Unit 3 MST 124

Unit 3
Functions
Introduction
Introduction
In mathematics you often work with situations in which one quantity
depends on another. For example:
The distance walked by a woman at a particular speed depends on the

time that shes been walking.
The height of a gondola on a Ferris wheel depends on the angle

through which the wheel has rotated since the gondola was in its
lowest position.
The number of 5-litre tins of a particular type of paint needed by a

decorator depends on the area that he intends to paint.
Whenever one quantity depends on another, we say that the rst quantity
is a function of the second quantity. The idea of a function is
fundamental in mathematics, and in particular it forms the foundation for
calculus, which youll begin to study in Unit 6.
In this unit youll be introduced to the terminology and notation that are
used for functions. Youll learn about some standard, frequently-arising
types of functions, and how to use graphs to visualise properties of
functions. Youll also learn how you can use your knowledge about a few
standard functions to help you understand and work with a wide range of
related functions. Later in the unit youll revise exponential functions and
logarithms, and practise working with them. In the nal section youll
revise inequalities, and see how working with functions and their graphs
can help you understand and solve some quite complicated inequalities.
A Ferris wheel
This is a long unit. The study calendar allows extra time for you to
study it.
1 Functions and their graphs

This section introduces you to the idea of a function and its graph, and
shows you some standard functions. Youll start by learning about sets,
which are needed when you work with functions and also in many other
areas of mathematics.
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Unit 3
Functions
1.1 Sets of real numbers

In mathematics a set is a collection of objects. The objects could be
anything at all: they could be numbers, points in the plane, equations or
anything else. For example, each of the following collections of objects
forms a set:
all the prime numbers less than 100
all the points on any particular line in the plane
all the equations that represent vertical lines
the solutions of any particular quadratic equation.
A set can contain any number of objects. It could contain one object, two
objects, twenty objects, innitely many objects, or even no objects at all.
Each object in a set is called an element or member of the set, and we
say that the elements of the set belong to or are in the set.
There are many ways to specify a set. If there are just a few elements,
then you can list them, enclosing them in curly brackets. For example, you
can specify a set S as follows:
S = {3, 7, 9, 42}.
Another simple way to specify a set is to describe it. For example, you can
say let T be the set of all even integers or let U be the set of all real
numbers greater than 5. We usually denote sets by capital letters.
The set that contains no elements at all is called the empty set, and is
denoted by the symbol .
Its often useful to state that a particular object is or is not a member of a
particular set. You can do this concisely using the symbols and ., which
mean is in and is not in, respectively. For example, if S is the set
specied above, then the following statements are true:
7S
and
Activity 1
10 . S.
Understanding set notation
Let X = {1, 2, 3, 4} and let Y be the set of all odd integers. Which of the
following statements are true?
(a) 1 X
(b) 1 Y
(c) 2 . X
(d) 2 . Y
Its often useful to construct new sets out of old sets. For example, if A
and B are any two sets, then you can form a new set whose members are
all the objects that belong to both A and B. This set is called the
intersection of A and B, and is denoted by A B. For instance, if
A = {1, 2, 3, 4}
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and
B = {3, 4, 5},
Functions and their graphs
then
A B = {3, 4}.
Similarly, if A and B are any two sets, then you can form a new set whose
members are all the objects that belong to either A or B (or both). This
set is called the union of A and B, and is denoted by A B. For example,
if A and B are as specied above, then
A B = {1, 2, 3, 4, 5}.
You might nd it helpful to visualise intersections and unions of sets by
using diagrams like those in Figure 1, which are known as Venn
diagrams. The Venn diagrams in the gure show the intersection and
union of the particular sets A and B above.
A
B
1
(a)
B
1
(b)
Figure 1 (a) The intersection (shaded) and (b) the union (shaded) of two
sets
Venn diagrams are named after the logician John Venn, who used
them in publications starting in 1880. However, the idea of using
diagrams in this way did not originate with Venn. The prolic Swiss
mathematician Leonhard Euler (pronounced oiler) used them in his
Letters to a German Princess (176062). Venn acknowledged Eulers
inuence by calling his own diagrams Eulerian circles. He extended
Eulers idea, using the diagrams to analyse more complex logical
problems. As well as working on logic at Cambridge University, Venn
was for some time a priest and later a historian. There is more about
Euler on page 214.
John Venn (18341923)

You can form intersections and unions of more than two sets in a similar
way. In general, the intersection of two or more sets is the set of all
objects that belong to all of the original sets, and the union of two or
more sets is the set of all objects that belong to any of the original sets.
For example, if A and B are as specied above and
C = {20, 21},
then
ABC =
and A B C = {1, 2, 3, 4, 5, 20, 21}.
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Unit 3
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Activity 2
Understanding unions and intersections of sets
Let P = {1, 2, 3, 4, 5, 6}, let Q = {2, 4, 6, 8, 10, 12} and let R be the set of
all integers divisible by 3. Specify each of the following sets.
(a) P Q
(b) Q R
(c) P Q R
(d) P Q
The set membership symbol is a stylised version of the Greek letter

(epsilon). The Italian mathematician Giuseppe Peano (18581932),
the founder of symbolic logic, used to indicate set membership in a
text published in 1889. He stated that it was an abbreviation for the
Latin word est, which means is. The symbol was then adopted by
the logician Bertrand Russell (18721970) in a text published in 1903,
but it was typeset in a form that looks like the modern symbol , and
this form has remained in use to the present day. Peano also
introduced the symbols and for intersection and union.
The empty set symbol was introduced in 1939 by the inuential
French mathematician Andre Weil (19061998). It was inspired by
the letter in the Norwegian alphabet.
Sometimes every element of a set A is also an element of a set B. In this

case we say that A is a subset of B, and we write A B. For example:
1
3
{1, 3} is a subset of {1, 2, 3} (as shown in Figure 2)
the set of integers is a subset of the set of real numbers.
Every set is a subset of itself, and the empty set is a subset of every set.
In this module, and particularly in this unit, youll mostly be working with
sets whose elements are real numbers. In the rest of this subsection, youll
meet some useful ways to visualise and represent sets of this type.
Figure 2 A subset of a set

(shaded)
The set of all real numbers is denoted by R. You can handwrite this as:
You saw in Unit 1 that you can visualise the real numbers as points on an
innitely long straight line, called the number line or the real line. Part
of the number line is shown in Figure 3. Although only the integers are
marked in the diagram, every point on the line represents a real number.
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
Figure 3 The number line
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You can use the number line to visualise sets of real numbers. For example:
Figure 4(a) shows the set {1, 0, 1}.
Figure 4(b) shows the set of real numbers that are greater than or
equal to 2 and also less than or equal to 6.
Figure 4(c) shows the set of real numbers that are greater than 5.
Figure 4(d) shows the set of real numbers that are less than
greater than or equal to 3.
1
2
or
In these kinds of diagrams, a solid dot indicates a number thats included

in the set, and a hollow dot indicates a number that isnt included. A
heavy line that continues to the left or right end of the diagram indicates
that the set extends indenitely in that direction.
321 0 1 2 3
(a)
7654321
0 1 2 3 4 5 6 7 8
(b)
321 0 1 2 3 4 5
1
2
(c)
(d)
Figure 4 Sets of real numbers

The sets in Figure 4(b) and (c) are examples of a special type of set of real
numbers, called an interval. An interval is a set of real numbers that
corresponds to a part of the number line that you can draw without lifting
your pen from the paper. The sets in Figure 4(a) and (d) arent intervals,
as they have gaps in them. In fact, the set in Figure 4(a) is the union of
three intervals (each containing a single number), and the set in
Figure 4(d) is the union of two intervals.
A number that lies at an end of an interval is called an endpoint of the
interval. For example, the interval in Figure 4(b) has two endpoints,
namely 2 and 6, and the interval in Figure 4(c) has one endpoint,
namely 5. The whole set of real numbers, R, is an interval with no
endpoints.
An interval that includes all of its endpoints is said to be closed, and one
that doesnt include any of its endpoints is said to be open. For example,
the interval in Figure 4(b) is closed (since it includes both its endpoints),
and the one in Figure 4(c) is open (since it excludes its single endpoint).
An interval that includes one endpoint and excludes another, such as the
interval in Figure 5, is said to be half-open (or half-closed). Since the
interval R has no endpoints, its both open and closed! This fact may seem
strange at the moment, but it will make more sense if you go on to study
pure mathematics at higher levels.
21 0 1 2 3 4
Figure 5 A half-open (or

half-closed) interval
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Unit 3
Functions
Activity 3
Recognising intervals
State whether each of the sets below is an interval. For each set that is an
interval, state whether its open, closed or half-open.
(a)
321 0 1 2 3
(c)
321 0 1 2 3
(e)
321 0 1 2 3
(g)
321 0 1 2 3
(b)
321 0 1 2 3
(d)
321 0 1 2 3
(f)
321 0 1 2 3
(h)
321 0 1 2 3
A convenient way to describe most intervals is to use inequality signs.

These are listed below, with their meanings. (Note that some texts use
slightly dierent inequality signs: ! and " instead of and .)
Inequality signs
<
>
is
is
is
is
less than
less than or equal to
greater than
greater than or equal to
For example, the interval in Figure 6(a) is the set of real numbers x such
that x > 2 (that is, such that x is greater than 2).
Similarly, the interval in Figure 6(b) is the set of real numbers x such that
x > 1 and x 4 (that is, such that x is greater than 1 and x is less than or
equal to 4). We usually write this description slightly more concisely, as
follows: the interval is the set of real numbers x such that 1 < x 4 (that
is, such that 1 is less than x, which is less than or equal to 4).
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0 1 2 3 4 5 6 7
(a)
1 0 1 2 3 4 5 6
(b)
Figure 6 Intervals
It might help you to remember the meanings of the inequality signs if you
notice that when you use either of the signs < or >, the lesser quantity is
on the smaller, pointed side of the sign. The same is true for the signs
and , except that one quantity is less than or equal to the other, rather
than denitely less than it.
The statement x > 2 is called an inequality. In general, an inequality is
a mathematical statement that consists of two expressions with an
inequality sign between them. A statement such as 1 < x 4 is called a
double inequality. The two inequality signs in a double inequality
always point in the same direction as each other.
Activity 4
Using inequality signs to describe intervals
(a) Draw diagrams similar to those in Figure 6 to illustrate the intervals

described by the following inequalities and double inequalities.
(i) 0 < x < 1
(ii) 3 x < 2
(iii) x 5
(iv) x > 4
(b) For each of the following diagrams, write down an inequality or double
inequality that describes the interval illustrated.
(i)
321 0 1 2 3 4 5 6
(iii)
4321 0 1
(v)
1 0 1 2 3 4 5 6 7
(ii)
54321 0
(iv)
54321 0 1
(vi)
2 3 4 5 6 7 8
Another useful way to describe intervals is to use interval notation. For

example, the interval described by the double inequality 4 x < 7 is
denoted in interval notation by [4, 7). The square bracket indicates an
included endpoint, and the round bracket indicates an excluded one. An
interval that extends indenitely is denoted by using the symbol (which
is read as innity), or its negative, (which is read as minus
innity), in place of an endpoint. For example, the interval described by
the inequality x 5 is denoted by [5, ), and the interval described by the
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Unit 3
Functions
inequality x < 6 is denoted by (, 6). We always use a round bracket
next to or in interval notation. Heres a summary of the notation.
Interval notation
Open intervals
(a, b)
a<x<b
Closed intervals
[a, b]
!
axb
(a, )
(, b)
x>a
x<b
[a, )
(, b]
xa
xb
(, )
(, )
{a}
x=a
Half-open (or half-closed) intervals

[a, b)
(a, b]
!
ax<b
a<xb
Notice that youve now seen two dierent meanings for the notation (a, b),
where a and b are real numbers. It can mean either an open interval, or a
point in the coordinate plane. The meaning is usually clear from the
context.
Activity 5
Using interval notation
Write each of the intervals below in interval notation.

(a)
321 0 1 2 3 4 5 6
(c)
4321 0 1
(e)
1 0 1 2 3 4 5 6 7
(b)
54321 0
(d)
54321 0 1
(f)
2 3 4 5 6 7 8
Sometimes you need to work with sets of real numbers that are unions of
intervals, like those in Figure 7.
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21 0 1 2 3 4 5
(a)
1 0 1 2 3 4 5 6 7 8 9
(b)
Figure 7 Two unions of intervals

You can denote a union of intervals in interval notation by using the usual
notation for intervals together with the union symbol . For example, the
sets in Figure 7 can be written as
(, 1] [2, 4)
and [0, 3) (3, 5] [7, 8],
respectively.
Activity 6
Denoting unions of intervals
For each of the following diagrams, write the set illustrated in interval
notation.
(a)
(b)
87654321 0 1 2 3
0 1 2 3 4 5 6 7
(c)
321 0 1 2 3
Its often useful to state that a particular number lies in, or doesnt lie in, a
particular interval or union of intervals. You can do this concisely using the
symbols and . in the usual way. For example, as illustrated in Figure 8,
1 [0, 4]
and
1 . [0, 4].
In the next subsection youll begin your study of functions.
1 0 1 2 3 4 5
Figure 8 The interval [0, 4]
1.2 What is a function?

As mentioned in the introduction to this unit, whenever one quantity
depends on another, we say that the rst quantity is a function of the
second quantity. Here are some more examples.
If a car is driving along a straight road, then its displacement s (in

km) from some reference point depends on the time t (in hours) that
has elapsed since the start of its journey. So s is a function of t.
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Unit 3
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The formula
C = 2r
expresses the circumference C of a circle in terms of its radius r (with
both C and r measured in the same units). So the value of C depends
on the value of r, and hence C is a function of r.
Figure 9
An electrocardiogram
(each high peak in voltage
corresponds to a heartbeat)
The electrical voltage between two points on a persons skin either

side of his or her heart (which can be measured using electrodes)
changes rhythmically with every heartbeat. So the voltage V (in volts,
say) depends on the time t (in seconds, say) that has elapsed since
some point in time, and hence V is a function of t. Theres no simple
formula for the relationship between t and V , but its often displayed
as an electrocardiogram (ECG), like the one in Figure 9.
In each of these examples, theres a rule that converts each value of one
variable (such as t, in the car example) to a value of the other variable
(such as s, in the car example). You can think of the rule as a kind of
processor that takes input values and produces output values, as
illustrated in Figure 10.
input value
processor
output value
Figure 10 A processor that takes input values and produces output

values
For instance, in the car example, an input value of 1.2 (a time, in hours)
might be converted by the processor to an output value of 60 (a
displacement, in kilometres). Similarly, in the circle example, an input
value of 3 (a radius, in centimetres) would be converted by the processor to
an output value of 2 3 = 6 (a circumference, in centimetres).
Sometimes the rule associated with a function can be expressed using a
formula, and sometimes it cant.
In each of the three examples in the list above, theres also a set of allowed
input values, and a set of values within which every output value lies. For
instance, with the car example, if the journey lasts three hours, then the
allowed input values are the real numbers between 0 and 3 inclusive (the
possible elapsed times, in hours), and the output values lie in the set R of
real numbers (they are displacements of the car from the reference point,
in kilometres).
A function is a mathematical object that describes a situation like those
listed above. Its dened as follows.
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A function consists of:
a set of allowed input values, called the domain of the function
a set of values in which every output value lies, called the

codomain of the function
a process, called the rule of the function, for converting each

input value into exactly one output value.
Its often useful to denote a function by a letter. If a function is denoted

by f , say, then for any input value x, the corresponding output value is
denoted by f (x), which is read as f of x.
For example, suppose that we denote the function associated with the car
example by f . If the rule of this function converts the input value 1.2 to
the output value 60, then we write
f (1.2) = 60.
Similarly, suppose that we denote the function associated with the circle
example by g. The rule of this function converts the input value 3 to the
output value 6, so we write
g(3) = 6.
This type of notation is known as function notation.
One use of function notation is for specifying the rule of a function, when
this can be done using a formula. For example, suppose that h is the
function whose domain and codomain each consist of all the real numbers,
and whose rule is square the input number. Then, for example,
h(2) = 4,
h(5) = 25
and h(1) = 1,
and the rule of h can be written as

h(x) = x2 .
Similarly, the rule of the function associated with the circle example can
be written as g(r) = 2r.
When you write down the rule of a function, it doesnt matter what letter
you use to represent the input value. So the rule of the function h above
could also be written as, for example,
h(t) = t2
or
h(u) = u2 .
The variable used to denote the input value of a function is sometimes

called the input variable.
Its traditional to use the letters f , g and h for functions, and the letters
x, t and u for input variables. Although you can use any letters, these ones
are often used in general discussions about functions. The most standard
letters are f for a function and x for an input variable.
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Unit 3
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Activity 7
Understanding function notation
(a) Suppose that f is the function whose domain and codomain each
consist of all the real numbers, and whose rule is f (t) = 4t. Write
down the values of f (5) and f (3).
(b) Suppose that g is the function whose domain and codomain each
consist of all the real numbers, and whose rule can be written in words
as multiply the input number by 2 and then subtract 1. Write down
the rule of g using the notation g(x).
Its important to appreciate that every value in the domain of a function

must have a corresponding output value, given by the rule of the function.
So, for example, a function f that has the rule
f (x) = x
cant
have any negative numbers in its domain, because if x is negative,
then x isnt dened.
As another example, if a function f describes how the displacement in
kilometres of a car from a particular point depends on the time in hours
since it started a 3-hour journey, then wed normally take the domain of f
to be the interval [0, 3].
In contrast, not every value in the codomain of a function actually has to
occur as an output value. For instance, with the car example, wed
normally take the codomain to be the whole set of real numbers, R. Its
good enough that this set contains every possible output value: it doesnt
matter that it also contains many values that couldnt be output values.
The set of values in the codomain of a function that do occur as output
values is called the image set of the function. For example, if f is the
function whose domain and codomain are each the whole set of real
numbers, R, and whose rule is f (x) = x2 , then the image set of f is the
interval [0, ).
Heres another fact about functions that its important to appreciate. Not
only must every value in the domain of a function have a corresponding
output value, given by the rule of the function, but it must have exactly
one output
value. For example, a function f cant have the rule
f (x) = x, because this rule assigns two output values to every input
value (except zero).
You can visualise the facts about functions described above by using a
type of diagram known as a mapping diagram, which is based on Venn
diagrams. (The word mapping is another name for function.) For
example, the mapping diagram in Figure 11 illustrates the function f that
has domain {1, 2, 3}, codomain {2, 4, 6, 8, 10} and rule f (x) = 2x. The
arrows indicate which input value goes to which output value. Notice that
exactly one arrow comes out of each input value. This corresponds to the
fact that each input value has exactly one output value. Notice also that
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the image set consists of all the values that have arrows going in to them,
and that (for this particular function f ) the codomain contains other
values too.
codomain
domain
1
2
3
f
2
4
6
8
10
image set
image of x under f,
value of f at x
Figure 11 A function f illustrated by a mapping diagram

Heres some more terminology associated with functions. If f is a function,
and x is any value in its domain, then the value f (x) is called the image
of x under f , or the value of f at x. This is illustrated in Figure 12. We
also say that f maps x to f (x).
For example, f (2) = 4 for the function f in Figure 11 above, so we can say
that the image of 2 under f is 4, or f takes the value 4 at 2, or f maps 2
to 4.
Activity 8
Understanding function terminology
f (x)
Figure 12 The image of a

value x under a function f
Suppose that f is the function whose domain and codomain each consist of
all the real numbers, and whose rule is f (t) = 4t. Write down the following
numbers.
(a) The image of 2 under f
(b) The image of 1 under f
(c) The value of f at 0.5
(d) The value of f at 0.2
(e) The number whose image under f is 44
(f) The number whose image under f is 1
(g) The number to which f maps 4
(h) The number that is mapped by f to 8
In this module youll be working only with functions whose domains and
codomains are sets of real numbers. Such functions are called real
functions. You can also have other types of functions, such as a function
whose domain and/or codomain is a set of another type of numbers
(complex numbers, for instance), or a set of points in the plane. Youll
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Unit 3
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meet many more types of functions if you go on to study mathematics
beyond this module.
Since well be working only with real functions in this module, well make
some simplifying assumptions.
In this module:
we use the word function to mean real function
we take the codomain of every function to be the whole set of

real numbers, since this set contains every possible output value.
These assumptions allow you to specify any function by stating its domain
and its rule. Its important to remember that to specify a function, a
domain must be stated, as well as a rule. Two functions with the same rule
but dierent domains are dierent functions.
The concept of a function was rst formally dened by the Swiss
mathematician Johann Bernoulli (16671748) in 1718. But the
mathematician who gave prominence to the concept, and who was
responsible for the notation f (x), was Bernoullis compatriot
Leonhard Euler. Euler was one of the most talented and productive
mathematicians of all time. He became blind in the early 1770s but
his output, rather than stopping, actually increased. His work covers
almost every area of mathematics, and his collected works run to
over 70 volumes, with further volumes still to appear.
Leonhard Euler (170783)
1.3 Specifying functions

Youve seen that to specify a function you have to state its domain and its
rule. There are various ways to state the domain and rule of a function.
Heres the format that well usually use in this module. For example, to
specify the function f whose rule is f (x) = x2 + 1 and whose domain is the
interval consisting of the real numbers between 0 and 6, inclusive, well
write either
f (x) = x2 + 1
(0 x 6)
f (x) = x2 + 1
(x [0, 6]).
or
Well be even more concise when we want to specify a function whose

domain is the largest possible set of real numbers for which its rule is
applicable. For example, the function
g(x) = x (x [0, ))
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is such a function: its domain is as large as it can be, because x is

dened only for non-negative values of x. Well usually specify a function
like this by stating just its rule. This is because of the following
convention, which is widely used in mathematics.
Domain convention
When a function is specied by just a rule, its understood that the
domain of the function is the largest possible set of real numbers for
which the rule is applicable.
For example, if you read the function h(x) = 1/x, and no domain is
stated, then you can assume that the domain of h is the set of all real
numbers except 0.
Notice that we say, for example, the function h(x) = 1/x, when we really
mean the function h with rule h(x) = 1/x. This is another convenient
convention, which is used throughout this module and throughout
mathematics in general.
Activity 9
Using the domain convention
Describe the domain of each of the following functions, both in words and
using interval notation.
1
1
(b) g(x) =
(c) h(x) = x 1
(a) f (x) =
x4
(x 2)(x + 3)
Another situation where we sometimes specify a function by giving just a

rule, rather than a rule and a domain, is where the domain is clear from
the context. For example, if the function f is such that f (t) is the
displacement in kilometres of a car at time t (in hours) after it began a
3-hour journey, then we assume that the domain of f is the interval [0, 3],
since in this context these are the values that t can take.
Functions specied by equations for one variable in terms

of another
Functions dont have to be specied using function notation. Sometimes
its convenient to express a function using an equation that expresses one
variable in terms of another variable. For example, as mentioned earlier,
the circumference C of a circle is given in terms of its radius r by the
formula
C = 2r.
(1)
Here C is a function of r, and, as youve seen, we can write this function as

g(r) = 2r
(r > 0).
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Unit 3
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But theres no need to use function notation: equation (1) is a perfectly
good specication of the rule of the function.
In general, any equation that expresses one variable in terms of another
variable species the rule of a function. If we wish to specify a domain
thats not the largest possible set of real numbers for which the equation is
applicable, then we can do so in the usual way. For example, we can write
C = 2r
(r > 0).
When a function is specied using an equation that expresses one variable

in terms of another variable, the output variable is called the dependent
variable, because its value depends on the value of the input variable.
The input variable is called the independent variable. For example, for
the function discussed above whose rule is C = 2r, the dependent variable
is C and the independent variable is r.
We often refer to an equation that species a function as a function. For
example, we might say the function y = x2 + 1, when we really mean the
function specied by the rule y = x2 + 1. This is another convenient
convention.
Both types of notation for functions function notation and equations
relating input and output variables are used throughout this module.
Piecewise-dened functions
Sometimes its useful to specify the rule of a function by using dierent
formulas for dierent parts of its domain. For example, you can specify a
function f as follows:
, 2
x
(x 0)
f (x) =
x + 5 (x < 0).
To nd the image of a number x under this function f , you use the rule
f (x) = x2 if x is greater than or equal to zero, and the rule f (x) = x + 5
if x is less than zero. For example,
f (2) = 22 = 4
and f (2) = 2 + 5 = 3.
A function dened in this way is called a piecewise-dened function.

Such piecewise-dened functions can be used to construct curves with a
great variety of shapes, so they are used extensively in the design of
objects such as car bodies and roads.
1.4 Graphs of functions

A convenient way to visualise many of the properties of a function is to
draw or plot its graph. The graph of a function f is the graph of the
equation y = f (x), for all the values of x that are in the domain of f . In
other words, its the set of points (x, y) in the coordinate plane such that x
is in the domain of f and y = f (x).
216
For example, the graph of the function f (x) = x2 + 1 is the graph of the
equation y = x2 + 1, which is shown in Figure 13(a).
Similarly, the graph of the function g(x) = x2 + 1 (0 < x 2) is the graph
of the equation y = x2 + 1 for values of x in the interval (0, 2], which is
shown in Figure 13(b).
y
4
3
3 2 1
y = x2 + 1
1
1
y = x2 + 1
(0 < x 2)
3 x
3 2 1
(a)
3 x
(b)
Figure 13 The graphs of (a) f (x) = x2 + 1 (b) f (x) = x2 + 1 (x (0, 2])

Notice that when we draw graphs we use similar conventions to those that
we use for illustrations of sets on the number line. For example, we use
solid and hollow dots to indicate whether points at the ends of a graph do
or dont lie on the graph. In Figure 13(b) the graph is labelled with its
rule and also with its domain, but we often omit the latter.
A function whose rule you cant express using a formula still has a graph.
For example, Figure 14(a) shows the graph of a function f that describes
the displacement of a car along a road from its starting point during a
3-hour journey. Similarly, the electrocardiogram that you saw earlier,
which is repeated in Figure 14(b), is the graph of a function (with the axes
omitted). This graph represents the changing voltage between two points
on a persons skin over a time period of about two seconds.
s
200
150
s = f (t)
100
50
(a)
(b)
Figure 14 (a) The graph of s = f (t), where f (t) is the displacement of a

car in kilometres at time t (in hours) (b) an electrocardiogram
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Unit 3
Functions
The graph of a function is normally drawn with the input numbers on the
horizontal axis and the output numbers on the vertical axis. (So, if the
axes are labelled with variables, then the variable on the horizontal axis is
the independent variable, and the variable on the vertical axis is the
dependent variable.) In this module well assume that graphs of functions
are always drawn like this.
You can read o the output number corresponding to any particular
input number by drawing a vertical line from the input number on the
horizontal axis to the graph and then a horizontal line across to the
vertical axis. For example, for the function f whose graph is shown in
Figure 15, the value of f (3) is about 5.
y
12
10
8
6
4
2
4 3 2 1 2
y = f (x)
7 x
Figure 15 The graph of a function f

One way to produce a graph of a function is to use a table of values, in the
way that you saw in Unit 2. Youre asked to do this in the next activity.
Activity 10
Plotting the graph of a function using a table of values
Consider the function f (x) = x3 .

(a) Use your calculator to complete the following table of values for this
function.
x
2 1.5 1 0.5 0 0.5 1 1.5 2
x3
(b) Plot the points given by your completed table on a pair of axes.
(c) Draw a smooth curve through the points.
If you use a table of values to plot the graph of a function whose domain
isnt the whole set of real numbers, remember to choose input values that
lie in the domain, and to make sure that you dont extend the graph
beyond the endpoints of the domain. Where appropriate, you should mark
the ends of the graph with solid or hollow dots.
A quicker way to obtain a graph of a function is to use a computer. You
can learn how to do that in the next activity.
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Activity 11
Plotting graphs of functions on the computer
Work through Section 4 of the Computer algebra guide.
As mentioned in Unit 2, one disadvantage of using a table of values to plot

a graph is that you cant be entirely sure about the shape of the graph
between the values in the table, or to the right or left of them. A graph
produced by a computer has similar disadvantages, as in essence its
plotted using a large table of values. In general, its useful to become
familiar with the shapes of the graphs of a variety of functions and
function types, and to learn how to sketch such graphs. Youll have
opportunities to do that throughout this module.
You saw in Unit 2 how to sketch the graphs of equations of the form
y = ax + b and y = ax2 + bx + c, where a, b and c are constants. So you
already know how to sketch the graphs of functions of the form
f (x) = ax + b and f (x) = ax2 + bx + c. The next example illustrates how
to adapt the methods in Unit 2 in order to sketch the graph of a function
whose rule has one of these forms, but whose domain isnt the largest
possible set of real numbers for which the rule is applicable.
Sketching the graph of a function whose domain is not

the largest set of numbers for which its rule is applicable
Example 1
Sketch the graph of the function

f (x) = 41 x2 2x + 6
(5 x < 7).
Solution
First sketch the graph of y = 41 x2 2x + 6, by using any of the
methods from Unit 2. Also include on the sketch the points
corresponding to the endpoints of the domain of f , plotted as solid or
hollow dots as appropriate, and labelled with their coordinates.
The required graph is part of the graph of y = 14 x2 2x + 6, which is
a u-shaped parabola. Completing the square gives
f (x) = 41 x2 2x + 6
= 41 (x2 8x) + 6
= 14 ((x 4)2 16) + 6

= 41 (x 4)2 4 + 6
= 14 (x 4)2 + 2.
The least value taken by (x 4)2 is 0, so the least value taken by

(x 4)2 + 2 is 2. This occurs when x 4 = 0, that is, when x = 4.
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Unit 3
Functions
So the parabola has vertex (4, 2). Also, since the expression
1
2
4 (x 4) + 2 is always positive, the parabola has no x-intercepts. Its
y-intercept is 6.
The endpoints of the domain of f are 5 and 7. We have
f (5) =
f (7) =
1
4
1
4
52 2 5 + 6 =
72 2 7 + 6 =
9
4
17
4 .
and
1 2
So the points (5, 94 ) and (7, 17
4 ) lie on the graph of y = 4 x 2x + 6.
These features give the following graph.

y
(0; 6)
(7; 17
4 )
(4; 2)
(5; 94 )
x
Erase the parts of the graph that dont lie between the points
(5, 94 ) and (7, 17
4 ) (or draw a new graph).
So the graph of f is as follows.
y
y = 41 x2 2x + 6
(5 x < 7)
(7; 17
4 )
(5; 94 )
x
With a little practice, you should be able to sketch the graph of a function
like the one in Example 1 without having to sketch a larger graph rst. Its
straightforward to do this for a simple graph, such as a straight line.
220
Activity 12 Sketching the graphs of functions whose domains are

not the largest sets of numbers for which their rules are applicable
Sketch the graphs of the following functions.
(a) f (x) = 3 2x (1 < x < 4)
(b) f (x) = 21 x2 2x 5
(x 5)
You can use the graph of a function to visualise its domain on the
horizontal axis. The domain consists of all the possible input numbers of
the function, that is, all points on the horizontal axis that lie directly
below or above a point on the graph, as illustrated in Figure 16.
y
Figure 16 The domain of a function marked on the horizontal axis
Activity 13
Identifying the domains of functions
Write down the domains of the functions whose graphs are shown below,
using interval notation. All the endpoints of the intervals involved are
integers, and in part (b) the graph continues indenitely to the left and
right.
(a)
(b)
4 x
5 4 3 2 1
2 x
As youve seen, a function has exactly one output number for every input
number. So if you draw the vertical line through any number in the
domain of a function on the horizontal axis, then it will cross the graph of
221
Unit 3
Functions
the function exactly once, as illustrated in Figure 17(a). If you can draw a
vertical line that crosses a curve more than once, then the curve isnt the
graph of a function. For example, the curve in Figure 17(b) isnt the graph
of a function.
y
x
(a)
x
(b)
Figure 17 (a) The graph of a function (b) a curve that isnt the graph of
a function
Activity 14
Identifying graphs of functions
Which of the following diagrams are the graphs of functions?

(a)
(b)
(c)
y
x
(e)
(h)
y
x
(k)
y
x
(j)
(g)
222
(f)
(i)
(d)
(l)
y
x
y
x
Increasing and decreasing functions

Figure 18 shows the graph of a function with domain [1, 9]. As x
increases, the graph rst slopes up, then slopes down, then slopes up
again. It changes from sloping up to sloping down when x = 2, and it
changes from sloping down to sloping up again when x = 6.
y
6
5
4
3
2
1
3 21 1
1 2 3 4 5 6 7 8 9 10 11 12 x
Figure 18 The graph of a function

To express these facts about the function f , we say that f is increasing on
the interval [1, 2], decreasing on the interval [2, 6], and increasing again
on the interval [6, 9]. Here are the formal denitions of these terms. The
denitions are illustrated in Figure 19.
Functions increasing or decreasing on an interval

A function f is increasing on the interval I if for all values x1 and
x2 in I such that x1 < x2 ,
f (x1 ) < f (x2 ).
A function f is decreasing on the interval I if for all values x1 and
x2 in I such that x1 < x2 ,
f (x1 ) > f (x2 ).
(The interval I must be part of the domain of f .)
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Unit 3
Functions
y
y = f (x)
f (x2 )
f (x1 )
y = f (x)
f (x1 )
f (x2 )
I
x
x 1 x2
y
4
3
2
1
3 2 1
(a)
x1 x2
(b)
Figure 19 (a) A function increasing on an interval I (b) a function

decreasing on an interval I
1 2 3 x
Figure 20 The graph of the

function f (x) = x2
For example, the function f (x) = x2 , whose graph is shown in Figure 20, is
decreasing on the interval (, 0] and increasing on the interval [0, ).
Activity 15
Identifying increasing functions
Which of the following graphs show functions that are increasing on their
whole domains?
(a)
(b)
(c)
y
x
(d)
y
x
y
x
1.5 Image sets of functions

Remember that the image set of a function is the set consisting of all the
values in its codomain that occur as output numbers. For example, the
image set of the function f (x) = x2 is the interval [0, ), because all
non-negative numbers occur as output numbers of this function, but no
negative numbers do. You saw in the last subsection that if you have a
graph of a function, then you can visualise the domain of the function on
the horizontal axis, as illustrated in Figure 21(a). In the same way, you
can visualise the image set of the function on the vertical axis.
224
The image set consists of all the possible output numbers, that is, all the
points on the vertical axis that lie directly to the right or left of a point on
the graph, as illustrated in Figure 21(b).
y
domain
image set
x
x
(b)
(a)
Figure 21 (a) The domain of a function marked on the horizontal axis

(b) the image set marked on the vertical axis
So you can use the graph of a function to help you nd its image set, as
demonstrated in the next example. Remember that a play button icon in
the margin next to a worked example indicates that a tutorial clip is
available for the example.
Example 2
Finding the image set of a function
Find the image set of the function

f (x) = x2 + 6x + 14
(6 < x < 2).
Solution
Obtain a sketch, plot or computer plot of the graph of the
function. Remember to stop the graph at the endpoints of its
domain, and to mark the resulting ends of the graph with solid or
hollow dots, as appropriate. Theres no need to nd the
intercept(s).
The parabola is u-shaped. Completing the square gives
f (x) = x2 + 6x + 14
= (x + 3)2 9 + 14
= (x + 3)2 + 5.
The least value taken by (x + 3)2 is 0, so the least value taken by
(x + 3)2 + 5 is 5. This occurs when x + 3 = 0, that is, when
x = 3.
So the vertex is (3, 5).
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Unit 3
Functions
Also
f (6) = (6)2 + 6 (6) + 14 = 14 and
f (2) = 22 + 6 2 + 14 = 30.
So the graph stops at the points (6, 14) and (2, 30), both of which
are excluded.
These features give the following graph.
y
30
(2; 30)
y = x2 + 6x + 14
( 6; 14)
( 3; 5)
5
x
The graph shows that the smallest value in the image set is the
y-coordinate of the vertex, and that the image set contains all the
values larger than this number, up to but not including f (2).
The graph shows that the image set of f is [5, 30).
You might have expected that if the domain of a function f is the interval
(6, 2), then its image set is the interval (f (6), f (2)). Example 2 shows
that this isnt necessarily true.
Activity 16
Finding image sets of functions
Find the image sets of the following functions.

(a) f (x) = x2 + 10x 24
(3 x < 6)
(b) f (x) = 2 2x (2 < x < 0)

(c) f (x) = x2 1
(d) f (x) = 1/x2
Hint for part (d): try to work out the answer by thinking about what
output numbers are possible for this function. If youre still not sure, try
plotting the graph of the function on a computer.
226
1.6 Some standard types of functions

As mentioned earlier, its useful to become familiar with some standard
types of functions and their graphs. Youll meet some types of functions in
this subsection, and further types later in the unit and in Unit 4. If you
study mathematics beyond this module, then youll meet many more types
of functions.
Linear functions
First consider any function whose rule is of the form
f (x) = mx + c,
where m and c are constants. Its graph is the graph of the equation
y = mx + c, which, as you saw in Unit 2, is the straight line with
gradient m and y-intercept c. For this reason, any function of the form
above is called a linear function.
Figure 22 shows the graphs of some linear functions.
y
3
2
1
1 1
2
3
y
3
2
1
1 2 3 x
(a)
3 21 1
y
3
2
1
1 2 3 x
2
3
(b)
3 21 1
1 2 3 x
2
3
(c)
Figure 22 The graphs of the linear functions (a) f (x) = 2x 3

(b) f (x) = 12 x + 1 (c) f (x) = 3
A linear function whose rule is of the form
f (x) = c,
where c is a constant, is called a constant function. Its graph is a
horizontal line. For example, the function f (x) = 3, whose graph is shown
in Figure 22(c), is a constant function.
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Unit 3
Functions
Quadratic functions
From what you saw in Unit 2, you also know that the graph of any
function of the form
f (x) = ax2 + bx + c,
(2)
where a, b and c are constants with a .= 0, is a parabola. You saw how to

nd various features of the parabola, such as its vertex and intercepts,
from the values of a, b and c. Any function whose rule is of form (2) is
called a quadratic function. The graphs of some quadratic functions are
shown in Figure 23.
y
3
2
1
21 1
2
3
(a)
y
3
2
1
y
3
2
1
1 2 3 x
5 4 3 21 1
1 2 3 x
2
3
(b)
1 1
2
3
1 2 3 x
(c)
Figure 23 The graphs of the functions (a) f (x) = x2

(b) f (x) = 15 x2 + 25 x 3 (c) f (x) = 4x2 + 8x 1
Polynomial functions
Linear functions and quadratic functions are particular types of polynomial
functions. Here are some more polynomial functions:
f (x) = 2x4 5x3 + x2 + 2x 2
g(x) = x3
h(x) = 71 x7 + 13 x6 + x5 52 x4 43 x3 + 4x2 + 1.
In general, if an expression is a sum of nitely many terms, each of which
is of the form axn where a is a number and n is a non-negative integer,
then the expression is called a polynomial expression in x. If the
right-hand side of the rule of a function is a polynomial expression in x,
then the function is called a polynomial function.
The word polynomial appears to be a hybrid word meaning many
names that is a mixture of Greek and Latin.
Polly, no meal
228
Note that the terms of a polynomial expression

must all have powers that
are non-negative integers; for example, x (which is the same as x1/2 ) is

not a polynomial expression.
The highest power of the variable x in a polynomial expression or function

is called the degree of the polynomial expression or function. For
example, the highest power of x in the rule of the polynomial function f
above is x4 , so the degree of this polynomial function is 4. Similarly, the
polynomial functions g and h above have degrees 3 and 7, respectively.
Quadratic functions are polynomial functions of degree 2, since the highest
power of x in the rule of a quadratic function is x2 .
Linear functions are polynomial functions of degree 1, 0 or no degree at all.
If a linear function is of the form f (x) = ax + b where a .= 0, then the
highest power of x is x1 , so the degree is 1. If it is of the form f (x) = c
where c .= 0, then the highest power of x is x0 (since the function can be
expressed as f (x) = cx0 ), so the degree is 0. The particular linear function
f (x) = 0 is usually regarded as not having a degree at all (or sometimes as
having degree ), for technical reasons.
Polynomial functions of degrees 3, 4 and 5 are called cubic, quartic and
quintic functions, respectively.
Figure 24 shows the graphs of the three polynomial functions above.
y
3
2
1
y
3
2
1
21 1
2
3
4
(a)
1 2 3 x
21 1
2
3
4
(b)
y
3
2
1
1 2 x
3 21 1
1 2 3 x
2
3
4
(c)
Figure 24 The graphs of the functions

(a) f (x) = 2x4 5x3 + x2 + 2x 2 (b) g(x) = x3
(c) h(x) = 17 x7 + 13 x6 + x5 52 x4 43 x3 + 4x2 + 1
Activity 17
Investigating graphs of polynomial functions
Use the module computer algebra system (CAS) to experiment with

plotting the graphs of some polynomial functions, to obtain a general idea
of the sorts of shapes that they have.
For example, you might like to try plotting the following polynomial
functions: f (x) = x3 , f (x) = x4 , f (x) = x4 4x3 , f (x) = x5 4x3 .
In Activity 17 you should have seen evidence of the following. Every

polynomial function has a graph thats a smooth, unbroken curve (or a
229
Unit 3
Functions
straight line; we normally consider a straight line to be a particular type of
curve). The graph often has a wiggly section, but (unless the function is
a constant function) if you trace your pen tip along the graph towards the
right, then eventually the wiggles stop and your pen either keeps moving
uphill or keeps moving downhill. The same happens if you trace your
pen tip towards the left.
In fact, the graph of every polynomial function (with domain R) that isnt
a constant function tends to innity or tends to minus innity at the
left and right. In other words, no matter how large a positive number you
choose, as you trace your pen tip along the graph, the y-values of the
graph either eventually exceed your chosen number (if the graph tends to
innity) or are eventually less than the negative of your chosen number (if
the graph tends to minus innity).
You can tell whether the graph of a polynomial function tends to innity
or tends to minus innity at each end by looking at the term in its rule
that has the highest power of x. This term is called the dominant term,
because for large values of x, the value taken by the dominant term
outweighs (dominates) the sum of the values taken by all the other terms.
For example, the dominant term in the rule
f (x) = 2x4 5x3 + x2 + 2x 2
is 2x4 . If the dominant term has a plus sign and
contains an even power of x, then the graph tends to innity at both

ends
contains an odd power of x, then the graph tends to minus innity at

the left and to innity at the right.
If the dominant term has a minus sign, then similar facts hold, but with
innity replaced by minus innity and vice versa.
To see examples of these facts, look at the graphs in Figure 24 above, and
at the graphs that you plotted in Activity 17.
The modulus function

All the functions that youve met so far have graphs that are smooth
curves. The modulus function has a graph thats smooth except at one
point, where it turns a corner!
As you saw in Unit 2, the modulus of a real number (also known as its
magnitude or absolute value) is its distance from zero, or its value
without its sign. For example, the modulus of 3 is 3, and the modulus of
3 is also 3. The modulus of a real number x is denoted by |x|. So, for
example, |3| = 3.
The modulus function is
f (x) = |x|.
230
It follows from the denition of modulus that

,
x, if x 0,
|x| =
x, if x < 0.
So the graph of the modulus function is the same as the graph of y = x
when x 0, and the same as the graph of y = x when x < 0. Its shown
in Figure 25. It has a corner at the origin, and the image set is [0, ).
y
3
2
1
1 2 3 x
3 2 1
Figure 25 The graph of the modulus function, f (x) = |x|
The reciprocal function

Remember that if x is any non-zero number, then the reciprocal of x is
1/x. The reciprocal function is the function
1
.
x
Its domain consists of all real numbers except 0. That is, its domain is the
set (, 0) (0, ). The graph of the reciprocal function is shown in
Figure 26.
f (x) =
y
4
3
2
1
4 3 21 1
1 2 3 4 x
2
3
4
Figure 26 The graph of the reciprocal function, f (x) = 1/x

The graph consists of two separate pieces, each of which gets closer and
closer to one of the coordinate axes at each end. To see why this is, rst
think about the piece of the graph to the right of the y-axis, that is, the
piece that shows the values of 1/x for positive values of x. Its shape can be
explained as follows. As x gets larger and larger, the value of 1/x gets
closer and closer to zero. Similarly, as x gets closer and closer to zero, the
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Unit 3
Functions
value of 1/x gets larger and larger. The shape of the piece of the graph to
the left of the y-axis, for negative values of x, has a similar explanation:
you might like to think it through.
If a curve has the property that, as you trace your pen tip along it further
and further from the origin, it gets arbitrarily close to a straight line, then
that line is called an asymptote of the curve. The phrase arbitrarily
close here has the following meaning: no matter how small a distance you
choose, if you trace your pen tip along the curve far enough, then
eventually the curve lies within that distance of the line, and stays within
that distance of the line.
So the coordinate axes are asymptotes of the graph of the reciprocal
function. Asymptotes are often drawn as dashed lines on graphs, when
they dont coincide with the coordinate axes.
The word asymptote comes from a Greek word meaning not
coinciding, used to describe a straight line that a curve approaches
arbitrarily closely but doesnt meet.
Rational functions
The reciprocal function and all polynomial functions are particular
examples of rational functions. In general, a rational function is a
function whose rule is of the form
p(x)
f (x) =
,
q(x)
where p and q are polynomial functions. If q is a constant function, then f
is a polynomial function, and if p(x) = 1 and q(x) = x, then f is the
reciprocal function. Here are some more examples of rational functions:
f (x) =
x2 + 1
,
2x + 4
f (x) =
2x2 6x 8
,
x2 x 6
f (x) =
7x + 5
.
x2 + 1
The graphs of these rational functions are shown in Figure 27. The dashed
lines are asymptotes. The third graph has the x-axis as an asymptote.
232
y
8
6
4
2
8 6 42 2
New functions from old functions
y
8
6
4
2
2 4 6 8 x
4
6
8
(a)
Figure 27 The graphs of (a) f (x) =
8 6 42 2
y
8
6
4
2
2 4 6 8 x
8 6 42 2
4
6
8
(b)
2 4 6 8 x
4
6
8
(c)
x2 + 1
2x2 6x 8
(b) f (x) = 2
2x + 4
x x6
7x + 5
x2 + 1
Every rational function has a graph that consists of one or more pieces,
each of which is a smooth curve. The graphs of many rational functions
have asymptotes, which can be horizontal, vertical or slant. For example,
the graph in Figure 27(a) has one vertical asymptote and one slant
asymptote.
(c) f (x) =
A detailed study of the graphs of rational functions is beyond the scope of

this module, but you can learn more about them in the follow-on module
to this one, Essential mathematics 2 (MST125).
2 New functions from old functions

In this section, youll learn how to use your knowledge about the graphs of
a few functions to deduce facts about the graphs of many more functions.
For example, you can use the graph of the function f (x) = x2 , which is
shown in Figure 28, to deduce the appearance of the graphs of the
functions g(x) = x2 + 1 and h(x) = 3x2 . The rules of these functions are
obtained from the rule for f simply by adding 1 and by multiplying by 3,
respectively.
2.1 Translating the graphs of functions
y
4
3
2
1
3 2 1
1 2 3 x
Figure 28 The graph of

f (x) = x2
Lets start by considering what happens to the graph of a function when

you add a constant to the right-hand side of its rule. You can see some
instances of this in the next activity.
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Unit 3
Functions
Activity 18 Investigating graphs of equations of the form

y = f (x) + c
Open the Translating and scaling graphs applet. Make sure that the
y = f (x) + c option is selected, and that the original function is f (x) = x2 .
Change the value of c to display the graph of y = x2 + c for various values
of c, and observe how the new graphs are related to the original graph. In
particular, notice the eect of positive values of c, and the eect of
negative values of c.
Now change the original function to a dierent function of your choice, and
repeat the process above.
The eects that you saw in Activity 18 are examples of translations of

graphs. Translating a shape means sliding it to a dierent position,
without rotating or reecting it, or distorting it in any way.
You saw that if you add any constant c to the right-hand side of the rule of
a function, then its graph is translated vertically. Specically, its
translated up by c units (the translation is down if c is negative). This is
because when you add the constant c, the y-values all increase by c.
Theres another fairly simple change that you can make to the rule of a
function, which causes its graph to be translated horizontally. To do this,
you replace each occurrence of the input variable x in the right-hand side
of the rule of the function by an expression of the form x c, where c is a
constant. For example, if you start with the function f (x) = x2 , then you
can replace x by x 3, say, to obtain the new function g(x) = (x 3)2 .
Youre asked to investigate changes of this sort in the next activity.
Activity 19 Investigating graphs of equations of the form

y = f (x c)
In the Translating and scaling graphs applet, select the y = f (x c)
option, and make sure that the original function is f (x) = x2 .
Change the value of c to display the graph of y = (x c)2 for various
values of c, and observe how the new graphs are related to the original
graph. In particular, notice the eect of positive values of c, and the eect
of negative values of c.
Now change the original function to a dierent function of your choice, and
repeat the process above.
In Activity 19 you saw that if you replace every occurrence of the input
variable x in the right-hand side of the rule of a function by the expression
x c, where c is a constant, then the graph of the function is translated
horizontally. Specically its translated to the right by c units (the
234
translation is to the left if c is negative). For example, if you replace x by

x 3, then the graph is translated to the right by 3 units (here c = 3).
Similarly, if you replace x by x + 3, then the graph is translated to the left
by 3 units (here c = 3).
To see why this happens, lets think about what happens when you
translate the graph of a particular function to the right by c units (where c
might be positive, negative or zero).
For example, Figure 29 shows the graph of the equation y = x2 (in black),
and the graph thats obtained by translating it to the right by 3 units (in
green). Lets try to work out the equation of this second graph. To do this,
we have to nd a relationship between x and y that holds for every point
(x, y) on the second graph. Now, whenever the point (x, y) lies on the
second graph, the point (x 3, y) lies on the original graph, so the second
coordinate, y, is the square of the rst coordinate, x 3. Hence the
following equation holds:
y = (x 3)2 .
This equation expresses a relationship between x and y, so it is the
equation of the second graph.
y
3
(x 3; y)
(x; y)
2
1
2 1
Figure 29 The graph of y = x2 , and the graph obtained by translating it

by 3 units to the right
More generally, consider any function f , and suppose that you translate its
graph to the right by c units, as illustrated in Figure 30 in a case where
c > 0. Then whenever the point (x, y) lies on the second graph, the point
(x c, y) lies on the original graph, so the following equation holds:
y = f (x c).
235
Unit 3
Functions
So this equation is the equation of the second graph.
y
(x c; y)
(x; y)
Figure 30 The graph of an equation of the form y = f (x), and the graph
obtained by translating it by c units to the right, where c > 0
This reasoning explains the eects that you saw in Activity 19.
Heres a summary of what youve seen so far in this subsection.
Translations of graphs
Suppose that f is a function and c is a constant. To obtain the graph
of:
y = f (x) + c, translate the graph of y = f (x) up by c units (the

translation is down if c is negative)
y = f (x c), translate the graph of y = f (x) to the right by c

units (the translation is to the left if c is negative).
These eects are illustrated in Figure 31.

y
y = f (x c); c < 0
y = f (x) + c; c > 0
y = f (x)
y = f (x c); c > 0
x
y = f (x) + c; c < 0
Figure 31 Pieces of graphs of equations of the form y = f (x) + c and

y = f (x c)
236
Activity 20
Understanding translations of graphs
You saw the graph of y = 1/x in the previous section, and its repeated in
Figure 32. Using this graph, and without using a computer, match up the
equations below with their graphs.
1
1
1
1
(a) y =
(b) y = 2
(c) y = + 2
(d) y =
x2
x
x
x+2
Graphs:
y
4
2
B
2 4 x
422
4
y
4
2
422
y
4
2
C
2 4 x
422
D
2 4 x
2
4 2
2
4 x

1
y=
x
y
4
2
422
2 4 x
Now suppose that you change the rule of a function in such a way that its
graph is translated horizontally, and then you change the rule of the new
function in such a way that its graph is translated vertically. The nal
result is that the graph of the original function is translated both
horizontally and vertically. For example, consider the equation y = x2 ,
whose graph is shown in Figure 33(a). If you replace x by x 4, then you
obtain the equation
y = (x 4)2 ,
and the graph is translated to the right by 4 units, as shown in
Figure 33(b). If you now add the constant 2, then you obtain the equation
y = (x 4)2 + 2,
and the original graph is now translated to the right by 4 units and up by
2 units, as shown in Figure 33(c).
y
8
6
4
2
4 2
(a)
y = (x 4)2
y
8
6
4
2
y = x2
2 4 6 x
4 2
(b)
2 4 6 x
y = (x 4)2 + 2
y
8
6
4
2
4 2
2 4 6 x
(c)
Figure 33 The graphs of (a) y = x2 (b) y = (x 4)2 (c) y = (x 4)2 + 2

237
Unit 3
Functions
In general, suppose that you start with an equation y = f (x). If you rst
replace x by x c, where c is a constant, then you obtain the equation
y = f (x c), and the graph is translated to the right by c units. If you
then add the constant d to the right-hand side, then you obtain the
equation
y = f (x c) + d,
and the original graph is translated to the right by c units and up by d
units.
In fact, the order in which you make the two changes doesnt matter. One
way to see this is to think about the situation geometrically. If you
translate a graph to the right by c units and then up by d units, then the
overall eect will be the same as if you had translated it up by d units and
then to the right by c units. You can also conrm it algebraically, as
follows. Suppose that you carry out the two changes to the equation
y = f (x) in the opposite order to the order used above. Adding d to the
right-hand side of the equation y = f (x) gives the equation y = f (x) + d,
and then replacing x in this equation by x c gives the nal equation
y = f (x c) + d, which is the same as the nal equation obtained above.
y
4
2
4 2
2
4 x
Activity 21 Understanding successive horizontal and vertical

translations of graphs
You saw the graph of y = |x| in the previous section, and its repeated in
(a) y = |x 2| + 1

y = |x|
(c) y = |x 2| 1
(d) y = |x + 2| 1
Graphs:
y
4
2
42 2
4
238
(b) y = |x + 2| + 1
y
4
2
2 4 x 42 2
4
y
4
2
2 4 x 42 2
4
y
4
2
2 4 x 42 2
4
2 4 x
2.2 Scaling the graphs of functions vertically

In this subsection youll see what happens to the graph of a function when
you multiply the right-hand side of its rule by a constant. The new
function that you obtain is called a constant multiple of the original
function. For example, the function g(x) = 3x2 is a constant multiple of
the function f (x) = x2 .
Activity 22
Investigating graphs of equations of the form y = cf (x)
In the Translating and scaling graphs applet, select the y = cf (x) option,
and make sure that the original function is f (x) = x2 1.
Change the value of c to display the graph of y = c(x2 1) for various
values of c, and observe how the new graphs are related to the original
graph. In particular, notice the eect of positive values of c, and the eect
of negative values of c. Also notice the eect of values of c such that
|c| < 1, and the eect of values of c such that |c| > 1.
Now change the original function to y = x3 , and repeat the process above.
If you wish, also try another function of your choice as the original
function.
The eects that you saw in Activity 22 are called vertical scalings.
Scaling a graph vertically by a factor of c means the following.
If c is positive, then move each point on the graph vertically, in the

direction away from the x-axis, until its c times as far from the x-axis
as it was before.
If c is negative, then move each point on the graph vertically, in the

direction away from the x-axis, until its |c| times as far from the
x-axis as it was before, and then reect it in the x-axis.
If c is zero, then move each point on the graph vertically until it lies
on the x-axis.
(In each of the rst two cases, if |c| is less than 1, then each point is
actually moved closer to the x-axis than it was before.)
Informally, when you scale a graph vertically by a factor of c, you stretch
or squash it parallel to the y-axis (depending on whether |c| is greater than
or less than 1), and if c is negative, you also reect it in the x-axis.
In Activity 22 you should have seen evidence of the following.
Vertical scalings of graphs

Suppose that c is a constant. To obtain the graph of y = cf (x),
scale the graph of y = f (x) vertically by a factor of c.
239
Unit 3
Functions
These eects are illustrated in Figure 35. They occur because when you
multiply the right-hand side of the rule of a function by the constant c, the
y-value corresponding to each x-value is multiplied by c.
y
y = cf (x); c > 1
y = f (x)
y = cf (x); 0 < c < 1
x
y = cf (x); 1 < c < 0
y = f (x)
y = cf (x); c < 1
Figure 35 Pieces of graphs of equations of the form y = cf (x)

Notice in particular what happens when c = 1. For any function f , the
graph of y = f (x) is the same shape as the graph of y = f (x), but
reected in the x-axis. The function that results from multiplying the
right-hand side of the rule of a function f by 1 is called the negative of
the function f .
y
3
2
1
3 211
2
3
1 2 3 x

y = x3
Activity 23
Understanding vertical scalings of graphs
You saw the graph of y = x3 in Subsection 1.6, and its repeated in

(a) y = 2x3
(b) y = 12 x3
(c) y = x3
(d) y = 21 x3
Graphs:
y
3
2
1
3 211
2
3
240
1 2 3 x
y
3
2
1
3 211
2
3
1 2 3 x
y
3
2
1
3 211
2
3
1 2 3 x
y
3
2
1
3 211
2
3
1 2 3 x
You can combine vertical scalings of graphs with vertical and/or horizontal
translations of graphs, in the same way that horizontal and vertical
translations of graphs were combined in the previous subsection. For
example, suppose that you start with the equation y = x3 , whose graph is
shown in Figure 37(a). If you multiply the right-hand side of this equation
by 4, then you obtain the equation y = 4x3 , and the graph is scaled
vertically by a factor of 4, as illustrated in Figure 37(b). If you then add
the constant 1 to the right-hand side of this new equation, then you obtain
the nal equation y = 4x3 + 1, and the original graph is rst scaled
vertically by a factor of 4, then translated up by 1 unit, as illustrated in
Figure 37(c).
y
4
2
42
2
4
y
4
2
y = x3
2 4 x
(a)
42
2
4
y
4
2
y = 4x3
2 4 x
(b)
42
2
4
y = 4x3 + 1
2 4 x
(c)
Figure 37 The graphs of (a) y = x3 (b) y = 4x3 (c) y = 4x3 + 1

When you combine vertical scalings and translations in this way, the order
in which you carry out the changes to the rule of the function does often
matter. Dierent orders can give dierent results. For example, suppose
that you start with the equation y = x3 , as above, and you make the same
two changes as above, but in the opposite order. Adding the constant 1 to
the right-hand side of the equation y = x3 gives the intermediate equation
y = x3 + 1, and the graph is translated up by 1 unit, as illustrated in
Figure 38(b). Then multiplying the right-hand side by 4 gives the nal
equation y = 4(x3 + 1), and the original graph is rst translated up by 1
unit and then scaled vertically by a factor of 4, as illustrated in
Figure 38(c). You can see that the nal equation and graph are dierent
from those obtained above.
y
4
2
42
2
4
(a)
y
4
2
y = x3
2 4 x
42
2
4
(b)
y
4
y = 4(x3 + 1)
2
y = x3 + 1
2 4 x
42
2
4
2 4 x
(c)
Figure 38 The graphs of (a) y = x3 (b) y = x3 + 1 (c) y = 4(x3 + 1)

241
Unit 3
Functions
In general, you can make any number of successive changes to the rule of a
function to scale and translate its graph in various ways, but you have to
be careful about the order in which you carry out the changes. Sometimes
the order matters, and sometimes it doesnt.
If you change the rule of a function to carry out a horizontal translation, a
vertical translation and a vertical scaling, then you can make the changes
in any order, except that the changes for the vertical translation and
vertical scaling must be made in the correct order relative to each other.
Example 3
of graphs
Understanding successive scalings and translations
For each of the following functions, describe how you could obtain its
graph by applying scalings and translations to the graph of the
function f (x) = x3 .
(a) g(x) = 12 (x + 3)3
(b) h(x) = 21 (x + 3)3 2
Solution
(a)
Try to work out how the equation y = 12 (x + 3)3 is obtained

from the equation y = x3 by making two or more changes of the
types that youve seen, one after another.
Consider the equation y = x3 . If you multiply the right-hand side
by 12 , you obtain the equation y = 12 x3 . If you then replace x by
x + 3, you obtain the equation y = 21 (x + 3)3 .
So the graph of the equation y = 12 (x + 3)3 is obtained by
starting with the graph of y = x3 , scaling it vertically by the
factor 21 , and then translating it to the left by 3 units.
(b)
Use the same method as in part (a). Here you can recognise
that the given equation is obtained from the equation in part (a)
by making a simple change.
The equation y = 12 (x + 3)3 2 is obtained from the nal
equation in part (a) by adding 2 to the right-hand side.
So the graph of the equation y = 21 (x + 3)3 2 is obtained by
starting with the graph of y = x3 , scaling it vertically by the
factor 21 , then translating it to the left by 3 units, and nally
translating it down by 2 units.
242
Figure 39 shows the graph of the function f (x) = x3 and the results of
applying the scalings and translations in Example 3 to this graph.
y
4
2
42
2
4
y = 12 x3
y
4
2
y = x3
2 4 x
(a)
42
2
4
y = 12 (x + 3)3 2
y
4
2
y = 21 (x + 3)3
y
4
2
2 4 x
42
2
4
(b)
2 4 x
(c)
42
2
4
2 4 x
(d)
Figure 39 Three graphs obtained by scaling and/or translating the graph

of y = x3
Activity 24
graphs
For each of the following functions, describe how you could obtain its
graph by applying scalings and translations to the graph of the function
f (x) = |x| (which is shown in Figure 40).
(a) g(x) = 2|x| + 3
(b) h(x) = 2|x + 2| + 3
(c) j(x) = 21 |x 3| 4
(d) k(x) = |x 1| + 1
In the nal activity of this subsection, youll see how the new ideas that
youve met can give you a deeper understanding of the shapes of the
graphs of quadratic functions.
Activity 25
y
3
2
1
Understanding successive scalings and translations of
3 21 1
2
3

y = |x|
y
4
3
2
1
Understanding the graph of a quadratic function
Consider the quadratic function f (x) = 2x2 + 12x + 19.

(a) Complete the square in the quadratic expression on the right-hand
side.
(b) Hence describe how you could obtain the graph of this function by
applying scalings and translations to the graph of the function
f (x) = x2 (which is shown in Figure 41).
1 2 3 x
3 2 1
1 2 3 x

y = x2
243
Unit 3
Functions
You can see that the method that you were asked to use in Activity 25 can
be applied to any quadratic function. You just need to start by completing
the square in the quadratic expression that forms the right-hand side of its
rule.
This tells you the following enlightening fact: the graph of any quadratic
function is the same basic shape as the graph of y = x2 , but scaled
vertically, and then translated horizontally and/or vertically.
2.3 Scaling the graphs of functions horizontally

In the rst activity of this subsection, youre asked to investigate a change
to the rule of a function that results in its graph being scaled horizontally.
The change is that you replace each occurrence of the input variable x in
the right-hand side of the rule of the function by an expression of the form
x/c, where c is a constant.
Activity 26
Investigating graphs of equations of the form y = f
*x(
c
In the Translating and scaling graphs applet, select the y = f (x/c) option,
and make sure that the original function is y = x3 .
Change the value of c to display the graph of y = (x/c)3 for various
non-zero values of c, and observe how the new graphs are related to the
original graph. In particular, notice the eect of positive values of c, and
the eect of negative values of c. Also notice the eect of values of c such
that |c| < 1, and the eect of values of c such that |c| > 1.
Now change the original function to y = x2 , and repeat the process above.
If you wish, also try another function of your choice as the original
function.
The eects that you saw in Activity 26 are called horizontal scalings.
Scaling a graph horizontally by a factor of c means the following.
If c is positive, then move each point on the graph horizontally, in the

direction away from the y-axis, until its c times as far from the y-axis
as it was before.
If c is negative, then move each point on the graph horizontally, in the

direction away from the y-axis, until its |c| times as far from the
y-axis as it was before, and then reect it in the y-axis.
If c is zero, then move each point on the graph horizontally until it lies
on the y-axis.
(In each of the rst two cases, if |c| is less than 1, then each point is
actually moved closer to the y-axis than it was before.)
244
Informally, when you scale a graph horizontally by a factor of c, you stretch

or squash it parallel to the x-axis (depending on whether |c| is greater than
or less than 1), and if c is negative, you also reect it in the y-axis.
In Activity 26 you should have seen evidence of the following.
Horizontal scalings of graphs

Suppose
( c is a non-zero constant. To obtain the graph of
* xthat
,
y=f
c
scale the graph of y = f (x) horizontally by a factor of c.
These eects are illustrated in Figure 42.
y=f
y=f
x
; 1 < c < 0
c
y=f
y = f ( x)
x
; c < 1
c
x
; 0<c<1
c
y = f (x)
y=f
x
; c>1
c
Figure 42 Pieces of graphs of equations of the form y = f (x/c)

To see why these eects occur, lets suppose that you scale the graph of a
particular function horizontally by a factor of c (where c might be positive
or negative), and lets try to work out how this aects the rule of the
function.
For example, Figure 43 shows the graph of the equation y = x2 (in black),
and the graph thats obtained by scaling it horizontally by a factor of 3 (in
green). Lets try to work out the equation of this second graph.
To do this, we have to nd a relationship between x and y that holds for
every point (x, y) on the second graph. Now whenever the point (x, y) lies
on the second graph, the point (x/3, y) lies on the original graph, so the
following equation holds:
* x (2
.
y=
3
245
Unit 3
Functions
This equation expresses a relationship between x and y, so it is the
equation of the second graph.
y
x
;y
3
(x; y)
2
1
6 5 4 3 2 1
Figure 43 The graph of y = x2 , and the graph obtained by scaling it

horizontally by a factor of 3
More generally, consider any function f , and suppose that you scale its
graph horizontally by a factor of c, as illustrated in Figure 44 in a case
where c > 1. Then whenever the point (x, y) lies on the second graph, the
point (x/c, y) lies on the original graph, so the following equation holds:
*x(
y=f
.
c
So this equation is the equation of the second graph.
y
x
;y
c
(x; y)
Figure 44 The graph of an equation y = f (x), and the graph obtained by

scaling it horizontally by a factor of c, where c > 1
Notice in particular what happens when c = 1. For any function f , the
graph of y = f (x) is the same shape as the graph of y = f (x), but
reected in the y-axis.
246
For convenience, the two facts that youve seen (in this subsection and the
previous one) about reections of graphs in the coordinate axes are
summarised below, and illustrated in Figure 45.
Reections of graphs in the coordinate axes

To obtain the graph of y = f (x), reect the graph of y = f (x) in the
x-axis.
To obtain the graph of y = f (x), reect the graph of y = f (x) in the
y-axis.
y
y = f (x)
y = f ( x)
y = f (x)
x
y = f (x)
(a)
(b)
Figure 45 Pieces of graphs of equations of the form (a) y = f (x) and

(b) y = f (x)
In the nal two activities of this section, youll need to put together all the
facts and skills about the eects of changing the equations of graphs that
youve learned in this section.
247
Unit 3
Functions
y
4
2
4 2 2
Activity 27 Understanding horizontal and vertical translations and

scalings of graphs
2 4 x
Figure
46 The graph of
y= x
The graph of the function f (x) = x is shown in Figure 46. Using this
graph, and without using a computer, match up the equations below with
their graphs.
(b) y = x
(c) y = 2 x 2
(a) y = x
(d) y = 12 x + 2
(e) y = 12 x
(f) y = x + 2
(g) y = x
(h) y = 12 x + 2
(i) y = 2 x + 2
Graphs:
y
4
2
4 2 2
B
2 4 x
4 2 2
E
2 4 x
248
H
2 4 x
2 4 x
F
2 4 x
y
4
2
4 2 2
2 4 x
y
4
2
4 2 2
y
4
2
4 2 2
y
4
2
4 2 2
2 4 x
y
4
2
4 2 2
y
4
2
4 2 2
y
4
2
I
2 4 x
y
4
2
4 2 2
4
2 4 x
More new functions from old functions
Activity 28 Understanding horizontal and vertical translations and

scalings of graphs, again
By considering graphs of functions that you met in Subsection 1.6, and
without using a computer, draw sketch graphs of the following functions.
In particular, mark the values of the intercepts, and in part (c) draw the
asymptotes, and label the vertical one with its equation. (You met the
idea of an asymptote on page 232.)
1
(a) f (x) = 2|x| + 1
(b) h(x) = (x 1)3
(c) g(x) =
x+3
Hint: in part (a), think about the graph of y = |x|; in part (b), think
about the graph of y = x3 ; in part (c), think about the graph of y = 1/x.
3 More new functions from old

functions
In this section youll meet some ways in which you can combine the rules
of two or more functions to obtain the rule of a new function. Youll also
see that for some functions theres a related function, called the inverse
function of the original function, which reverses the eect of the original
function.
3.1 Sums, dierences, products and quotients of

functions
In this short subsection youll see how to combine functions by forming
sums, dierences, products and quotients of them. These combinations
mean just what the names suggest.
Suppose that f and g are functions. The sum of f and g has the rule
h(x) = f (x) + g(x).
There are two dierences of f and g, with rules
h(x) = f (x) g(x) and h(x) = g(x) f (x).
The product of f and g has the rule
h(x) = f (x)g(x).
There are two quotients of f and g, with rules
h(x) =
f (x)
g(x)
and h(x) =
g(x)
.
f (x)
249
Unit 3
Functions
For example, if f (x) = x2 and g(x) = x, then the sum of f and g has the
rule h(x) = x2 + x. The domain of each of the combined functions above is
the intersection of the domain of f and the domain of g, with the
additional requirement for the rst quotient of f and g that the numbers x
such that g(x) = 0 are removed, since its not possible to divide by zero,
and a similar additional requirement for the second quotient.
Activity 29 Understanding sums, dierences, products and

quotients of functions
Consider the functions f (x) = 2x 1 and g(x) = x + 3. Find the rules of
the sum, the two dierences, the product and the two quotients of f and g,
and state the domain of each of these functions.
You can form sums and products of three or more functions. For example,
the sum of the functions f (x) = x2 , g(x) = x and h(x) = 1 is the function
s(x) = x2 + x + 1.
There isnt much more to be said about sums, products, dierences and
quotients of functions, at this stage. Its usually not easy to deduce the
shape of the graph of any one of these functions from the shapes of the
graphs of the original functions. These types of combinations of functions
will be important later in the module.
3.2 Composite functions

Theres another useful way to combine two functions to obtain a new
function. This is to apply one function after the other.
For example, suppose that f and g are functions. Consider any value that
lies in the domain of f . If you input this value to the function f , then you
obtain an output value. If this output value is in the domain of the
function g, then you can, in turn, input it to the function g, to obtain a
nal output value. This process is illustrated in Figure 47.
input
value
output
value
nal
output
value
Figure 47 An input value processed by two functions f and g one after

the other
The function whose rule is given by this process, and whose domain is the
largest set of real numbers to which you can apply the process, is called a
composite function, or just composite, of f and g. Its denoted by
g f (the symbol is read as circle). Note that the function thats
applied rst is written second in this notation youll see why shortly.
250
The largest set of real numbers for which you can apply the process is the
set of all numbers in the domain of f such that f (x) lies in the domain of
g. For example, Figure 48 illustrates the process of nding the image of
the number 3 under the composite function g f , where f (x) = x2 and
g(x) = x + 1. It shows that (g f )(3) = 10.
3
10
Figure 48 The image of 3 under g f , where f (x) = x2 and g(x) = x + 1
Activity 30
Understanding composite functions
Suppose that f (x) = x2 and g(x) = x + 1, as in the paragraph above. Find

the image of 5 under the composite function g f .
In general, for any two functions f and g, the process of nding the image
of an input value x under the composite function g f is as shown in
Figure 49.
x
f (x)
g(f (x))
Figure 49 The image of x under g f , for any functions f and g

So a composite function can be dened concisely as follows.
Composite functions
Suppose that f and g are functions. The composite function g f
is the function whose rule is
(g f )(x) = g(f (x)),
and whose domain consists of all the values x in the domain of f such
that f (x) is in the domain of g.
Its important to remember that g f means f followed by g, not the other

way round, as you might at rst expect. To understand why the notation
is this way round, consider the equation in the box above:
(g f )(x) = g(f (x)).
It would be confusing if f and g were in dierent orders on the two sides of
the equation.
251
Unit 3
Functions
The process of forming a composite function from two functions is called
composing the functions. You can compose two functions f and g in
either order, so they have two composite functions: g f , which means
rst f and then g, and f g, which means rst g and then f . These two
composite functions are illustrated in Figure 50.
x
f (x)
g(x)
g(f (x))
f (g(x))
Figure 50 The image of x under g f and f g, respectively

The next example shows you how to work out the rules of composite
functions.
Example 4
Composing functions
Suppose that f (x) = x2 and g(x) = x + 1. Find the rules of the

composite functions g f and f g.
Solution
(g f )(x) = g(f (x)) = g(x2 ) = x2 + 1
and
(f g)(x) = f (g(x)) = f (x + 1) = (x + 1)2 .
Figure 51 illustrates the composite functions in Example 4.

x
x2
x+1
x2 + 1
(x + 1)2
Figure 51 The image of x under particular composite functions g f and

f g
252
Notice that the composite functions g f and f g in Example 4 are

dierent functions. If f and g are two dierent functions, then the
composite functions g f and f g are usually dierent from each other.
Activity 31
Composing functions
(a) Suppose that f (x) = x 3 and g(x) =

following composite functions.
(i) g f
(ii) f g
x. Find the rules of the
(iii) f f
(iv) g g
(b) Determine the domain of the function g f .
You can compose more than two functions. For example, if f , g and h are
functions, then you can form a composite function whose rule is given by
rst applying f , then g, then h. This composite function is denoted by
h g f , and its rule can be stated as
(h g f )(x) = h(g(f (x))).
Example 5
Composing three functions
Suppose that f (x) = x + 2, g(x) = 1/x and h(x) =

of the composite function h g f .
x. Find the rule
Solution
(h g f )(x) = h(g(f (x)))
= h(g(x + 2))
'
+
1
=h
x+2
"
1
=
x+2
1
.
=
x+2
Activity 32
Composing three functions
Suppose that f (x) = x + 2, g(x) = 1/x and h(x) = x, as in Example 5.

Find the rules of the following composite functions.
(a) f g h
(b) g h f
(c) f h g
(d) f g f
253
Unit 3
Functions
3.3 Inverse functions

In this subsection youll learn whats meant by the inverse of a function.
To illustrate the idea, lets consider the function f (x) = 2x. The mapping
diagram in Figure 52 shows some of the inputs and corresponding outputs
of this function, linked by arrows.
4
1
0
8
2
0
1
2
10
1
2
5
inputs
outputs
Figure 52 Some inputs and outputs of the function f (x) = 2x

Imagine a full version of the mapping diagram in Figure 52, which shows
all the inputs and outputs of the function f . You cant actually draw such
a diagram, of course, because f has innitely many inputs and outputs.
The inverse function, or simply inverse, of the function f , which is
denoted by f 1 , is the function whose mapping diagram is obtained by
reversing the directions of all the arrows in this full version of the diagram.
For example, Figure 53 shows some of the inputs and outputs of f 1 .
4
1
0
8
2
0
1
2
10
1
2
5
outputs
y
f 1
Figure 54 A mapping
diagram illustrating a
function f and its inverse
function f 1
254
inputs
Figure 53 Some inputs and outputs of the function f 1 , where f (x) = 2x
f
x
f 1
This gure shows that, for example,

f 1 (8) = 4,
f 1 (1) =
1
2
and
f 1 (10) = 5.
You can imagine a full version of the mapping diagram in Figure 53, which
shows all the inputs and outputs of f 1 . You can think of the inverse
functions of other functions in the same way. Essentially, the inverse
function f 1 of a function f is the function that has the reverse eect
of f . That is, if inputting a number x to f gives the number y, then
inputting the number y to f 1 gives the original number x, as illustrated
in Figure 54. For example, if f is the function f (x) = 2x, as above, then
inputting 5 to f gives 10, and inputting 10 to f 1 gives 5.
Another way to think of the inverse function f 1 of a function f is that

f 1 undoes the eect of f . For example, if f is the function f (x) = 2x, as
above, and you input the number 5 to f , then you get 10; and if you then
take this output 10 and input it to the inverse function f 1 , then you get 5
back again.
You can sometimes write down the rule of an inverse function by thinking
about it in this way. For example, consider once more the function
f (x) = 2x. This function doubles numbers, so the function that undoes its
eect halves numbers. So the rule of the inverse function of this function
f is
f 1 (x) = 21 x.
In the next activity youre asked to write down the rules of the inverse
functions of some other simple functions.
Activity 33
Finding the rules of inverse functions of simple functions
(a) Write down the rules of the inverse functions of the following
functions.
(i) f (x) = x + 1
(iii) f (x) = 13 x
(ii) f (x) = x 3
(b) Can you think of a function with an inverse function that has the same
rule as the original function? Can you think of another such function?
Some functions dont have inverse functions. For example, consider the
function f (x) = x2 . Some of the inputs and outputs of this function are
shown in Figure 55.
0
21
1
4
1
2
3
inputs
outputs
Figure 55 Some inputs and outputs of the function f (x) = x2

If you reverse the directions of all the arrows in the full version of this
mapping diagram, then the new diagram that you get isnt the mapping
diagram of a function. Thats because, in the new diagram, some of the
input numbers have more than one output number, as illustrated in
Figure 56. Remember that, for a function, every input number must have
exactly one output number.
255
Unit 3
Functions
0
12
1
2
3
3
outputs
0
1
4
9
inputs
Figure 56 After reversing the directions of the arrows in Figure 55

You can see that this problem will arise whenever, for the original
function f , there are two or more dierent input values that have the same
output value. A function for which this doesnt happen, and which
therefore does have an inverse function, is said to be one-to-one. In other
words, we have the following denition.
One-to-one functions
A function f is one-to-one if for all values x1 and x2 in its domain
such that x1 .= x2 ,
f (x1 ) .= f (x2 ).
For example, the function f (x) = x3 is one-to-one, because no two dierent

numbers have the same cube. On the other hand, the function f (x) = x2
isnt one-to-one, because, for instance,
f (3) = 32 = 9
and f (3) = (3)2 = 9,
so f (3) = f (3).
Activity 34
Recognising whether functions are one-to-one
Which of the following functions are one-to-one? For each function that
isnt one-to-one, state two input numbers that have the same output
number.
(a) f (x) = |x|
(b) f (x) = x + 1
(e) f (x) = x
(f) f (x) = 1
(c) f (x) = x4
(d) f (x) = x5
A useful way to recognise whether a function is one-to-one is to look at its

graph. Youve seen that, for a one-to-one function, every output number is
obtained from exactly one input number. So if you draw any horizontal
line that crosses the graph of the function, then it crosses it exactly once,
as illustrated in Figure 57(a). If you can draw a horizontal line that crosses
a graph more than once, then the graph isnt the graph of a one-to-one
256
function. For example, the function whose graph is shown in Figure 57(b)
isnt one-to-one, since the dashed horizontal line shows that the two input
numbers marked as x1 and x2 have the same output number.
y
x
(a)
x1
x2
(b)
Figure 57 The graphs of (a) a one-to-one function and (b) a function

that isnt one-to-one
Activity 35
Recognising the graphs of one-to-one functions
For each of the following diagrams, state whether its the graph of a
one-to-one function, the graph of a function that isnt one-to-one, or not
even the graph of a function.
(a)
(b)
(c)
y
x
(e)
y
x
(f)
(g)
(h)
y
x
(k)
y
x
(j)
y
x
(i)
(d)
(l)
y
x
y
x
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Unit 3
Functions
The following important fact summarises the ideas that youve just met.
Only one-to-one functions have inverse functions.
A function that has an inverse function is said to be invertible. So

invertible function means the same as one-to-one function.
B
f
y
x
f 1
Figure 58 A mapping
diagram illustrating a
function f and its inverse
function f 1 . The set B is
the image set of f , not
necessarily its whole
codomain.
Now lets consider the domains of inverse functions. If f is any one-to-one

function, then the domain of f 1 is the image set of the original
function f . To see this, think of the mapping diagram for the original
function f . Now imagine reversing the directions of all the arrows, to
obtain the mapping diagram for the inverse function f 1 . The numbers
that have arrows starting from them are the numbers in the image set of
the original function f . So the domain of f 1 is the image set of f .
You can also see, by thinking about these diagrams, that the image set of
an inverse function f 1 is the domain of the original function f . Heres a
concise denition of an inverse function, which summarises what youve
seen so far. Its illustrated in Figure 58.
Inverse functions
Suppose that f is a one-to-one function, with domain A and image
set B. Then the inverse function, or simply inverse, of f , denoted
by f 1 , is the function with domain B whose rule is given by
f 1 (y) = x,
where
f (x) = y.
The image set of f 1 is A.
The next example illustrates how you can use this denition to nd an
inverse function, even when the rule of the original function is more
complicated than those that youve seen so far.
Example 6
Finding an inverse function
Find the inverse function of the function

f (x) = 2x + 1.
Solution
To nd the rule of f 1 , rearrange the equation f (x) = y to
express x in terms of y.
258
The equation f (x) = y gives

2x + 1 = y
2x = y 1
x = 21 (y 1).
Hence the rule of f 1 is
f 1 (y) = 12 (y 1).
Usually, change the input variable from y to x, as this is the letter
normally used for the input variable of a function.
That is, it is
f 1 (x) = 12 (x 1).
The domain of f 1 is the image set of f , which is R. By the
domain convention, you dont need to write down this domain.
The domain of f 1 is the whole of R, so the rule above completely
species f 1 .
The method used to nd the rule of the inverse function in Example 6 is

summarised below.
Strategy:
To nd the rule of the inverse function of a one-to-one
function f
Write y = f (x) and rearrange this equation to express x in terms

of y.
Use the resulting equation x = f 1 (y) to write down the rule

of f 1 . (Usually, change the input variable from y to x.)
If its possible to rearrange an equation y = f (x) to express x in terms of y

(so that each value of y gives exactly one value of x), then this shows that
the function f has an inverse function, whose rule is given by the
rearranged equation. Remember that to fully specify the function f 1 , you
also have to indicate its domain, which, as youve seen, is the image set
of f . (As always, if the domain of f 1 is the largest set of real numbers for
which its rule is applicable, then theres no need to state its domain
explicitly.)
259
Unit 3
Functions
Activity 36
Finding inverse functions
Find the inverse functions of the following functions. (Remember to nd

the domain of the inverse function in each case.)
1
(c) f (x) = 5 +
(a) f (x) = 3x 4
(b) f (x) = 2 12 x
x
In cases that are trickier than those in Activity 36, it often helps to obtain
a graph of the original function f . For example, this can be useful if the
domain of the function f isnt the largest set of real numbers for which its
rule is applicable, or if youre not sure whether f is one-to-one. Heres an
example.
Example 7
Finding another inverse function
Does the function

f (x) = x2 4x + 1
(x [1, 1])
have an inverse function? If so, nd it.
Solution
Obtain a sketch or computer plot of the graph of f .
The graph of f is shown below.
y
6
2
y = x 4x + 1 5
4
3
2
1
21
1
2
260
1 2 x
Think about whether every horizontal line that crosses the graph
of f does so exactly once.
The graph shows that f is one-to-one and therefore has an inverse
function.
Try to nd the rule of f 1 in the usual way, by rearranging the
equation f (x) = y. For a quadratic function like f , it helps to begin
by completing the square.
The equation f (x) = y gives
x2 4x + 1 = y
(x 2)2 4 + 1 = y
(x 2)2 3 = y
(x 2)2 = y + 3
&
x2= y+3
&
x=2 y+3
Decide whether the + or the applies. Remember that the nal
equation above is a rearrangement of the equation f (x) = y, so x is
an element of the domain of f , which is [1, 1]. Now 2 plus the
positive square root of something cant be equal to a number in this
interval, but 2 minus the positive square root of something can, so
the correct sign is .
Since the domain of f is [1, 1], each input value x of f is less than 2.
So
&
x = 2 y + 3.
&
f 1 (y) = 2 y + 3;
that is,
f 1 (x) = 2
x + 3.
To nd the domain of f 1 , nd the image set of f , using the

graph to help you.
The domain of f 1 is the image set of f . The graph shows that this is
[f (1), f (1)] = [2, 6].
Finally, specify f 1 by stating its domain and rule.
So the inverse function of f is the function
f 1 (x) = 2 x + 3 (x [2, 6]).
261
Unit 3
Functions
Activity 37
Finding more inverse functions
In each of parts (a)(c), determine whether the function has an inverse

function. If it does, then nd the inverse function.
(a) f (x) = x2 + 2x + 2
(x (2, 2))
(b) f (x) = x2 + 2x + 2
(x (0, 2))
(c) f (x) = 1 x (x [3, 1])
Heres a useful fact that sometimes gives you a quick way of conrming
that a function has an inverse function.
If a function is either increasing on its whole domain, or decreasing on
its whole domain, then it is one-to-one and so has an inverse function.
This fact holds because if a function is either increasing on its whole

domain or decreasing on its whole domain, then any horizontal line drawn
on its graph will cross the graph at most once, as illustrated in Figure 59.
x
(a)
x
(b)
Figure 59 (a) A function thats increasing on its whole domain (b) a

function thats decreasing on its whole domain
Heres another useful property of inverse functions, which you can
understand by thinking about mapping diagrams. If a function f has an
inverse function f 1 , then f 1 also has an inverse function, namely f . In
other words, f and f 1 are inverses of each other. This is because the
mapping diagram for each of these functions is obtained by reversing the
directions of the arrows in the mapping diagram for the other function. In
particular, each of the functions f and f 1 undoes the eect of the other.
So if you take any value x in the domain of f , input it to f , and then input
the resulting output value to f 1 , then you get the value x back again; and
similarly if you take any value x in the domain of f 1 , input it to f 1 , and
then input the resulting output value to f , then you get the value x back
262
again. These two facts can be stated concisely as follows, using the
notation for composite functions.
For any pair of inverse functions f and f 1 ,
(f 1 f )(x) = x,
(f f
)(x) = x,
for every value x in the domain of f , and

for every value x in the domain of f 1 .
A warning
When youre working with the notation f 1 , where f is a function, its
important to appreciate that it doesnt mean the function g with rule
1
.
g(x) = (f (x))1 ; that is, g(x) =
f (x)
This function g is called the reciprocal of the function f , and its never
denoted by f 1 . For example, consider the function f (x) = x + 5. Its
inverse function is
f 1 (x) = x 5,
whereas its reciprocal is
1
g(x) =
.
x+5
Graphs of inverse functions

Theres a useful geometric connection between the graph of a function and
the graph of its inverse function. Figure 60(a) shows the graphs of the
function f
(x) = x2 1 (x 0) and its inverse function
f 1 (x) = x + 1 (x 1), drawn on axes with equal scales. Similarly,
Figure 60(b) shows the graphs of the function f (x) = 2x + 1 (x [2, 1])
and its inverse function f 1 (x) = 21 (x 1) (x [3, 3]), again drawn on
axes with equal scales.
y
3
2
1
321
1
2
3
(a)
y = x2 1
y y = 2x + 1
3
2
y = 21 (x 1)
1
y = x + 1
1 2 3 x
321
1
2
3
1 2 3 x
(b)
Figure 60 Graphs of pairs of inverse functions

In each case the graphs of f and f 1 are the reections of each other in
the line y = x, which is shown as a green dashed line.
263
Unit 3
Functions
This happens for every pair of inverse functions, when their graphs are
drawn on axes with equal scales. To see why, lets start by considering any
point on the graph of the function f (x) = x2 1 (x 0). For example, the
point (2, 3) lies on this graph, because inputting 2 to this function f gives
the output 3. It follows that inputting 3 to the inverse function f 1 gives
the output 2, and so the point (3, 2) lies on the graph of f 1 . You can see
that, for any pair of inverse functions f and f 1 , if you swap the
coordinates of any point on the graph of f , then youll get the coordinates
of a point on the graph of f 1 , and vice versa.
y
3
2
1
1
1
Now when you swap the coordinates of a point, the resulting point is the
reection of the original point in the line y = x (provided the axes have
equal scales). This is illustrated in Figure 61, for the example discussed
above. This reasoning explains the connection between the graphs of a
function and its inverse function, which is summarised below.
Graphs of inverse functions
Figure 61 The points (2, 3)

and (3, 2) are reections of
each other in the line y = x
The graphs of a pair of inverse functions are the reections of each

other in the line y = x (when the coordinate axes have equal scales).
Activity 38
Sketching graphs of inverse functions
Each of the following diagrams shows the graph of a function, drawn on

axes with equal scales. The line y = x is shown as a green dashed line. The
vertical dashed lines in graph (c) are asymptotes.
For each graph, sketch the graph of the inverse function.
(a)
(b)
(c)
(d)
(4; 4)
1
( 2; 2)
264
Functions that arent one-to-one

Finally in this subsection, lets consider functions that dont have inverses,
because theyre not one-to-one. An example of such a function is
f (x) = x2 , whose graph is shown in Figure 62.
When you have a function like this, its sometimes useful to consider some
sort of inverse ofthe function. For example, youd probably consider the
function h(x) = x to be some sort of inverse of the function f (x) = x2 .
Heres the approach that we take in situations like this. Starting with the
function f thats not one-to-one, we specify a new function that has the
same rule as f , but a smaller domain. We choose the new domain to make
sure that the following two conditions are satised:
the new function is one-to-one and therefore has an inverse
the image set of the new function is the same as the image set of the
original function.
y
3
2
1
3 2 1
1 2 3 x
Figure 62 The graph of the

function f (x) = x2
For example, for the function f (x) = x2 , we could take the new function to
be the function
g(x) = x2
(x [0, )),
whose graph is shown in Figure 63(a). This function g is one-to-one, and

therefore has an inverse function, namely
g 1 (x) = x,
whose graph is shown in Figure 63(b).
y
3
2
1
3 2 1
(a)
y
3
2
1
1 2 3 x
3 2 1
1 2 3 x
(b)
Figure 63 The graphs of (a) thefunction g(x) = x2 (x [0, )) and

(b) its inverse function g 1 (x) = x
A function thats obtained from another function f by keeping the rule the
same but removing some numbers from the domain is called a restriction
of the original function f . The process of obtaining such a function is called
restricting the domain of f , or restricting f . So the function g above is
a restriction of the function f (x) = x2 . The graph of a restriction of a
function is obtained by erasing part of the graph of the original function.
When you want to restrict the domain of a function that isnt one-to-one
to enable you to nd an inverse function, theres always more than one
possibility for the new domain. For example, for the function f (x) = x2 ,
you could have chosen the new domain (, 0] instead of [0, ). This
would have given you the new function g whose graph is shown in
265
Unit 3
Functions
Figure 64(a). The inverse function of this function is g 1 (x) = x,

whose graph is shown in Figure 64(b).
y
3
2
1
3 21 1
2
3
(a)
y
3
2
1
1 2 3 x
3 21 1
1 2 3 x
2
3
(b)
Figure 64 The graphs of (a) the function

g(x) = x2 (x (, 0]) and
(b) its inverse function g 1 (x) = x

Usually when we restrict the domain of a function to enable us to nd an
inverse function, we choose the new domain that seems to be the most
convenient.
Activity 39
y
3
2
1
2 1
Restricting a function to nd an inverse function
Consider the function f (x) = (x 1)2 , whose graph is shown in Figure 65.
Specify a one-to-one function g that is a restriction of f and has the same
image set as f . Find the inverse function g 1 of g, and sketch its graph.
1 2 3 4 x

y = (x 1)2
In Unit 4 youll see some more examples of this process of restricting the
domain of a function to enable you to nd an inverse function. Its useful
in particular for trigonometric functions, which youll meet in that unit.
4 Exponential functions and logarithms

In this section youll revise exponential functions and logarithms. In
particular, youll have the opportunity to practise working with
logarithms. Its important that you can do this uently and correctly.
266
Exponential functions and logarithms
4.1 Exponential functions

An exponential function is a function whose rule is of the form
f (x) = bx ,
where b is a positive constant, not equal to 1. (Note that in some other
texts the case b = 1 is not excluded.) The number b is called the base
number, or just base,% of# the exponential function. For example,
x
f (x) = 2x and g(x) = 21 are exponential functions, with base numbers 2
and 21 , respectively. The graphs of these two functions are shown in
Figure 66. The y-intercept is 1 in each case.
y
y = 2x
y=
1
2
y
4
4 2
4 2
(b)
(a)
Figure 66 The graphs of (a) y = 2x (b) y =
% 1 #x
2
To see why these graphs have the shapes that they do, rst consider the
function f (x) = 2x . Some values of this function are given in Table 1. The
corresponding points are shown in Figure 67 (except that the nal two
points are o the scale).
Table 1 Values of 2
x
2x
1
8
1
4
4
2
4 3 2 1 0 1 2 3
1
16
1
2
4 2
1 2 4 8 16
Notice that f (0) = 20 = 1, which explains why the y-intercept is 1.

Notice also that each time the value of x increases by 1 unit to the next
integer up, the value of f (x) doubles. So as x takes values that are further
and further along the number line to the right, the value of f (x) = 2x
increases, and increases more and more rapidly. This explains the shape of
the graph as x increases.
Figure 67 Points on the

graph of y = 2x with integer
values of x
Similarly, each time the value of x decreases by 1 unit to the next integer
down, the value of f (x) halves. So, as x takes values that are further and
further along the number line to the left, the value of f (x) = 2x gets closer
and closer to zero, but never reaches zero. This gives the shape of the
graph as x decreases.
% #x
The shape of the graph of the function g(x) = 21 can be explained in a
similar way, and you might like to think it through for yourself.
267
Unit 3
Functions
Alternatively, you can deduce it from the shape of the graph of the
function f (x) = 2x . Notice that, for any number x,
% 1 #x
= (21 )x = 2x .
2
% #x
So the rule of the function g(x) = 21 can be written as g(x) = 2x , and
hence, by what you saw in Subsection 2.3, its graph is the same shape as
the graph of f (x) = 2x , but reected in the y-axis.
In the next activity youre asked to investigate the shapes of the graphs of
some more exponential functions.
Activity 40
Investigating the graphs of exponential functions
Open the Exponential functions applet. Display the graph of y = bx for

various positive values of b, and observe the shapes of the graphs.
In particular, notice how the graphs obtained when 0 < b < 1 dier from
those obtained when b > 1.
In Activity 40 you should have observed the following facts, which are
illustrated in Figure 68.
Graphs of exponential functions

The graph of the function f (x) = bx , where b > 0 and b .= 1, has the
following features.
b>1
The graph lies entirely above the x-axis.
If b > 1, then the graph is increasing, and it gets steeper as x

increases.
If 0 < b < 1, then the graph is decreasing, and it gets less steep
as x increases.
The x-axis is an asymptote.
The y-intercept is 1.
The closer the value of b is to 1, the atter is the graph.
y
b > 1,
b close to 1
1
1
x
Figure 68
268
Graphs of equations of the form y = bx
0 < b < 1,
b close to 1
x
0<b<1
A helpful way to remember the nal feature listed in the box above is to
notice that when the value of b is exactly 1, the function f (x) = bx is the
function f (x) = 1x , that is, f (x) = 1, and hence its graph is the horizontal
line with y-intercept 1, as shown in Figure 69. Remember, though, that
this function f isnt an exponential function its a constant function.
y
1
x
The exponential function

Theres a particular exponential function thats crucially important in
mathematics. Youve seen that the graph of every exponential function
f (x) = bx passes through the point (0, 1). The steepness of the graph at
this point depends on the value of b, as you can see from the examples in
Figure 70.

y=1
y = 5x y = 2x y = 1:5x
5
4
3
2
1
4 3 2 1
4 x
Figure 70 Graphs of exponential functions

In the next activity youre asked to investigate the steepness of the graphs
of exponential functions at the point (0, 1).
Activity 41
(0, 1)
Investigating the gradient of the graph of y = bx at
Use the Exponential functions applet to do the following.

(a) Reset the applet and choose the option to show a grid. Zoom in on the
point (0, 1) until the graph looks like a straight line. Keep the scales
on the axes the same as each other.
(b) Change the value of b until the gradient of the graph at (0, 1) appears
to be exactly 1; that is, until the graph goes up by the same distance
vertically as it goes along horizontally to the right. What value of b
seems to achieve this?
In Activity 41 you should have found that the value of b that gives a
gradient of 1 at (0, 1) seems to be about 2.7. In fact, the precise value is a
269
Unit 3
Functions
special number, usually denoted by the letter e, whose rst few digits are
2.718 28 . . . . The number e is irrational, like , so its digits have no
repeating pattern, and it cant be written down exactly as a fraction or a
terminating decimal. It occurs frequently in mathematics, and youll learn
more about it, and why its so important, later in the module.
So the exponential function with the rule f (x) = ex has the special
property that its gradient is exactly 1 at the point (0, 1). Its graph is
shown in Figure 71. This function is important both in applications of
mathematics and in pure mathematics, and because of its importance its
sometimes referred to as the exponential function. The expression ex is
sometimes written as exp x, or exp(x). An approximate value for e is
available from your calculator keypad, just as for , and you can also work
out values of ex by using a function button on your calculator.
y
5
4
3
2
1
4 3 2 1
4 x
Figure 71 The graph of y = ex

The use of the letter e for the base of the exponential function was
introduced by Leonhard Euler (see page 214).
4.2 What is a logarithm?

In this subsection, youll revise logarithms, which are closely related to
exponential functions. Logarithms are sometimes called logs, for short.
The rst thing to remember about logarithms is that whenever youre
working with them, youre always using logarithms to a particular base
(also called base number ). Lets start by considering logarithms to base 10,
which are known as common logarithms. These are dened as follows.
The logarithm to base 10 of a number x is the power to which 10 must
be raised to give the number x.
270
For example,
the logarithm to base 10 of 100 is 2, because 100 = 102 .
Similarly,
the logarithm to base 10 of 1000 is 3, because 1000 = 103 , and
1
1
is 1, because 10
= 101 .
the logarithm to base 10 of 10
The logarithm to base 10 of a number x is denoted by log10 x, so the three
logarithms found above can be written as follows:
log10 100 = 2
log10 1000 = 3
%1#
log10 10
= 1.
You can see that if you can easily write a number as a power of 10, then
its straightforward to nd its logarithm to base 10. For other numbers,
you can use your calculator to nd an approximate value. For example, a
calculator gives
log10 42 = 1.623 249 2 . . . ,
which is the same as saying that
101.623 249 2... = 42.
The button on a calculator for nding common logarithms is usually
labelled log.
Activity 42
Understanding logarithms to base 10
(a) Find the following numbers, without using your calculator.

(i) log10 10 000
(ii) log10
1
100
(iii) log10 10
(iv) log10 1
(b) If the number x is such that log10 x = 21 , what is x?

(c) Use your calculator to nd the following numbers, to three decimal
places.
(i) log10 3700
(ii) log10 370
(iii) log10 37
(iv) log10 3.7
(v) log10 0.37
(vi) log10 0.037
Notice that only positive numbers have logarithms to base 10. For
example, the negative number 2 has no logarithm to base 10, because
theres no power to which 10 can be raised to give 2. Similarly, 0 has no
logarithm to base 10, because theres no power to which 10 can be raised
to give 0.
However, logarithms themselves can be positive, negative or zero. For
example, youve seen that
%1#
= 1 and log10 1 = 0.
log10 100 = 2, log10 10
271
Unit 3
Functions
Now lets consider logarithms to other bases. Like the base of an
exponential function, the base of a logarithm can be any positive number
except 1. Logarithms to other bases work in the same way as logarithms to
base 10. Heres a general denition of logarithms, to any base.
Logarithms
The logarithm to base b of a number x, denoted by logb x, is the
power to which the base b must be raised to give the number x. So
the two equations
and x = by
y = logb x
are equivalent.
Remember that:
the base b must be positive and not equal to 1
only positive numbers have logarithms, but logarithms

themselves can be any number.
For example, log6 36 = 2 because 36 = 62 .
Activity 43
Understanding logarithms to any base
(a) Find the following numbers without using your calculator.

(i) log3 9
(vi) log8 2
(x) log8
1
8
(xiv) log5 5
(ii) log2 8
(vii) log2
(iii) log4 64
1
2
(viii) log2
(iv) log5 25
1
8
(ix) log3
(v) log4 2
1
27
(xii) log4 41
(xiii) log6 6
3
(xv) log7 7
(xvi) log2 1
(xvii) log15 1
(xi) log3 3
(b) Find the solution of each of the following equations in x.

(i) log2 x = 5
(ii) log8 x =
1
3
(iii) log7 x = 1
Youve seen that its straightforward to write down the logarithm to base b
of a number if you can express the number as a power of b. In particular,
for any base b, its straightforward to write down the logarithm to base b
of 1, and the logarithm to base b of b itself, because
1 = b0
and b = b1 .
This gives the following useful facts.
272
Logarithm of the number 1 and logarithm of the base

For any base b,
logb 1 = 0
and
logb b = 1.
Logarithms were invented by the Scottish mathematician John Napier

for the purpose of easing the labour involved in astronomical and
navigational calculations. Napiers rather awkward initial
formulation, published in 1614, was greeted enthusiastically by the
English mathematician Henry Briggs (15611630), who immediately
set about trying to improve it. The following year Briggs visited
Napier, and together they invented logarithms to base 10, the rst
publication of which appeared in 1617 shortly after Napiers death.
The most common choices for the base of logarithms are 10, 2 and e.
(Remember that e is the important constant whose value is approximately
2.718.) Usually, once youve chosen a base, you use the same base for all
the logarithms in your calculations (otherwise, your calculations may be
wrong!).
John Napier (15501617)
In university-level mathematics, the number most commonly used as the

base for logarithms is e. As youll see later in the module, logarithms to
base e turn out to be easier to work with in many ways than logarithms to
any other base.
Logarithms to base e are called natural logarithms. The notation ln is
often used in place of loge , and this is the notation that will be used in
this module. For example, the natural logarithm of 5 is written as
ln 5 rather than
loge 5.
(The rst symbol in ln is the letter l, not the digit 1.)

Theres no consensus about how the notation ln should be pronounced,
but some common pronunciations are log, ell enn, linn and lawn. The
box below summarises the denition of a natural logarithm, using this
notation.
Natural logarithms
The natural logarithm of a number x, denoted by ln x, is the power
to which the base e must be raised to give the number x. So the two
equations
y = ln x
and x = ey
are equivalent.
273
Unit 3
Functions
Scientic calculators have a button for nding natural logarithms, usually
labelled ln.
The properties that logb 1 = 0 and logb b = 1 for any base b give the
following two useful properties of natural logarithms.
ln 1 = 0
and
ln e = 1.
Here are some calculations involving natural logarithms for you to try.
Activity 44
Understanding natural logarithms
(a) Find the following numbers without using your calculator.
(i) ln e4
(ii) ln e2
(iii) ln e3/5
(iv) ln e
' +
' +
1
1
(v) ln
(vi) ln 3
e
e
(b) If the number x is such that ln x = 12 , what is x?
(c) Use your calculator to nd the following numbers to three decimal
places.
(i) ln 5100
(iv) ln(51e)
(ii) ln 510
(v)
(iii) ln 51
ln(51e2 )
The natural logarithm of a number x is sometimes denoted by log x, rather

than by ln x or loge x. For example, this notation is used by some
computer algebra systems. Confusingly, the same notation, log x, is also
sometimes used to denote log10 x, the common logarithm of x. For
example, as mentioned earlier, the button on a calculator for nding
common logarithms is usually labelled log. Even more confusingly, the
notation log x is also sometimes used to denote the logarithm of x with no
specic base (but the same base for each use of the notation). So, wherever
you see the notation log x used, its important to check its meaning. The
notation log x isnt used at all in the main books of this module, but you
may have to use it with your calculator or computer.
274
Logarithms to base e were rst described as natural logarithms by

the Danish mathematician Nicolaus Mercator (162087), in his
treatise Logarithmotechnica published in 1668. Mercators treatise was
dicult to follow, so later writers tended to refer to the exposition of
it given by the English mathematician John Wallis (16161703),
published in the same year. The rst published use of the notation
ln for a natural logarithm was in a book written by an American
mathematician, Irving Stringham, which was published in 1893. He
explained his choice, in a textbook published a little later, as follows:
In place of e log we shall henceforth use the shorter symbol ln, made
up of the initial letters of logarithm and of natural or Napierian.
4.3 Logarithmic functions

A logarithmic function is a function whose rule is of the form
f (x) = logb x,
where b is a positive constant, not equal to 1. The number b is called the
base or base number of the logarithmic function. For example,
f (x) = log2 x and g(x) = ln x are logarithmic functions, with bases 2
and e, respectively.
For any positive constant b not equal to 1, the logarithmic function
f (x) = logb x is the inverse function of the exponential function g(x) = bx .
This is because, as youve seen, the two equations
y = logb x
and x = by
are rearrangements of each other.

In particular, the function f (x) = ln x is the inverse function of the
function g(x) = ex , which explains why these two functions usually share
the same button on a calculator.
So the graphs of f (x) = logb x and g(x) = bx are reections of each other
in the line y = x (provided that the coordinate axes have equal scales).
275
Unit 3
Functions
For example, Figure 72 shows the graphs of y = ln x and y = ex .
y
4
y = ex
y = ln x
2
4 2
Figure 72 The graphs of y = ln x and y = ex

You can deduce the following general properties of the graphs of
logarithmic functions from the properties of the graphs of exponential
functions, using the fact that theyre reections of each other in the line
y = x. These properties are illustrated in Figure 73.
Graphs of logarithmic functions

The graph of the function f (x) = logb x, where b > 0 and b .= 1, has
the following features.
276
The graph lies entirely to the right of the y-axis.
If b > 1, then the graph is increasing, and it gets less steep as x

increases.
If 0 < b < 1, then the graph is decreasing, and it gets less steep as
x increases.
The y-axis is an asymptote.
The x-intercept is 1.
The closer the value of b is to 1, the steeper is the graph.
y 0 < b < 1;
b close
to 1
b>1
b > 1;
b close
to 1
0<b<1
Figure 73 Graphs of equations of the form y = logb x

The properties in the box below are often useful when youre working with
logarithms. Theyre just the two facts about composing a pair of inverse
functions that are stated in the box on page 263, in the particular case
when the functions are f (x) = bx and f 1 (x) = logb x.
For any base b,
logb (bx ) = x and
blogb x = x.
In particular,
ln(ex ) = x
and eln x = x.
These properties hold for all appropriate values of x.
Activity 45
Simplifying expressions involving e and ln
Simplify the following expressions.

(a) eln(7x)
(b) ln(e8x )
(e) ln(ex/2 ) + 3 ln 1
(c) ln(e2x ) + ln(e3x )
(f) e2 ln c
(h) ln(ey+2 ) + 2 ln(ey1 )
(d) ln(e2 ) ln e
(g) eln(3a) + 4e0
(i) e3 ln B
(j) e2+ln x
Hint: in some parts you may need to use the index laws from Unit 1.
277
Unit 3
Functions
4.4 Logarithm laws

In this subsection youll revise three laws for logarithms, which are often
useful when youre working with logarithms. These laws are really just the
same as the following three index laws that were given in Unit 1, rewritten
using logarithm notation.
Three index laws from Unit 1

bm bn = bm+n
bm
= bmn
bn
(bm )n = bmn
Here are the three logarithm laws.
Three logarithm laws

logb x + logb y = logb (xy)
' +
x
logb x logb y = logb
y
r logb x = logb (xr )
As with the index laws, these logarithm laws apply to all appropriate
numbers. So the base b of the logarithms can be any positive number
except 1, the numbers x and y must be positive (since only positive
numbers have logarithms), and r can be any number (in particular, it can
be fractional and/or negative).
To see how these three logarithm laws are deduced from the three index
laws above, lets write m = logb x and n = logb y. This is the same as
saying that x = bm and y = bn .
So,
xy = bm bn = bm+n ,
from which it follows that
logb (xy) = m + n = logb x + logb y.
This is the rst logarithm law.
Also,
x
bm
= n = bmn ,
y
b
' +
x
logb
= m n = logb x logb y.
y
This is the second logarithm law.
278
Finally,
xr = (bm )r = bmr ,
logb (xr ) = mr = r logb x.
This is the third logarithm law.
Example 8
Using the logarithm laws
Write the expression 3 ln 6 2 ln 2 as the logarithm of a single number.
Solution
3
'
3 ln 6 2 ln 2 = ln 6 ln 2 = ln
63
22
+
= ln 54
Here are some examples for you to try.
Activity 46
Using the logarithm laws
(a) Write each of the following expressions as the logarithm of a single

number or expression.
(i) ln 5 + ln 3
(ii) ln 2 ln 7
(iv) ln 3 + ln 4 ln 6
(iii) 3 ln 2
(v) ln 24 2 ln 3
(vii) 3 log2 5 log2 3 + log2 6
(viii)
1
2
(vi)
1
3
log10 27
ln(9x) ln(x + 1)
(b) Simplify the following expressions.

(i) ln c3 ln c
(ii) 3 ln(p2 )
(iv) ln(3u) ln(2u)
(iii) ln(y 2 ) + 2 ln y
(v) ln(4x) + 3 ln x ln(e6 )
1
2
ln(y 3 )
(vi)
1
2
ln(u8 )
(c) Can you explain the pattern in the answers to Activity 42(c)? (This
activity is on page 271.)
Hint: notice that each number in Activity 42(c) is of the form
log10 (37 10n ), for some integer n.
(d) Suppose that the multiplication button on your scientic calculator
doesnt work. Can you use the remaining buttons to nd the value of
1567 2786, at least approximately?
Hint: start by writing 1567 2786 = eln(15672786) .
279
Unit 3
Functions
Solving exponential equations

An exponential equation is one in which the unknown is in the
exponent, such as
5x = 130.
Equations like this often arise when you use exponential functions to
model real-life situations. Youll see some examples in the next subsection.
You can solve exponential equations by using the third of the three
logarithm laws given earlier in this subsection. For this purpose its best to
think of the law with its left and right sides swapped:
logb (xr ) = r logb x. In particular,
Example 9
ln(xr ) = r ln x.
Solving an exponential equation
Solve the equation 2 1.53x = 45, giving the solution to three

signicant gures.
Solution
The equation can be solved as follows.
2 1.53x = 45
Rearrange it into the form (number)(expression in x) = number.
1.53x = 22.5
Take the natural logarithm of both sides.
ln(1.53x ) = ln 22.5
Use the fact that ln(xr ) = r ln x.
3x ln 1.5 = ln 22.5
Divide both sides by the coecient of the unknown.
x=
ln 22.5
3 ln 1.5
Use your calculator to evaluate the answer.

x = 2.56 (to 3 s.f.).
The solution is approximately x = 2.56.
(Check: when x = 2.56,
LHS = 2 1.532.56 = 45.020 557 . . . 45 = RHS.)
280
The crucial step in the method demonstrated in Example 9 is to take logs

of both sides of the exponential equation. Then you can use the property
in the box above to turn the exponent into a factor, which makes it
straightforward to solve the equation. You dont have to use natural
logarithms in this method you can use logarithms to any base, as long as
youre consistent.
In fact, if your calculator has a button for nding logarithms to any base,
then you can make the working in Example 9 slightly shorter by
proceeding as follows, starting from the second equation in the solution:
1.53x = 22.5
3x = log1.5 22.5
x = 31 log1.5 22.5
x = 2.56 (to 3 s.f.).
Activity 47
Solving exponential equations
Solve the following exponential equations, giving your answers to three

signicant gures.
(a) 5x = 0.5
(b) 4e7t = 64
(c) 5 2u/2 + 30 = 600
(d) 23x5 = 100
Remember that you can always check a solution that youve found for an
equation by substituting it into the equation.
In the nal activity of this subsection you can learn how to use the
computer to work with expressions involving exponentials and logarithms.
281
Unit 3
Functions
Activity 48
computer
Working with exponentials and logarithms on the
Work through Section 5 of the Computer algebra guide.
4.5 Alternative form for exponential functions

Youve seen that an exponential function is one whose rule is of the form
f (x) = bx ,
where b is a positive constant, not equal to 1.
Theres a useful alternative way to express a rule of this form. If you let k
be the number such that ek = b (in other words, if you take k = ln b), then
f (x) = bx = (ek )x = ekx .
For example, another way to express the rule
f (x) = 5x
is, approximately,
f (x) = e1.609 438x ,
because ln 5 = 1.609 437 912 . . . .
In general, we have the following fact.
Any exponential function f (x) = bx , where b is a positive constant
not equal to 1, can be written in the alternative form
f (x) = ekx ,
where k is a non-zero constant. The constant k is given by k = ln b.
Activity 49
function
Understanding the alternative form of an exponential
Consider the exponential function f (x) = 3x . Write its rule in the form
f (x) = ekx , giving the constant k to seven signicant gures. Use each
form of the rule in turn to work out f (1.5) to three signicant gures, and
check that you get the same answer.
The fact in the box above gives us the following alternative denition of an
exponential function.
282
An exponential function is a function whose rule is of the form

f (x) = ekx ,
where k is a non-zero constant. This alternative form is the one thats
usually used in university-level mathematics, as it turns out to be easier to
work with, for reasons that youll see later in this module.
The properties of the graphs of exponential functions that you saw in
Subsection 4.1 can be stated in terms of this alternative form as in the box
below. These properties are illustrated in Figure 74.
Graphs of exponential functions

The graph of the function f (x) = ekx , where k .= 0, has the following
features.
The graph lies entirely above the x-axis.
If k > 0, then the graph is increasing, and it gets steeper as x

increases.
If k < 0, then the graph is decreasing, and it gets less steep as x

increases.
The x-axis is an asymptote.
The y-intercept is 1.
The closer the value of k is to 0, the atter is the graph.
k>0
y
k > 0,
k close to 0
1
1
x
k < 0,
k close to 0
k<0
Figure 74 Graphs of equations of the form y = ekx

The fact that every exponential function can be written in the form
f (x) = ekx also tells you the following enlightening fact.
The graph of every exponential function is a horizontal scaling of the
graph of the exponential function f (x) = ex .
Thats because, from what you saw in Subsection 2.3, if f is any function,
then the graph of y = f (kx) is a horizontal scaling of the graph of
y = f (x) by the factor c = 1/k. So the graphs of all exponential functions
283
Unit 3
Functions
have the same basic shape, just stretched or squashed horizontally by
dierent amounts, and possibly reected in the y-axis.
It follows that the graphs of all logarithmic functions are vertical scalings
of each other, since the graphs of logarithmic functions are reections of
the graphs of exponential functions in the line y = x.
4.6 Exponential models

Functions with rules of the form f (x) = abx , where a is a non-zero number
and b is a positive number not equal to 1, are useful for modelling some
types of real-life situations. Models of this type are called exponential
models.
From what you saw in the previous subsection, rules of this form are the
same as rules of the form f (x) = aekx , where a and k are non-zero
numbers. Well use this alternative form in this subsection.
From your work in Subsection 2.2, you know that the graph of the function
f (x) = aekx is obtained by vertically scaling the graph of the function
g(x) = ekx by a factor of a. Also, as you saw in Subsection 4.5, the graph
of the function g(x) = ekx is itself obtained by horizontally scaling the
graph of the function h(x) = ex by a factor of 1/k. So the graph of any
function of the form f (x) = aekx is obtained by scaling the graph of the
function h(x) = ex both horizontally and vertically. In particular, since the
graph of h(x) = ex has y-intercept 1, the graph of f (x) = aekx has
y-intercept a. Figure 75 shows the graphs of some functions of this form.
y
4
y=
3e0:5x
2
4 2
2
4 x
4 y = 3e 0:5x
2
4 2
2
4
(a)
4 x
2
4 2
2
4
4
(b)
4 x
y = 4e0:3x
(c)
Figure 75 The graphs of three functions of the form f (x) = aekx

A quantity that changes in a way that can be modelled by a function
whose rule is of the form f (x) = aekx , where a and k are non-zero
constants, is said to change exponentially. If a and k are both positive,
then the graph of f looks like the graph in Figure 76(a), or a part of it. In
this case the quantity is said to grow exponentially, the function is
called an exponential growth function, and the graph is called an
exponential growth curve.
Similarly, if a is positive as before but k is negative, then the graph of f
looks like the graph in Figure 76(b), or a part of it. In this case the
284
quantity is said to decay exponentially, the function is called an

exponential decay function, and the graph is called an exponential
decay curve.
y
x
(a)
x
(b)
Figure 76 (a) An exponential growth curve (b) an exponential decay

curve
concentration (g/ml)
An example of a real-life situation that can often be modelled by an

exponential decay function is the concentration of a prescription drug in a
patients bloodstream. The concentration always peaks shortly after the
drug is administered, and then falls, quickly at rst but more slowly later,
as the drug is metabolised or eliminated from the body. For example,
Figure 77 shows such a model. The model covers the period of time after
the concentration of the drug peaks. The unit g/ml is micrograms per
millilitre.
14
12
10
8
6
4
2
Prescription drugs
4 5 6 7 8 9 10 11 12 13
time since drug administered (hours)
Figure 77 An exponential decay curve modelling the concentration of a

particular prescription drug in a patients bloodstream
The next example is about a model of this type. The solution to part (a) of
the example involves solving a pair of simultaneous exponential equations.
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Unit 3
Functions
Example 10
Using an exponential model
A drug is administered to a patient, and blood tests show that after

one and a quarter hours the concentration of the drug in the patients
bloodstream is 105 ng/ml (nanograms per millilitre), and after two
and three quarter hours it is 86.0 ng/ml. Assume that the
concentration of the drug in the patients bloodstream can be
modelled by an exponential decay function f , where f (t) is the
concentration (in ng/ml) at time t (in hours) after the drug was
administered, for t 1.
(a) Find the function f , giving each of the two constants in it to
three signicant gures.
(b) What is the predicted concentration of the drug after 10 hours?
Solution
(a) Let f (t) = aekt , where a and k are constants.
Use the information that you know about f to nd the values
of a and k.
From the information given in the question, f (1.25) = 105 and
f (2.75) = 86.0, so
ae1.25k = 105
and
ae2.75k = 86.0.
(3)
These are simultaneous exponential equations in a and k. To

solve them, rst eliminate a, by dividing one equation by the
other.
Hence
ae2.75k
86
=
,
ae1.25k
105
which gives
e2.75k1.25k =
e1.5k =
86
105
'
86
105
+
86
1.5k = ln
105
ln(86/105)
k=
1.5
k = 0.133 075 . . . .
Now nd a, by substituting into one of equations (3).
286
Substituting k = 0.133 075 . . . into the rst of equations (3) gives

ae1.25(0.133 075... ) = 105
105
a = 1.25(0.133 075... )
e
a = 124.002 . . . .
So a = 124 and k = 0.133, both to three signicant gures.
Hence the required function f is given, approximately, by
f (t) = 124e0.133t
(t 1).
(b) The predicted concentration after 10 hours, in ng/ml, is

f (10) = (124.002 . . . )e(0.133 075... )10 = 32.7712 . . . .
That is, the concentration is predicted to be about 33 ng/ml.
There are many other types of real-life situations that can be modelled by
exponential growth and decay functions. These include the level of
radioactivity in a sample of radioactive material, which decreases over
time, and, sometimes, the size of a population of organisms, such as
bacteria, plants, animals or even human beings, which often increases over
a period of time. The next activity is about an exponential model for the
growth of a population of bacteria.
Activity 50
Using an exponential model
The uid in a test tube was inoculated with a sample of bacteria, which
began to divide after 8 hours. Tests after 9 hours and 12 hours showed
that the test tube contained about 300 and 4200 bacteria per millilitre,
respectively. Assume that the number of bacteria per millilitre can be
modelled by an exponential growth function f , where f (t) is the number of
bacteria per millilitre at time t (in hours), for 8 t 24.
Radioactive waste
(a) Find the function f , giving each of the two constants in it to three
signicant gures.
(b) What is the predicted number of bacteria per millilitre after 24 hours?
Give your answer in scientic notation, to two signicant gures.
Exponential growth and decay functions have an interesting characteristic

property. If f is such a function, and you start with any value of x and add
a number to it, then the value of f (x) is multiplied by a factor. This factor
doesnt depend on the value of x that you started with, but only on the
number that you added.
Bacteria dividing
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Unit 3
Functions
For example, if f is such a function, and f (5) happens to be 3 times larger
than f (1), then also f (20) will be 3 times larger than f (16) (because you
need to add the same number of units, namely 4, to get from 16 to 20 as
from 1 to 5). Similarly, f (154) will be 3 times larger than f (150), and f (2)
will be 3 times larger than f (2), and so on.
To see why this happens, consider any exponential growth or decay
function f (x) = aekx . When you start with a particular value of x, and
add p units, say, the value of f (x) changes from
f (x) = aekx
to f (x + p) = aek(x+p) .
Now,
aek(x+p) = aekx+kp = aekx ekp = f (x) ekp .
So the value of f (x) is multiplied by the factor ekp .
Heres a concise statement of the fact discussed above.
A characteristic property of exponential growth and decay

functions
If f (x) = aekx , then whenever p units are added to the value of x, the
value of f (x) is multiplied by ekp .
For any exponential function f , the factor by which f (x) is multiplied

when a particular amount is added to the value of x is called a growth
factor or a decay factor, according to whether f is an exponential
growth or decay function. Growth factors are greater than 1, and decay
factors are between 0 and 1, exclusive. The next example illustrates how to
nd growth or decay factors.
Example 11
Finding growth factors for exponential growth
Suppose that the number f (t) of bacteria per millilitre of uid in a

test tube at time t (in hours) after it was inoculated is modelled by
the exponential growth function
f (t) = 3e0.4t
(9 t 30).
By what factor is the number of bacteria predicted to multiply

(a) every hour?
(b) every two and a half hours?
Give your answers to three signicant gures.
288
Solution
Use the property in the box above.
(a) In every hour, the number of bacteria is predicted to multiply by
the factor
e0.41 = e0.4 = 1.49 (to 3 s.f.).
(b) In every period of 2.5 hours, the number of bacteria is predicted
to multiply by the factor
e0.42.5 = e1 = 2.72 (to 3 s.f.).
Here are some examples of exponential growth and decay for you to
analyse.
Activity 51
Finding growth factors for exponential growth
Suppose that the number of trees of a particular variety in a region, at

time t (in decades) after the variety was introduced, is modelled by the
exponential growth function
f (t) = 700e0.06t
(10 t 50).
By what factor is the size of the tree population predicted to multiply

(a) every decade?
(b) every century?
(c) every ve years?
Give your answers to three signicant gures.
The phrase grown exponentially is used frequently in the media.

However the intended meaning is nearly always that a quantity has
grown a lot, or has grown quickly, rather than the true meaning. Its
possible for a quantity to grow slowly but exponentially. For example,
this is true of the size of the population of trees in Activity 51.
Modelling tree populations is

a long-term project
Activity 52
Finding decay factors for exponential decay
Suppose that the radioactivity level r(t) (in becquerels) of a sample of

radioactive material at time t (in years) after the level was rst measured
is modelled by the function
r(t) = 2800e0.035t
(t 0).
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Unit 3
Functions
By what factor is the radioactivity level predicted to multiply
(a) every year?
(b) every 25 years?
(c) every century?
Give your answers to two signicant gures.
The characteristic property of exponential growth and decay functions in

the box above gives us a useful way to describe how quickly a particular
instance of exponential growth or decay is taking place. For example,
consider again the situation in Example 11, where the number f (t) of
bacteria per millilitre of uid in a test tube at time t (in hours) after it was
inoculated is modelled by the exponential growth function
f (t) = 3e0.4t
(9 t 30).
You can see that, by the property in the box, if p is the number such that
e0.4p = 2,
(4)
then whenever you add p to t, the value of f (t) is multiplied by exactly 2.

This value of p is called the doubling period for the function f . Solving
equation (4) gives
0.4p = ln 2;
that is,
p = (ln 2)/0.4 = 1.732 867 . . . .
So the number of bacteria per millilitre doubles every 1.7 hours,

approximately. In general we make the following denitions.
Doubling and halving periods

Suppose that f is an exponential growth function. Then p is the
doubling period of f if whenever you add p to x, the value of f (x)
doubles.
Similarly, suppose that f is an exponential decay function. Then p is
the halving period of f if whenever you add p to x, the value of
f (x) halves.
Heres a summary of how to nd doubling or halving periods.
Strategy:
To nd a doubling or halving period
If f (x) = aekx is an exponential growth function (so k > 0), then the
doubling period of f is the solution p of the equation ekp = 2; that is,
p = (ln 2)/k.
Similarly, if f (x) = aekx is an exponential decay function (so k < 0),
then the halving period of f is the solution p of the equation ekp = 21 ;
that is, p = (ln 21 )/k = (ln 2)/k.
290
Inequalities
In many exponential models, such as the ones in Example 11 and

Activities 51 and 52, the input variable represents time. When this is the
case, the doubling or halving period is more usually called the doubling
or halving time. In the particular case of radioactive decay, the halving
time is usually called the half-life.
Activity 53
Finding doubling and halving periods
Find the doubling time of the exponential growth in Activity 51, and the
half-life of the exponential decay in Activity 52. Give your answers to
three signicant gures.
Radiocarbon dating is a method of estimating the age of material

that originates from a living organism, such as an animal or a plant.
A living organism absorbs the radioactive isotope carbon-14 from the
atmosphere. When it dies, the amount of carbon-14 in its remains
decays exponentially, with a half-life of 5730 years. A measurement of
the amount of carbon-14 thats left can be used to estimate when the
organism was alive, up to about 60 000 years ago. For example,
radiocarbon dating of organic material at Stonehenge was used in
2008 to determine that the monument was built in about 2300 BC.
Radiocarbon dating was developed by Willard Libby at the
University of Chicago in 1950. He was awarded the Nobel Prize for
Chemistry in 1960.
Willard Libby (190880)
5 Inequalities
In the module so far youve worked with equations of various types.
However, sometimes you need to work not with equations, but with
inequalities. Whereas an equation expresses the fact that two quantities
are equal, an inequality expresses the fact that one quantity is greater
than, less than, greater than or equal to, or less than or equal to, another
quantity.
Stonehenge, Wiltshire
291
Unit 3
Functions
5.1 Terminology for inequalities

You saw some examples of inequalities in Section 1 of this unit. Here are a
few more:
x 5,
3a 2 > b + 1,
p2 5p + 6 0,
2t > 10.
In general, an inequality is the same as an equation, except that instead

of an equals sign it contains one of the four inequality signs, <, , >
and . In other words, an inequality is made up of two expressions, with
one of the four inequality signs between them.
Inequality signs
<
>
is
is
is
is
less than
less than or equal to
greater than
greater than or equal to
Much of the terminology that applies to equations also applies to

inequalities. For example:
an inequality in x is one that contains the variable x and no other

variables (the rst inequality above is an example)
the solutions of an inequality are the values of its variables that

satisfy it in other words, they are the real numbers for which it is
true (for instance, the values x = 6 and t = 4 are solutions of the rst
and fourth inequalities above, respectively, and there are many other
solutions)
solving an inequality means nding all its solutions
two inequalities are equivalent if they contain the same variables and
are satised by the same values of those variables
rearranging an inequality means transforming it into an equivalent

inequality.
Most inequalities have either innitely many solutions or no solutions. A

useful way to specify all the solutions of an inequality is to state the set
that they form. This set is called the solution set of the inequality. For
example, the solution set of the simple inequality x 5 is the interval
[5, ), and the solution set of the inequality x2 < 0 is the empty set .
In this section youll learn how to rearrange inequalities, and youll see how
you can use this technique to help you solve some types of inequality in
one variable. Youll also see how the graphs of functions can help you
visualise the solution sets of inequalities, and youll learn some further
techniques that you can use to extend the range of inequalities in one
variable that you can solve.
292
Inequalities
5.2 Rearranging inequalities

You can rearrange inequalities using methods similar to those that you use
for rearranging equations. However there are some important dierences.
Here are the three main ways to rearrange an inequality.
Rearranging inequalities
Carrying out any of the following operations on an inequality gives an
equivalent inequality.
Rearrange the expressions on one or both sides.
Swap the sides, provided you reverse the inequality sign.
Do any of the following things to both sides:
add or subtract something
multiply or divide by something thats positive
multiply or divide by something thats negative, provided you

reverse the inequality sign.
To understand why these rules make sense, consider, for example, the
simple, true inequality 1 < 2.
You can swap the sides of this inequality to obtain another true
inequality, provided you reverse the inequality sign. This gives 2 > 1.
You can multiply both sides of the original inequality 1 < 2 by the
positive number 3, say, to obtain another true inequality. This gives
3 < 6.
You can multiply both sides of the original inequality 1 < 2 by the
negative number 3, say, to obtain another true inequality, provided
you reverse the inequality sign. This gives 3 > 6.
When youre rearranging an inequality, you should not multiply or divide

both sides by a variable, or by an expression containing a variable, unless
you know that the variable or expression takes only positive values or takes
only negative values. Thats because in other cases you cant follow the
rule about when to reverse the inequality sign, so usually the inequality
that you obtain wont be equivalent to the original one.
5.3 Linear inequalities

As youd expect, a linear inequality in one unknown is the same as a
linear equation in one unknown, but with one of the four inequality signs
in place of the equals sign. You can solve such an inequality by using the
same methods that you use to solve a linear equation in one unknown (see
Subsection 5.3 of Unit 1), except that you need to use the rules for
rearranging inequalities instead of the rules for rearranging equations.
Heres an example.
293
Unit 3
Functions
Example 12
Solving a linear inequality
Solve the inequality

5x
7
1 > 4x + .
2
2
Give your answer as a solution set in interval notation.
Solution
The inequality
5x
7
1 > 4x +
2
2
can be rearranged as follows.
Clear the fractions, by multiplying through by 2. This is a positive
number, so leave the direction of the inequality sign unchanged.
5x 2 > 8x + 7
Get all the terms in the unknown on one side, and all the other
terms on the other side. Collect like terms.
5x 8x > 7 + 2
3x > 9
Obtain x by itself on one side, by dividing through by 3. This is
a negative number, so reverse the inequality sign.
x < 3
The solution set is the interval (, 3).
Activity 54
Solving linear inequalities
(a) Solve the following linear inequalities. Give your answers as solution
sets in interval notation.
x
(i) 5x + 2 < 3x 1
(ii) 6 3x 1
2
(b) An employee has achieved 54%, 69% and 72% in the rst three of her
four assignments in a workplace training course. She has to achieve an
average of at least 60% over all four assignments (which are
equally-weighted) to pass the course. Let x% be the score that she will
achieve for her nal assignment. Write down an inequality that x must
satisfy if the employee is to pass the course, and solve it to nd the
acceptable values of x.
294
Inequalities
5.4 Quadratic inequalities

Sometimes you have to solve a quadratic inequality. As youd expect,
this is an inequality thats the same as a quadratic equation, but with one
of the inequality signs in place of the equals sign.
The rst step in solving a quadratic inequality is to simplify it, if possible,
in the same ways that youd simplify a quadratic equation (this was
covered in Subsection 4.3 of Unit 2). In particular, you should get all the
terms on one side of the inequality sign, leaving just the number zero on
the other side. Of course, you have to simplify the inequality using the
rules for rearranging inequalities.
Once youve simplied the inequality, you can solve it by considering the
graph of the function whose rule is given by the quadratic expression on
one side of the inequality, as illustrated in the next example.
Example 13
Solving a quadratic inequality

x2 + 3 4x.
Solution
Get all the terms on one side. Simplify the inequality in other
ways if possible in this case theres no further simplication to be
done.
Rearranging the inequality x2 + 3 4x gives
x2 4x + 3 0.
Roughly sketch the graph of y = x2 4x + 3. The only features
that you need to show are the x-intercepts and whether the parabola
is u-shaped or n-shaped. In particular, theres no need to nd the
vertex.
The x-intercepts of the graph of f (x) = x2 4x + 3 are given by
x2 4x + 3 = 0;
that is,
(x 1)(x 3) = 0,
so they are 1 and 3.
Also, the graph is u-shaped. So the graph is as follows.
295
Unit 3
Functions
y
y = x2 4x + 3
From the sketch you can see that the value of the expression
x2 4x 3 is greater than or equal to zero precisely when x 1 or
x 3 (since the parabola lies above or on the x-axis for these values
of x).
The solution set is (, 1] [3, ).
As you become more familiar with the method in Example 13, you might
nd that you dont need to sketch the graph instead you can just work
out the intercepts, note whether the graph is u-shaped or n-shaped, and
picture the sketch in your head. You might like to try this in the later
parts of the next activity.
If you prefer, you can always make sure that the coecient of x2 is positive
by, if necessary, multiplying the inequality through by 1 and reversing
the inequality sign. Then you dont need to think about whether the
parabola is u-shaped or n-shaped, as it will always be u-shaped.
Activity 55
Solving quadratic inequalities
Solve the following inequalities.

(a) x2 + x < 2
(b) x2 + 7x < 10
(c) x2 2x
The next example illustrates an alternative method for solving a quadratic

inequality. This method starts in the same way as the method that youve
just seen you rearrange the inequality to obtain a quadratic expression
on one side. Then, instead of using a graph to determine the values of x
that make the value of the quadratic expression greater than, less than or
equal to zero, you obtain the same information by constructing a type of
table known as a table of signs. You can use this alternative method
whenever you can factorise the quadratic expression. It might seem a little
more complicated than the method that youve just practised, but its
worth learning, as you can use it to solve more complicated inequalities.
Youll see this in the next subsection.
296
Example 14
Inequalities
Solving a quadratic inequality using a table of signs

2x2 + x 6 0.
Solution
Make sure that all the terms are on one side, and simplify the
inequality in other ways if possible here theres no simplication to
be done. Next, factorise the quadratic expression on the left-hand
side, and nd the values of x for which the resulting factors are equal
to zero.
Factorising gives
(x + 2)(2x 3) 0.
The factors x + 2 and 2x 3 are equal to zero when x = 2 and
when x = 23 , respectively.
Construct a table, as follows. In the top row, write, in increasing
order, the values of x for which the factors are equal to zero, and also
the largest open intervals to the left and right of, and between, these
values. In the left-most column, write the factors x + 2 and 2x 3,
and then their product (x + 2)(2x 3).
We have the following table.
x
(, 2) 2 (2, 23 )
3
2
( 32 , )
x+2
2x 3
(x + 2)(2x 3)
The factor x + 2 is zero when x = 2, negative when x < 2 and
positive when x > 2, so ll in its row appropriately. Use similar
thinking to ll in the row for the factor 2x 3. Finally, use the signs
of x + 2 and 2x 3 to nd the signs of (x + 2)(2x 3) for the various
values of x, and enter these in the bottom row. For example, if x + 2
and 2x 3 are both negative, then their product is positive.
x
(, 2) 2 (2, 23 )
3
2
( 23 , )
x+2
2x 3
+
0
+
+
(x + 2)(2x 3)
Use the entries in the bottom row to help you solve the inequality.
Remember that youre looking for the values of x such that
(x + 2)(2x 3) is positive or zero.
The solution set is (, 2] [ 32 , ).
297
Unit 3
Functions
Activity 56
Solving quadratic inequalities using tables of signs
Solve the following inequalities using tables of signs.

(a) 2x2 5x 3 < 0
(b) 2x2 + 4x + 16 0
5.5 More complicated inequalities

In this nal subsection youll see how to use tables of signs to solve more
complicated inequalities. In particular, this method is useful for some
inequalities that contain algebraic fractions, such as
3
2x + 3.
x1
When youre trying to solve an inequality that contains an algebraic
fraction, remember that youre not allowed to multiply it through by a
variable or expression, unless you know that the variable or expression
takes only positive values or takes only negative values. Thats because
otherwise you cant follow the rule about when to reverse the inequality
sign. So, for example, you cant simplify the inequality above by
multiplying through by x 1. (You could consider the two cases x 1 > 0
and x 1 < 0 separately, but its more straightforward to use the method
illustrated in the following example.)
Example 15
Solving an inequality containing algebraic fractions

3
2x + 3.
x1
Solution
Get all the terms on one side, leaving only 0 on the other side.
3
(2x + 3) 0
x1
Combine the terms into a single algebraic fraction, and
simplify it.
298
Inequalities
3
(2x + 3)(x 1)
0
x1
x1
3 (2x + 3)(x 1)
0
x1
3 (2x2 + x 3)
0
x1
2x2 x + 6
0
x1
2x2 + x 6
0
x1
Factorise the numerator and denominator, where possible. Find
the values of x for which the factors are equal to 0.
(2x 3)(x + 2)
0
x1
A factor is equal to 0 when x = 2, x = 1 or x = 23 .
Construct a table of signs to help you nd the values of x for
which the whole fraction is positive, negative or zero. You need a row
for each of the three factors.
x
(, 2) 2 (2, 1) 1 (1, 32 )
3
2
( 23 , )
2x 3
x+2
x1
(2x 3)(x + 2)
x1
Fill in the row for each factor. Then use the signs of the factors to
nd the signs of the whole fraction, and enter these in the bottom
row. Note that where a factor in the denominator takes the value 0,
the fraction is undened. Use the symbol to indicate this.
x
(, 2) 2 (2, 1)
(1, 32 )
3
2
( 23 , )
2x 3
x+2
x1
+
0
+
+
0
+
+
+
+
+
(2x 3)(x + 2)
x1
Use the entries in the bottom row to help you solve the inequality.
Remember that youre looking for the values of x such that
(2x 3)(x + 2)/(x 1) is positive or zero.
The solution set is [2, 1) [ 32 , ).
299
Unit 3
Functions
Activity 57
Solving inequalities containing algebraic fractions
Solve the following inequalities by using tables of signs.

(a)
3x 4
1
2x + 1
(b)
2x2 + 5x 8
2
x3
You can check the solution set that youve found for an inequality by
obtaining the graph of an appropriate function. For example, consider
again the inequality in Example 15. It was rearranged into the form
(2x 3)(x + 2)
0.
x1
Figure 78 shows the graph of the equation
y=
(2x 3)(x + 2)
,
x1
as a computer would plot it. The expression in x here is the left-hand side
of the inequality above. The graph shows that this expression takes values
greater than or equal to zero roughly when x is in the set [2, 1) [ 23 , ).
This accords with the solution set found in Example 15.
y
30
20
10
4 3 2 110
4 x
20
30
Figure 78 The graph of y = (2x 3)(x + 2)/(x 1)

You can also use graphs to check the solution set of an inequality directly
from the original, un-rearranged version of the inequality. This provides a
more thorough check on your working. To do this, you usually have to
obtain two graphs on the same axes.
This method is illustrated in the next example, in which the solution set of
the original version of the inequality in Example 15 is estimated from a
graph.
300
Example 16
Inequalities
Estimating solutions from a graph
Use the graph below to estimate the solution set of the inequality
3
2x + 3.
x1
y
30
20
10
4 3 2 110
20
30
y = 2x + 3
1
2
y=
4 x
3
x 1
Solution
Estimate the values of x for which the graph of y = 3/(x 1) lies
below or on the graph of y = 2x + 3.
The graph shows that 3/(x 1) is less than or equal to 2x + 3
roughly when x is in the set [2, 1) [ 23 , ). So this set is the
solution set of the inequality, at least approximately.
This agrees with the solution set found in Example 15.
Estimating solution sets from graphs in the way illustrated in Example 16

can be useful not only as a check on algebraic working, but also when you
need only an approximate solution set, or when you dont know a method
for solving an inequality or equation algebraically.
301
Unit 3
Functions
Activity 58
Estimating solutions from a graph
Use the graph below to estimate the following.

15
(a) The solutions of the equation x =
.
x2
15
.
(b) The solution set of the inequality x
x2
15
(c) The solution set of the inequality x >
.
x2
y
30
20
15
10
x 2
8 6 4 2
10
y=x
y=
8 x
20
30
Learning outcomes
After studying this unit, you should be able to:
302
understand and use the terminology and notation associated with

functions
work with graphs of functions
work with a range of standard types of functions, and understand

their properties and graphs
understand the changes to the rules of functions that cause their

graphs to be translated or scaled, horizontally or vertically
form sums, dierences, products, quotients and composites of functions
understand whats meant by the inverse function of a one-to-one

function, and nd the inverse in some cases
understand the properties and graphs of exponential and logarithmic

functions
work uently and correctly with logarithms
work with exponential models
solve some types of inequalities in one variable
construct tables of signs.
Solutions to activities
Solution to Activity 1
(a) True (b) True (c) False
(d) True
(a) P Q = {2, 4, 6}
(b) Q R = {6, 12}
(c) P Q R = {6}
(d) P Q = {1, 2, 3, 4, 5, 6, 8, 10, 12}
(a) This set is an open interval.
(b) This set is not an interval.
(c) This set is a closed interval.
(d) This set is a half-open interval.
(e) This set is not an interval.
(b) (i) 2 < x 5

(ii) x 4
(iii) x < 0
(iv) 3 x < 1
(v) 0 x 6
(vi) 3 < x < 7
(a) (2, 5]
(b) [4, )
(c) (, 0)
(d) [3, 1)
(e) [0, 6]
(f) (3, 7)
(f) This set is a closed interval. (It has only one

endpoint, and it includes it.)
(a) (, 5) [2, 1]
(g) This set is an open interval.
(b) [1, 2) [3, 4) [5, 6)
(h) This set is not an interval.
(c) (, 0) (0, )
(a) (i)
(a) f (5) = 4 5 = 20 and f (3) = 4 (3) = 12

(b) g(x) = 2x 1
21 0 1 2 3
(a) The image of 2 is 8, because f (2) = 8.
(ii)
(b) The image of 1 is 4, because f (1) = 4.

54321 0 1 2 3 4
(c) The value of f at 0.5 is 2, because f (0.5) = 2.

(d) The value of f at 0.2 is 0.8, because
f (0.2) = 0.8.
(e) The number 11 has image 44 under f , because
f (11) = 44.
(iii)
0 1 2 3 4 5 6 7
(f) The number

f ( 14 ) = 1.
1
4
has image 1 under f , because
(g) The function f maps 4 to f (4) = 16.
(iv)
(h) The function f maps 2 to 8 because

f (2) = 8.
1 2 3 4 5 6 7 8
303
Unit 3
Functions
(a) The domain of f is the set of all real numbers

except 4, that is, the set (, 4) (4, ).
(a) The graph is part of the straight-line graph of

y = 3 2x.
(b) The domain of g is the set of all real numbers

except 2 and 3, that is, the set
(, 3) (3, 2) (2, ).
( 1; 5)
(c) The domain of h is the set of all real numbers x

such that x 1 is non-negative. This set can be
described more concisely as the set of all real
numbers greater than or equal to 1. In interval
notation this set is denoted by [1, ).
(0; 3)
( 23 ; 0)
x
y = 3 2x
( 1 < x < 4)
(a)
1.5
8 3.375 1 0.125
x3
0.5
(4; 5)
0.5
1.5
(b) The graph is part of an n-shaped parabola.

Completing the square gives
f (x) = 12 x2 2x 5
0.125 1 3.375 8
(b) and (c)
= 12 (x2 + 4x) 5
= 12 ((x + 2)2 4) 5
= 21 (x + 2)2 + 2 5
8
6
y=
x3
4
2
3 2 1
2
3 x
= 12 (x + 2)2 3
So the vertex is (2, 3). Also
f (5) = 21 (5)2 2 (5) 5 = 15
2 .
So the graph stops at the point (5, 15

2 ) (which is
included). These features give the following sketch.
6
8
x
( 2; 3)
(0; 5)
( 5; 15
2 )
304
y = 21 x2 2x 5
(x 5)
(a) The domain is [1, 3].
(b) The domain is (, 2] [1, ).
Diagrams (a), (c), (f), (h), (i) and (j) are the graphs
of functions.
Graphs (a) and (b) show functions that are
increasing on their whole domains.
(For the function in graph (c), if you take x1 and x2
to be values slightly less than 0 and slightly greater
than 0, respectively, then the function takes a
smaller value at x2 than it does at x1 , so it is not
increasing on its whole domain.
For the function in graph (d), you can nd values
x1 and x2 with x1 < x2 such that the function takes
the same value at x2 as it does at x1 , so it is not
increasing on its whole domain.)
(a) The graph of f is part of an n-shaped parabola.
Completing the square gives
f (x) = x2 + 10x 24
= (x2 10x) 24
= ((x 5)2 25) 24
= (x 5)2 + 25 24
= (x 5)2 + 1.
So the vertex is (5, 1). Also
f (3) = 32 + 10 3 24 = 3
and
f (6) = 62 + 10 6 24 = 0.
So the graph stops at the points (3, 3) (which
is included) and (6, 0) (which is excluded).
These features give the graph below. The image

set is shown on the y-axis.
y
(5; 1)
(6; 0)
x
y=
x2
+ 10x 24
(3; 3)
The graph shows that the image set of f is the

interval [3, 1].
(b) The graph of f is below, with the image set
shown on the y-axis.
y
( 2; 6)
y = 2 2x
2
(0; 2)
x
The graph shows that the image set is (2, 6).

(c) The image set of f is the interval [1, ).
This is because the image set of the function
g(x) = x2 is [0, ), since every non-negative
number can be expressed as the square of a
number. Hence the image set of the function
f (x) = x2 1 is [1, ).
(The graph of f is shown below, with the image
set shown on the y-axis.)
y
y = x2 1
x
305
Unit 3
Functions
(d) The image set of f is the interval (0, ).

This is because the image set of f doesnt
contain any negative numbers, since the value
of 1/x2 cant be negative. Similarly, the image
set doesnt contain 0. However, the image set
does contain every positive number, because
every positive number can be expressed as 1/x2
for some number x.
(The graph of f is shown below, with the image
set shown on the y-axis.)
y
y=
1
x2
x
(The eects that you should have seen are described
in the text after the activity.)
1
x2
1
(b) y = 2
x
1
(c) y = + 2
x
1
(d) y =
x+2
(a) y =
is the equation of graph D.

is the equation of graph A.
is the equation of graph C.
is the equation of graph B.
(a) y = |x 2| + 1 is the equation of graph C.
(a) y = 2x3 is the equation of graph A.
(b) y = 12 x3 is the equation of graph D.
(c) y = x3 is the equation of graph B.
(d) y = 12 x3 is the equation of graph C.
(a) The graph of g(x) = 2|x| + 3 can be obtained
from the graph of f (x) = |x| by rst scaling it
vertically by the factor 2 and then translating it
up by 3 units.
(b) The graph of h(x) = 2|x + 2| + 3 can be
obtained from the graph of f (x) = |x| by rst
scaling it vertically by the factor 2, then
translating it to the left by 2 units, and nally
translating it up by 3 units.
(You can carry out the operations in any order,
except that you have to do the vertical scaling
before the vertical translation.)
(c) The graph of j(x) = 12 |x 3| 4 can be
scaling it vertically by the factor 12 , translating
it to the right by 3 units, and nally translating
it down by 4 units.
before the vertical translation.)
(d) The graph of k(x) = |x 1| + 1 can be
reecting it in the x-axis (that is, scaling it
vertically by the factor 1), then translating it
to the right by 1 unit, and nally translating it
up by 1 unit.
except that you have to do the reection before
the vertical translation.)
(b) y = |x + 2| + 1 is the equation of graph D.
(c) y = |x 2| 1 is the equation of graph B.
(a) Completing the square gives

f (x) = 2x2 + 12x + 19
= 2(x2 + 6x) + 19
= 2((x + 3)2 9) + 19
= 2(x + 3)2 18 + 19
= 2(x + 3)2 + 1.
(d) y = |x + 2| 1 is the equation of graph A.
306
(b) This equation is obtained from the equation
y = x2 by rst multiplying the right-hand side
by 2, then replacing x by x + 3, and nally
adding 1 to the right-hand side. So its graph is
obtained from the graph of y = x2 by rst
scaling vertically by a factor of 2, then
translating to the left by 3 units, and nally
translating up by 1 unit.
(a) The graph of the function f (x) = 2|x| + 1 can
be obtained from the graph of y = |x| by rst
scaling it vertically by a factor of 2 and then
translating it up by 1 unit.
y

before the vertical translation. The resulting
graph is shown below.)
y = 2x2 + 12x + 19
= 2(x + 3)2 + 1
y
4
3
2
1
4 3 21 1
y = 2jxj + 1
1
x
(b) The graph of the function h(x) = (x 1)3 can

be obtained from the graph of y = x3 by
translating it to the right by 1 unit.
y
1 x
y = (x 1)3

(a) y =
(b) y
(c) y
(d) y
(e) y
(f) y
(g) y
(h) y
(i) y
x is the equation of graph D.
= x is the equation of graph H.
= 2 x 2 is the equation of graph C.
= 21 x + 2 is the equation of graph E.
= 12 x is the equation of graph F.
= x + 2 is the equation of graph G.
= x is the equation of graph B.
= 12 x + 2 is the equation of graph I.
= 2 x + 2 is the equation of graph A.
(c) The graph of the function g(x) = 1/(x + 3) can

be obtained from the graph of y = 1/x by
translating it to the left by 3 units.
x = 3
y
1
3
x
y=
1
x+3
(You can nd the y-intercept of this graph

simply by substituting x = 0 into its equation.)
307
Unit 3
Functions
The sum of f and g has rule
h(x) = 2x 1 + x + 3,
which can be simplied to
h(x) = 3x + 2.
One dierence of f and g has rule
h(x) = 2x 1 (x + 3),
h(x) = x 4.
The other dierence of f and g has rule
h(x) = x + 3 (2x 1),
h(x) = x + 4.
The product of f and g has rule
h(x) = (2x 1)(x + 3),
which can also be expressed as
h(x) = 2x2 + 5x 3.
The two quotients of f and g have rules
2x 1
h(x) =
x+3
and
x+3
h(x) =
.
2x 1
All of these functions have domain R, except the
nal two functions, which have domains
(, 3) (3, ) and (, 12 ) ( 21 , ),
respectively.
The function f maps 5 to 25, and the function g
maps 25 to 26, so (g f )(5) = 26.
(g f )(x) = g(f (x)) = g(x 3) = x 3
(ii) (f g)(x) = f (g(x)) = f ( x) = x 3
(a) (i)
(iii)
(iv)
(f f )(x) = f (f (x)) = f (x 3)
= (x 3) 3 = x 6
!
(g g)(x) = g(g(x)) = g( x) =
x
= (x1/2 )1/2 = x1/4
308
(b) If x is in the domain of f , then f (x) = x 3.

For x 3 to be in the domain of g, the value of
x 3 must be greater than or equal to zero,
which means that the value of x must be
greater than or equal to 3. That is, the domain
of g f is [3, ).
(a)
(f g h)(x) = f (g(h(x)))
= f (g( x))
'
+
1
=f
x
1
= +2
x
(b)
(g h f )(x) = g(h(f (x)))

= g(h(x + 2))
= g( x + 2)
1
=
x+2
(f h g)(x) = f (h(g(x)))
' ' ++
1
=f h
x
)" $
1
=f
x
+
'
1
=f
x
1
= +2
x
(c)
(d)
(f g f )(x) = f (g(f (x)))

= f (g(x + 2))
+
'
1
=f
x+2
1
=
+2
x+2
1 + 2(x + 2)
=
x+2
2x + 5
=
x+2
(a) (i)
The rule of the inverse function of

f (x) = x + 1 is f 1 (x) = x 1.
(ii) The rule of the inverse function of

f (x) = x 3 is f 1 (x) = x + 3.
(iii) The rule of the inverse function of
f (x) = 13 x is f 1 (x) = 3x.
(b) Some possible answers are f (x) = x, f (x) = x
and f (x) = 1/x. There are many others, such
as f (x) = 3 x and f (x) = 8/x.
(a) The function f (x) = |x| is not one-to-one. For
example, f (1) = f (1) = 1.
(b) The function f (x) = x + 1 is one-to-one.
(c) The function f (x) = x4 is not one-to-one. For
example, f (1) = f (1) = 1.
(d) The function f (x) = x5 is one-to-one.
(e) The function f (x) = x is one-to-one.
(f) The function f (x) = 1 is not one-to-one. For
example, f (0) = f (1) = 1.
Diagrams (a), (b), (e), (f), (h), (i) and (k) are the
graphs of one-to-one functions.
Diagrams (c), (d) and (j) are the graphs of
functions that arent one-to-one.
Diagrams (g) and (l) are not the graphs of functions.
(a) The equation f (x) = y gives
3x 4 = y
3x = y + 4
x = 13 (y + 4).
Since the equation f (x) = y can be rearranged
to express x as a function of y, the function f
has an inverse function f 1 , with rule
f 1 (y) = 13 (y + 4);
that is,
f 1 (x) = 13 (x + 4).
(b) The equation f (x) = y gives

2 12 x = y
2 y = 12 x
x = 2(2 y).
Since the equation f (x) = y can be rearranged
to express x as a function of y, the function f
has an inverse function f 1 , with rule
f 1 (y) = 2(2 y);
that is,
f 1 (x) = 2(2 x).
The domain of f 1 is the image set of f , which
is R. So the inverse function f 1 is given by
f 1 (x) = 2(2 x).
(c) The equation f (x) = y gives

1
5+ =y
x
1
=y5
x
1
x=
.
y5
Hence f has an inverse function f 1 , with rule
1
;
f 1 (y) =
y5
that is,
1
.
f 1 (x) =
x5
is (, 5) (5, ). This is the largest set of
real numbers for which the rule of f 1 is
applicable. So the inverse function f 1 is given
by
1
f 1 (x) =
.
x5

is R. So the inverse function f 1 is given by
f 1 (x) = 13 (x + 4).
309
Unit 3
Functions
that is,
f 1 (x) = 1 + x 1.
The domain of f 1 is the image set of f . The
graph shows that this is
(f (0), f (2)) = (2, 10).
So the inverse function of f is the function
f 1 (x) = 1 + x 1 (x (2, 10)).
(a) The graph of f is shown below.

y
10
8
6
y = x2 + 2x + 2
4
2
2 x
The graph shows that f isnt one-to-one. So it

doesnt have an inverse function.
(b) The graph of f is shown below.
y
10
8
6
y = x2 + 2x + 2
(c) The equation f (x) = y gives

1x=y
x = 1 y.
Hence f has an inverse function f 1 , with rule
f 1 (y) = 1 y;
that is,
f 1 (x) = 1 x.
is
[f (1), f (3)] = [0, 4].
Hence the inverse function of f is
f 1 (x) = 1 x (x [0, 4]).
(The graph of f , shown below, might help you nd
the image set of f .)
y
4
3
2
1
2
2
2 x
y =1 x
The graph shows that f is one-to-one and

therefore has an inverse function. The equation
f (x) = y gives
x2 + 2x + 2 = y
(x + 1)2 1 + 2 = y
(x + 1)2 + 1 = y
(x + 1)2 = y 1
&
x+1= y1
&
x = 1 y 1.
Since the domain of f is (0, 2), each input
value x of f is greater than 1. So
&
x = 1 + y 1.
&
f 1 (y) = 1 + y 1;
310
1 2 x
4321
(a)
y
2
( 2; 2)
(b)
y
Hence the rule of g 1 is
g 1 (y) = 1 + y;
that is,
g 1 (x) = 1 + x.
The domain of g 1 is the image set of g, which is
[0, ).
So the inverse function of g is the function
g 1 (x) = 1 + x (x [0, )).

The graph of the function g is as shown below.
(c)
y
3
2
1
Hence the graph of g 1 is as shown below.

y
3
2
1
(d)
y
1 2 3 x
2 1
(4; 4)
2 1
1 2 3 x
(Alternatively, you can take the domain of g to be

(, 1]. Then g 1 is the function
g 1 (x) = 1 x (x [0, ).)
The image set of f is [0, ).
A one-to-one function g that is a restriction of f
and has the same image set as f is
g(x) = (x 1)2 (x [1, )).
The equation g(x) = y gives
(x 1)2 = y
x1= y
x = 1 y.
Since the domain of g is [1, ), each input value x
of f is greater than or equal to 1. So
x = 1 + y.
(The answer is given in the text after the activity.)
311
Unit 3
Functions
(a) (i)
(b) (i)
log10 10 000 = 4, since 10 000 = 104 .
(ii) log10
1
100
= 2, since
1
100
(ii) If log8 x = 31 , then x = 81/3 = 2.
= 102 .
(iii) If log7 x = 1, then x = 71 = 7.
(iii) log10 10 = 1, since 10 = 101 .
(iv) log10 1 = 0, since 1 = 100 .

(b) If the number x is such that log10 x =
101/2 = x; that is, x = 10.

(c) (i)
1
2,
then
(a) (i)
(iii) ln e3/5 = 35
(iv) ln e = ln(e1/2 ) = 12
' +
1
(v) ln
= ln(e1 ) = 1
e
' +
1
(vi) ln 3 = ln(e3 ) = 3
e
(ii) log10 370 = 2.568 (to 3 d.p.)

(iii) log10 37 = 1.568 (to 3 d.p.)
(iv) log10 3.7 = 0.568 (to 3 d.p.)
(v) log10 0.37 = 0.432 (to 3 d.p.)
(vi) log10 0.037 = 1.432 (to 3 d.p.)
(Youll be asked to explain the pattern in the
answers to part (c) later in this section.)
(a) (i)
(b) If the number x is such that ln x = 12 , then
x = e1/2 = 1/ e.
(c) (i)
log3 9 = 2, since 3 = 9.
(iii) ln 51 = 3.932 (to 3 d.p.)
(ii) log2 8 = 3, since 2 = 8.
(iv) ln(51e) = 4.932 (to 3 d.p.)
(iii) log4 64 = 3, since 4 = 64.

(iv) log5 25 = 2, since 52 = 25.
(v) log4 2 = 12 , since 41/2 = 2.
(vi) log8 2 =
(vii) log2
(viii)log2
(ix) log3
(x) log8
since 8
1/3
= 2.
1
1
= 21 .
2 = 1, since 2
1
3
= 81 .
8 = 3, since 2
1
1
3
= 27
.
27 = 3, since 3
1
1
= 81 .
8 = 1, since 8
1
(xi) log3 3 = 1, since 3 = 3.

(xii) log4
1
4
= 1, since 41 = 41 .
(xiii)log6 6 = 1, since 61 = 6.
(xiv)log5 5 = 21 , since 51/2 = 5.
(xv) log7 3 7 = 31 , since 71/3 = 3 7.

(xvi)log2 1 = 0, since 20 = 1.
(xvii)log15 1 = 0, since 150 = 1.
312
ln 5100 = 8.537 (to 3 d.p.)
(ii) ln 510 = 6.234 (to 3 d.p.)
1
3,
ln e4 = 4
(ii) ln e2 = 2
log10 3700 = 3.568 (to 3 d.p.)
If log2 x = 5, then x = 25 = 32.
(v) ln(51e2 ) = 5.932 (to 3 d.p.)
(a) eln(7x) = 7x
(b) ln(e8x ) = 8x
(c) ln(e2x ) + ln(e3x ) = 2x + 3x = 5x
(d) ln(e2 ) ln e = 2 1 = 1
x
x
(e) ln(ex/2 ) + 3 ln 1 = + 3 0 =
2
2
% ln c #2
2
2 ln c
(ln c)2
=c
(f) e
=e
= e
(g) eln(3a) + 4e0 = 3a + 4 1 = 3a + 4
ln(ey+2 ) + 2 ln(ey1 )
= y + 2 + 2(y 1)
= y + 2 + 2y 2
= 3y
%
#3
(i) e3 ln B = e(ln B)3 = eln B = B 3
(h)
(j) e2+ln x = e2 eln x = e2 x
(a) (i)
(v)
ln(4x) + 3 ln x ln(e6 ) = ln(4x) + ln x3 6

= ln(4x x3 ) 6
= ln(4x4 ) 6
(Other acceptable nal answers include
ln 4 + ln(x4 ) 6 and ln 4 + 4 ln x 6.)
(vi)
1
2
ln 5 + ln 3 = ln(5 3) = ln 15
(ii) ln 2 ln 7 = ln( 72 )
(iii) 3 ln 2 = ln(23 ) = ln 8
'
(iv) ln 3 + ln 4 ln 6 = ln
(vii)
(viii)
(b) (i)
+
= ln 2
ln 24 2 ln 3 = ln 24 ln 32
' +
24
= ln
32
= ln( 38 )
(v)
(vi)
34
6
1
3
log10 27 = log10 271/3 = log10 3

3 log2 5 log2 3 + log2 6
= log2 53 log2 3 + log2 6
+
' 3
5 6
= log2
3
= log2 250
1
2
(The answer 4 ln u is just as acceptable.)

(c) The pattern can be explained as follows. It
follows from the logarithm laws that, for any
value of n,
log10 (37 10n ) = log10 37 + log10 (10n )
= log10 37 + n.
So if you multiply 37 by 10n , then its common
logarithm is increased by n.
(d)
(a)
(ii) 3 ln(p2 ) = ln(p6 )

(The answer 6 ln p is just as acceptable.)
(iii)
ln(y ) + 2 ln y
ln(y )
*
(
= ln(y 2 ) + ln(y 2 ) ln (y 3 )1/2
= ln(y 2 ) + ln(y 2 ) ln(y 3/2 )
+
' 2
y y2
= ln
y 3/2
= ln(y 5/2 )
(The nal answer
acceptable.)
5
2
(iv) ln(3u) ln(2u) = ln
ln y is just as
'
3u
2u
= ln( 23 )
(The nal answer ln 3 ln 2 is just as

acceptable.)
5x = 0.5
ln(5x ) = ln 0.5
x ln 5 = ln 0.5
ln 0.5
x=
ln 5
x = 0.430 676 . . .
The solution is x = 0.431 (to 3 s.f.).
(Alternatively, you can proceed as follows:
5x = 0.5
x = log5 (0.5)
x = 0.430 676 . . . .)
1
2
1567 2786 = eln(15672786)

= eln 1567+ln 2786
e7.356 918 242+7.932 362 154
e15.289 280 4
4 365 662
(This is the exact answer.)
ln(9x) ln(x + 1)
= ln(9x)1/2 ln(x + 1)
'
+
(9x)1/2
= ln
x+1
' +
3 x
= ln
x+1
' 3+
c
3
ln c ln c = ln
= ln(c2 )
c
(The nal answer 2 ln c is just as
acceptable.)
ln(u8 ) = ln(u4 )
(b)
4e7t = 64
e7t = 16
ln(e7t ) = ln 16
7t = ln 16
t = 17 ln 16
t = 0.396 084 . . .
The solution is t = 0.396 (to 3 s.f.).
313
Unit 3
(c)
Functions
5 2u/2 + 30 = 600
5 2u/2 = 570
2u/2 = 114
ln(2u/2 ) = ln 114
1
2 u ln 2 = ln 114
u ln 2 = 2 ln 114
2 ln 114
u=
ln 2
u = 13.665 780 . . .
The solution is u = 13.7 (to 3 s.f.).
(Alternatively, you can proceed as follows from
the third equation above:
2u/2 = 114
u
= log2 114
2
u = 2 log2 114
u = 13.665 780 . . . .)
(d)
23x5 = 100
ln(23x5 ) = ln 100
(3x 5) ln 2 = ln 100
ln 100
3x 5 =
ln 2
ln 100
3x =
+5
'ln 2
+
ln 100
1
+5
x= 3
ln 2
x = 3.881 285 396 . . .
The solution is x = 3.88 (to 3 s.f.).
(Alternatively, you can proceed as follows:
23x5 = 100
3x 5 = log2 (100)
3x = log2 (100) + 5
x = 13 (log2 (100) + 5)
x = 3.881 285 396 . . . .)
Since ln 3 = 1.098 612 (to 7 s.f.), the rule of f can be
written, approximately, as
f (x) = e1.098 612x .
Using the original form of the rule gives
f (1.5) = 31.5 = 5.20 (to 3 s.f.).
Using the alternative form gives
f (1.5) = e1.098 6121.5 = 5.20 (to 3 s.f.).
314
(a) Let f (t) = aekt , where a and k are constants.
Then f (9) = 300 and f (12) = 4200, so
ae9k = 300 and ae12k = 4200.
(5)
Hence
4200
ae12k
=
,
ae9k
300
which gives
e12k9k = 14
e3k = 14
3k = ln 14
k = 31 ln 14
k = 0.879 685 . . . .
The rst of equations (5) can be written as
a(e3k )3 = 300 and substituting e3k = 14 into
this equation gives
a 143 = 300,
so
75
300
a= 3 =
14
686
= 0.109 329 . . . .
So a = 0.109 and k = 0.880, both to three
signicant gures.
Hence the required function f is given,
approximately, by
f (t) = 0.109e0.880t (8 t 24).
(b) The predicted number of bacteria per millilitre
after 24 hours is
f (24) = (0.109 329 . . . )e(0.879 685... )24
= 1.6 108 (to 2 s.f.).
(a) Every decade the size of the tree population is
predicted to multiply by the factor
e0.061 = 1.06 (to 3 s.f.).
(b) Every century (10 decades) the size of the tree
population is predicted to multiply by the factor
e0.0610 = 1.82 (to 3 s.f.).
(c) Every ve years (0.5 decades) the size of the
tree population is predicted to multiply by the
factor
e0.060.5 = 1.03 (to 3 s.f.).
(a) Every year the level of radioactivity is predicted

to multiply by the factor
e0.0351 = 0.97 (to 2 s.f.).
(a) (i)
(b) Every 25 years the level of radioactivity is

predicted to multiply by the factor
e0.03525 = 0.42 (to 2 s.f.).
(c) Every century (100 years) the level of
radioactivity is predicted to multiply by the
factor
e0.035100 = 0.030 (to 2 s.f.).
(a) The exponential growth function in Activity 51
is
f (t) = 700e0.06t (10 t 50),
where f (t) is the number of trees at time t (in
decades) after the variety was introduced.
The doubling time p (in decades) for this
exponential growth is given by
p = (ln 2)/0.06 = 11.6 (to 3 s.f.).
So the number of trees doubles every 11.6
decades (116 years), approximately.
(b) The exponential decay function in Activity 52 is
r(t) = 2800e0.035t (t 0),
where r(t) (in becquerels) is the level of
radioactivity of the sample of radioactive
material at time t (in years) after the level was
rst measured.
The half-life p (in years) for this exponential
decay is given by
p = (ln 21 )/(0.035) = 19.8 (to 3 s.f.).
So the level of radioactivity halves every 19.8
years, approximately.
Rearranging the inequality gives

5x + 2 < 3x 1
2x < 3
x < 32 .
The solution set is the interval (, 23 ).
(ii) Rearranging the inequality gives

x
6 3x 1
2
12 6x x 2
7x 14
x 2.
The solution set is the interval (, 2].
(b) Each acceptable value of x satises the
inequality
54 + 69 + 72 + x
60.
4
Solving this inequality gives
195 + x
60
4
195 + x 240
x 45.
So the employee must score at least 45% in her
nal assignment to pass the course.
(a) The inequality can be rearranged as follows:
x2 + x < 2
x2 + x 2 < 0.
The graph of f (x) = x2 + x 2 is u-shaped. Its
intercepts are given by
x2 + x 2 = 0;
that is,
(x + 2)(x 1) = 0.
So they are x = 2 and x = 1.
315
Unit 3
Functions
Hence the graph is as shown below.

y
y = x2 + x 2
The graph of f (x) = x2 + 2x is u-shaped. Its

x-intercepts are given by
x2 + 2x = 0;
that is,
x(x + 2) = 0
So they are x = 0 and x = 2. Hence the graph
is as shown below.
y
y = x2 + 2x
The solution set is (2, 1).

(b) The inequality can be rearranged as follows:
x2 + 7x < 10
x2 + 7x 10 < 0.
The graph of f (x) = x2 + 7x 10 is n-shaped.
Its intercepts are given by
x2 + 7x 10 = 0;
that is,
x2 7x + 10 = 0,
or
(x 5)(x 2) = 0.
So they are x = 5 and x = 2. Hence the graph
is as shown below.
y
y = x2 + 7x 10
The solution set is (, 2) (5, ).

(c) The inequality can be rearranged as follows:
x2 2x,
x2 2x 0
x2 + 2x 0.
316
The solution set is [2, 0].
(a) The inequality is
2x2 5x 3 < 0,
which can be factorised as
(2x + 1)(x 3) < 0.
A factor is equal to 0 when x = 12 or x = 3.
A table of signs for the expression on the
left-hand side of the inequality is given below.
x
(, 12 ) 12
( 21 , 3)
(3, )
2x + 1
x3
+
0
+
+
(2x + 1)
(x 3)
The solution set is ( 12 , 3).

(b) The inequality is
2x2 + 4x + 16 0;
that is,
x2 2x 8 0.
Factorising gives
(x + 2)(x 4) 0.
x
(, 2) 2 (2, 4) 4 (4, )
x+2
x4
+
0
+
+
(x + 2)
+
0
(x 4)
The solution set is (, 2] [4, ).
(Heres a slightly dierent way to solve the

inequality in part (b).
Unlike the working above, this alternative
working doesnt involve multiplying through by
a negative number to simplify the inequality, so
the inequality sign stays the same way round as
in the original inequality. You might nd that
this approach helps you to avoid errors.
The inequality is
2x2 + 4x + 16 0.
Factorising gives
A factor of the numerator or denominator of

the expression on the left-hand side is equal to
0 when x = 5 or x = 12 .
A table of signs for the expression follows.
x
(, 21 ) 12
( 12 , 5)
(5, )
x5
2x + 1
0
+
+
+
x5
2x + 1
The solution set is ( 21 , 5].
2(x + 2)(x 4) 0.
(b) The inequality can be rearranged as follows:
2x2 + 5x 8
2
x3

x
3x 4
1
2x + 1
3x 4
10
2x + 1
3x 4 2x + 1
0
2x + 1 2x + 1
3x 4 2x 1
0
2x + 1
x5
0.
2x + 1
(, 2) 2 (2, 4)
(4, )
2
x+2
x4
+
0
+
+
2(x + 2)
(x 4)
The solution set is

(, 2] [4, ).)
(a) The inequality can be rearranged as follows:
2x2 + 5x 8
20
x3
2x2 + 5x 8 2(x 3)
0
x3
x3
2x2 + 5x 8 2x + 6
0
x3
2x2 + 3x 2
0
x3
(2x 1)(x + 2)
0.
x3
A factor of the numerator or denominator of
the expression on the left-hand side is equal
to 0 when x = 2, x = 12 or x = 3.
317
Unit 3
Functions
A table of signs for the expression follows.

To save space, the last row uses the notation
(2x 1)(x + 2)
f (x) =
.
x3
x
(, 2) 2 (2, 12 )
1
2
( 21 , 3)
(3, )
2x 1
x+2
x3
0
+
+
+
+
+
0
+
+
+
f (x)
The solution set is [2, 12 ] (3, ).
(a) The solutions of the equation x =
roughly 3 and 5.
15
are
x2
(b) The solution set of the inequality x

roughly (, 3] (2, 5].
(c) The solution set of the inequality x >
roughly (3, 2) (5, ).
(The answers given here are in fact exact.)
318
15
is
x2
15
is
x2
Acknowledgements
Acknowledgements
Grateful acknowledgement is made to the following sources:
Page 201: Erik De Graaf / Dreamstime.com
Page 210: Madhero88. This le is licensed under the Creative Commons
Attribution-Share Alike Licence
http://creativecommons.org/licenses/by-sa/3.0/
Page 285: Taken from: http://focustreatmentcenters.com/chronic-painaddiction-to-prescription-drugs/
Page 287: Taken from:
http://blogs.dallasobserver.com/unfairpark/2012/04/dallasowned west texas nuclea.php
http://www2.estrellamountain.edu/faculty/farabee/biobk/biobookmito.html
http://en.wikipedia.org/wiki/File:Willard Libby.jpg
Page 291: Gareth Wiscombe. This le is licensed under the Creative
Commons Attribution Licence
http://creativecommons.org/licenses/by/3.0/
Every eort has been made to contact copyright holders. If any have been
inadvertently overlooked the publishers will be pleased to make the
necessary arrangements at the rst opportunity.
319

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What is a function and how is the idea of a function fundamental in mathematics?

What is a function and how is the idea of a function fundamental in mathematics?

What are some standard types of functions that are introduced?

What are some standard types of functions that are introduced?

Unit 3

The distance walked by a woman at a particular speed depends on the

The height of a gondola on a Ferris wheel depends on the angle

The number of 5-litre tins of a particular type of paint needed by a

1 Functions and their graphs

1.1 Sets of real numbers

all the prime numbers less than 100

all the points on any particular line in the plane

all the equations that represent vertical lines

the solutions of any particular quadratic equation.

Understanding set notation

Functions and their graphs

John Venn (18341923)

and A B C = {1, 2, 3, 4, 5, 20, 21}.

Understanding unions and intersections of sets

The set membership symbol is a stylised version of the Greek letter

Sometimes every element of a set A is also an element of a set B. In this

{1, 3} is a subset of {1, 2, 3} (as shown in Figure 2)

the set of integers is a subset of the set of real numbers.

Figure 2 A subset of a set

Figure 3 The number line

Functions and their graphs

Figure 4(a) shows the set {1, 0, 1}.

In these kinds of diagrams, a solid dot indicates a number thats included

Figure 4 Sets of real numbers

Figure 5 A half-open (or

A convenient way to describe most intervals is to use inequality signs.

Functions and their graphs

Using inequality signs to describe intervals

(a) Draw diagrams similar to those in Figure 6 to illustrate the intervals

Another useful way to describe intervals is to use interval notation. For

Half-open (or half-closed) intervals

Using interval notation

Write each of the intervals below in interval notation.

Functions and their graphs

Figure 7 Two unions of intervals

and [0, 3) (3, 5] [7, 8],

Denoting unions of intervals

In the next subsection youll begin your study of functions.

Figure 8 The interval [0, 4]

1.2 What is a function?

If a car is driving along a straight road, then its displacement s (in

The electrical voltage between two points on a persons skin either

Figure 10 A processor that takes input values and produces output

Functions and their graphs

A function consists of:

a set of allowed input values, called the domain of the function

a set of values in which every output value lies, called the

a process, called the rule of the function, for converting each

Its often useful to denote a function by a letter. If a function is denoted

and the rule of h can be written as

The variable used to denote the input value of a function is sometimes

Understanding function notation

Its important to appreciate that every value in the domain of a function

Functions and their graphs