2 2y x x y= − ⇒ − = ±

Since 2x ≥ , 2 2x y x y− = ⇒ = + .

1f ( ) 2 , 0x x x− = + ≥ , [ )1 ff

R D 2,− = = ∞ .

(b) g( ) 2 ,x x x= − ∈�

The line 1y = cuts the graph at two points, 1,3x = . Hence g is not 1-1.

(c) { }2 3

h( ) , \ 11

xx x

x

+= ∈

−�



Since every horizontal line , 2y k k= ≠ cuts the graph at most once, hence h is 1-1.

2 3

1

2 3

2 3

3

2

xy

x

xy y x

xy x y

yx

y

+=

−

⇒ − = +

⇒ − = +

+⇒ =

−

Hence { }1 3h ( ) , \ 2

2

xx x

x

− += ∈

−� ,

{ }1 hhR D \ 1− = = � .

3. ( ( ) 2g f( ) 3 5x x= +

( )23 1 2x= + +

( )3 f( ) 2x= +

Let f( )y x= , then g( ) 3 2y y= + .

Hence g : 3 2x x +� .

Alternatively, let 2

1 1y x x y= + ⇒ = ± − .

Therefore

1f ( ) 1x x− = ± − . Note that 1f − is a relation and not a function. Then,

( ) ( )( )

21 1g( ) g f f ( ) 3 f ( ) 5

3 1 5 3 2

x x x

x x

− −= = +

= − + = +

Tutorial Questions

1.(i)

2 2

2

4 3

ff ( )

f ( 6 )

f ( 5 )

25 30

λ

λ λ

λ

λ λ

= −

= −

= +

(ii)

2

2 2

f ( ) 6

( 3 ) 9

x x x

x

λ

λ λ

= −

= − −

( )2

fR 9 ,λ= − ∞

Since f (0) 0 f (6 )λ= = ,



f is not a one-to-one function and hence

inverse of f does not exist.

Minimum value of f occurs when

3x λ= , hence 3k λ= .

2 2

2 2

2

( 3 ) 9

9 ( 3 )

3 9

y x

y x

x y

λ λ

λ λ

λ λ

= − −

+ = −

= ± +

For 23 3 9x x yλ λ λ< ⇒ = − +

1 2 2f ( ) 3 9 , 9x x xλ λ λ−∴ = − + ≥ −

For 23 3 9x x yλ λ λ> ⇒ = + +

1 2 2f ( ) 3 9 , 9x x xλ λ λ−∴ = + + ≥ −

2. ( ) ( )f fD 0, , R 0,1= ∞ =

[ )

( ) ( )

g g

h h

D ,R 5,

D , 5 , R 0,

= = − ∞

= −∞ − = ∞

�

Since [ ) ( )g fR 5, 0, D= − ∞ ⊄ ∞ = , fg does not exist. For fg to exist, ( )gR 0,= ∞ ,

therefore 2 15

33 5 0 | |x x− > ⇒ > .

Since ( )f gR 0,1 D= ⊆ =� , gf exists. ( )2 2gf( ) 3 f( ) 5 3e 5, 0x

x x x−= − = − > .

( ) ( )g

f gfR 0,1 5, 2 R= → − − = .

Since ( )h gR 0, D= ∞ ⊆ =� , gh exists. ( ) ( )( )2 5

3gh( ) 3 h( ) 5 3 5

10, 5

xx x

x x

− += − = −

= − − < −

.

( ) ( )g

h ghR 0, 5, R= ∞ → − ∞ = .

Since [ ) ( )g hR 5, , 5 D= − ∞ ⊄ −∞ − = , hg does not exist. Furthermore g hR D {}∩ = , it is

not possible to find a restriction of g such that hg exist.

3.(i) f( ) ( 2)( 8)x x x= − −

( )22 10 16 5 9x x x= − + = − −



[ )

( ) ( )

f f

g g

D , R 9,

D 0, ,R 1,

= = − ∞

= ∞ = − ∞

�

Since [ ) ( )f gR 9, 0, D= − ∞ ⊄ ∞ = , gf does not exist.

(ii) Since f(4) 8 f(6)= − = , f is not 1-1. Therefore 1f − does not exist. From the expression

of f, domain of f must be restricted to ( ,5]−∞ . Hence

h : ( 2)( 8), , 5x x x x x− − ∈ ≤� � and [ )hR 9,= − ∞ .

(iii) Now, 1 1h h ( ) h h( )x x x− −= = for all x such that both functions are defined,

[ ) ( ]

[ ]

1 1 1 hhh h h h

h h

D D D D

R D 9, ,5

9,5

x − − −∈ ∩ = ∩

= ∩ = − ∞ ∩ −∞

= −

4. ( )22g( ) 4 4 2 1 1x x x x= − = − −

(i) Since g(2) 8 g( 1)= = − , therefore g is not 1-1. Hence 1g− does not exist.

(ii) From the expression of g, 12

k = .

(iii)

1hh( ) h( ) h ( )x x x x−= ⇒ = . Hence we are looking for the intersection of these two

graphs. From diagram, they intersect at the origin, which is therefore where P is.

5. (i)

fR ( ,2]= −∞

-1 ffD R ( , 2]= = −∞



Set of values of x = [ 0 , 2 ]

(ii)

g : e , , 0x a

x x x+ ∈ <� �

( )gR 0,ea=

e lnx ay x y a

+= ⇒ = −

Hence 1g : ln , 0 e

ax x a x

− − < <� .

( ] ( )f gR , 2 ,0 D= −∞ ⊄ −∞ =

For gf to exist, ( )fR ,0= −∞ , i.e.

332 0 8 2

4

xx x− < ⇒ > ⇒ > . Therefore maximal domain is ( )2,∞ . Clearly

( )gf gR R 0,ea= = .

6. .(i) 2f ( ) 6x x x= − −

( )

( )

2 1 14 4

225 14 2

6 x x

x

= − + + −

= − +

( )

( )

225 14 2

2251

2 4

2512 4

y x

x y

x y

= − +

⇒ + = −

⇒ + = ± −

Since 12

x ≤ − , 25 14 2

x y= − − − .

1 25 2514 2 4

f : , ,x x x x− − − − ∈ ≤� �

(ii) ( )gR 2,= − ∞ .

(iii) ( ) ( )g hR 2, 1, D= − ∞ ⊄ ∞ = , hence hg does not exist.

(iv) For hg to exist, ( )gR 1,= ∞ , therefore e 2 1 e 3 ln 3x xx− > ⇒ > ⇒ > .

Hence ln 3α = .

( )

( )( )

hg( ) ln ln g( )

ln ln e 2 , ln 3x

x x

x

=

= − >



(v) ( )( )hg( ) ln ln e 2 ln 2x

x = − =

( )

( )

2

2

ln e 2 2 e e 2

ln e 2

x x

x

− = ⇒ = +

⇒ = +.

7.(i)

[ )fR 0,= ∞

(ii) From graph [ )1,1S = − .

(iii) 1 1

1 1

x xy y

x x

+ += ⇒ ± =

− − Let z y= ±

1 1 21

1 1 1

x zz x

x z z

+ +⇒ = ⇒ = = +

− − −

Therefore,2

11

xy

= +−

or 2

11

xy

= +− −

.

Since 1x < and 0y ≥ , 2

11

xy

= +− −

.

1

1

2f ( ) 1 , , 0

1x x x

x

− = + ∈ ≥− −

� .

8.(i)

( )

( )-1f ,

axy

bx a

bxy ay ax

by a x ay

ayx

by a

ax ax x

bx a b

=−

− =

− =

=−

= ≠−



( )2

2

2

f

,

axa

bx ax

axb a

bx a

a x

abx abx a

ax x

b

− =

− −

=− +

= ≠

2fR \

a

b

=

�

8.(ii) fg does not exist, because fD \

a

b

=

� but

{ }gR \ 0= � . i.e. g fR D⊄ at a

xb

= .

8.(iii)

( )

( )

-1

2

2

f

2 0

2 0

20,

x x

axx

bx a

ax bx ax

bx ax

x bx a

ax x

b

=

=−

= −

− =

− =

= =

9.(i)

( ) ( ) ( ) ( )

( ) ( )

( )( ) ( )( )2

f 27 f 45 f 3+4 6 f 1+4 11

f 3 f 1

2 3 1 7 1

11

+ = +

= +

= − + −

=

9.(ii)

7

3

2 4 10 -7




1.

The idea here is to draw a horizontal line y k= .

If the line cuts the graph of f, then fRk ∈ .

Hence we are finding all the values of k, such

that the equation f( )k x= has solutions.

Consider

( )2

2 211 1 1 0

1

xk k x x x kx k

x

+= ⇒ − = + ⇒ + + − =

−. For this equation to have solutions, since it is a

quadratic equation, we have

Discriminant = ( )24 1 0k k− − ≥ .

Solving the inequality, we get 2 2 2k ≥ + or 2 2 2k ≤ − .

Therefore, ( )fR ,2 2 2 2 2 2, = −∞ − ∪ + ∞ .

2.

[Only consider the part of graph for 0x > - given domain.]

First, use the GC to graph 1

y xx

= − for 0x > as follows:

In the Y= window, enter 1

1Y (X X ) /(X 0)−−−−= − >= − >= − >= − > .

Press WINDOW, then enter suitable values such as

Xmin = -10, Xmax = 10, Ymin = -10, Ymax = 10.

Press GRAPH.

x

y

y k=

( )1 2,2 2 2+ +

( )1 2,2 2 2− −

x

y

g( )y x=

y = b

Alternatively,

( )2

2f '( ) 1 0

1

1 2

f (1 2) 2 2 2

xx

x

= − =−

= ±

± = ±

Hence, stationary pts:

( )1 2,2 2 2± ±



From the graph of g, a horizontal line y = b, where b ∈� , cuts the graph of g( )y x= at

exactly one point, hence g is a 1-1 function and thus 1−g exists.

Let 1

g( )y x xx

= = − . Then 2

2

2

1

1 0

4

2

xy

x

x yx

y yx

−=

⇒ − − =

± +⇒ =

Since x+∈� ,

2

0

4

2

x

y yx

>

+ +⇒ =

But 2

1 1 4g ( ) g ( )

2

y yx y y

− − + += ⇒ =

Replacing all y by x, 2

1 4g ( )

2

x xx

− + +=

Hence 2

1 4g : ,

2

x xx x

− + +∈� � . (since RRDx gg

==∈ −1 )

Assignment Questions

1. (i) Let f(x) = y where y = 1 − (x − 2)

2

⇒ x = 2 ± 1 y−

= 2 − 1 y− (∵ x < 2)

∴ f −1

: x � 2 − 1 x− , x < 1

(ii)

0 1 2 3 x

y

-1 -2 -3

y = f(x)

y = f-1

(x)

y = x

-3



2. (i)

[ ) { }g fR 0, \ 3 =D fg does not exist= ∞ ⊄ ⇒�

{ }f gR \ 0 D gf exists= ⊂ = ⇒� �

( )

2

1gf : , , 3

3x x x

x∴ ∈ ≠

−� � ( )gfR 0,= ∞

(ii) Every horizontal line y = k, { }\ 0k ∈� cuts the graph of ( )fy x= exactly once.

Hence, f is 1-1. 1f − exists.

(iii)

y

x 3

( )fy x=y k=

y

x O

( )gy x=

x

y

O



3.

Min point is (3, 1)

Any horizontal line ,y k= where 1,k > cuts the graph

more than once. This implies that f is not a 1-1

function and thus 1f −

does not exist.

(i) From graph, we can deduce that the smallest possible

value of b is 3.

3

e ln 3x

y y x−

= ⇒ = −

3 lnx y⇒ − = ±

Therefore,

3 lnx y= + or 3 lnx y= − .

Since 3x ≥ , 3 lnx y∴ = + . 1f : 3 ln , 1x x x

− + ≥� .

(ii) For -1f g to exist,

-1g fR D⊂

[ )-1 ffD R 1,= = ∞

[ )gR ,a= ∞

Hence, smallest value of a is 1.

( ) [ ) [ )

-1g g f D R R

, 1, 3,

g

−∞ ∞ → ∞ → ∞

{ }

{ }

f

f

D \ 3

R \ 0

=

=

�

� [ )

g

g

D

R 0,

=

= ∞

�

topic 6: functions · pdf filenational junior college mathematics department 2010 2010 / sh1 /...

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