What is a function

A func­tion is a rule that takes each mem­ber of one set and gives it exactly one part­ner in another set. The idea runs through the whole of higher math­e­mat­ics: every graph you will draw in this chap­ter, every for­mula such as $f(x) = x^2 + 3x + 5$, and every log­a­rithm and mod­u­lus you meet later is a func­tion. This les­son sets out what a func­tion is, how to recog­nise one from an arrow dia­gram, and the words used to describe it: image, pre-image, domain, codomain and range.

Rela­tions and func­tions#

In the chap­ter on sets and rela­tions you met a rela­tion from a set $A$ to a set $B$: a col­lec­tion of ordered pairs $(a, b)$ with $a \in A$ and $b \in B$. A rela­tion can be shown as an arrow dia­gram, with an arrow from $a$ to $b$ for every pair $(a, b)$.

A func­tion is a spe­cial kind of rela­tion.

Def­i­n­i­tion. A rela­tion $f$ from $A$ to $B$ is called a func­tion if each and every ele­ment of $A$ is asso­ci­ated with a unique (exactly one) ele­ment of $B$. We write

$$f : A \to B$$

and read it as "$f$ maps $A$ to $B$".

So two con­di­tions must both hold:

1. Every ele­ment of $A$ has an arrow leav­ing it. No ele­ment of $A$ is left out.
2. No ele­ment of $A$ has two or more arrows leav­ing it.

Noth­ing is said about $B$. An ele­ment of $B$ may receive sev­eral arrows, or none at all.

Exam­ple 1#

Let $A = \{1, 2, 3, 4\}$ and $B = \{a, b, c, d\}$, and let $f$ be the rela­tion with arrows $1 \to a$, $2 \to a$, $3 \to b$, $4 \to c$. Is $f$ a func­tion?

Check the two con­di­tions. Every ele­ment of $A$ (that is $1$, $2$, $3$ and $4$) has an arrow. Each has only one arrow. So $f : A \to B$ is a func­tion.

Notice two things that do not spoil it. Both $1$ and $2$ go to the same ele­ment $a$; that is allowed. The ele­ment $d$ of $B$ receives no arrow; that is allowed too.

As a set of ordered pairs, $f = \{(1, a), (2, a), (3, b), (4, c)\}$.

Exam­ple 2#

Let $B = \{1, 2, 3\}$ and $C = \{a, b, c, d\}$, and let $g$ have arrows $1 \to a$ and $2 \to a$ only. Is $g$ a func­tion from $B$ to $C$?

The ele­ment $3$ of $B$ has no arrow. Con­di­tion 1 fails, so $g$ is not a func­tion from $B$ to $C$.

Exam­ple 3#

Let $C = \{1, 2, 3\}$ and $D = \{a, b, c, d\}$, and let $h$ have arrows $1 \to a$, $1 \to b$, $2 \to c$, $3 \to d$. Is $h$ a func­tion from $C$ to $D$?

The ele­ment $1$ has two arrows, one to $a$ and one to $b$. Its part­ner is not unique, so con­di­tion 2 fails. $h$ is not a func­tion.

A quick way to remem­ber both tests: in an arrow dia­gram of a func­tion, every ele­ment on the left sends out exactly one arrow.

Image and pre-image#

If $f : A \to B$ is a func­tion and $(a, b) \in f$, that is, the arrow from $a$ lands on $b$, then

• $b$ is called the image of $a$, and we write $b = f(a)$;
• $a$ is called a pre-image of $b$.

In Exam­ple 1, the image of $3$ is $b$, so $f(3) = b$. The ele­ment $a$ of $B$ has two pre-images, $1$ and $2$. The ele­ment $d$ has no pre-image at all.

An ele­ment of the domain always has exactly one image. An ele­ment of the codomain may have one pre-image, sev­eral, or none.

Domain, codomain and range#

If $f : A \to B$ is a func­tion, then

• $A$ is called the domain of $f$;
• $B$ is called the codomain of $f$;
• the set of all images of the ele­ments of $A$ is called the range of $f$. It is writ­ten $f(A)$.

The range is always a sub­set of the codomain: $f(A) \subseteq B$.

Exam­ple 4#

For the func­tion of Exam­ple 1, write the domain, codomain and range.

$$\text{Domain} = A = \{1, 2, 3, 4\}$$

$$\text{Codomain} = B = \{a, b, c, d\}$$

The images are $f(1) = a$, $f(2) = a$, $f(3) = b$, $f(4) = c$. Col­lect­ing them, and writ­ing each only once,

$$\text{Range} = f(A) = \{a, b, c\}$$

The codomain is the set the arrows are allowed to land in; the range is the set they actu­ally land in.

Func­tions given by a for­mula#

Most func­tions in this chap­ter are given by a rule writ­ten as a for­mula. The rule tells you how to find the image of any ele­ment of the domain.

Exam­ple 5#

Let $A = \{-2, -1, 0, 1, 2\}$ and let $f : A \to \mathbb{Z}$ be given by $f(x) = x^2$. Find the range of $f$, the pre-images of $4$ and the pre-images of $3$.

Work out each image:

$$f(-2) = 4, \quad f(-1) = 1, \quad f(0) = 0, \quad f(1) = 1, \quad f(2) = 4$$

Each ele­ment of $A$ has exactly one image, so $f$ is a func­tion. The range is $\{0, 1, 4\}$.

The pre-images of $4$ are the ele­ments whose image is $4$: these are $-2$ and $2$.

No ele­ment of $A$ has image $3$, so $3$ has no pre-image. That is no prob­lem: $3$ belongs to the codomain $\mathbb{Z}$ but not to the range.

Exam­ple 6#

Let $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x) = 3x - 2$. Find the image of $4$ and the pre-image of $13$.

The image of $4$ is $f(4) = 3 \times 4 - 2 = 10$.

For the pre-image of $13$, solve $f(x) = 13$:

$$\begin{aligned} 3x - 2 &= 13 \\ 3x &= 15 \\ x &= 5 \end{aligned}$$

So the pre-image of $13$ is $5$.

Exam­ple 7#

Let $A = \{1, 4, 9\}$ and $B = \{-3, -2, -1, 1, 2, 3\}$. A rela­tion sends each $x$ in $A$ to every num­ber in $B$ whose square is $x$. Is this a func­tion?

The num­ber $4$ is sent to $2$ and also to $-2$, since both square to $4$. One ele­ment has two part­ners, so the rela­tion is not a func­tion.

Prac­tice#

1. Let $A = \{1, 2, 3\}$ and $B = \{x, y, z\}$. Is $\{(1, x), (2, y), (3, z)\}$ a func­tion from $A$ to $B$?
2. With the same sets, is $\{(1, x), (2, x), (3, x)\}$ a func­tion from $A$ to $B$?
3. With the same sets, is $\{(1, x), (1, y), (2, z), (3, z)\}$ a func­tion from $A$ to $B$? Give a rea­son.
4. With the same sets, is $\{(1, y), (3, z)\}$ a func­tion from $A$ to $B$? Give a rea­son.
5. Let $f$ be defined on $A = \{0, 1, 2, 3\}$ by $f(x) = 2x + 1$. Find the range of $f$.
6. Let $f$ be defined on $A = \{-2, -1, 0, 1, 2\}$ by $f(x) = x^2 - 1$. Find the range of $f$ and the pre-images of $3$.
7. For $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 3x - 2$, find the image of $-1$ and the pre-image of $-8$.
8. A func­tion $f$ has domain $\{1, 2, 3, 4\}$ and codomain $\{5, 6, 7, 8\}$, with $f(1) = 5$, $f(2) = 5$, $f(3) = 6$, $f(4) = 8$. Write its range, and name the ele­ment of the codomain that has no pre-image.
9. Let $A = \{1, 4, 9\}$ and $B = \{1, 2, 3\}$. Each $x$ in $A$ is sent to the num­ber in $B$ whose square is $x$. Is this a func­tion? If so, write it as a set of ordered pairs.

Answers#