MODUL AM015 2021 INTAKE

AM015 – Functions And Graphs
Example 5.6
Sketch the graph of the following functions. Hence, find the domain and range.

(a) f(x) = x  3

(b) f(x) =  x  5

(c) f(x) = 3  x

vii) Absolute Value Function f(x)

f(x) f(x) = |x| 0x

x f(x) = –|x|

0 Domain :  ,

Domain :  , Range : (, 0]

Range : [0, )

Page 10 of 28

AM015 – Functions And Graphs

f(x) f(x)

f(x) = |x+a|

x f(x) = |x| + a

a
0x

–a 0

Domain :  , Domain :  ,
Range : [0, )
Range : [a, )

Example 5.7
Sketch the graph of the following functions. Hence, find the domain and range.
(a) f(x) = |x+3|

(b) f(x) = |x| – 2

viii) Piecewise Function

(a) f ( x)   x 2, x0
 x 5, x0


Page 11 of 28

AM015 – Functions And Graphs

2x  4,  4  x  1
(b) f (x)  3  4x, -1 x 1

 x, x 1

Exercise
Sketch the graph of the following functions. Hence, state the domain and range.
a) f (x)  2  x 1
b) f (x)  x 1

LECTURE 3 OF 5
At the end of the lecture, you should be able to

i. represent a composite function by an arrow diagram.
ii. find composite function.
iii. find one of the functions when the composite and the other functions are given.

Definition of Composite Functions:
It is also possible to take the output values from one function and use them as the input values
for another function. The functions which are composed in this way are called composite
functions or function of a function.

Consider two functions f(x) and g(x).
We define f  g(x) = f [g(x)] meaning that the output values of the function g are used as the
input values for the function f.

Page 12 of 28

This can be represented in an arrow diagram: AM015 – Functions And Graphs

f og C
f[g(x]
AB

gf
x g(x)

Similarly, we define g  f(x) = g [f(x)] meaning that the output values of the
function f are used as the input values for the function g.

This can be represented in an arrow diagram. C
g[f(x)]
g of

AB

fg
x f(x)

Note :

i. ( f  g)(x)  f(x). g(x)

ii. (g  f)(x)  g(x). f(x)
iii. ( f  f)(x)  f(x)2
iv. f 2 (x)  [f(x)]2
v. f 2 (x) = ff(x)

Page 13 of 28

AM015 – Functions And Graphs

Example 5.8
If f(x) = 3x + 1 and g(x) = 2 – x, compose the

(a) (f  g)(x)

(b) (g  f)(x)

Example 5.9
If f(x) = 2x – 1 and g(x) = x3, find the values of

(a) gf(3)

(b) fg(3)
(c) f 2(3)

Example 5.10 (x  1).

The functions f and g are defined by f (x)  2  x and g(x)  3
x 1

(a) Show that f 2(x) = x.
(b) Find an expression for g2(x)

Page 14 of 28

AM015 – Functions And Graphs

Example 5.11
Given that f(x) = 2x + 3 and f0g(x) = 10x – 9, find g(x).

Example 5.12
Given the function f (x)  x2  3x 1, x   3 . Find the function g(x) if f [g(x)]  2x 3.

2

Example 5.13
If g(x) = 2x + 3 and fg(x) = 10x – 9, find f(x).

Page 15 of 28

AM015 – Functions And Graphs
Example 5.14
If fg(x) = 4x2 – 2x + 1 and g(x) = 2x + 1, find gf(x). Subsequently, find the values of x that
satisfy fg(x) = gf(x).

Example 5.15

Find the function of g(x) , given f (x)  x2 1, x  0 and ( f  g)(x)  x2  2x  2, x  1.

Exercise

Given fg(x)  x2 1. Find
a) g(x) if f (x) 1 2x
b) f (x) if g(x)  x  3

Page 16 of 28

AM015 – Functions And Graphs

LECTURE 4 OF 5
At the end of the lecture, you should be able to

i. use algebraic approach or horizontal line test to determine whether a function is one to
one.

ii. determine the inverse of a function if it exists.
iii. find the domain and range of an inverse function.
iv. sketch the graph of the function and its inverse on the same axes.

E) Inverse Functions
i) The inverse function exists if and only if a function is one to one function. If f is a function

from A to B then and inverse function for f is a function in the opposite direction, from B to A.
Thus, if an input x into the function produces an output y, then inputting y into the inverse
function f -1 produces the output x.

A fB

xy

f -1
ii) There are two methods to determine whether the function is one-to-one function,

by using
(a) Algebraic method
(b) Graphical method

iii) Properties of inverse
(a) Domain of f = Range of f -1
(b) Range of f = Domain of f -1
(c) (f -1)-1 = f
(d) f -1[f(x)] = x or f[f -1(x)] = x
(e) (fog) -1 = g -1of -1

iv) A many-to-one function can have an inverse by restricting the domain of the function so that
it is one-to-one.

(a) Algebraic Method
A function f with a domain is called a one-to-one function if no two elements of x have the
same image, that is f(x1) ≠ f(x2) for x1 ≠ x2. To prove that a function f is one-to-one, we must
show that f(x1) = f(x2) implies that x1 = x2.

Page 17 of 28

AM015 – Functions And Graphs

Example 5.16

By using the algebraic method, determine whether f is a one-to-one function or not.

a) f(x) = 2x + 3, xR b) f(x) = x2, x R

c) f(x) = x2, x ≥ 0

(b) Graphical method
The graph of a one-to-one function does not have the same y-coordinate for two different
x-coordinates on the graph. Consequently, if a horizontal line intersects the graph y = f(x) at
more than one point, then f is not a one-to-one function since there are two different values
of x, namely x1 and x2 such that f(x1) = f(x2)

Example 5.17
Use graphical method to determine whether each of the following functions is one-to-one function.
a) f (x)  2x 1, x  R.

b) f(x) = (x – 2)2 + 3

Page 18 of 28

AM015 – Functions And Graphs

Inverse of a Function

The function g is inverse of f if ( f g)(x)  x . The function f and g are inverses of each other

if ( f g)(x)  x and (g f )(x)  x

Example 5.18
Given the function f (x)  3x and g(x)  x , find

3
a) fg(x)

b) gf (x)

Conclude your answer in part (a) and (b).

Note: Range f(x) = Domain f 1 ( x)

 f [ f 1(x)]  x and f 1[ f (x)]  x

Domain f(x) = Range f 1 ( x) ,

Example 5.19
Find the inverse of f(x) = 3 – x if exist.

Note:
Such a function f is called self-inverse, i.e. it is its own inverse.

Page 19 of 28

AM015 – Functions And Graphs

Example 5.20

Given the function f(x) = x  2 , (x  3) , find f -1.

x3

Example 5.21
A function f is given by f (x)  x2  4x 1, if the domain of f is set of all real numbers, show that

the function f has no inverse. Restate the domain so that f(x) is one-to-one function.

Example 5.22
A function f is defined by f (x)  x 1 , x 1

a) find an expression for f 1
b) find the domain and range for f 1

Page 20 of 28

AM015 – Functions And Graphs

Example 5.23
The functions f and g are defined by f : x  2x  3 and g : x  x 1. Find
(a) f -1 and g -1
(b) (g  f )-1
(c) f -1  g -1

Sketch the Graph of the Function and its Inverse.

y f(x)
y=x
f -1(x)

b x
a 0b

a

The graph of y = f(x) and f -1 (x) are reflection of each other in the line y = x.
Page 21 of 28

AM015 – Functions And Graphs
Example 5.24
For each the functions below, Sketch the graph of f and f -1 on the same axes. State the domain

and the range of f 1(x) .

a) f (x)  2x  5
b) f (x)  (x  9)2 , x  9
c) f (x)  x  2, x  2

Exercise
Sketch the following functions. Hence, without finding their inverses, sketch their inverses on the
same axes.

a) f (x)  x2  2, x  0
b) f (x)  x  3, x  3
c) f (x)  x3

LECTURE 5 OF 5
At the end of the lecture, you should be able to

i. use algebraic approach or horizontal line test to determine whether a function is one to
one.

ii. determine the inverse of a function if it exists.
iii. find the domain and range of an inverse function.
iv. sketch the graph of the function and its inverse on the same axes.
F) Exponential and Logarithmic Functions

Relationship between Exponential and Logarithmic Functions
The inverse of exponential function is a logarithmic functions and vice versa.
The exponential function, f(x) = ax and its inverse f 1(x)  log a. x.
The logarithmic function, g(x)  log a x and its inverse g 1(x)  a x.
The domain of exponential function is  ,.

Page 22 of 28

y f (x) = ax AM015 – Functions And Graphs

1 y=x
f (x) = loga x
01 x

Since loga x is the inverse function of ax, it follows that the domain of loga x is the range of ax.

Example 5.25

Given f (x)  e2x 1.

a) Show that f(x) is one-to-one function.
b) Find f -1(x)
c) Determine the domain and the range of f -1(x)

Example 5.26
Given f(x) = 2(3x) – 1
a) Find f -1(x)

b) Determine the domain and range of f(x) and f -1(x)

Page 23 of 28

AM015 – Functions And Graphs

Example 5.27

Given f(x) = ln (3x + 2)

a) Show that the function is one-to-one function.
b) Find f -1(x).
c) Determine the domain and range of f -1(x).

Example 5.28

Given f(x) = log3 (2x + 1).
a) Find f -1(x).
b) Determine the domain and range of f -1(x).

Sketching the graph of the Exponential Function

(a) f(x) = ax, xR , a > 1 f(x) = ax , a > 1

f(x)

1
x

Page 24 of 28

AM015 – Functions And Graphs

Basic properties :

i) f(x) > 0 for xR.
ii) When x = 0, f(x) = 1
iii) When x   , f(x)  
iv) When x  - , f(x)  0

(b) f(x) = ax, xR , 0 < a < 1

f(x)
f(x) = ax , 0 < a < 1

1
x

Basic properties :

i) f(x) > 0 for x  R .

ii) When x = 0, f(x) = 1
iii) When x   , f(x)  0
iv) When x  - , f(x)  

Example 5.29
Sketch the graph of the function below and state the domain and range of f(x).
a) y = e2x + 1

b) y = 3 – e2x

Page 25 of 28

AM015 – Functions And Graphs

c) y = 1 + e– 2x

Sketching the graph of the Logarithmic Function
A logarithmic function is defined as f(x) = log b x, x  R, b > 0, b  1
(a) f(x) = logb x , xR , b > 1

f(x) f(x) = logb x , b > 1

1 x

Basic properties:
i) When x = 1, f(x) = 0
ii) When x  0 , f(x)  -
iii) When x   , f(x)  

(b) f(x) = logb x, xR , 0 < b < 1

f(x)

x
1

f(x) = logb x , 0< b < 1

Basic properties:
i) When x = 1, f(x) = 0
ii) When x  0 , f(x)  
iii) When x   , f(x)  -

Page 26 of 28

AM015 – Functions And Graphs

Example 5.30

Sketch the graph of the following function.
a) y = ln (2x – 1)

b) y = – ln (2x – 1)

c) y = ln (1 – 2x)

Example 5.31
Given f(x) = 2 + ln (x + 1), sketch the graph of f(x) and the f -1 (x) on the same axes. State the
domain and range of f(x) and f -1(x).

Page 27 of 28

AM015 – Functions And Graphs

Example 5.32

Given f(x) = 22x + 3
a) Sketch the graph of f and f -1 on the same axes.
b) Determine the domain and range of f(x) and f -1(x).

Example 5.33

1

The functions f and g are given by f (x)  3  e3x and g(x)  ln( x  3) 3 . Show that f and g

are inverse of each other. Hence, sketch graph f and g .

Exercise
The functions f and g are defined as follows.

f (x)  4  x2

1x

g(x)  e 2
a) Determine whether f is one-to-one function. Give reason for your answer.
b) Find g 1 and state the domain.
c) Obtain the composite function g 1 f (x) .

Page 28 of 28

AM015/5.Functions and Graph

CHAPTER 5: FUNCTIONS AND GRAPH

TUTORIAL 1 OF 8

1. The function f is given f : x 3x  6, x  R. Find
a) f (4)
b) The value of x whose image is 18.
c) The value of x such that f (x)  2x 1

2. If f (x)  1 , x  1, find
x 1

a) f  1 
a

b) The value of a if f  1   2
a

3. If f (x)  2x  3,  3  x  2. Sketch the graph and state the domain and range.

4. Sketch the graph of the following functions and state its domain and range.

a) f (x)  x2  2x  3
b) f (x)  2x2  4x  5

c) f (x)  x  32, x  3.

5. Sketch the graph of the following function. Hence, state the domain and range.

a) f x  x3  4

b) f (x)  x3  2x2  5x  6
c) f (x)  3x2 (4  x)

ANSWERS

1. a) 6 b) 8 c) 7

2. a) a b) 2
1a 3

3. Df  [3,2], Rf   3, 7
4. a) Df   ,, Rf  [4, )

b) Df   ,, Rf  (,  3]

c) Df  (,  3] Rf  [0, )

5. a) Df   ,, Rf  (, )
b) Df   ,, Rf  (, )
c) Df   ,, Rf  (, )

Page 1 of 10

AM015/5.Functions and Graph

TUTORIAL 2 OF 8

1. Sketch the graph of the following function. Hence, state the domain and range.

(a) f (x)  | x  4 | (b) f  x  3 x 1

(c) f  x  x  3 (d) f (x)  2 1 x  3

2. Given f (x)  2x 1 and g(x)  x  4.

a) Write f (x) as a piecewise function. Hence, find the function h(x)  2 f (x) 3g(x).
b) Sketch the graph of h(x).

3. Sketch the graph of the following function. Hence, state the domain and range.

 1 x0

f x   x 2 1 0 x 1

3x 1 x 1

4 x  1
 x 1
4. Given f x   x2 3 x 1

 2  x

a) Sketch the graph of f (x) and find the range.

b) Find the value of f (3), f (0),and f (1) .

 x, -4  x  0
 x0
5. A piecewise function g is given as g(x)   2, x0

 x,


Sketch the graph of g(x) . Hence, determine the domain and range.

ANSWERS (b) Df   ,, Rf  [3, )

1. (a) Df   ,, Rf  [0, ) (d) Df  (, 1] Rf  [3, )

(c) Df  [3, ) Rf  [0, )

 2x 1, x1 7x 10, x1
 2  2
2. a) f (x)   h(x) 
b) x 1 x 1
2x 1, 2 x 14, 2

Graph

3. Df   ,0  (0, ) Rf  [1, )

4. a) Rf  (, 1)  [3, 4] b) 4, 3, 4

5. Dg  [4, ), Rg  0, 

Page 2 of 10

AM015/5.Functions and Graph

TUTORIAL 3 OF 8

1. Given g(x)  x2  x 12 and h(x)  x  3, where x 1 .
2

a) Find f (x)  g(x) , hence state the domain.
3h(x)

b) Find the value of x such that 2 f (x)  x .
3

2. If f x  2x , g x  x 1 and h x  x2 , find fg  x and gh x .

3. If f (x)  2x 3 and g(x)  x2  5 , find

(a) gf 2 (b) fg 3

(c) fg a 1

4. Find the function of g(x) , given

(a) f (x)  2x 3 and  f g(x)  2x 1.

(b) f (x)  x2 1, x  1 and ( f  g)(x)  x2  2x  2, x  1.

(c) f  x  x 1 and g f x  3 2x  x2 .

5. Given f  x  6x  p, find the function g for each of the composite function below,
(a)  f g  x  8x2  5
(b)  g f  x  36x2 12 px  p2

6. If (g f )(x)  1 2x , x  2 and f (x)  3 2x. find g(x) .
2x

ANSWERS

1. a) f (x)  x  4 , x  3, Df  [ 1 , ) b) x  8,
3 2

2. 2(x 1), x2 1

3. a) 6 b) 25 c) 2a2  4a  9

4. a) g(x)  x  2 b) g(x)  x 1 c) g(x)  4  x2

5. a) g(x)  8x2 5 p b) g(x)  x2

6 ,

6. g(x)  2x  4 , x  1
x 1

Page 3 of 10

AM015/5.Functions and Graph

TUTORIAL 4 OF 8

1. For each of the following function, determine whether f is one to one function or not.

(a) f (x)  3x 1 (b) f (x)  x2  4x  3

(c) f x  x 5 (d) f (x)  4(x  2)2 1, x  2.

2. Find f 1  x . In the same axes sketch the graph of f  x and f 1  x . State its domain and

range.

(a) f x  2x 1, x R

(b) f  x  x2  3x  4, x   3

2

(c) f  x  x3 3, x  R

3. Given a function f (x)  x2  3
(a) Sketch the graph f (x) and state its domain and range.
(b) Hence verify that f (x) is one-to-one function or not.
(c) Determine the existence of f 1(x)
(d) Hence, sketch the f (x), x  0 and f 1(x) in the same axes.

4. Given f (x)  x2  2x  3, x  1. Find f 1(x) and hence, evaluate f 1(6).

5. Consider the function f (x)  5x where x  1 .
4x 1 4

Write an expression for f 1(x) and state its domain.

6. Find the value of q for the given functions.

a) f (x)  4(x  2)2  q, x  2 and f 1(5)  3.

b) f (x)  q(x  3)2  2, x  3 and f 1(2)  2.

ANSWERS (b) not one-to-one function
(b) one-to-one function
1. (a) one-to-one function
(c) not one-to-one function

2. (a) f 1(x)  x  1 , D  R  (,), Rf  D  (,)
2 f f
f 1 1

(b) f 1(x)  x  25  3 , Df  R 1  (,), Rf  D 1  (,)
4 2 f f

(c) f 1(x)  3 x  3, Df  R 1  [ 32 , ), Rf  D 1  [ 25 , )
f f 4

Page 4 of 10

AM015/5.Functions and Graph
3. (a) Graph,

Df  (,) Rf  (,3]

(b) f(x) is not one-to-one function for that domain, but one-to-one function for domain.

x  0 or x  0.
(c) f 1(x) exist for where the function is one to one function.

(d) graph

f 1(x) 1 x  2

4. f 1 6  3

5. f 1(x)   x  x , D f 1   ,\ 5
5 4x 4x 5 4

6. q  1, q = 4

TUTORIAL 5 OF 8

1. Find the inverse function of the following function and state its domain and range.

(a) f  x  2x (b) f (x)  e2x 1

(c) f  x  3 ex (d) f (x)  logx 1

(e) hx  ln3x 1 (f) f  x  ln  x 1
 2 
(g) f (x) = 5x +1
(h) g(x) = 2 – 10x

2. The functions f is defined by f : x  1 ln x, x  0. Find f 1  x and f 1(2)

2

3. Given the function f (x)  ln  px q  and the inverse function f 1 ( x)  5ex 7 . Find the
 2x   rex 
 

possible values of p, q and r.

ANSWERS

1. (a) f 1(x)  log2 x. Df 1   0,  , R 1   , 
f

(b) f 1( x)  1 ln  x  1 , D f 1  1, ,Rf 1  ,
2

(c) f 1( x )  ln3  x , D 1  ,3 ,Rf 1  ,
f

(d) f 1(x)  10x 1, D 1   ,  , R 1  1,
f f

(e) h1  ex 1  ,, Rh1   1 ,  
3 , Dh1  3 

(f) f 1  2ex 1, D  ,, Rf 1  1,
f
1

Page 5 of 10

AM015/5.Functions and Graph

(g) f 1  log 5 (x 1), Df 1  (1, ), Rf 1  (, )

(h) g 1  log (2  x), Dg1   , 2, Rg1  (, )

2. f 1(x)  e2x , 54.5982
3. q  5,r  2, p  7 or q  5,r  2, p  7

TUTORIAL 6 OF 8

2 x

1. Functions f and g are defined as f (x)  5(2  ln3x) and g(x)  e 5 respectively.
3

Write down an expression for ( f  g)(x). Hence, find g 1(x).

2. Given f(x) = 4(3 – ln 2x) and g(x)  e3bx , where b is a constant.
2

(a) Write down an expression for  f  gx.

(b) Find the value of b such that f and g are inverse of each other.

3. The functions f and g are given by f (x)  3  e3x and g(x)  ln(3x  5).
(a) Find the value of x such that ( f g)(x)  4 .
(b) If ( f  g 1)(b)  4, find the value of b .

4. Given the function f (x)  log x .
1 log x

a) Find f 1(x) .
b) If g(x) 101x , find f g(x).

ANSWERS

1. x , g 1(x)  5(2  ln 3x)

2. (a) ( f g)(x)  4bx b1
(b) 4

3. (a) x   4 (b) b  ln5  1.6094
3

x

4. a) f 1 (x)  101x

b) f  g(x)  1 x .
x

Page 6 of 10

AM015/5.Functions and Graph

TUTORIAL 7 OF 8

1. Find f 1  x . In the same axes sketch the graph of f  x and f 1  x . State its domain and

range.

(a) f  x  e2x  5

(b) f  x  log3 5  x
(c) f x  lnx 3

(d) f  x  3 ex

(e) f  x  ln  x 1
 2 

2. Given f (x)  22x , find f 1 (x). Hence, sketch f (x) and f 1(x) on the same graph.

3. Given the function as follows g  x  ex3 and f  x  2  ln x 1
(a) In a separate diagram, sketch the graph g  x and f  x . Hence, state the domain and

range for each of the function g and f .

(b) Find  f gx. Hence, solve for x such that  f g  x  2.

ANSWERS

1. (a) f 1 ( x)  1 ln( x  5), Df  R 1  (,), Rf  D 1  (5,)
2 f f

(b) f 1(x)  5  3x , Df  R 1  (,5), Rf  D 1  (,)
f f

(c) f 1(x)  ex  3, Df  Rf 1  (3,), Rf  Df 1  (,)

(d) f 1(x)  ln | 3  x |, Df  R 1  (,), Rf  D 1  (,3)
f f

(e) f 1(x)  2ex 1, Df  R 1  (1,), Rf  D 1  (,)
f f

2. f 1 ( x)  1 log 2 x
2

3. (a) Df : 1, 
Rf : , 
Dg : , 
x  3ln2  3.693
Rg : 0, 

 (b)  f g  x  2  ln ex3 1 ,

Page 7 of 10

AM015/5.Functions and Graph

TUTORIAL 8 OF 8

1. Given g(x)  x  4. Find the function k (x) if g k(x)  x  2 where x  2.Hence, determine
the value of x such that k 1(x)  11.

2. Given f (x)  log2 x and g(x)  2x.
a) Show that f and g are one to one functions. Without finding the inverse functions, show
that functions f and g are inverse of each other.
b) On the same axes, sketch the graphs of f and g . Label all the intercepts and line of

symmetry. Hence, state the domain and range of each function.

3. Given two function f (x)  74xm and g(x)  13 log7 x . If f (x) and g(x) are inverse of
n

each other, express x in terms of m and n.

4. Given g(x)  5x . By defining g2 (x)  (g g)(x), determine the function g2 (x). Hence,
x5

obtain the inverse of g(x).

 5. Given f (x)  x2 1, x  0. Determine g(x) if fg(x)  1 e23x1 1 .
55

ANSWERS

1. k(x)  4  x  2, k 1(x)  (x  4)2  2 for x  4, x  1.

2. a) Shown.
b) Graph.

Df  Rg  0, 

Rf  Dg  , 

3. x  13  m or mn52
n4
x  7 4n .

4. g2(x)  x, g1(x)  5x .
x5

5. g(x)  e2(3x1) .

Page 8 of 10

AM015/5.Functions and Graph

EXERCISE

1. Find the intersection point between f (x)  2x2 16 and g(x)  4x . Hence, sketch both
graphs f (x) and g(x) on the same axes. Then, state its domain and range.

2. Given the function f (x)  p  qx2 such that f ( 13)  5 and f (3)  17 . Find the exact

values of p and q.

3. Given a quadratic function where its domain and range are  ,  and [2, ) . It is also

known that the symmetrical axis of the function is x  3 and one of the root is 4. Express the

quadratic function in its general form completely.

4. A piecewise function f (x) is given as

f ( x)   2x2, x0
 x  2, 4  x  0.


a) Sketch the graph of f (x) .

b) Hence, find the domain and range of f (x) .

5. The function f is defined by f : x  x2  2x  5, x  R, x  1.

Find the function g if the composite function f  g is given by

f  g : x  4x2  20x  29, x  R, x  5 .
2

6. Functions f and g are defined by f (x)  x , x  1 and g(x)  ax2  bx  c, x  R
x 1

where a, b and c are constants.

(a) Find f  f , hence find f 1.

(b) Find the values of a, b and c if g f  2x2  5x  2 .
(x 1)2

7. Find the value of h(2) given that h  gx  4x2  4x  4 and g(x) =2x + 1.

8. Consider the functions f (x)  x2  4x  3.

(a) Express f (x) in the form of (x  h)2  k, such that x  h. Hence, find the value of h.
(b) Sketch the graph of f (x) for x  h and explain why f is a one-to-one function.
(c) State the range of f (x).

9. Given g(x)  5  ax and h(x)  2x2  5. Find g 1(x) and the value of a so that
2

2g 1(x2 )  h(x).

Page 9 of 10

AM015/5.Functions and Graph

ANSWERS

1. Intersection point at (4,16) and (-2,-8), Df  Rf  (, )
Dg  (,), Rg  [16,)

2. p  1,q  2
3. Minimum point (3,-2) subs (4,0) into f (x)  a(x  3)2  2, a  2 , f (x)  2x2 12x 16

4. a) Graph

b) Df  [4, ), Rf  [2, )
5. g(x)  2x  4, x  R, x  5 .

2

6. a) x , f 1(x)  x , x  1 b) a  1, b 1, c  2
x 1

7. h(2)  7

8. (a) f (x)  (x  2)2 1, h  2.

(b) graph, from the graph horizontal line cross the function at one point.
(c) [1, ).

9. 2x  5 , a = 2
a

PSE 2020/2021 [2 marks]
[6 marks]
1. Given f (x)  4x2  4, g(x)  1 5x.
[6 marks]
a) State the domain and range of g(x). [7 marks]

b) Find g 1(x). Hence, determine ( f  g 1)(x) in the simplest form.

2. Given f (x)  ln(2x  5) and g(x)  ex  3.Find
a) (g  f )1(4).
b) The value k if (g 1  f )(k)  0.

ANSWERS

1. a) Dg  (, 15], Rg  [0, )
b) g 1(x)  1 x2 , x  0. ( f  g 1)(x)  4(x4  2x2  24) .
5 25

2. a) (g  f )1(x)  1  1  5, (g  f )1(4)  2.
2x3 

b) (g 1  f )(k)   ln[ln 2k  5  3], k  e4  5 @ 24.8

2

Page 10 of 10

AM 015/6. Polynomials

CHAPTER 6: POLYNOMIALS

SUBTOPIC
6.1 Polynomials
6.2 Remainder Theorem, Factor Theorem and Zeroes of Polynomials
6.3 Partial Fractions

LECTURE 1 OF 5

Learning Objective:
By the end of this lesson, student should be able to
a) Perform addition, subtraction and multiplication of polynomials.
b) Perform division of polynomials.

6.1 Polynomials

The word polynomial is derived from two Greek words, ‘poly’ means many and
‘nomial’ means terms or names. Many real-life situations can be modeled by
polynomial functions. For examples, the observed pitch of a train whistle is a
polynomial function of the speed of the train and the shape of an Astro disc is
parabolic, which is a polynomial quadratic function of its radius.

Definition of Polynomials

A polynomial in a finite sum of terms in which all the variables have whole number

exponents and no variable appears in denominator. A polynomial P(x) of degree n is

defined as
P(x)  anxn  an1xn1 ... a1x  a0; an  0

where nZand a0, a1, a2,..., an are called the coefficient of the polynomial .
The coefficient of the highest power of x, an, is the leading coefficient.
The constant term is a0 .
The degree of the polynomial is determined by the highest power of x .

For Example
The table below shows the degree, leading coefficient, constant term and the name
of the polynomial. Complete the table.

Polynomials Degree Leading Constant Name of
coefficient term polynomial
1 P(x)  7 1 -7 constant
2 P(x)  5x  6 2 0 -6 linear
3 P(x)  7x2  x 3 quadratic
4 P(x)  2x3  7x2  x  3 7 3
cubic

Examples of non-polynomial expressions

1  4x , 5  3x1 , x2  3x  3 and also the expression that contains the non-
x
x3

positive power of x .

Page 1 of 20

AM 015/6. Polynomials

Monomials, Binomials and Trinomials

For Example:

Name Example Number of terms
Monomial x3 one
Binomial two
Trinomial 3x3  2x three
7x3  2x2 1

A) The Algebraic Operations on Polynomials

Addition and subtraction

The addition and subtraction of the polynomial P(x) and Q(x) can be performed by
collecting like terms.

Example 1

Given P(x)  5x3  4 and Q(x)  x3  3x2  4x . Determine

(a) P(x) Q(x) (b) P(x) Q(x)

Multiplication

Note that every term in one polynomial is multiplied by each term in the other
polynomial.

Example 2

Given P(x)  x2 1and Q(x)  2x3  x2 1.

Determine: (a) 4Q(x) (b) P(x) Q(x)

If P( x) is a polynomial of degree m and Q( x) is a polynomial of degree n ,

then product P(x) Q(x) is a polynomial of degree m  n .

Page 2 of 20

AM 015/6. Polynomials
Division

In the division of integer, 11  5  1
22

 the quotient is 5
 the remainder is 1
 the divisor is 2
The statement could be expressed as

11 5(2) 1

= divisor (quotient) + remainder
In the same way, the division of polynomials can be expressed in the form

P(x)  Q(x)  R(x) or P(x)  D(x) Q(x)  R(x)
D(x) D(x)

When dividing polynomials, the quotient and remainder can be found by using long
division.

Q(x)
D(x) P(x)

R(x)

The division operation will be continued until the degree of the
remainder is less than the degree of the divisor.

B) Long Division
Example 3

Find the quotient and the remainder for polynomial P(x)  2x2  3x  6 when
divided by D(x)  x 1 . Hence, express P(x) in terms of Q(x) and R(x) .

Page 3 of 20

AM 015/6. Polynomials

Example 4

Given that 3x3  4x2  x  7 is divided by 3x  4 . Write your answer in the form of
P(x)  D(x) Q(x)  R(x)

Example 5

Use long division to find x3  27 .

3 x

When the remainder is zero; we say that x3  27 is exactly divisible
by 3 x

Page 4 of 20

AM 015/6. Polynomials
Miscellaneous Exercise

1. Divide x2  2x 5 by x
2. Find the remainder when x3 1 is divided by x  2

x3  2x  5
3. Find the quotient and remainder for x2  3

x3  4x2  5x  2

4. Use long division to find

x2
5. Given that 4x3 3x 5 is divided by 2x  3. Write your answer in the form of

P(x)  D(x) Q(x)  R(x)

Answer Miscellaneous Exercise:

1. x2  2x 5  xx  2 5

 2. x3 1 x  2 x2  2x  4 9
 3. x3  2x 5  x x2  3 5x  5
 4. x3  4x2  5x  2  x  2 x2  2x 1
 5. 4x3 3x  5  2x 3 2x2 3x 3 14

Page 5 of 20

AM 015/6. Polynomials

LECTURE 2 OF 5

Learning Objective:
By the end of this lesson, student should be able to
a) Apply the remainder theorem to solve problems.

6.2 Remainder Theorem, Factor Theorem and Zeros of Polynomials.

By using long division polynomial P(x)  2x3  x2  3x 1 is divided by x  2 will give

remainder 19.

2x2  3x  9
x  2 2x3  x2  3x 1

 (2x3  4x2)

3x2  3x
 (3x2  6x)

9x 1
 (9x 18)

19

We can still get the remainder without going through the long division. This is called

the remainder theorem. Substitute x  2 in the above polynomial, we obtain
P(2)  2(2)3  22  3(2) 1
 19

The Remainder Theorem

A) The Divisor is a Linear Factor

When a polynomial P( x) is divided by a linear factor x  a , then the remainder is
P(a)

Proof

Let P(x) be a polynomial of degree n where n  2 .
Then P(x)  Q(x)(x  a)  R (from polynomial division)

When x  a ,

P(a)  Q(a)(a a)  R

Since (a  a)  0 , then the remainder R(a)  P(a)

Note:

1. If P(x) is divided by ax – b = a  x  b  , then R = P  b
 a   a 

2. If P(x) is divided by ax +b = a  x  b  , then R = P   b 
 a   a 

3.
4. Page 6 of 20

AM 015/6. Polynomials

Example 1

Find the remainder when P(x)  x3  2x2  4x  5 is divided by:

(a) x  2 (b) 3x 1

Example 2

When P(x)  2x3  3x2  kx3  6 is divided by 2x 1 the remainder is 16. Determine

k.

Example 3

The expression x3  5x2  qx  9, where q is a constant, gives a remainder of 13
when divided by (x  4). Find the remainder when the same expression is divided by
(x  3) .

Page 7 of 20

AM 015/6. Polynomials
Example 4

Given that P(x)  2x3  ax2  6x 1. When P(x) is divided by x  2 , the remainder
is twice of the remainder when P(x) is divided by x 1. Find a.

B) The Divisor is a Quadratic Expression

When a polynomial P(x) is divided by a quadratic expression ax2  bx  c , then the
remainder is R(x)  ax b .

P(x)  Q(x) D(x)  R(x)
P(x)  Q(x) D(x)  ax  b

Example 5
By using Remainder Theorem, determine the remainder when

P(x)  x3  3x2  3x  4 is divided by x2  4.

Example 6

The polynomial P(x) leaves a remainder of 1 on division by x  3, a remainder of 3
on division by x  3 . Find the remainder when P(x) is divided by x2  9 .

Page 8 of 20

AM 015/6. Polynomials
Example 7

The polynomial P(x)  x3  px2  qx  8 gives a remainder is (2  x)when divided
by (x2  x  2) . Find the values of p and q .

Miscellaneous Exercise

The polynomial x3  ax2  bx1 left a remainder of 5 when it is divided by x 1
and a remainder of 7 when divided by x  2 . Find the values of a and b .

Answer: a  4,b  7

Page 9 of 20

AM 015/6. Polynomials

LECTURE 3 OF 5

Learning Objective:
By the end of this lesson, student should be able to
a) Apply factor theorems to solve problems
b) Find the roots of the equations and the zeroes of a polynomials.

The Factor Theorem

If the remainder obtained from dividing the polynomial P(x) by ( x  a ) is zero, then
the linear term (x  a) is called a factor of the polynomial P(x) .

If P(a)  0 then (x  a) is a factor of P(x)

x 1
x 1 x2  2x 1

 (x2 1)

x 1
 (x 1)

0

By using remainder theorem,

P(1)  (1)2  2(1) 1
0

If (x  a) is a factor of P(x) then P(a)  0.

If (ax  b) is a factor of P(x) then P   b  = 0.
 a 

Example 1

(a) Show that (x 3) is a factor of P(x)  3  7x  5x2  x3 .

(b) By using Remainder Theorem, determine whether the following linear
functions are factor of the given polynomials:

P(x)  2x3  3x  6; Dx  (x 1)

Page 10 of 20

AM 015/6. Polynomials
Example 2

If 2x 1 is a factor of polynomial P x  2x3  px2 - 5 , find the values of p .

Example 3

Given that  x  2 and  x 1 are both factors of the polynomial
P x  2x3  ax2  bx  5. Find the values of the constant a and b .

Example 4

When the polynomial P(x) is divided by (x2 1) , the remainder is (mx  n). Find the
constants m and n given that (x 1) is a factor of P(x) and when P(x) is divided
by (x 1) , the remainder is 4.

Page 11 of 20

AM 015/6. Polynomials
Example 5

Given that the expression P(x)  3x3  ax2  bx 12 is exactly divisible by x2  2x 3

.

(a) Determine the values of a and b .

(b) Hence, factorize the expression completely.

Roots and Zeros of a Polynomial
Definition

 A zero of a polynomial P(x) is a number a such that P(a)  0
 x  a is called a root of the polynomial equation P(x)  0
 In general, if x  a is a root of a polynomial equation P(x)  0 then

(x  a) is a factor of P(x) .

 Every polynomial equation of degree n has exactly n roots. Some of

these roots may be repeated.

Example 6

(a) Show that x 1 is a factor of P(x)  2x3  5x2  4x  3

(b) Factorize the expression completely.
(c) Hence, find all the zeros of the expression.
(d) State all the roots of the expression.

Page 12 of 20

AM 015/6. Polynomials

Example 7

Find all the zeros and roots of P(x)  6x3 13x2  4 .

Miscellaneous Exercise

1. Show that  4 is a zero of 6x3  23x2  5x  4.
2. Determine whether the following linear functions are factors of the given

polynomials:

P(x)  2x3  3x2 8x  3; Dx  (2x 1)
3. Prove that  x  2 is a factor of P(x)  2x3  x2  8x  4 . Hence, factorize

P(x) completely.
4. Factorize P(x)  2x3  9x2  3x  4 completely and write all the zeros.
5. Prove that 3x  2 is a factor of the polynomial 6x3  7x2  x  2 . Hence, find

the roots of the equation 6x3  7x2  x  2  0
6. Determine all the roots of x2  2x2 5x 6
7. Find the roots for the equation x3  3x2  x 1  0

Answer Miscellaneous Exercise:

1.  4 is a zero of 6x3  23x2  5x  4.
2. 2x 1is a factor of P(x)  2x3  3x2  8x  3

3.  x  2 is a factor of P(x)  2x3  x2  8x  4 .
P(x)  x  22x 1x  2

4. P(x)   x 12x 1 x  4 .

Zeroes are  1 , 1, 4
2

5. 3x  2 is a factor of P(x)  6x3  7x2  x  2 .
Roots: x   1 , x  2 , x 1
23

6. Roots : x  1, x  2, x  3
7. Roots: x  1, x  0.414, x  2.414

Page 13 of 20

AM 015/6. Polynomials

LECTURE 4 OF 5

Learning Objective:
By the end of this lesson, student should be able to
a) Construct partial fractions decomposition when the denominators are in the form
of

(i) Linear factor , ax  b

(ii) A repeated linear factor, ax  bn

6.3 Partial Fractions

Two or more proper fractions can be combined to give a single fraction, for example

(i) 1  2  11 (ii) 3  2  x  6
3 5 15 x (x  2) x(x  2)

Equally, a single proper fraction can be expressed as a sum or difference of two or
more algebraic fractions, for example

(i) 11  1  2 1 and 2 are known as partial fractions of 11
15 3 5 35 15

(ii) x  6  3  2 x6
x(x  2) x (x  2)
are known as partial fractions of
3 and 2
x (x  2) x(x  2)

Definition
A fraction can be expressed or decomposed as the sum of two or more separate
proper fraction.
This is known as partial fractions.

P(x)  a(x)  c(x)  e(x)
Q(x) b(x) d(x) f (x)

Partial fractions

Page 14 of 20

Summary AM 015/6. Polynomials
Denominator of Proper fractions

Linear Repeated Quadratic
factor linear factor factor

Type 1: Linear Factors in the Denominator

If Q(x) is a product of linear factors, thus P(x) can be

a1x  b1 a2 x  b2 .....ar x  br 

expressed as A  B  C and A, B,C are constants.
a1x  b1 a2x  b2 a3x  b3

For Example:

x x5 6  x A  x B 6 and x 7x  4  A  B
4  x3 x5
 4x   3x  5

Example 1

Express 2x  3 in partial fractions.

x 13x  2

Page 15 of 20

AM 015/6. Polynomials

Example 2

5x 1

Express (x 1)(2x 1)1 x in partial fractions.

Type 2: Repeated Linear Factors in the Denominator
The numerators are just constant terms

(a) 6x2 15x  8  A x B 5  C .
(x  3)(x  5)2 x3  (x  5)2

Repeated factors
(b) x 1  A  B  C

x2 (3x  5) x x2 3x  5

Repeated factors

Example 3 in partial fractions.

2x 1
Express x2 (2  x)

Page 16 of 20

AM 015/6. Polynomials

Miscellaneous Exercise

5

1. Express x  x 1 in partial fractions.

2. Express 7x in partial fractions.
(x  2)2

7

3. Express  x  3 (x 1)2 in partial fractions.

4. Express x3 in partial fractions.

 x  22  x 1

Answer Miscellaneous Exercise:

1. x  5   5   x 5
x
x 1 1

2. 7  7  14

 x  22 x  2 x  22

3. 7  7  16 7  7

 x  3(x 1)2 16x 3 x 1 4 x 12

4.   x3 1    x 2 2  x 1   2

x 22 x  22 x 1

Page 17 of 20

AM 015/6. Polynomials

LECTURE 5 OF 5

Learning Objective:
By the end of this lesson, student should be able to
a) Construct partial fractions decomposition when the denominators are in the form
of

(iii) A quadratic factor, ax2 bx  c

Type 3: Quadratic Factors in the Denominator

(i) Denominator with quadratic factors that can be factorized

Example 1 in partial fractions.

Express x
4x2 1

Example 2

3

 Express x2 1  x 1 in partial fractions.

Page 18 of 20

AM 015/6. Polynomials

(ii) Denominator with Quadratic factors that cannot be factorized

For Example

(i) 7x 10  A  Bx  C

(x  2)(x2  7x  3) x  2 x2  7x  3

(ii) 6x2  8  A  Bx  C
x(5x2  6x  2) x 5x2  6x  2

Example 3

Express 4x 1 in partial fraction

(x 1)(x2  5x 1)

Example 4

3

x  2 x2 1
 Express in partial fractions:

Page 19 of 20

AM 015/6. Polynomials

Miscellaneous Exercise

x 1

 1. Express  x 1 x2  5x  6 in partial fractions

2x2  x 1 in partial fractions.
 2. Express
 x  3 2x2 1

Answer

1.  x 1 x 1  6  6  1  3 1 2  2  1 3
x x
x2 5x x 1

2x2  x 1  22  6x 1
2x2 
 x  3 2x2 1 x
   2. 19 3 19 1

Page 20 of 20

AM015/6. POLYNOMIAL
TUTORIAL CHAPTER 6 : POLYNOMIALS

TUTORIAL 1 OF 9

1. Functions f, g and h are defined by: f  x  2x  3 , g  x 10  x2 and

h x  x3  5x2  3 . Simplify the following,

a) hx  g x  f x b) 3 f x  2g x c) f x.g x

2. Find the quotient and remainder of the following functions by long division. Write the

answers in the form Px  Qx Dx  Rx .

4x3  2x2  x 1 2x3  7x2  7x 15
b) x2  3
a)

x2

3. Find the quotient and remainder of the following functions by long division. Write the

answers in the form P(x)  Q(x)  R(x)
D(x) D(x)

2x3  4x2  x  3 x3  27
(a) x2  2x  3
(b)

x3

1. a) x3  6x2  2x 10 b) 2x2  6x  29 c) 30 20x 3x2  2x3

 2. (a) 4x3  2x2  x 1 4x2 10x  21 x  2  43

 (b) 2x3 7x2  7x 15  2x  7 x2 3  13x  6

 3. (a) 2x3  4x2  x 3  2x x2  2x 3  7x 3

 (b) x3  27  x2  3x  9 x 3

TUTORIAL 2 OF 9

1. By using the remainder theorem, find the remainder when the following functions are
divided by the linear factors indicated.

(a) x3  2x  4; x 1 (b) 2x3  x2 13x  6,x  2 (c) x3 3x2  5x ;2x 1

2. Given the remainder is 9 when P x  x3  px2  qx 1 is divided by  x  2 and 19
when divided by  x  3 , find the values of p and q .

3. The polynomial Sx  5x3  px2  qx 1 gives a remainder of 26 on division by
 x  1 and remainder of 8 on division by  x 1 . Find the values of the constants p

and q .

Page 1 of 8