6.1 What is a function?

Mathematicians use functions to describe relationships between quantities. Examples of functions you may seen before are:

•

quadratic functions, e.g. $f(x)=3x^{2}-1$ ;
•

trigonometric functions: e.g. $\cos$ , $\sin$ and $\tan$ ;
•

exponential functions, e.g. $f(x)=2^{x}$ and $\exp$ ; and
•

logarithmic functions, e.g. $\log_{10}$ and $\ln$ .

We give a more formal definition of a function and then discuss how this aligns with the intuitive ideas you will have learned in your previous studies.

Definition 6.1 (Function, domain, codomain, graph).

A function is an object consisting of three sets:

(i)

a set $D$ called its domain;
(ii)

a set $C$ called its codomain; and
(iii)

a subset $G$ of $D\times C$ called its graph.

These three sets must have the properties:

(a)

for every $x\in D$ , there exists $y\in C$ such that $(x,y)\in G$ ; and
(b)

if $(x,y)\in G$ and $(x,z)\in G$ then $y=z$ .

Note, we typically give the function a name such as $f$ . Writing ‘ $f(x)=y$ ’ means that $(x,y)\in G$ . The two defining properties (a) and (b) of a function in Definition 6.1 state that each element of the domain is matched with exactly one element of the codomain. [Check that you see why.]

Example 6.2.

You may have seen statements like ‘ $f(x)=3x^{2}-1$ ’ in your previous studies. Let’s discuss what this means in terms of the formal definition above.

The variable $x$ in ‘ $f(x)=3x^{2}-1$ ’ can be any real number. So the domain could be $D=\mathbb{R}$ .

The result of calculating $f(x)$ for a real number $x$ will also be a real number. So in this example the codomain could be $C=\mathbb{R}$ .

Here we define the graph as the set $G=\{(x,3x^{2}-1):x\in\mathbb{R}\}$ in Definition 6.1. Since elements of $G$ are pairs of real numbers, these can be plotted on a pair of coordinate axes to create a geometric object in the plane.

This geometrical representation of the graph is consistent with how you will have used the term in the past.

Note from the graph that $f(x)\geq-1$ for all real $x$ , and so we could define the codomain of the function $f$ to be $C=[-1,\infty)$ instead of the whole of $\mathbb{R}$ .

Definition 6.3 (Function notation).

The notation ‘ $f:D\to C$ ’ means $f$ is a function with domain $D$ and codomain $C$ . We say ‘ $f$ maps $D$ to $C$ ’. In this situation the graph is not yet specified.

The notation ‘ $f(x)$ ’ means the element of the codomain corresponding to $x$ in the domain under $f$ . In other words, $y=f(x)$ if and only if $(x,y)\in G$ . We say ‘ $y$ is the value of the function $f$ at $x$ ’.

When $x$ is a variable, $f(x)$ can be thought of as a formula that specifies how to calculate the element of the codomain corresponding to any $x$ in the domain.

The notation ‘ $f:x\mapsto y$ ’ can be used to mean the same thing as $f(x)=y$ . We say ‘ $f$ sends (or maps) $x$ to $y$ ’.

When referring to a function, you should use its name (e.g. $f$ ) and not the value at a point, e.g. $f(x)$ . If you want to refer to the value of a function evaluated for a specific element of the domain $a\in D$ then you should use $f(a)$ .

There are several ways of visualising what a function is; Figure 6.1 shows the two most common ones. In Figure 6.1(a), we visualise a function $f$ as a machine that takes an element $a\in D$ as an input, processes it, and then outputs $f(a)$ . In Figure 6.1(b), we visualise a function $f$ as a mapping from its domain to its codomain.

(a) A function as a machine

(b) A function as a mapping

Figure 6.1: Two ways of picturing a function

f

You will sometimes see functions defined loosely using just a formula with no mention of a domain or codomain. For example, you may see phrases such as ‘let $f(x)=x^{2}$ ’. This informality rarely causes problems, but you should remember that a function always has a domain and codomain.

When a domain is not stated explicitly, you can infer the implied domain from the context. For now, we define it to be the largest subset of $\mathbb{R}$ for which the formula makes sense. The implied codomain is usually $\mathbb{R}$ . However, to avoid ambiguity, it is good practice always to state the domain and codomain explicitly.

Exercise 6.4.

What is the implied domain of the function $f(x)=1/x$ ? Following Definition 6.1, write down a domain, codomain and graph of $f$ . Sketch the graph of $f$ .

Solution (please try for yourself before looking)

The expression $1/x$ is undefined when $x=0$ and so $0$ cannot be in the domain of $f$ defined by $f(x)=1/x$ . The (implied) domain of $f$ is $\mathbb{R}\setminus\{0\}$ .

The codomain could be $\mathbb{R}$ . The codomain could also be $\mathbb{R}\setminus\{0\}$ since there is no $x\in\mathbb{R}$ such that $f(x)=0$ .

With $f:\mathbb{R}\setminus\{0\}\to\mathbb{R}$ the graph is the set

\bigl{\{}(x,1/x):x\in\mathbb{R}\setminus\{0\}\bigr{\}}.

Exercise 6.5.

What is the implied domain of the function $\tan(x)=\sin(x)/\cos(x)$ ?

Solution (please try for yourself before looking)

The fraction $\sin(x)/\cos(x)$ is undefined whenever $\cos(x)=0$ , i.e. when $x\in\bigl{\{}n\pi-\frac{\pi}{2}:n\in\mathbb{Z}\bigr{\}}$ . So the implied domain of $\tan$ is $\mathbb{R}\setminus\bigl{\{}n\pi-\tfrac{\pi}{2}:n\in\mathbb{Z}\bigr{\}}$ .

Two functions are said to be equal when their domains, codomains and graphs are equal as sets.

Example 6.6.

(a)

Functions $f:\mathbb{R}\to\mathbb{R}$ with $f(x)=x^{2}$ and $g:\mathbb{Z}\to\mathbb{R}$ with $g(x)=x^{2}$ are not equal, even though they are defined by the same formula, because their domains (and therefore also their graphs) are different.
(b)

If we consider the functions $p(x)=x+1$ and $q(x)=\dfrac{(x+1)(x-1)}{x-1}$ with their implied domains, these functions are not equal because the implied domain of $q$ cannot contain the value $1$ , whereas the implied domain of $p$ is $\mathbb{R}$ which does contain 1. However, the functions $p:\mathbb{R}\setminus\{1\}\to\mathbb{R}$ and $q:\mathbb{R}\setminus\{1\}\to\mathbb{R}$ with the above definitions are equal because their domains, codomains and graphs

$\{(x,x+1):x\in\mathbb{R}\setminus\{1\}\}$

and

$\left\{\left(x,\frac{(x+1)(x-1)}{x-1}\right):x\in\mathbb{R}\setminus\{1\}\right\}$

are equal.

When defining a function, it is our job to choose an appropriate domain for whatever purpose the function is serving. For example, if we wish to prove something about integers, then it may make more sense to define function $g$ rather than function $f$ in the example above.

The importance of the codomain may not yet be clear. For now, you can think of the codomain as specifying the sort of object that a function outputs. Paying attention to the codomain becomes more important when we start to think about composing functions in Section 6.4.

Exercise 6.7.

Let $C=D=\mathbb{R}$ and consider the set

G:=\{(x,y):x\in C,y\in D,x^{2}+y^{2}=1\}.

1.

Sketch the set $G$ . [You may wish to start by listing some elements of $G$ .]
2.
The definition of a function $f:D\to C$ with graph $G$ in Definition 6.1 has two conditions on the sets, namely:
1. (a)
  
  for every $x\in D$ , there exists $y\in C$ such that $(x,y)\in G$ ; and
2. (b)
  
  if $(x,y)\in G$ and $(x,z)\in G$ then $y=z$ .
Explain why $C$ , $D$ and $G$ in this example do fit the definition of a function.

Solution (please try for yourself before looking)

1.

You may recognise the set $G$ as a unit circle centred at $(0,0)\in\mathbb{R}\times\mathbb{R}$ .
2.
Consider the conditions that must be satisfied by a function:
1. (i)
  
  for every $x\in D$ , there exists $y\in C$ such that $(x,y)\in G$ ; and
2. (ii)
  
  if $(x,y)\in G$ and $(x,z)\in G$ then $y=z$ .
Condition (i) is not satisfied because for any $x<-1$ or $x>1$ there is no $y$ such that $(x,y)\in G$ . This can be seen from the sketch of the graph.
Condition (ii) is not satisfied because, for example, $(0,-1)$ and $(0,1)$ both belong to $G$ , but $-1\neq 1$ .