MST121 Chapter A3 Functions

MST121
Using Mathematics
Chapter A3
Functions
About this course

This course, MST121 Using Mathematics, and the courses MU120 Open
Mathematics and MS221 Exploring Mathematics provide a flexible means of
entry to university-level mathematics. Further details may be obtained from the
address below.
MST121 uses the software program Mathcad (MathSoft, Inc.) and other
software to investigate mathematical and statistical concepts and as a tool in
problem solving. This software is provided as part of the course.
This publication forms part of an Open University course. Details of this and
other Open University courses can be obtained from the Student Registration
and Enquiry Service, The Open University, PO Box 197, Milton Keynes
MK7 6BJ, United Kingdom: tel. +44 (0)845 300 6090, email
general-enquiries@open.ac.uk
Alternatively, you may visit the Open University website at
http://www.open.ac.uk where you can learn more about the wide range of
courses and packs offered at all levels by The Open University.
To purchase a selection of Open University course materials visit
http://www.ouw.co.uk, or contact Open University Worldwide, Walton Hall,
Milton Keynes MK7 6AA, United Kingdom, for a brochure: tel. +44 (0)1908
858793, fax +44 (0)1908 858787, email ouw-customer-services@open.ac.uk
The Open University, Walton Hall, Milton Keynes, MK7 6AA.

First published 1997. Second edition 2001. Third edition 2008. Reprinted 2008.
c 1997, 2001, 2008 The Open University
Copyright $
All rights reserved. No part of this publication may be reproduced, stored in a
retrieval system, transmitted or utilised in any form or by any means, electronic,
mechanical, photocopying, recording or otherwise, without written permission from
the publisher or a licence from the Copyright Licensing Agency Ltd. Details of such
licences (for reprographic reproduction) may be obtained from the Copyright
Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS;
website http://www.cla.co.uk.
Edited, designed and typeset by The Open University, using the Open University
TEX System.
Printed in the United Kingdom by Cambrian Printers, Aberystwyth.
ISBN 978 0 7492 2939 9
3.2
Contents
Study guide 4
Introduction 5
1 What is a function? 6
1.1 Function notation 6
1.2 Graphs of functions 9
2 Quadratic functions 15
2.1 The exhibition hall problem 15
2.2 Graphs of quadratic functions 17
3 Trigonometric and exponential functions 27
3.1 Trigonometric functions 27
3.2 Exponential functions 32
4 Inverse functions 35
4.1 What is an inverse function? 35
4.2 Inverse trigonometric functions 40
4.3 Logarithms 42
5 Functions, graphs and equations on the computer 47
Summary of Chapter A3 48
Learning outcomes 48
Summary of Block A 49
Solutions to Activities 50
Solutions to Exercises 57
Index 62
3
Study guide
There are five sections in this chapter. They are

intended to be studied consecutively in five study
sessions. Subsection 2.2 requires the use of an
audio CD player, and Section 5 requires the use
of the computer and Computer Book A.
The pattern of study for each session might be as
follows.
Study session 1: Section 1
Each of the sections should take two and a half to
three hours, the longest being Section 4.
Before studying this chapter, you should be
familiar with the following topics, which are
covered in the Revision Pack and Chapter A0:
% basic properties of powers ax , where a > 0
and x is a real number;
% the notion of a function, and, in particular,
trigonometric and exponential functions.
The optional Video Band A(v) Algebra

workout – Powers and logarithms could be
viewed at any stage during your study of
this chapter.
4
Introduction
One of the major concerns of mathematics (and science) is the way in

which variables are related. For example, the formula
C = 2πr
expresses the circumference C of a circle in terms of its radius r. In this
formula, r is an independent variable whereas C is a dependent variable.
In more complicated formulas, there may be several independent variables.
For example, in the formula A =
21 ab sin θ for the area of a triangle,
a
, b and θ are independent variables, and A is the dependent variable;
see Figure 0.1. In order to study the relationship between dependent and
independent variables systematically, the concept of a function is
introduced.
Figure 0.1 Area of a triangle
Section 1 introduces functions as ‘processors’ and describes their notation

and their representation by graphs. It also introduces the concept of an
interval of the real line; roughly speaking, this is an unbroken set of real
numbers.
In Section 2, you will see how problems about areas lead to equations
whose solutions can be estimated by using the graph of an appropriate
function. We discuss techniques of translation and scaling for sketching
graphs of functions, particularly quadratic functions.
Section 3 is concerned with trigonometric functions and exponential
functions. You will see how fundamental properties of these functions are
related to the shapes of their graphs.
If the effect of a given function can be reversed, then we obtain its inverse
function. Section 4 defines this concept, explains how to identify inverse
functions, and describes various standard inverse functions.
Section 5 uses the computer to plot accurate graphs of functions and to
obtain numerical solutions of equations for which there is no convenient
formula. It also describes how the computer can solve some equations by
performing algebraic manipulations.
5
1 What is a function?
1.1 Function notation

In this chapter, we concentrate on functions which represent the
relationship between a dependent variable and a single independent
variable. In this case, it is common to take x as the independent variable
(though other variables may be used, such as t when time is involved) and
y as the dependent variable. So y and x are typically related by formulas
such as
√
y = 2x − 1, y = x2 , y = x,
y = tan x, y = 2x .
In each case, for each value of x, the formula gives exactly one value of y
– a unique value. If y and x are related by such a formula, then we say
informally that ‘y is a function of x’. The concept of a function is,
however, more general than this, covering cases in which there is no simple
formula relating y to x.
In general, a function can be thought of as a processor which converts
inputs (values of x) into outputs (values of y). Figure 1.1 illustrates this
concept in relation to the formula y = x
2 .
Figure 1.1 A processor for squaring
In general, a function is specified by giving:

(a) a set of allowed input values, called the domain of the function;
(b) a process, called the rule of the function, for converting each input
The word unique is taken to value into a unique output value.
mean ‘one and only one’ in
For a function called f , we use the notation f (x) = . . . to specify the rule
mathematics.
of f , so y = f (x). For example, for the function in Figure 1.1, the rule is
Pronounce f (x) as ‘f of x’.
f (x) = x2 . (1.1)
It is important to note that a function f is independent of the variable
names, which could be changed without changing the function itself. For
example, the formula s

= t2 also leads to the function f , with s = f (t) = t2 .
Various notations can be used to specify both the rule of a function and its
domain, the commonest being of the form
Notice the resemblance here f (x) = x
2 + 1 (0 ≤ x ≤ 6). (1.2)
to the way we specify a
sequence in closed form; see Equation (1.2) specifies a function f with rule f (x) = x2 + 1, whose
Chapter A1, Subsection 1.1. domain is the set of real numbers x satisfying 0 ≤ x ≤ 6.
Activity 1.1 Specifying functions
(a) Describe the domain of the function f given by the specification

f (t) = −t (−1 < t < 2).
6
SECTION 1 WHAT IS A FUNCTION?
(b) For each of the following functions f , write down a specification of the
function and find the value f (1), the output corresponding to the
input 1.
(i) The function f which converts any non-negative real number x
into x2 .
(ii) The function f which converts any real number x which satisfies
−1 ≤ x ≤ 1 into 2x + 1.
Solutions are given on page 50.
For any x in the domain of f , we call f (x) the image of x under f , the
value of f at x, or simply an image value. For example, for the function We usually use ‘image’, but
in Figure 1.1, the image of −3 under f is f (−3) = (−3)2 = 9. as you will see, there are
contexts where the
The set of allowed input values of a function, the domain, depends on: alternatives are more natural.
% the nature of the rule of the function – for example, the rule in
equation (1.1) can be applied to any real number x, whereas the rule
√
f (x) = x
can be applied only to non-negative real numbers;
% the context in which the function is applied – for example, if we are
using the function with rule f (x) = x
2 in a modelling context, to
determine the area of a square of side x, say, then it is inappropriate
to include non-positive values of x in the domain.
Thus the domain of a function may be restricted by the nature of the rule
of the function or by the context in which the function is applied.
In describing domains of functions, it is often convenient to use interval
notation. Roughly speaking, an interval is an unbroken set of real numbers
– that is, a ‘subset’ of the real line which can be ‘drawn without lifting A subset of the real line that
your pen from the paper’. is not an interval is the
‘broken’ subset comprising
the digits 0, 1, . . . , 9.
Intervals
Let a and b be real numbers, with a < b.
The closed interval [a, b] is the set of real numbers x such
that a ≤ x ≤ b.
The open interval (a, b) is the set of real numbers x such Some texts use the notation
that a < x < b. ]a, b[ for open intervals.
The closed interval [a, b] includes its endpoints a and b, whereas the open
interval (a, b) excludes them. For example, in part (ii) of Activity 1.1(b),
the domain of the function is the closed interval [−1, 1].
Interval notation may be used when specifying functions. For example,
rather than writing
f (x) = 2x + 1 (−1 ≤ x ≤ 1)
to specify the function in part (ii) of Activity 1.1(b), we could write
f (x) = 2x + 1 (x in [−1, 1]).
We can also use this notation to represent intervals which include one
endpoint but exclude the other. Also, by using the symbols ∞ and −∞, Read ∞ and −∞ as ‘infinity’
we can represent intervals which extend indefinitely far to the right or left; and ‘minus infinity’.
see Table 1.1 (overleaf).
7
CHAPTER A3 FUNCTIONS
Table 1.1 Interval notation

Inequality Interval
Closed intervals a≤x≤b [a, b]
Remember that the symbols a≤x [a, ∞)
∞ and −∞ do not represent x≤a (−∞, a]
real numbers, so they are not
Open intervals a<x<b (a, b)
considered to be endpoints of
a<x (a, ∞)
intervals.
x<a (−∞, a)
The real line R is also an Half-open (or half-closed) a≤x<b [a, b)
interval, sometimes written as intervals a<x≤b (a, b]
(−∞, ∞).
Activity 1.2 Using interval notation
For each of the following inequalities, write down the corresponding

interval and describe it as closed, open or half-open (half-closed).
(a) 0 < x < 1 (b) −3 ≤ x ≤ 2 (c) −2 < x ≤ 2 (d) x ≥ 0
When the domain of a function is restricted only by the nature of the rule
of the function, we usually do not state the domain explicitly, but rely on
the following convention.
Domain convention
When a function is specified by just a rule, it is understood that the
domain of the function is the largest possible set of real numbers for
which the rule is applicable.
Consider, for example, the two functions g and h specified as follows:

√
1
g(x) = x and h(x) =
.
x
√
By the domain convention, the function g has domain [0, ∞), since
x is
defined only for x ≥ 0. Similarly, the function h has domain R excluding 0,

which consists of the two open intervals (−∞, 0) and (0, ∞).
Activity 1.3 Using the domain convention
Specify the domain of each of the following functions, as given by the
domain convention. In each case, explain your answer.

√
(a) f (x) = x − 1
1 1
(b) f (x) = +
x−2 x+3
8
Here are some general remarks about function notation.

1. As you have seen, it is common practice to use the name f for the
function currently being discussed. Other names like g and h are used
in order to distinguish between functions.
2. Alternative names for ‘function’ are mapping and, in geometric
contexts, transformation.
3. There is an alternative notation for the rule of a function. Rather than
writing f (x) = x2 , for example, we can write
f : x )−→ x
2 . Read f : x )−→ x2 as ‘the
function f maps x to x2 ’.
This notation aims to suggest the sense of x being converted
(‘mapped’) to x
2 by the function.
It is also common practice to write, for example, ‘the function
f (x) = x2 ’ rather than ‘the function f with rule f (x) = x 2 ’.

4. For all the functions in this chapter, both the inputs and the outputs
are real numbers. Such functions are called real functions. More
general functions may have inputs or outputs of a very different nature,
such as points in the plane, words or even people. When dealing with
such general functions, it is common to add to the function
specification information about a set in which the output values lie.
We call such a set the codomain
√ of the function. For example, to
specify the function f (x) = x in this detailed way, we could write
f : [0, ∞) −→ R Here [0, ∞) is the domain and
√ R is the codomain. For a real
x )−→ x. function, the codomain can
Notice that two different types of arrow are used here. always be chosen to be R.
1.2 Graphs of functions

For any real function f , the points (x, y) in the Cartesian plane which
satisfy y = f (x) form the graph of the function, often referred to as ‘the
graph of y = f (x)’. Such a graph gives a means of visualising the function
geometrically, often illustrating key features of the function. For example,
in Chapter A2 you met straight-line graphs with equations of the form
y = mx + c. These are the graphs of linear functions of the form
f (x) = mx + c. Here the domain convention
is being used.
Figure 1.2 Graph of y = mx + c
The fact that the graphs of these functions are (straight) lines means that
they have constant slope, and this in turn means that if x increases by a
fixed amount, then so does y; see Figure 1.2.
9
Graphs of linear functions can be drawn accurately by plotting just two

points and drawing a line passing through them. For most other functions,
however, we can sketch graphs only approximately.
A basic method is to construct a table of values of f (x) for several values
of x, plot the corresponding points (x, f (x)), and attempt to sketch a
For most of the functions smooth curve through these points. The more points we can plot, the
considered in this chapter, the better the sketch will be, and the more confident we shall be that the
graph will be smooth. function does not behave wildly between consecutive plotted points. One
possibility is to use a graphics calculator or computer to do this, but first
it is important to develop some familiarity with the graphs of a number of
common types of functions.
We begin with the ‘squaring function’
f (x) = x2 ,
and sketch the corresponding graph of y = x2 by constructing a table of

values such as the following.
x −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
2
x 4 2.25 1 0.25 0 0.25 1 2.25 4
Where convenient, we use the

same scale on both axes of a
graph, as here.
Figure 1.3 Graph of y = x2
In Figure 1.3 we have plotted 9 points and joined them by a smooth

U-shaped curve. This characteristic smooth U-shape can be confirmed by
plotting more points if necessary. The curve is symmetric in the y-axis;
For example, the points (2, 4) that is, the part to the left of the y-axis is the mirror image, or reflection,
and (−2, 4) both lie on the in the y-axis of the part to the right. This is because if the point (u, v) lies
graph. on the graph, then v = f (u) = u2 , so we can also write
v = (−u)2 = f (−u).
Hence the point (−u, v) also lies on the graph.
Next, we sketch the graph of the reciprocal function
1
Recall that, for x ,= 0, the f (x) = .
x
reciprocal of x is 1/x.
By the domain convention in Subsection 1.1, the domain of the reciprocal
function is the real line R excluding the point 0.
As before, we construct a table of values for f (x) = 1/x, and try to sketch
a smooth curve (or curves) through the corresponding points (x, f (x)).
x −2 −1.5 −1 −0.5 0.5 1 1.5 2
These values of 1/x are

correct to two decimal places. 1/x −0.5 −0.67 −1 −2 2 1 0.67 0.5
10
Observe from the table that points on the graph of f lie in either the first
or the third quadrant. Since 0 is not in the domain of f , no point in the
first quadrant is joined by the graph to a point in the third quadrant. The
graph is shown in Figure 1.4.
In this case, using the same

scale on both axes suggests
that the graph is symmetric
in the line y = x, as discussed
below.
Figure 1.4 Graph of y = 1/x

Figure 1.4 shows that the graph consists of two smooth curves, each of
which approaches – that is, becomes arbitrarily close to – an axis at each
end. This behaviour of the graph occurs because, for x > 0,
1
if x is large, then f (x) = is small, For example, if x = 1000,
x
and then 1/x = 10−3 .
1
if x is small, then f (x) = is large,
x
with similar properties for x < 0.
Any line which a curve approaches ‘far from the origin’ is called an
asymptote. Thus both the axes are asymptotes of the graph of the
reciprocal function.
The graph of y = 1/x also has some symmetry properties, which can be
seen in the points plotted in Figure 1.4. For example, the points (2, 0.5),
(−2, −0.5), (0.5, 2) and (−0.5, −2) all lie on the graph. In general, if the
point (u, v) lies on the graph, then so do the points (−u, −v), (v, u) and
(−v, −u); see Figure 1.5.
As in Figure 1.5, it is often

sufficient to show just two
axis scale markers to indicate
the scales being used for a
graph, and to omit the label
for the origin unless the origin
has a special role (such as an
intercept) in the graph under
consideration.
Figure 1.5 Symmetry properties of the graph of y = 1/x
11
The idea of restricting the If the domain of a function f has been restricted due to the context in
domain of a function was which the function is applied, then the graph of f is restricted in a
introduced on page 7. corresponding way. For example, the graphs of the functions
g(x) = 1/x (1 ≤ x ≤ 2) and h(x) = 1/x (−∞ < x < 0)
are both parts of the graph of the function f (x) = 1/x; see Figure 1.6.
Here, in each case, the

domain on the x-axis has
been emphasised. To indicate
whether a point is included or
excluded from a domain or
graph, we can draw it as a
solid dot (included) or an
open circle (excluded).
Figure 1.6 Graphs of g and h
One feature of a function f which can often be read off from the graph
of f is the image set, or image, of f . The image set is the complete set
of image values of f – that is, all possible values f (x), where x lies in the
domain of f . In terms of the graph of f , the image set of f is the set of
points on the y-axis which are at the same height as a point of the graph.
Figure 1.7 shows the image sets of two of the functions considered above.
The image set of the function f (x) = x
2 is the interval [0, ∞), and the
image set of the function g(x) = 1/x (1 ≤ x ≤ 2) is the interval [
12 , 1].
Figure 1.7 Image sets of f and g
Activity 1.4 Sketching graphs
Calculate image values By constructing a suitable table of values, sketch the graphs of each of the
correct to two decimal places. following functions, and hence state their image sets.
(a) f (x) = x
3 (−1 ≤ x ≤ 1) (b) f (x) = 1/x2
12
So far, the graphs considered have consisted of a smooth curve (or curves).
The next graph is smooth except at one point, where it turns a corner!
This is the graph of the modulus function.
The modulus, or absolute value, of a real number x is the

magnitude of x, regardless of its sign; it is denoted by |x|.
For example:
the modulus of −3 is |−3| = 3;
the modulus of 0 is |0| = 0;
the modulus of
π is |π| = π.
From this definition, it follows that For example,
"
x, if x ≥ 0, |−3| = −(−3) = 3.
|x| =
−x, if x < 0.
This observation enables us to sketch the graph of the modulus function
f (x) = |x|
without constructing a table of values. We just use the part of the graph of
y = x for which x ≥ 0, and the part of the graph of y = −x for which
x < 0, as shown in Figure 1.8.
Figure 1.8 Constructing the graph of the modulus function
The resulting graph of y = |x| has a corner at the origin, and its image set
is [0, ∞).
Activity 1.5 Graphs involving the modulus
Without constructing a table of values, sketch the graphs of each of the

following functions.
1
(a) f (x) = |x|3 (−1 ≤ x ≤ 1) (b) f (x) =
|x|
13
Summary of Section 1
This section has introduced:
% notations for specifying the rule and domain of a function;
% interval notation, including [a, b] for a closed interval and (a, b) for an
open interval;
% a domain convention for functions;
% the graph and image set of a function;
% the graphs of some common functions;
% the modulus, or absolute value, of a real number.
Exercises for Section 1

Exercise 1.1
For each of the following functions, write down the domain of the function
using interval notation.
1
√
(a) f (x) = x
4 (x < 1) (b) f (x) =
√ (c) f (x) = 9 + x
x
Exercise 1.2
Calculate image values By constructing tables of values, or otherwise, sketch the graphs of each of
correct to two decimal places. the following functions, and state their image sets.
1 1
(a) f (x) = x4 (b) f (x) = √ (c) f (x) = '
x |x|
14
2 Quadratic functions
To study Subsection 2.2, you will need an audio CD player and CDA5505.
In Section 1, we discussed the idea of a (real) function, and considered
some examples of common functions. Now we see how functions and their
graphs can be applied in a particular modelling situation.
2.1 The exhibition hall problem Stages in the modelling cycle

for this problem are given in
A hall is to be used for an exhibition in such a way that half the floor the margin.
space is devoted to exhibits, the rest being clear. After measuring the hall
and making simplifying assumptions (such as taking convenient
approximate dimensions), the following problem is stated. Specify
purpose
The Exhibition Hall Problem

A rectangular hall of dimensions 16 metres by 12 metres is to have a
uniform border around the walls for exhibits (as shown in Figure 2.1).
The border is to take half the area of the hall. Determine the width
of the border.
In Figure 2.1, the variable x has been used to denote the width of the From now on we assume that
border, and the variable A denotes the area of the clear space. all lengths are in metres.
Create
model
Figure 2.1 Introducing variables for the exhibition hall problem
To solve the exhibition hall problem, we need to find the value of x for
which A is half the total area of the hall. You are now asked to find a
formula for A in terms of x.
Activity 2.1 Relating variables
(a) Write down the largest closed interval consisting of values of x which
make sense for the exhibition hall problem.
(b) For values of x in this interval, write down expressions for the length
and width of the clear space.
(c) Deduce an expression for the area of the clear space A, in terms of x, Do
multiplying out any brackets. mathematics
15
In Activity 2.1, you should have found that x lies in the interval [0, 6], and
that, for such x,
A = 4x
2 − 56x + 192.
The exhibition hall problem is to find a value of x such that A is half the
area of the hall; that is, A = 12 × 16 × 12 = 96. Therefore we have to solve
the equation 4x
2 − 56x + 192 = 96; that is,
4x
2 − 56x + 96 = 0. (2.1)
Then we have to choose a solution in the interval [0, 6], if possible.
Activity 2.2 Solving the equation
(a) Solve equation (2.1).

(b) Hence solve the exhibition hall problem.
Comment
Interpret You can check whether your answer to part (b) is reasonable by studying
results Figure 2.1, which has been drawn (to scale) with A = 96.
We have now solved the exhibition hall problem, and without any
reference to functions or graphs! What makes this problem straightforward
is the existence of a convenient method for solving equation (2.1). For
many problems, however, we arrive at equations for which no such method
is available. In such cases, it is helpful to express the problem in terms of
functions and their graphs.
For example, a function which arises in the exhibition hall problem is
f (x) = 4x2 − 56x + 192 (x in [0, 6]).
This function represents the area of the clear space in the exhibition hall,
when the border has width x. The graph of y = f (x) is given in Figure 2.2.
(A way of sketching this graph is given in Subsection 2.2.)
For this graph, it is not

convenient to use the same
scale on both axes, because
the domain of f is [0, 6] and
the image set of f is [0, 192].
Figure 2.2 Graph of f (x) = 4x2 − 56x + 192 (x in [0, 6])
The graph of f has the following features:
% f (0) = 192 (if x = 0, then the hall is completely clear);
% f (6) = 0 (if x = 6, then there is no clear space);
% as x increases from 0 to 6, the clear area f (x) decreases from 192 to 0.
16
SECTION 2 QUADRATIC FUNCTIONS
The key point to notice from Figure 2.2 is that it provides some indication
of the answer to the exhibition hall problem. The answer to this problem
is a solution of the equation f (x) = 96; that is, it is the value of x at the
point P where the horizontal line y = 96 meets the graph of y = f (x).
From Figure 2.2, we can see that this value of x is approximately 2, which
is the value found in the solution to Activity 2.2(b).
The graph of the same function f can also be used to obtain an
approximate value of the solution to a modified exhibition hall problem.
Suppose that the area of the clear space is required to be
43 of the total
area (192 square metres); that is, A =

43 × 192 = 144. Then the border
width x must satisfy

f (x) = 144. (2.2)
It can be seen from Figure 2.3 that in this case x is approximately 1.
Figure 2.3 Modified exhibition hall problem
Activity 2.3 The modified exhibition hall problem
(a) Solve equation (2.2).

(b) Hence solve the modified exhibition hall problem.
So, expressing a problem in terms of a function and its graph can provide
an approximation to the solution to the problem. A graph can also convey
more general information. For example, if the area of clear space in the
exhibition hall is required to be any particular value, A say, in the image
set [0, 192], then there is a unique corresponding value of x for which this
area occurs, namely, the solution in [0, 6] of the equation f (x) = A.
In Section 5, we use graphs in this way to help solve equations for which
there is no convenient formula.
2.2 Graphs of quadratic functions

A quadratic function is one whose rule is of the form
f (x) = ax2 + bx + c (where a =
, 0).
In the audio which follows, techniques of translation and scaling are
introduced. These techniques can be used to sketch the graph of any
quadratic function, without constructing a table of values, by relating the
required graph to the familiar graph of y = x
2 in a simple geometric way. See Subsection 1.2.
17
To prepare for these new graph-sketching techniques, you are reminded of

See Chapter A2, a technique, ‘completing the square’, which is based on the identity
Subsection 2.3.
x2 + 2px = (x + p)2 − p2 .
(2.3)
Using equation (2.3), any quadratic expression ax2 + bx + c can be
rearranged in completed-square form as a(x + p)2 + q .
Example 2.1 Completing the square
Rearrange −2x2 + 4x − 1 in completed-square form.

Solution
In order to use equation (2.3), we first take out the factor −2:
−2x2 + 4x − 1 = −2(x2 − 2x + 12 ).
Next, we complete the square for the expression in the bracket:
Since the coefficient of x
x
2 − 2x + 1
= (x − 1)2 − 1 + 1
= (x − 1)2 − 12 .
is −2, we take p = −1 in 2 2
equation (2.3). Thus we obtain

−2x2 + 4x − 1 = −2((x − 1)2 − 12 ) = −2(x − 1)2 + 1,
which is in the required form.
The following examples of completed-square form are used on the audio.
Activity 2.4 Completing the square
Rearrange each of the following expressions in completed-square form.

3 1 2
(a) x2 + 4x + 3 (b)
x2 + x + 4
(c) 4x
2 − 56x + 192 (d) 2
x +x
/>E ;9BC6= C> ,-*((&( "1A34:B 'F)#$ +3=5 '$ 2.A3?8B >7 @D35A3C94
7D=4C9>=B!%
Frame 1
The graph of the function f (x) = x 2
x –2 – 1.5 –1 – 0.5 0 0.5 1 1.5 2
x2 4 2.25 1 0.25 0 0.25 1 2.25 4
y y axis of symmetry
4 4
y = x2 y = x2
3 3
parabola
2 2
1 1 vertex
at (0, 0)
–2 –1 0 1 2 x –2 –1 0 1 2 x
18
Frame 2
1
The graph of the function f (x) = x 2 + 2
x –2 – 1.5 –1 – 0.5 0 0.5 1 1.5 2
x2 4 2.25 1 0.25 0 0.25 1 2.25 4
x 2 + 21
y
y = x2 Which curve is
4
1
y = x2 + 2 ,
B
3 A or B ?
A
2
1 Vertex:
2
Axis of
–2 1 1 2 x symmetry:
–2
Frame 3
2
The graph of the function f (x) = (x + 21 )
x –2 – 1.5 –1 – 0.5 0 0.5 1 1.5 2
x2 4 2.25 1 0.25 0 0.25 1 2.25 4
x + 21 – 1.5
2
(x + 21 ) 2.25
y
4 y = x2 Which curve is
y = (x + 1 ) ,
2
A 2
3 A or B ?
B
2
1
4 Vertex:
Axis of
–2 1 1 1 2 x symmetry:
–2 2
19
Frame 4
Translations
2
f (x) = x 2 + q f (x) = (x + p)
y Translate y Translate
up by to left
q by p
0 x –p 0 x
Down if To right if
Vertex: (O, q) q negative Vertex: (– p, O) p negative
Match each function to its graph.

2
(1) f (x) = x 2 + 1 (2) f (x) = (x + 1) (3) f (x) = x 2 – 1 (4) f (x) = (x – 1) 2
y y y y
A B C D
1 1
1 x 1
–1
–1 x 1 x –1 –1 1 x
Frame 5
2
The function f (x) = (x + 2) – 1
2 2
f (x) = x 2 f (x) = (x + 2) f (x) = (x + 2) – 1
y y y
4 4 4
Translate Translate
to left down
by 2 by 1 –2
–2 2 x –2 2 x 2 x
–1
Vertex: (0, 0) Vertex: (– 2, 0) Vertex: ( – 2 , – 1 )
Frame 6
2 1
The function f (x) = (x + 21 ) + 2 2
f (x) = (x + 1 ) + 1
2
f (x) = (x + 1 )
2 2 2
f (x) = x 2 Translate y y
Translate
y
1 by 1 by 1
–1 1 x –1 1 x –1 1 x
Vertex: (0, 0) Vertex: Vertex:
20
Frame 7
The function f (x) = a x 2
x –2 – 1.5 –1 – 0.5 0 0.5 1 1.5 2
x2 4 2.25 1 0.25 0 0.25 1 2.25 4
2x 2
1 2
2x
– x2
y y
2
y = 2x
y
y = 1 x2 d
2
x
d
y = x2 2d d
1
d 2 d
x x y = –x 2
y -scaling y -scaling y -scaling

1
with factor 2 with factor 2 with factor – 1
Frame 8
Graph of the parabola y = a x 2
y
a>1 a = 1 (basic parabola)
0<a<1
y - scaling
by factor a
–1<a<0
a < –1 a= –1
21
Frame 9
The function f (x) = – 2 x 2 + 4 x – 1
f (x) = x 2
y
4
Complete the square
(Example 2.1):
f (x) = – 2 x 2 + 4 x – 1
2
1 x = – 2 (x – 1) + 1
y -scaling
with factor – 2
2
f (x) = – 2 x 2 f (x) = – 2(x – 1)2 f (x) = – 2 (x – 1) + 1
y y y
1 1 1
1 x Translate 1 x Translate 1 x
to right up
by 1 by 1
Vertex: (0, 0) Vertex: (1, 0) Vertex: (1, 1)
Frame 10
1
The function f (x) = 2
x2+ x
f (x) = x 2
y
4 Complete the square
(Activity 2.4 (d)):
f (x) = 1 x 2 + x
2
2
1 x = 1 (x + 1) – 1
2 2
y -scaling
with factor
1 1 2 1 2 1
f (x) = 2 x 2 f (x) = 2 (x + 1) f (x) = 2 (x + 1) – 2
y y y
4 Translate 4 Translate 4
by by
1 x 1 x 1 x
Vertex: Vertex: Vertex:
22
Frame 11
2
Summary : f (x) = a (x + p) + q
a positive a negative
a=1 Vertex:
( – p, q)
a small
( – 1 < a < 1) Axis of
symmetry:
a large
x =–p
(a < – 1, a > 1)
Here are some further remarks about sketching quadratic graphs.

1. Another useful technique to use when sketching the graph of a function
is to check where the graph meets the axes – that is, to find the x- and
y-intercepts. For a quadratic function f (x) = ax
2 + bx + c, By the domain convention,
% the y-intercept is found by putting x = 0 to give f has domain R.
f (0) = c,
% the x-intercepts (if any) are found by solving the equation A quadratic equation may
2 have 0, 1 or 2 solutions.
f (x) = ax + bx + c = 0.
For example, in Frame 9, we sketched the graph of the function
f (x) = −2x2 + 4x − 1. In this case:

% f (0) = −1, so the y-intercept is −1;
% the equation −2x

2 + 4x − 1 = 0 has solutions x = 1 ± 1
2
2, so the
x-intercepts are x 0 0.29 and x 0 1.71.
These x- and y-intercepts are shown in Figure 2.4.
2. In both Frames 9 and 10, we first performed a y-scaling, followed by a

horizontal and then a vertical translation. You may prefer, however, to
reverse the order of the y-scaling and the horizontal translation. This
approach is illustrated in Figure 2.5 for the function
Figure 2.4 Graph of
f (x) = −2x
2 + 4x − 1 = −2(x − 1)2 + 1. (2.4)
y = −2x2 + 4x − 1
Figure 2.5 Horizontal translation before y-scaling
23
The vertical translation by 1 unit upwards then completes the process,

as in Frame 9.
3. Even when the graph of the function f that you wish to sketch requires
different scales on the axes in order to represent it, the translation and
scaling techniques can still be used. Such situations may occur when
large coefficients exist in the expression for f (x), as in the case of the
function
f (x) = 4x
2 − 56x + 192.
Some trial and error may be In such cases, it is necessary to choose scales so that any intercepts can
needed when finding suitable be represented. Sketching the function f above is the subject of the
axis scales. next activity.
Activity 2.5 Sketching a quadratic graph
Use the techniques introduced in the audio to sketch the graph of the
quadratic function
You rearranged this f (x) = 4x2 − 56x + 192,
quadratic expression into including the x- and y-intercepts.
completed-square form in
Activity 2.4(c). A solution is given on page 51.
Comment
The graph of the function introduced in connection with the exhibition
hall problem is part of the graph sketched in this activity; see Figure 2.2.
The techniques of translation and scaling can be used to help sketch the
graphs of functions other than quadratic ones. For example, the graph of
the function
f (x) = 2|x − 1| − 3
can be obtained from the graph of the modulus function f (x) = |x| by
performing:
% a y-scaling with factor 2;
% a horizontal translation by 1 unit to the right;
% a vertical translation by 3 units downwards.
The stages in this process are shown in Figure 2.6. It is convenient to
perform the two translations together, as in Figure 2.6(b).
Here the graph of y = |x| is

shown in colour.
Figure 2.6 Constructing the graph of y = 2|x − 1| − 3
24
The graph of the function f (x) = 2|x − 1| − 3 in Figure 2.6(b) shows the
y-intercept as
f (0) = 2|0 − 1| − 3 = 2 − 3 = −1,
and the x-intercepts as x = − 21 and x =
52 . The x-intercepts are found by

solving the equation

f (x) = 2|x − 1| − 3 = 0;
that is,
|x − 1| =
23 .
3
The equation |x − 1
| = 2
is equivalent to the two equations Another way to write these
3 two equations is
x−1= 2
and

x−1= − 32 .
5
x − 1 = ± 32 .
Thus the solutions are x = 2
and x = − 12 , confirming the x-intercepts
given in the figure.
In addition to horizontal and vertical translations and y-scalings of graphs,
there are also x-scalings which correspond to squashing or stretching a
graph in the x-direction by a given factor. For example, consider the effect
of squashing the graph of the function f (x) = x
2 in the x-direction by the
factor
21 . Let (u, v) be a point on the graph of f ; see Figure 2.7(a). On
what curve does the point ( 21 u, v) lie? Since
v = f (u) = u
2 ,
we have
v = u2 = 4( 21 u)2 .
Thus the point ( 21 u, v) lies on the graph of y = 4x2 .
So if g is the function g(x) = 4x2 , then its graph can be obtained by
squashing the graph of f in the x-direction by the factor
21 ; see
Figure 2.7(b).
The function g can be expressed in terms of f as
g(x) = 4x
2 = (2x)2 = f (2x),
and the graph of g can be obtained from that of f by applying an x-scaling Alternatively, the graph of g
with factor 21 . can be obtained from that of
f by means of a y-scaling
with factor 4.
Figure 2.7 An x-scaling
More generally, the graph of y = f (bx), where b ,= 0, is obtained from the

graph of y = f (x) by an x-scaling with factor 1/b. You will meet some An x-scaling with factor −1
examples of the x-scaling of graphs in Section 3. corresponds to reflection in
the y-axis.
25
The scalings and translations that you have met in this section
(summarised in the table below) can, with two exceptions, be applied in
any order with the same result. The exceptions are that the result of
applying both a horizontal translation and an x-scaling depends in general
on the order in which these are applied, and similarly for both a vertical
translation and a y-scaling.
% the idea of finding information about the solution to a problem by
using the graph of an appropriate function;
% a technique for sketching the graph of a quadratic function using the
completed-square form;
% techniques for translating and scaling a known graph y = f (x).
Graph Translation or scaling of y = f (x)
y = f (x + p) Horizontal translation by p units to the left
(right if p is negative)
y = f (x) + q Vertical translation by q units upwards
(downwards if q is negative)
y = af (x) y-scaling with factor a
y = f (bx) x-scaling with factor 1/b

Exercise 2.1
A square garden with side length 8 metres is to have uniform borders
around three sides, the rest being lawn. The borders are to take half the
area of the garden. Let x denote the width of the border (in metres), and
let A denote the area of the lawn (in square metres).
Solve this problem by determining
(a) the largest closed interval consisting of values of x which make sense
for this problem;
(b) the area A of the lawn, in terms of x;
(c) the value of x for which A is half the area of the garden.
Exercise 2.2
(a) Sketch the graph of the function
f (x) = 2x2 − 24x + 64,
by using translations and scalings of the graph of y = x2 , and by
finding the x- and y-intercepts.
(b) What is the relevance of the graph in part (a) to the ‘garden problem’
in Exercise 2.1?
Exercise 2.3
Sketch the graph of the function
2
f (x) = − + 1,
x+1
by using translations and scalings of the graph of y = 1/x, and by finding
the x- and y-intercepts.
26
3 Trigonometric and exponential functions
In this section, two important classes of functions are introduced, both

based upon ideas that you have seen before. Trigonometric functions
follow from the introduction of the sine, cosine and tangent of an angle,
and exponential functions relate to powers of numbers and to
geometric sequences. Both of these classes of functions have very See Chapter A1, Section 3.
significant applications within mathematical models of various real-life
phenomena. Examples include the oscillating level of a tide (sine and
cosine functions) and radioactive decay (exponential functions).
3.1 Trigonometric functions

We start by recalling the definitions of cosine and sine. If P (x, y) is any
point on the unit circle, as shown in Figure 3.1, and the line segment OP Recall that the unit circle has
is at an angle t measured from the positive x-axis, then centre at the origin O and
radius 1.
cos t = x and sin t = y.
Here the angle t may be of any size (positive, negative or zero); that is, t Here, and later in this
may take any real value. chapter, it is assumed that
t is measured in radians.
Recall that π radians = 180◦.
The symbol θ rather than t
was used in Chapter A2,
Subsection 3.1.
Figure 3.1 Defining cos and sin
Now cos and sin, as defined above, are functions in the sense described in
Section 1. There is no simple formula to calculate the values of these
cosine and sine functions, but the geometric definitions of cos t and sin t
provide the rules for a pair of functions nonetheless. For each input
value t, there are unique output values x = cos t and y = sin t. Since t may
take any real value,
the domain of each of the functions cos and sin is R.
In the next activity you are asked to describe what image values can be
obtained from these two functions.
27
Activity 3.1 Set of images under cos and sin
By referring to the definitions of cos and sin in terms of the unit circle,
write down the image set – that is, the complete set of image values – for
each of these functions.
We seek next to sketch the graphs of the trigonometric functions cos and
sin. A first step might be, as with the functions considered previously, to
draw up a table of values. A calculator could be used to obtain values for a
table directly, but in the absence of a simple formula for the rules of cos
and sin, these cannot be checked independently except in a few special
See also Chapter A2, cases. We look at these special cases first.
Subsection 3.1.
Activity 3.2 Images of cos and sin in special cases
(a) By referring to the unit circle or to an appropriate right-angled

You may like to check that triangle in each case, find the values obtained when each of cos and sin
your calculator gives the same is applied to the inputs 0, 61 π, 41 π, 13 π, 12 π.
values with these inputs.
(b) Use your results from part (a) and the trigonometric formulas
Remember that your

calculator should be in radian cos(
π − t) = − cos t, sin(π − t) = sin t
mode. to find the values obtained when each of cos and sin is applied to the
inputs 32 π, 43 π, 56 π, π.
You met the trigonometric (c) Use your results from parts (a) and (b) and the trigonometric formulas
formulas in parts (b) and (c)
cos(−t) = cos t, sin(−t) = − sin t
in Chapter A2,
Subsection 3.1. If you are to find the values obtained when each of cos and sin is applied to the
confident about finding values inputs − 16 π, − 41 π, − 13 π, − 12 π, − 32 π, − 34 π, − 56 π, −π.

of trigonometric functions,
then you may prefer to use an
alternative method in
parts (b) and (c).
The table below is based on the values obtained in Activity 3.2.
t −π − 56 π − 34 π − 23 π − 21 π − 31 π − 14 π − 16 π 0
The values of cos t and sin t in cos t −1 −0.87 −0.71 −0.5 0 0.5 0.71 0.87 1
the table are given correct to sin t 0 −0.5 −0.71 −0.87 −1 −0.87 −0.71 −0.5 0
two decimal places.
1 1 1 1 2 3 5
t 0 6π 4π 3π 2π 3π 4π 6π π
cos t 1 0.87 0.71 0.5 0 −0.5 −0.71 −0.87 −1
sin t 0 0.5 0.71 0.87 1 0.87 0.71 0.5 0
This table covers a selection of values for t within the interval [−π, π].
Outside this range, we can apply the formulas
See Chapter A2, cos(t + 2π) = cos t, sin(t + 2π) = sin t,
Subsection 3.1.
which describe the fact that each of the graphs of the functions cos and sin
repeats itself after every

2π radians; see Figure 3.2 below. We say that a
function f is periodic, with period p, if f (t + p) = f (t) for all t in the
domain of f . Thus the cosine and sine functions are both periodic, with
period 2π
.
28
SECTION 3 TRIGONOMETRIC AND EXPONENTIAL FUNCTIONS
Figure 3.2 cos and sin are periodic
We have put the graphs of cos and sin on separate sets of axes, since the
dependent variables are different: x for the cosine function and y for the
sine function. However, if we concentrate on cos and sin as functions, and
leave in the background their definitions in terms of the unit circle, then
both can be expressed in the form y = f (x), with x as the independent Remember that a function is
variable and y as the dependent variable. With this choice of variables, independent of the variables
both graphs can be placed on the same set of axes, as shown in Figure 3.3 which are chosen to label its
below. inputs and outputs and to
display its graph.
Function notation would lead

us to write cos(x) and sin(x),
but where there is no
ambiguity we often omit the
brackets.
Figure 3.3 Graphs of cos and sin on the same axes
As with the quadratic functions in Section 2, it is possible to translate or

scale these graphs of cos and sin in either the x- or the y-direction; see
Figure 3.2. For example, the graph of the function f (x) = cos(x + p) is
obtained by translating the graph of cos x by p units to the left. This and Similar effects can be
other transformations of the graph of cos are illustrated in Figure 3.4 demonstrated starting from
overleaf. the graph of the sine function.
29
Figure 3.4 Transforming the graph of cos
Note that the graph of y = cos(x − 21 π) in Figure 3.4(a) is the same as the
graph of y = sin x. This reflects the fact that, for all values of x, we have
cos(x − 12 π) = sin x. Similarly, the graphs in Figures 3.4(b) and 3.4(e) are
identical, since cos(x + π) = − cos x for all x.
Activity 3.3 Transforming the graph of cos
Graphs that can be obtained By considering the effect of an appropriate translation or scaling on the
from that of the sine function graph of cos, sketch the graph of each of the following functions.
(or cosine function) by 1
(a) y = −3 + cos x (b) y = cos x (c) y = cos(3x) (d) y = cos(−x)
translation and/or scaling are 2
often called ‘sinusoidal’. Solutions are given on page 53.
30
This subsection concludes by considering briefly the tangent function.

This is defined by the rule See Chapter A2,
sin x Subsection 3.1.
tan x = ,
cos x
which produces an output for all real numbers x except when cos x = 0.
Figure 3.3 shows that cos x = 0 occurs for each of the values
x = ± 12 π, ± 23 π, ± 52 π, . . . . Hence the domain of tan is all of R excluding
these values, so it is composed of the open intervals
. . . , (− 25 π, − 32 π), (− 32 π, − 12 π), (− 21 π, 12 π), ( 12 π, 32 π), ( 32 π, 52 π), . . . .
Let us concentrate first on the central interval, (− 21 π, 12 π).
Activity 3.4 Images of tan in special cases
(a) By using the definition of tan and your answers to Activity 3.2(a), find You may like to check that
the values obtained when tan is applied to the inputs 0, 61 π, 14 π, 13 π. your calculator gives the same
values with these inputs.
(b) Use your results from part (a) and the trigonometric formula
tan(−x) = − tan x See Chapter A2,
to find the values obtained when tan is applied to the inputs
Subsection 3.1.
− 16 π, − 41 π, − 13 π.
We could now draw up a table of values for the function tan within the
central interval (− 21 π, 12 π). However, to cut a long story short, the graph of
tan is as shown in Figure 3.5.
Figure 3.5 Graph of tan
The behaviour of tan x within the central interval as x approaches 12 π

(from the left) can be deduced from the rule tan x = sin x/cos x, by noting
that sin x is close to sin( 21 π) = 1 whereas cos x approaches zero through

positive values. Similarly, as x approaches − 21 π from the right, sin x is
close to sin(− 21 π) = −1 while cos x again approaches zero through positive
values. So as x approaches
12 π from the left, tan x becomes large through It can be seen from the graph
positive values. Similarly, as x approaches − 21 π
from the right, tan x that the image set of tan is R.
becomes large through negative values.
Outside the central interval, the graph is determined by the fact that the This fact holds because
function tan is periodic with period π; that is, tan(x + π) = tan x. sin(x + π) − sin x
Translating the graph of tan horizontally by any integer multiple of π = ;
cos(x + π) − cos x
(either to left or right) gives the same graph once more. For this graph,
see Chapter A2,
each of the vertical lines x = ± 12 π, ± 32 π, ± 52 π, . . . is an asymptote. Exercise 3.2(a).
31
3.2 Exponential functions

If a is a positive real number and n is a positive integer, then an represents
a multiplied by itself n times. A meaning can also be assigned in a natural
See Chapter A0, way to ax , where x is any real number. When x is not a positive integer,
Subsection 3.1. this is achieved as follows.
Here x = 0. (i) a0 = 1.
Here x = −n. (ii) If n is a positive integer, then a−n = 1/an .
√ p
Here x = p/q. (iii) If p, q are integers, with q > 0, then ap/q = ( q a ) .
(iv) If x is an irrational number, then the value of ax can be found to any

desired accuracy by approximating x more and more closely by
rational numbers p/q, for which ap/q is defined by (iii) above.
With these definitions, we have the following rules for powers, where
x and y are any real numbers:
y
ax+y = ax ay and (ax ) = axy . (3.1)
For each positive real number a, the process just described for finding
values of ax for any real number x is a rule for the function
There is one such function for f (x) = ax ,
each positive real number a.
with domain R. Each such function is called an exponential function,
and the number a is called the base of the function.
We can draw up tables of values for such functions. Below is a table for
Note that ( 12 )x can also be the exponential functions with base 2 and base 21 , for values of x within
written as 2−x or as 1/2x . the interval [−2, 2].
x −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
x
The values of 2 and ( 12 )x
in 2x
0.25 0.35 0.5 0.71 1 1.41 2 2.83 4
the table are given correct to ( 12 )x 4 2.83 2 1.41 1 0.71 0.5 0.35 0.25
two decimal places.
Using these values, we sketch the graphs of the two functions. Because
( 12 )x = 2−x , the graph of y = ( 12 )x is the reflection of the graph of y = 2x in

the y-axis.
Figure 3.6 Graphs of two exponential functions
32
Activity 3.5 Sketching graphs of exponential functions
(a) Use your calculator, where necessary, to find image values of the
function f (x) = 5x for x = −1, −0.75, −0.5, . . . , 1.
(b) Use your results from part (a) to sketch the graph of the function
f (x) = 5x for x in [−1, 1].
(c) Explain how, without performing any further calculations, you could
use the result from part (b) to sketch the graph of the function
g(x) = ( 15 )x .
These graphs may bring to mind the graphs which arise from geometric
sequences. The closed forms of geometric sequences with initial term 1 can See Chapter A1, Section 3.
be written as yn = an (n = 0, 1, 2, . . .).
You may recall from the study of geometric sequences that:
(i) if a > 1, then an → ∞ as n → ∞; The graphs of geometric
(ii) if 0 < a < 1, then an → 0 as n → ∞. sequences in Chapter A1
consisted of isolated points,
Similar properties hold for the exponential functions: whereas the graphs of
(i) if a > 1, then ax → ∞ as x → ∞; exponential functions are
(ii) if 0 < a < 1, then ax → 0 as x → ∞. smooth curves.
These forms of behaviour for large positive values of x are evident in the
graphs of the two exponential functions sketched in Figure 3.6. In general,
the graph of f (x) = ax , where a > 0, has one of the three forms shown in
Figure 3.7 below.
The image set of any

exponential function
f (x) = ax , for a > 1 or
0 < a < 1, is the open
interval (0, ∞).
Also, the x-axis is an
asymptote of the graph of f .
Figure 3.7 Graphs of exponential functions f (x) = ax : three cases
While an exponential function of the form f (x) = ax is defined for each

positive number a, one value of a is worthy of particular mention. This is
the number e = 2.
718 281
. . . . Like π, the number e is an irrational number
which can be determined to as many decimal places as desired. There are
various equivalent ways of defining e. One of these is that e is the unique Why this feature is singled
number such that the graph of y = ex has a tangent line at (0, 1) whose out to define e will become
slope is 1; see Figure 3.8, overleaf. more apparent when you
study the calculus in Block C.
In that block the notion of a
tangent line to a curve is
discussed.
33
Figure 3.8 Graph of the function f (x) = ex
The function f (x) = ex is often referred to as the exponential function,

Your calculator should have and is denoted also by exp. Thus exp x = ex and exp(−x) = e−x .
an ex or exp button, which
calculates values of this
exponential function. In
particular, calculating
e1 = exp 1 will give you an
approximate value for e.
% the trigonometric functions cos, sin and tan, together with their
graphs;
% periodic functions;
% the exponential functions f (x) = ax (where a > 0), together with their
graphs;
% the exponential function f (x) = ex (or exp x), where e = 2.718 281 . . . .

Exercise 3.1
By considering the effect of an appropriate translation or scaling on the
graph of sin, sketch each of the following graphs.
(a) y = 1 + sin x (b) y = − sin x (c) y = sin(4x)
Exercise 3.2
By considering the effect of an appropriate translation on the graph of the
function f (x) = ex (see Figure 3.8), sketch the graph of the function
g(x) = ex+1 .
34
4 Inverse functions
In Section 1, functions were introduced as a way to study a particular type

of relationship between a dependent variable and an independent variable.
Often, however, we wish to reverse the roles of dependent and independent
variables. For example, the formula C = 2πr for the circumference C of a
circle of radius r can be rearranged as
1
r =
C,
2
π
to express the radius in terms of the circumference. This idea leads to the
concept of an inverse function.
4.1 What is an inverse function?

Consider the two ‘squaring functions’
g(x) = x2 (x in [0, ∞)) and h(x) = x
2 . (4.1)
These functions have the same rule, but the domain of g is [0, ∞), whereas
the domain of h is R, by the domain convention; see Figure 4.1.
Figure 4.1 Two squaring functions
The functions g and h have the same image set, namely [0, ∞). Recall that the image set of a
function is the set of all
If we now consider trying to reverse the effects of g and h, then we find a possible image values of the
significant difference between these two functions: function.
% for each value of y in the image set [0, ∞), there is exactly one value of
x in [0, ∞) such that g(x) = y;
% for each value of y in the image set [0, ∞), apart from 0, there are two
values of x in R such that h(x) = y.
This difference is illustrated in Figure 4.1, with the value y = 4.
A consequence of this difference is that it is not possible to find a function
which reverses the effect of h. As you will see, however, the effect of g can Recall that a function has to
be reversed by a function. give a unique output for each
input.
35
To describe this difference, some new ideas are introduced. A function f is

said to be one-one, or one-to-one, if it has the following property:
for all x1 , x2 in the domain of f, if x1 =
, x2 , then f (x1 ) =
, f (x2 ).
The function h in
A function which is not one-one is said to be many-one, or many-to-one.
Figure 4.1(b) is many-one;
for example, h(−2) = h(2).
In terms of the graph of a function, being one-one means that each

horizontal line which meets the graph does so exactly once. The function g
defined in equations (4.1) has this property since, as can be seen in
Figure 4.1(a), this function is increasing. As you take greater values of x
(further to the right on the x-axis), the corresponding values of g(x) are
greater (further up the y-axis). Written in symbols, a function f is said to
In some texts, functions with

be increasing if it has the following property:
these properties are called
‘strictly increasing’ and
for all x1 , x2 in the domain of f, if x1 < x2 , then f (x1 ) < f (x2 ).
‘strictly decreasing’. Note Similarly, a function f is decreasing if it has the following property:
also that these definitions

provide formal meanings for for all x1 , x2 in the domain of f, if x1 < x2 , then f (x1 ) > f (x2 ).
everyday words. If a function is either increasing or decreasing, then it is certainly one-one.
For example, in Figure 4.2, the function on the left is increasing (so it is
one-one), the function in the middle is decreasing (so it is one-one), but the
function on the right is neither increasing nor decreasing (it is many-one).
Figure 4.2 Types of behaviour
Sometimes, it can be difficult to check whether a given function is

increasing or decreasing, but in this chapter we consider only functions
where this can be deduced immediately from the graph of the function.
Activity 4.1 Increasing functions and decreasing functions
For each of the following functions, state whether the function is:
increasing, decreasing, neither increasing nor decreasing, one-one,
many-one.
(a) f (x) = cos x (see Figure 3.3)
(b) g(x) = sin x (− 12 π ≤ x ≤ 12 π) (see Figure 3.3)
(c) h(x) = ( 12 )x (see Figure 3.6)
36
SECTION 4 INVERSE FUNCTIONS
For any one-one function f , it is possible to define an inverse function,

or inverse for short, denoted by f −1 . As shown in Figure 4.3, the domain The notation f −1 should not
of the inverse function f −1 is the image set of f , the set consisting of all be confused with 1/f , which
possible image values f (x), and the rule of the inverse function is is the function with rule
x )−→ 1/f (x). Shortly, you
f −1 (y) = x, where y = f (x). will see that particular inverse
This inverse function f −1 reverses, or undoes, the effect of f . functions have special names.
Also, f undoes the the effect of f −1 , which is itself one-one.

For example, the function h(x) = ( 21 )x is one-one (see Activity 4.1(c)), so it
has an inverse function h−1 , which undoes the effect of h. In particular:
h(1) = 21 , so h−1 ( 12 ) = 1;
h(2) = 14 , so h−1 ( 14 ) = 2.
For convenience, the domain

of f is taken to be a closed
interval in this figure.
Figure 4.3 A function and its inverse
In Figure 4.3(b) you can also see that the image set of the inverse
function f −1 is equal to the domain of f . Furthermore, as shown in
Figure 4.4, if f −1 is written as a function of x, in the usual way, then the
graph of y = f −1 (x) is obtained from the graph of y = f (x) by exchanging
the roles of x and y. This exchange of roles corresponds to reflecting the
graph of y = f (x) in the 45◦ line, as indicated by the double-headed arrow, In the case of equal scales on
and using the old x-scale on the new y-axis and vice versa. the axes, the 45◦ line is the
line y = x, as in Figure 4.5.
Figure 4.4 Obtaining the graph of y = f −1 (x) by reflection
For example, the function g(x) = x2 (x in [0, ∞)), considered earlier, is

one-one, and its inverse function g −1 is given by
√
g −1 (y) = y (y in [0, ∞)).
37
√
Exchanging the roles of x and y, the graph of y = g −1 (x) = x is the
reflection of the graph of g in the 45◦ line; see Figure 4.5, in which the
x-scale and the y-scale are the same.
√
Figure 4.5 Obtaining the graph of y = x
The example below shows you how to find an inverse function in a
particular case.
Example 4.1 Finding an inverse function
Show that the function

f (x) = 2x + 1 (x in [−1, 1])
has an inverse function f −1 . Find the rule of f −1 , and sketch its graph.
Solution
First we sketch the graph of f . This is the graph of y = 2x + 1, restricted
to the closed interval [−1, 1].
Figure 4.6 Graph of y = f (x)

From the graph we see that:
% the function f is increasing, and so one-one;
If the domain of f had been % the image set of f is [−1, 3].
the open interval (−1, 1),
Therefore f has an inverse function f −1 with domain [−1, 3] and image set
then the image set would
[−1, 1]. We can find the rule of f −1 by solving the equation
have been (−1, 3).
y = f (x) = 2x + 1
to obtain x in terms of y:
y = 2x + 1, so x = 21 (y − 1).
38
Thus the inverse function is
f −1 (y ) = 21 (y − 1) (y in [−1, 3]),
which, expressed in terms of x, is
f −1 (x) = 21 (x − 1) (x in [−1, 3]).
The graph of y = f −1 (x) is obtained by reflecting the graph of y = f (x) in

the 45◦ line; see Figure 4.7.
Figure 4.7
Graph of y = f −1 (x)
In the following activity, you have the opportunity to practise the above
method of finding inverse functions. In part (c) of this activity, the graph
of f requires different scales on the axes. So, when reflecting the graph
of f in the 45◦ line to obtain the graph of f −1 , remember to exchange the
scales as well.
Activity 4.2 Finding inverse functions
For each of the following functions f , find the inverse function f −1 and
sketch the graph of y = f −1 (x).
(a) f (x) = 3x − 1 (x in (0, 2))
(b) f (x) = x3 (x in [0, ∞))

2
(c) f (x) = 4x − 56x + 192 (x in [0, 6])
(Hint : In this part, make use of the solution to Activity 2.5. Also,
when solving the equation y = f (x) to obtain x in terms of y, consider
carefully which solution to choose.)
Comment
Note that we can use the inverse function found in part (c) to solve the See Subsection 2.1.
exhibition hall problem. Since the function
f (x) = 4x
2 − 56x + 192 (x in [0, 6])
has inverse function
'
f −1 (y) = 7 − 1 + 14 y (y in [0, 192]),
the solution of the equation f (x) = 96 which lies in [0, 6] is
' √
x = f −1 (96) = 7 − 1 + 14 × 96 = 7 − 25 = 2,
as expected.
39
4.2 Inverse trigonometric functions
When working with trigonometry, we often wish to find an angle θ for

See Chapter A2, Section 3. which we know the value of sin θ, cos θ or tan θ. It is always necessary to
be careful when finding such a θ, since equations of the form
sin θ = k, cos θ = k and tan θ = k (4.2)
do not have unique solutions. For example, the equation sin θ = 0 has
infinitely many solutions: θ = 0, ±π, ±2π, . . . . In particular, the sine
function is not one-one (and nor are the cosine and tangent functions).
In order to define inverse functions which give unique solutions to
equations (4.2), we restrict the domains of the sine, cosine and tangent
functions to obtain one-one functions. For the sine function, we restrict the
domain as follows:
f (x) = sin x
(− 21 π ≤ x ≤ 12 π).
Figure 4.8
Graph of f (x) = sin x (− 12 π ≤ x ≤ 12 π) and its inverse f −1
Figure 4.8(a) shows that:
% the image set of f is [−1, 1].
Therefore f has an inverse function f −1 with domain [−1, 1] and image set
[
− 12 π, 12 π]. The graph of f −1 , obtained by reflecting the graph of f in the
45◦ line, is shown in Figure 4.8(b).
The name ‘arcsine’ arises This inverse function is given the name arcsine. Thus, for −1 ≤ y ≤ 1,
from an old meaning of ‘arc’
as ‘angle’.
x = arcsin y means that y = sin x and − 21 π ≤ x ≤ 12 π.
In words, for −1 ≤ y ≤ 1,
arcsin y is that angle in [− 21 π, 12 π] whose sine is y.
We can calculate some particular values of the arcsine function, by using
our knowledge of the sine function. For example, the value of arcsin 0 is
that angle in [
− 12 π, 12 π] whose sine is 0. We know that this angle is 0, so
arcsin 0 = 0.
Similarly, since
sin( 16 π) = 1
2
and sin(− 21 π) = −1,
we have
arcsin( 12 ) = 16 π and arcsin(−1) = − 12 π.
40
In a similar way, we can restrict the domains of the cosine and tangent
functions to define the inverse functions arccosine and arctangent. The Alternative names for the
details are given in Figures 4.9 and 4.10. inverse trigonometric
functions are
sin−1 , cos−1 , tan−1 .
Calculators and computers
may also use the names
inv sin, inv cos, inv tan
or
asin, acos, atan.
Figure 4.9 Graph of f (x) = cos x (0 ≤ x ≤ π) and its inverse f −1
Notice that the graph of

arctan has two horizontal
asymptotes:
y = ± 12 π.
Figure 4.10 Graph of f (x) = tan x (− 12 π < x < 12 π) and its inverse f −1
Activity 4.3 Evaluating inverse trigonometric functions
(a) Which of the following are valid expressions?
(i) arcsin(
2π) (ii) arccos 0
(iii) arctan( 21 π)
(iv) arctan(tan( 45 π))
(b) Write down exact values for each of the following expressions, giving You should not need to use
angles in radians. your calculator in this
√ activity.
(i) arcsin( 21 3 ) (ii) arccos(−0.5) (iii) arctan 1
(iv) arcsin(sin( 51 π)) (v) cos(arccos(0.9))
(vi) arctan(tan( 45 π))
41
Finding a value of θ
If you know that sin

θ = 0.61, for example, then the function arcsin
gives a value of θ in the interval [
− 12 π, 12 π
] whose sine is 0.61, namely
0.656 (to 3 d.p.). But you may also know that
θ lies in the interval [
12 π, π].
See Chapter A2, By using the property sin x = sin(π − x), it can be deduced that the
Subsection 3.1. required value of θ is
π − 0.656 0 2.486.
Similar remarks apply to finding values of angles corresponding to a given
cosine or tangent value.
4.3 Logarithms
In Subsection 3.2, we saw that the graph of the function f (x) = ax , where
a > 0, a ,= 1, takes one of two forms, as shown in Figure 4.11.
Figure 4.11 Graphs of y = ax
The function g(x) = 1x = 1 is In Figure 4.11(a) the function f is increasing, whereas in Figure 4.11(b) it
a constant function, which is is decreasing. In both cases, therefore, f is one-one. Also in both cases, the
neither increasing nor domain of f is R and the image set is (0, ∞), the set of positive real
decreasing. numbers. Therefore, for a > 0, a ,= 1, the function f (x) = ax has an
inverse function with domain (0, ∞) and image set R. This inverse function
We read loga y as ‘log to the is called the logarithm to the base a, denoted by loga . Thus, for y > 0,
base a of y’.
x = loga y means that y = ax ;
Note that x = loga (ax ), for x in words,
in R, and y = aloga y , for
y > 0.
the logarithm to the base a of y is that power of a which equals y.
For example, the value of log2 8 is that power of 2 which equals 8. Since
8 = 23 , we have
log2 8 = 3.
Similarly, since
4 = 22 ,
2 = 21 ,
1 = 20 and

1
2
= 2−1 , (4.3)
we have
log2 4 = 2,
log2 2 = 1,
log2 1 = 0 and log2 ( 12 ) = −1. (4.4)
The graph of y = loga x can be obtained by reflecting the graph of y = ax
in the 45◦ line. Figure 4.12 shows the graphs of y = 2x and y = log2 x, with
the values in equations (4.3) and (4.4) plotted.
42
The graph of y = log2 x has

the y-axis as an asymptote.
Figure 4.12 Graphs of y = 2x and y = log2 x
Activity 4.4 Evaluating logarithms
(a) Find exact values for each of the following expressions.

(i) log10 (10 000) (ii) log3 ( 91 )
(b) Express the meaning of log2 10 in words, and hence give the exact
value of 2log2 10 .
Logarithms have a number of useful properties arising from their definition

as inverse functions of exponential functions. These are listed below.
Properties of logarithms (a > 0, a ,= 1)

(a) loga 1 = 0 and loga a = 1.
(b) For x > 0 and y > 0, A special case of
(i) loga (xy) = loga x + loga y, property (b)(ii) is
(ii) loga (x/y) = loga x − loga y. loga (1/y) = − loga y.
(c) For x > 0 and p in R,
loga (xp ) = p loga x.
For example, property (b)(i) can be checked as follows.

If we write p = loga x and q = loga y, then x = ap and y = aq . Thus
xy = ap × aq = ap+q , See Subsection 3.2.
and hence
loga (xy) = p + q = loga x + loga y,
as required.
These properties allow us to rearrange many expressions involving
logarithms; for example,
$ &1/3 $ &
xy 1 xy
loga = loga (property (c), with p = 13 )
z 3
z
= 13 (loga (xy) − loga z) (property (b)(ii))
= 13 (loga x + loga y − loga z) (property (b)(i)).
43
Activity 4.5 Using the properties of logarithms
Verify each of the following equations.

(a) loga 6 + loga 8 − loga 2 − loga 24 = 0
$ 4 3x &
x
4
(b) log2 = 4 log2 x + 6x − x2
2x2
Base 2 is also used, for The most commonly used bases for logarithms are 10 and
example in computer science. e = 2.718 281 . . . . Logarithms to the base 10 are called common
logarithms since they were used for many years (from the early 17th
Century until calculators were invented) to facilitate multiplication and
division. Briefly, the product xy can be evaluated by first finding log10 x
and log10 y from logarithm tables, and then finding
10log10 x+log10 y = xy,
using tables of powers of 10 (called antilogarithms). In this way the
problem of multiplying x and y is replaced by the simpler problem of
adding log10 x and log10 y. Often, log10 is called simply log.
The number e = 2.718 281 . . . is called the base of natural logarithms,
The symbol ‘ln’ is read as and the function loge is often called ln. As pointed out in Subsection 3.2,
‘ell en’. Both the functions the tangent line to the graph of y = ex at the point (0, 1) has slope 1, and
log and ln appear on a this implies that the tangent line to the graph of y = ln x at the point
scientific calculator. (1, 0) also has slope 1.
We can use logarithms to solve equations of the form
ax = k,
where k > 0 and a > 0, a ,= 1. The solution is x = loga k, but calculators
and computers evaluate logarithms only to base 10 and base e (at the time
of writing). To solve an equation such as
2x = 1000, (4.5)
Alternatively, apply log10 to we first apply the function ln to both sides:
both sides.
ln(2x ) = ln(1000).
By property (c), we have
ln(1000)
x ln 2 = ln(1000) and hence x = 0 9.966.
ln 2
In particular, we can use logarithms when working with geometric
See Chapter A1. sequences and linear recurrence sequences, to discover how far along the
sequence we need to go for the terms to reach a given value. For example,
suppose that a savings account contains (in
£)
sn = 1000 × (1.
05)n−1 (n = 1, 2, 3, . . .)
See Chapter A1, Activity 3.3. on 1 January of the
nth year, arising from an initial deposit of £1000 and
annual interest of 5%. If we leave the money to accumulate at this rate, at
the start of which year will the account first contain more than
£2000 ?
To answer this question, we begin by solving
1000 × (1.05)n−1 = 2000; that is, (1.

05)n−1 = 2.
44
Applying ln to both sides, as we did with equation (4.5), we obtain

ln 2
n−1= 0 14.2, so n 0 15.2.
ln(1.05)
Therefore, at the start of the 16th year, the account will (for the first time)
contain more than
£2000.
Activity 4.6 Trebling the deer population
Suppose that a deer population has size Pn at the start of year n, where Recall that Pn is a model of a
deer population; see
Pn = 2666.6̇ × (1.15)n−1 + 3333.3˙ (n = 1, 2, 3, . . .). Chapter A1, Exercise 4.3(a).
(a) Find the deer population at the start of year 1.
(b) At the start of which year will the population reach more than three
times its size at the start of year 1?
% the ideas of one-one, many-one, increasing and decreasing functions;
%
the fact that a one-one function f has an inverse function f −1 which
undoes the effect of f and whose domain is the image set of f ;
% the inverse trigonometric functions arcsine, arccosine and arctangent;
% the logarithm functions loga , where a > 0, a ,= 1, and their properties;
% the use of logarithms to solve equations of the form ax = k.

Exercise 4.1
For each of the following functions f , find the inverse function f −1 and
sketch the graph of y
= f −1 (x).
(a) f (x) = 4x + 3 (b) f (x) = 2x
2 − 24x + 64 (x ≥ 6) For part (b), see the solution
to Exercise 2.2(a).
Exercise 4.2
(a) Use your calculator to evaluate each of the following expressions,
giving your answers in radians (correct to six significant figures).
(i) arcsin(0.1) (ii) arccos(−0.85) (iii) arctan(0.1)
(b) Write down exact values (in radians) for each of the following
expressions.
√ √
(i) arcsin(− 21 2 ) (ii) arccos 1 (iii) arctan(− 3 )
45
Exercise 4.3
(a) Evaluate each of the following expressions, using your calculator where
appropriate, giving your answers correct to six significant figures.
(i) log2 64 (ii) log10 (0.001) (iii) ln 1 (iv) ln 10
x
(b) Solve the equation 10 = 2 for x, by applying ln to both sides. Hence
evaluate log2 10, correct to six significant figures.
(c) Verify the equation

$ &
ex+1
ln = x + 1 − 2 ln x − ln(x + 1).
x3 + x2
Exercise 4.4
The amount mn (in

£) owing at the start of year n of a mortgage is given
See Chapter A1, by
Exercise 4.3(b).
mn = −6048.6 × (1.
05)n−1 + 16 048.6 (n = 1, 2, . . . , 20).
At the start of which year will the amount owing be less than half the
amount owing at the start of year 1 ?
46
5 Functions, graphs and equations on the
computer
In Section 2, you met the exhibition hall problem, which was solved using
the quadratic equation formula. Many problems, however, lead to more
complicated equations. For example, a problem about volumes might lead
to a cubic expression, of the form
ax3 + bx2 + cx + d (where a =
, 0).
More generally, a polynomial of degree n is an expression of the form
When a polynomial has
an xn + an−1 xn−1 + · · · + a1 x + a0 (where an ,= 0).
general degree, as here, we
For example, the linear expression 2x + 1 is a polynomial of degree 1, the
often use sequence notation
quadratic expression 3x2 − 4 is a polynomial of degree 2, and the cubic for the coefficients.
expression x3 − x2 + x − 1 is a polynomial of degree 3.
A function whose rule involves a polynomial of degree n is called a
polynomial function of degree n, and an equation of the form
an xn + an−1 xn−1 + · · · + a1 x + a0 = 0 (where an ,= 0)
is called a polynomial equation of degree n. Polynomial equations of
degree 3 are called cubic equations.
There is no simple general rule for finding the solutions, or roots, of a
polynomial equation of degree n greater than 2. Instead, such equations
can be solved approximately by various methods; for example, we can find
approximate solutions by using the computer to plot the graph of the
corresponding polynomial function.
In this section the computer is used to plot graphs of functions, and to find
approximate solutions of equations using these graphs. You will also meet
two non-graphical ways of solving equations.
0676A C> ,><?DC6A +>>: * 7>A C86 E>A: 9= C89B B64C9>=%
In this section you saw several methods (including graphical and algebraic
methods) for finding solutions of equations, using the computer. You also
saw how the computer’s graphing capability can be used to find the
greatest or least values taken by a function in an interval.
47
Summary of Chapter A3
In this chapter you met (real) functions, introduced as processors for

turning inputs x into outputs y. The graphs of such functions provide a
geometric interpretation of them, and when functions are used to solve
problems, the shapes of their graphs may provide useful information.
In particular, you saw how to sketch the graphs of quadratic functions,
trigonometric functions, exponential functions and various related inverse
functions.
Learning outcomes
You have been working towards the following learning outcomes.
Terms to know and use

Function, domain, rule, image, image set, closed interval,
open interval, domain convention, real function, graph of a function,
asymptote, modulus, absolute value, quadratic function,
completed-square form, horizontal translation, vertical translation,
x-scaling, y-scaling, periodic function, sine, cosine and tangent
functions, exponential function, one-one, many-one, increasing
and decreasing functions, inverse function, arcsine, arccosine
and arctangent functions, logarithm, polynomial, cubic equation.
Symbols and notation to know and use

f (x) = x2 (x ≥ 0) or f : x )−→ x2 (x ≥ 0) for a function;
[a, b] and (a, b) for closed and open intervals;
|x| for the modulus of x;
f −1 for the inverse function of f ;
arcsin, arccos, arctan;
exp, loga , log, ln.
Mathematical skills
% Use function notation and sketch graphs of functions in simple cases.
% Use interval notation and the modulus function.
% Introduce appropriate functions when solving equations.
% Find inverse functions of given one-one functions and sketch their
graphs.
% Manipulate logarithms.
Modelling skills
% Introduce algebraic symbols to represent unknown or general
quantities.
% Be able to interpret problems concerning areas and volumes in terms
of equations, functions and their graphs.
48
SUMMARY OF CHAPTER A3
Mathcad skills
% Create a function and plot its graph.
% Find numerical solutions of equations by using the graph of an
appropriate function or by using a solve block.
% Solve quadratic equations symbolically.
Ideas to be aware of
% Equations may be solved by various techniques, which may be more or
less suitable in different cases.
Summary of Block A
This block has introduced several key topics:
% linear recurrence sequences;
% lines, circles and trigonometry;
% functions.
You saw in Chapter A1 that linear recurrence sequences can be used to
model various real-life phenomena, such as the height of a bouncing ball
and deer populations. These phenomena each involve a list of separate, or
discrete, values, so it is appropriate to model them with a sequence which
has a subscript, n say, taking integer values. Such models are called

discrete models, and the subscript n is called a discrete variable.
In Chapters A2 and A3, on the other hand, you saw several problems, such
as the location of a train or the area of free space in an exhibition hall,
which were modelled using formulas or functions involving a variable, x say,
taking any value in an interval of the real line. Such models are called
continuous models, and the variable x is called a continuous variable.
When constructing a model, it is usually clear whether a discrete model or
a continuous model is appropriate. For example, a discrete model seems
suitable for monthly car sales figures, which take separate values, whereas
a continuous model seems suitable for the temperature of a cooling cup of
tea, which takes a value at every instant of time during a given period.
In later blocks of the course, both types of model will be developed further
in various ways.
Later in the course, you will meet the word ‘continuous’ used in a different
but related way. A continuous function is, roughly speaking, one whose This informal definition is
graph can be drawn without lifting the pen from the paper. In this course, made precise, and more
most of the functions considered are continuous. For example, all general, in courses on pure
polynomial functions are continuous, and the modulus function, discussed mathematics.
in Section 1, is continuous (but not smooth, since it has a corner at the
origin).
49
Solutions to Activities
Solution 1.1 (b) x −2 −1.5 −1 −0.5 0.5 1 1.5 2

(a) The domain of f is the set of real numbers t 1/x2 0.25 0.44 1 4 4 1 0.44 0.25
satisfying −1 < t < 2.
(b) (i) f (x) = x2 (x ≥ 0) and f (1) = 12 = 1.
(ii) f (x) = 2x + 1 (−1 ≤ x ≤ 1) and

f (1) = 2 × 1 + 1 = 3.
Solution 1.2
(a) (0, 1) is open.
(b) [−3, 2] is closed.
(c) (−2, 2] is half-open.
(d) [0, ∞) is closed.
Solution 1.3
Figure S.2
(a) Since the square root applies only to a√
non-negative number, the rule f (x) = x − 1 is The image set is the open interval (0, ∞).
applicable only to those x for which x − 1 ≥ 0;
that is, x ≥ 1. Thus√the domain of the function Solution 1.5
f with rule f (x) = x − 1 is [1, ∞). (a) Since "
(b) The expression 1/(x − 2) gives a real number for x3 , if x ≥ 0,
|x|3 =
all x in R except x = 2, and the expression (−x)3 , if x < 0,
1/(x + 3) gives a real number for all x in R we can sketch the graph of the function
except x = −3. Thus the domain of the function f (x) = |x|
3 (−1 ≤ x ≤ 1) by using the right half
with rule f (x) = 1/(x − 2) + 1/(x + 3) is R of Figure S.1 and its reflection in the y-axis.
excluding 2 and −3. (This consists of the three

open intervals (−∞, −3), (−3, 2) and (2, ∞).)
Solution 1.4
The ranges of x-values used in this solution are
sufficient to indicate the overall shapes of the graphs.
(a) x −1 −0.75 −0.5 −0.25 0 0.25 0.5 0.75 1
x3 −1 −0.42 −0.13 −0.02 0 0.02 0.13 0.42 1
Figure S.3
(b) Since "

1 1/x, if x > 0,
=
|x| 1/(−x), if x < 0,
we can sketch the graph of f (x) = 1/|x| by using
the right half of Figure 1.4 and its reflection in
the y-axis.
Figure S.1
The image set is the closed interval [−1, 1].

Figure S.4
50
SOLUTIONS TO ACTIVITIES
Solution 2.1 (b) Of these solutions, only x 0 0.917 lies in the

interval [0, 6], so this gives the solution to the
(a) The width of the border can be any number
modified exhibition hall problem. The width of
between 0 and 12 × 12 = 6 metres. So the largest
the border is 0.917 metres.
closed interval in which x can lie is [0, 6].
(As predicted using the graph, the solution is
(The values x = 0 and x = 6 are, in a sense,
approximately 1.)
exceptional since x = 0 corresponds to
‘no border’, and x = 6 corresponds to
‘no clear space’. Such exceptional points are Solution 2.4
often included in the domain of a function which
(a) Using equation (2.3) with p = 2, we obtain
is part of a model, provided that the rule of the
function is applicable at the points.) x2 + 4x + 3 = (x + 2)2 − 4 + 3
(b) The length of the clear space is 16 − 2x, and the = (x + 2)2 − 1.
width is 12 − 2x. (b) Using equation (2.3) with p = 12 , we obtain
(c) We have
x2 + x + 3
4 = (x + 12 )2 − 1 3
4 + 4
A = (16 − 2x)(12 − 2x) = (x + 21 )2 + 1
2 2.
= 192 − 56x + 4x .
(c) First take out the factor 4, to give
4x2 − 56x + 192 = 4(x2 − 14x + 48),
Solution 2.2
and then use equation (2.3) with p = −7:
(a) The equation
4x2 − 56x + 192 = 4((x − 7)2 − 49 + 48)
4x2 − 56x + 96 = 0
= 4((x − 7)2 − 1)
is equivalent to
= 4(x − 7)2 − 4.
2
x − 14x + 24 = 0.
(d) First take out the factor 12 , to give
The solutions are therefore 1 2
( 2x + x = 12 (x2 + 2x),
14 ± (−14)2 − 4 × 24
x= and then use equation (2.3) with p = 1:
√ 2 1 2
= 12 (14 ± 100); 2x + x = 12 ((x + 1)2 − 1)
that is, = 12 (x + 1)2 − 21 .
x = 12 and x = 2.
These solutions could also be found by Solution 2.5
factorising the expression x2 − 14x + 24. By the solution to Activity 2.4(c),
(b) Of these solutions, only x = 2 lies in the interval f (x) = 4x2 − 56x + 192
[0, 6], so this gives the solution to the exhibition
hall problem: the border width is 2 metres. = 4(x − 7)2 − 4,
so the graph of y = f (x) can be obtained from the
Solution 2.3 graph of y = x2 by performing:
(a) Equation (2.2) can be written as % a y-scaling with factor 4;
4x2 − 56x + 192 = 144;

% a horizontal translation by 7 units to the right;
that is,
(The order of performing the y-scaling and the
2 horizontal translation may be reversed.)
4x − 56x + 48 = 0,
which is equivalent to The vertex of the parabola y = f (x) is at (7, −4).
2 The y-intercept is
x − 14x + 12 = 0.
The solutions are therefore f (0) = 192,
(
14 ± (−14)2 − 4 × 12 and the x-intercepts are x = 6 and x = 8. The
x=
x-intercepts are found by solving the equation
1
√ 2
= 2 (14 ± 148) f (x) = 4x2 − 56x + 192 = 0,
√
= 7 ± 37; which is equivalent to
that is, x2 − 14x + 48 = (x − 6)(x − 8) = 0.
x 0 13.083 and x 0 0.917.
51
The stages in the process, other than the y-scaling, For the input 14 π, we use the following
are shown in Figure S.5. right-angled triangle in a similar way.
Figure S.7
This gives
√ √
cos( 14 π) = sin( 14 π) = 1/ 2 = 21 2 0 0.71.
Figure S.5 (b) Use of the formulas indicated gives
cos( 23 π) = − cos( 13 π) = − 21 = −0.5,
Solution 3.1 √
sin( 23 π) = sin( 13 π) = 21 3 0 0.87,
As the point (x, y) rotates around the unit circle, √
cos( 34 π) = − cos( 14 π) = − 21 2 0 −0.71,
each of the x- and y-coordinates varies between −1 √
and 1 (both inclusive). Hence the complete set of sin( 34 π) = sin( 14 π) = 21 2 0 0.71,
√
image values for cos t = x and sin t = y is, in each cos( 56 π) = − cos( 16 π) = − 21 3 0 −0.87,
case, the interval [−1, 1].
sin( 56 π) = sin( 16 π) = 1
2 = 0.5,
cos π = − cos 0 = −1,
Solution 3.2
sin π = sin 0 = 0.
(a) For the inputs 0 and 12 π, it is easiest to refer
directly to the unit circle. If t = 0, then the (c) Use of the formulas indicated gives
√
corresponding point on the unit circle is cos(− 16 π) = cos( 61 π) = 21 3 0 0.87,
(x, y) = (1, 0), so
sin(− 16 π) = − sin( 61 π) = − 12 = −0.5,
cos 0 = 1, sin 0 = 0. √
cos(− 14 π) = cos( 41 π) = 21 2 0 0.71,
If t = 12 π, then (x, y) = (0, 1), so √
sin(− 14 π) = − sin( 14 π) = − 21 2 0 −0.71,
cos( 12 π) = 0, sin( 12 π) = 1. cos(− 13 π) = cos( 13 π) = 1
2 = 0.5,
√
For the inputs 1
6π
and 1
3 π,
it is easiest to refer to sin(− 13 π) = − sin( 13 π) = − 21 3 0 −0.87,
the following right-angled triangle, and to use cos(− 12 π) = cos( 21 π) = 0,
the expressions for cos and sin in terms of the
opposite and adjacent sides and the hypotenuse sin(− 12 π) = − sin( 12 π) = −1,
of the triangle (Chapter A2, Subsection 3.2). cos(− 23 π) = cos( 32 π) = − 21 = −0.5,
√
sin(− 23 π) = − sin( 23 π) = − 21 3 0 −0.87,
√
cos(− 34 π) = cos( 43 π) = − 21 2 0 −0.71,
√
sin(− 34 π) = − sin( 34 π) = − 21 2 0 −0.71,
√
cos(− 56 π) = cos( 65 π) = − 21 3 0 −0.87,
sin(− 56 π) = − sin( 65 π) = − 12 = −0.5,
cos(−π) = cos π = −1,
sin(−π) = − sin π = 0.
Figure S.6
This gives
√
cos( 16 π) = 1
2 3 0 0.87, sin( 16 π) = 21 = 0.5,
√
cos( 13 π) = 1
2 = 0.5, sin( 13 π) = 21 3 0 0.87.
52
Solution 3.3 Solution 3.4

(a) The graph of y = −3 + cos x is obtained from (a) From the definition of tan, we have
that of y = cos x by translating downwards by sin 0 0
3 units. This gives the graph in Figure S.8. tan 0 = = = 0,
cos 0 1
sin( 1
6 π)
1 √
tan( 61 π) = 1 = 1
2
√ = 13 3 0 0.58,
cos( 6 π) 2 3
1 1
√
1 sin( 4 π) 2 2
tan( 4 π) = = 1 √ = 1,
cos( 14 π) 2 2
√
1
1
sin( 3 π) 1
2 3
√
tan( 3 π) = 1 = 1 = 3 0 1.73.
cos( 3 π) 2
(b) Use of the formula tan(−x) = − tan x gives

√
tan(− 16 π) = − tan( 16 π) = − 13 3 0 −0.58,
tan(− 14 π) = − tan( 14 π) = −1,
√
tan(− 13 π) = − tan( 31 π) = − 3 0 −1.73.
Figure S.8 Solution 3.5

(b) The graph of y = 12 cos x is obtained from that (a) The required values (correct to two decimal
of y = cos x by performing a y-scaling with places) of 5x are shown in the following table.
factor 12 . This gives the graph in Figure S.9.
x −1 −0.75 −0.5 −0.25 0
5x 0.2 0.30 0.45 0.67 1
x 0 0.25 0.5 0.75 1

5x 1 1.50 2.24 3.34 5
(b) The graph is shown in Figure S.11.
Figure S.9
(c) The graph of y = cos(3x) is obtained from that

of y = cos x by performing an x-scaling with
Figure S.11
(c) We have g(x) = ( 15 )x = 5−x . The graph of this
function is obtained from that of f (x) = 5x by
reflection in the y-axis, which is equivalent to an
x-scaling with factor −1. Hence the graph of
g(x) = ( 15 )x is as in Figure S.12.
Figure S.10
(d) The graph of y = cos(−x) is obtained from that

of y = cos x by performing an x-scaling with
factor −1. This is the same as reflecting the
graph in the y-axis, which gives the graph of
y = cos x once more. This corresponds to the
fact that cos(−x) = cos x for all x.
Figure S.12
53

(a) (a) The graph of f is as in Figure S.16.
Figure S.13
The function f (x) = cos x is neither increasing Figure S.16

nor decreasing, and it is many-one.
(b) From the graph we see that:
% the image set of f is (−1, 5).
Therefore f has an inverse function f −1 with
domain (−1, 5) and image set (0, 2).
We can find the rule of f −1 by solving
y = f (x) = 3x − 1
y = 3x − 1, so x = 13 (y + 1).
Figure S.14 Thus the inverse function is
The function g(x) = sin x (− 21 π ≤ x ≤ 21 π) is f −1 (y) = 13 (y + 1) (y in (−1, 5)),
increasing, so it is one-one.
or, in terms of x,
(c)
f −1 (x) = 13 (x + 1) (x in (−1, 5)).
The graph of y = f −1 (x) is found by reflecting
that of y = f (x) (Figure S.16) in the 45◦ line,
and is shown in Figure S.17.
Figure S.17
Figure S.15
The function h(x) = ( 21 )x is decreasing, so it is

one-one.
54
(b) The graph of f is as in Figure S.18. From the graph we see that:
% the function f is decreasing, and so one-one;
% the image set of f is [0, 192].
domain [0, 192] and image set [0, 6].
y = f (x) = 4x2 − 56x + 192,
where 0 ≤ y ≤ 192,
4x2 − 56x + 192 − y = 0,
Figure S.18 so
(
56 ± (−56)2 − 16(192 − y)
From the graph we see that: x=
# 8 %
(
% the function f is increasing, and so one-one; = 18 56 ± 64 + 16y
% the image set of f is [0, ∞). '
Therefore f has an inverse function f −1 with = 7 ± 1 + 14 y.
domain [0, ∞) and image set [0, ∞). (Alternatively, on completing the square we
We can find the rule of f −1
by solving obtain
3
y = f (x) = x , where y ≥ 0, y = 4x2 − 56x + 192 = 4(x − 7)2 − 4,
to obtain x in terms of y: which can be rearranged as

√ (x − 7)2 = 1 + 41 y,
y = x3 , so x = 3 y.
'
Thus the inverse function is giving x = 7 ± 1 + 41 y, as before.)
√
f −1 (y) = 3 y (y in [0, ∞)), Now, we need to choose the solution of this
or, in terms of x, equation which lies in [0, 6], the domain of f .
√ Therefore we choose
f −1 (x) = 3 x (x in [0, ∞)). '
x = 7 − 1 + 14 y.
that of y = f (x) (Figure S.18) in the 45◦ line, Thus the inverse function is
and is shown in Figure S.19. '
f −1 (y) = 7 − 1 + 14 y (y in [0, 192]),
or, in terms of x,
'
f −1 (x) = 7 − 1 + 41 x (x in [0, 192]).

that of y = f (x) (Figure S.20) in the 45◦ line,
and is shown in Figure S.21.
Figure S.19
(c) The graph of f is as in Figure S.20; it can be

obtained from Figure S.5 by restricting the
domain to [0, 6].
Figure S.21
Figure S.20
55
Solution 4.3 (b) By properties (b)(i) and (b)(ii),

$ 4 3x &
(a) (i) The domain of arcsin is [−1, 1]. Since 2π x 4 2
log2 = log2 (x4 ) + log2 (43x ) − log2 (2x ).
does not lie in this interval, arcsin(2π) is not 2x 2
valid.
Now
(ii) The domain of arccos is [−1, 1]. Since 0 lies
log2 (x4 ) = 4 log2 x,
in this interval, arccos 0 is valid.
log2 (43x ) = 3x log2 4 = 3x × 2 = 6x,
(iii) and (iv) The domain of arctan is R. Since
1 5 1 by property (c), and
2 π and tan( 4 π) are real numbers, arctan( 2 π)
and arctan(tan( 54 π)) are both valid. 2
√ log2 (2x ) = x2 ,
(b) (i) arcsin( 21 3 ) = 31 π
so
(ii) arccos(−0.5) = 23 π $ &
x4 43x
(iii) arctan 1 = 14 π log2 = 4 log2 x + 6x − x2 .
2x 2
(iv) arcsin(sin( 15 π)) = 15 π
(v) cos(arccos(0.9)) = 0.9
Solution 4.6
(vi) The value 54 π does not lie in the domain of
the function (a) At the start of year 1, the deer population is
f (x) = tan x (− 12 π < x < 12 π), P1 = 2666.6̇ × 1 + 3333.3̇ = 6000.

(b) We need to solve the equation
but 14 π does lie in this domain, and
tan( 54 π) = tan( 41 π). So Pn = 3 × 6000 = 18 000;
arctan(tan( 54 π)) = arctan(tan( 14 π)) = 14 π. that is,

2666.6̇ × (1.15)n−1 + 3333.3̇ = 18 000,
or
Solution 4.4
18 000 − 3333.3̇
(a) (i) log10 (10 000) = log10 (104 ) = 4 (1.15)n−1 = = 5.5.
2666.6̇
(ii) log3 ( 19 ) = log3 (3−2 ) = −2 Applying ln to both sides and rearranging, we
(b) Since log2 10 is the power to which 2 must be obtain
raised to obtain 10, ln(5.5)
n−1= 0 12.2, so n 0 13.2.
ln(1.15)
2log2 10 = 10.
So the population will exceed 18 000 at the start
of year 14.
Solution 4.5
(a) By properties (b)(i) and (b)(ii),
loga 6 + loga 8 − loga 2 − loga 24
$ &
6×8
= loga = loga 1.
2 × 24
Hence, by property (a),
loga 6 + loga 8 − loga 2 − loga 24 = 0.
56
Solutions to Exercises
Solution 1.1 (c) Since this function has the property that
(a) (−∞, 1) 1 1
f (−x) = ( = ( = f (x),
√ √ |−x| |x|
(b) The expression x is defined for x ≥ 0, so 1/ x
is defined for x > 0. Thus, by the domain it is symmetric in the y-axis. Now
convention, the domain is (0, ∞). 1 1
√ ( =√ for x > 0,
(c) The expression 9 + x is defined for 9 + x ≥ 0, |x| x
which is equivalent to x ≥ −9. Thus, by the
domain convention, the domain is [−9, ∞). so we can use the solution to part (b), together
with its reflection in the y-axis.
Solution 1.2
(a) x −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
x4 16 5.06 1 0.06 0 0.06 1 5.06 16
Figure S.24
The image set is the open interval (0, ∞).
Solution 2.1
(a) The largest closed interval in which x can lie is
[0, 4].
Figure S.22
The image set is the closed interval [0, ∞).

(b) x 0.1 0.5 1 1.5 2
√
1/ x 3.16 1.41 1 0.82 0.71
Figure S.25
(b) The lawn has length 8 − x and width 8 − 2x, so

its area is
Figure S.23
A = (8 − x)(8 − 2x)
The image set is the open interval (0, ∞). = 64 − 24x + 2x2 .
(c) Half the area of the garden is 32, so x must
satisfy
64 − 24x + 2x2 = 32;
57
that is, (b) In the garden problem, we needed to find a

2 solution of the equation
2x − 24x + 32 = 0,
f (x) = 2x2 − 24x + 64 = 32
or, equivalently,
which lies in the interval [0, 4]. The graph
x2 − 12x + 16 = 0.
plotted in part (a) shows that the solution of
The solutions of this equation are this equation which lies in the interval [0, 4] is
( between 1 and 2, as found in the solution to
12 ± (−12)2 − 4 × 16
x= Exercise 2.1(c).
1
√ 2
= 2 (12 ± 80) Solution 2.3
√
= 6 ± 2 5; The graph of y = f (x) can be obtained from the
that is, graph of y = 1/x (see Figure 1.4) by performing:
x 0 10.47 and x 0 1.53.

% a y-scaling with factor −2;
% a horizontal translation by 1 unit to the left;
Of these solutions, only x 0 1.53 lies in the
interval [0, 4], so the width of the border is
% a vertical translation by 1 unit upwards.
1.53 metres. The stages in this process are shown in Figure S.27.
Solution 2.2 The y-intercept is

2
(a) First, we express the rule of f in f (0) = − + 1 = −1,
completed-square form: 0+1
and the x-intercept is x = 1. The x-intercept is found
f (x) = 2x2 − 24x + 64
by solving the equation
= 2(x2 − 12x + 32)
2
= 2((x − 6)2 − 36 + 32) f (x) = − + 1 = 0,
x+1
= 2(x − 6)2 − 8. which is equivalent to
Therefore the graph of y = f (x) can be obtained x + 1 = 2,
from the graph of y = x2 by performing:
giving x = 1.
% a y-scaling with factor 2;
% a horizontal translation by 6 units to the
right;
The stages in this process are shown in
Figure S.26.
The y-intercept is
f (0) = 64,
and the x-intercepts are x = 4 and x = 8. The
x-intercepts are found by solving the equation
f (x) = 2x2 − 24x + 64 = 0,
which is equivalent to
x2 − 12x + 32 = (x − 4)(x − 8) = 0.
Figure S.27
Figure S.26
58
SOLUTIONS TO EXERCISES

(a) The graph of y = 1 + sin x is obtained from that The graph of g(x) = ex+1 is obtained from that of
of y = sin x by translating upwards by 1 unit. f (x) = ex by applying a horizontal translation by
This gives the graph in Figure S.28. 1 unit to the left. This gives the graph in Figure S.31.
Figure S.28
(b) The graph of y = − sin x is obtained from that

Figure S.31
of y = sin x by performing a y-scaling with
factor −1. This is the same as reflection in the (Note that the graph of g may also be obtained from
x-axis, and gives the graph in Figure S.29. that of f by applying a y-scaling with factor e, since
ex+1 = e × ex .)
Solution 4.1
(a) The graph of f is as in Figure S.32.
Figure S.29
(c) The graph of y = sin(4x) is obtained from that

of y = sin x by performing an x-scaling with
Figure S.32
From the graph we see that:

% the image set of f is the whole of R.
domain R and image set R, the domain of f .
Figure S.30
y = f (x) = 4x + 3
y = 4x + 3, so x = 14 (y − 3).
Thus the inverse function is
f −1 (y) = 14 (y − 3),
or, in terms of x,
f −1 (x) = 41 (x − 3).
59
The graph of y = f −1 (x) is found by reflecting Thus the inverse function is

that of y = f (x) (Figure S.32) in the 45◦ line, as '
shown in Figure S.33. f −1 (y) = 6 + 4 + 21 y (y in [−8, ∞)),
or, in terms of x,
'
f −1 (x) = 6 + 4 + 12 x (x in [−8, ∞)).

that of y = f (x) (Figure S.34) in the 45◦ line, as
shown in Figure S.35.
Figure S.33
(b) The graph of f is as in Figure S.34 (see the

solution to Exercise 2.2).
Figure S.35
Solution 4.2
(a) (i) arcsin(0.1) 0 0.100 167
Figure S.34 (ii) arccos(−0.85) 0 2.586 78
From the graph we see that: (iii) arctan(0.1) 0 0.099 668 7

√
% the function f is increasing, and so one-one; (b) (i) arcsin(− 12 2 ) = − 14 π
% the image set of f is the interval [−8, ∞). (ii) arccos 1 = 0
√
Therefore f has an inverse function f −1 with (iii) arctan(− 3 ) = − 13 π
domain [−8, ∞) and image set [6, ∞), the
domain of f .
Solution 4.3
(a) (i) log2 64 = log2 26 = 6
y = f (x) = 2x2 − 24x + 64, where y ≥ −8,
(ii) log10 (0.001) = log10 (10−3 ) = −3
(iii) ln 1 = loge 1 = 0
2x2 − 24x + 64 − y = 0, (iv) ln 10 0 2.302 59
so (b) Applying ln to both sides of the equation
(
24 ± 242 − 8(64 − y) 10 = 2x gives
x=
# 4 % ln(2x ) = ln 10, so x ln 2 = ln 10,
1
(
= 4 24 ± 64 + 8y
giving
'
= 6 ± 4 + 12 y. ln 10
x= 0 3.321 93.
ln 2
Now, we need to choose the solution of this
Since 10 = 2x ,
equation which lies in [6, ∞), the image set
of f −1 . Therefore we choose log2 10 = log2 (2x )
'
= x 0 3.321 93.
x = 6 + 4 + 12 y.
60
SOLUTIONS TO EXERCISES
(c) By properties (b)(i) and (ii),

$ x+1 &
e
ln = ln(ex+1 ) − ln(x3 + x2 )
x3 + x2
= x + 1 − ln(x2 (x + 1))
= x + 1 − ln(x2 ) − ln(x + 1)
= x + 1 − 2 ln x − ln(x + 1),
as required.
Solution 4.4
At the start of year 1, the amount owing is
m1 = −6048.6 × (1.05)0 + 16 048.6
= 10 000.
Thus we need to solve the equation
1
mn = 2 × 10 000 = 5000;
that is,
−6048.6 × (1.05)n−1 + 16 048.6 = 5000,
or
5000 − 16 048.6
(1.05)n−1 = 0 1.826 637 6.
−6048.6
Applying ln to both sides and rearranging, we obtain
ln(1.826 637 6)
n−1= 0 12.3, so n 0 13.3.
ln(1.05)
So the amount owing will be less than £5000 at the
start of year 14.
61
Index
absolute value 13 one-one function 36

antilogarithm 44 one-to-one function 36
arccosine function 41 open interval 7
arcsine function 40
parabola
arctangent function 41
axis of symmetry 18
asymptote 11
vertex 18
base of a logarithm 42 period 28
base of an exponential function 32 periodic function 28
polynomial 47
closed interval 7 polynomial equation 47
codomain 9 polynomial function 47
common logarithm 44
completing the square 18 quadratic function 17
continuous function 49 sketching 17, 22
continuous model 49
real function 9
continuous variable 49
reciprocal function 10
cosine function 27
rule 6
cubic equation 47
sine function 27
decreasing function 36
degree of a polynomial 47 tangent function 31
discrete model 49 transformation 9
discrete variable 49 translation 20
domain 6
domain convention 8 value of a function 7
endpoint of an interval 7 x-scaling 25

exponential function 32
y-scaling 21
function 6
geometric sequence 33
graph of a function 9
half-closed interval 8
half-open interval 8
image 7, 12
image set 12
image value 7
increasing function 36
interval notation 8
inverse function 37
finding 37
sketching 37
linear function 9
logarithm 42
properties 43
many-one function 36
many-to-one function 36
mapping 9
modulus 13
modulus function 13
natural logarithm 44
62

MST121 Chapter A3 Functions

Uploaded by

Copyright:

Available Formats

MST121 Chapter A3 Functions

Uploaded by

Document Information

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

MST121 Chapter A3 Functions

Uploaded by

Copyright:

Available Formats

What mathematical concepts are covered in the document?

What mathematical concepts are covered in the document?

What software is used as part of the course?

What software is used as part of the course?

MST121

About this course

The Open University, Walton Hall, Milton Keynes, MK7 6AA.

There are ﬁve sections in this chapter. They are

The optional Video Band A(v) Algebra

One of the major concerns of mathematics (and science) is the way in

Figure 0.1 Area of a triangle

Section 1 introduces functions as ‘processors’ and describes their notation

1.1 Function notation

Figure 1.1 A processor for squaring

In general, a function is speciﬁed by giving:

It is important to note that a function f is independent of the variable

example, the formula s

Activity 1.1 Specifying functions

(a) Describe the domain of the function f given by the speciﬁcation

Solutions are given on page 50.

Table 1.1 Interval notation

Activity 1.2 Using interval notation

For each of the following inequalities, write down the corresponding

Consider, for example, the two functions g and h speciﬁed as follows:

deﬁned only for x ≥ 0. Similarly, the function h has domain R excluding 0,

Activity 1.3 Using the domain convention

Specify the domain of each of the following functions, as given by the

domain convention. In each case, explain your answer.

Here are some general remarks about function notation.

f (x) = x2 ’ rather than ‘the function f with rule f (x) = x 2 ’.

1.2 Graphs of functions

Figure 1.2 Graph of y = mx + c

Graphs of linear functions can be drawn accurately by plotting just two

and sketch the corresponding graph of y = x2 by constructing a table of

Where convenient, we use the

Figure 1.3 Graph of y = x2

In Figure 1.3 we have plotted 9 points and joined them by a smooth

These values of 1/x are

In this case, using the same

Figure 1.4 Graph of y = 1/x

As in Figure 1.5, it is often

Figure 1.5 Symmetry properties of the graph of y = 1/x

Here, in each case, the

Figure 1.6 Graphs of g and h

Figure 1.7 Image sets of f and g

Activity 1.4 Sketching graphs

The modulus, or absolute value, of a real number x is the

Figure 1.8 Constructing the graph of the modulus function

Activity 1.5 Graphs involving the modulus

Without constructing a table of values, sketch the graphs of each of the

% notations for specifying the rule and domain of a function;

Exercises for Section 1

2.1 The exhibition hall problem Stages in the modelling cycle

The Exhibition Hall Problem

Figure 2.1 Introducing variables for the exhibition hall problem

Activity 2.1 Relating variables

area of the hall; that is, A = 12 × 16 × 12 = 96. Therefore we have to solve

Activity 2.2 Solving the equation

(a) Solve equation (2.1).

For this graph, it is not