9.1 The Idea of a Function

SL 2.2

In this section you will learn

what a relation and a function are, using arrow (mapping) diagrams
the words domain, codomain and range
function notation: $f(x)$, $f : x \mapsto \ldots$, and other letters such as $v(t)$ and $C(n)$
how a function is shown as a graph, and how to read the domain and range from it
the vertical line test for deciding whether a graph is a function
how a function can be used as a mathematical model

Relations and functions

In real life, two quantities are often linked by a rule. For example:

the perimeter $P$ of a square depends on its side $s$: $\;P = 4s$;
the time $t$ hours for a $120$ km journey depends on the speed $v$ km/h: $\;t = \dfrac{120}{v}$.

Any rule that links the elements of one set to the elements of another set is called a relation. A relation is a function if every input gives exactly one output.

We can show a relation with an arrow diagram (also called a mapping diagram). In Figure 1 (left), $X = \{-2, 1, 2\}$, $Y = \{0, 1, 4, 9\}$ and the rule is “square the number”:

\[ -2 \mapsto (-2)^2 = 4, \qquad 1 \mapsto 1^2 = 1, \qquad 2 \mapsto 2^2 = 4 . \]

Every element of $X$ has exactly one arrow, so this is a function. It does not matter that two inputs ($-2$ and $2$) go to the same output.

In Figure 1 (right), the element $5$ has two arrows (to $3$ and to $7$). So this relation is not a function.

Figure 1: Left: a function. Right: a relation that is not a function.

Function, domain, codomain, range Not in the formula booklet

A function $f$ from a set $X$ to a set $Y$ is a rule that gives each element of $X$ exactly one element of $Y$.

The domain is the set of inputs, $X$.
The codomain is the set $Y$ that the outputs are taken from.
The range is the set of outputs that actually occur. It is part of the codomain.

In Figure 1 (left), the domain is $\{-2, 1, 2\}$, the codomain is $\{0, 1, 4, 9\}$ and the range is $\{1, 4\}$.

Example 1 Which of these relations, from $X = \{1, 2, 3\}$, are functions? For each function, state the range.

(a) $1 \mapsto 5,\; 2 \mapsto 6,\; 3 \mapsto 7$

(b) $1 \mapsto 4,\; 2 \mapsto 4,\; 3 \mapsto 4$

(c) $1 \mapsto 2,\; 1 \mapsto 3,\; 2 \mapsto 3,\; 3 \mapsto 1$

(d) $1 \mapsto 0,\; 3 \mapsto 2$

Solution

(a) Each input has exactly one output: a function. Range $\{5, 6, 7\}$.

(b) Each input has exactly one output (it is the same output, but that is allowed): a function. Range $\{4\}$.

(c) The input $1$ has two outputs: not a function.

(d) The input $2$ has no output: not a function from $X$.

Function notation

We usually give a function a letter, such as $f$, $g$ or $h$. There are two common ways to write the rule:

\[ f(x) = x^2 + 3 \qquad\text{or}\qquad f : x \mapsto x^2 + 3 . \]

Both mean “$f$ is the function that takes $x$ and gives $x^2 + 3$”. We read $f(x)$ as “$f$ of $x$”. $f(x)$ is the output (the image of $x$).

The domain is often written next to the rule, for example $f(x) = x^2 + 3,\; x \in \mathbb{R}$, or $x \geq 0$.

Other letters are used too, especially in applications: $v(t)$ for a velocity at time $t$, $C(n)$ for the cost of $n$ items, $A(r)$ for the area of a circle of radius $r$.

Example 2 Let $f(x) = 5x - x^2$.

(a) Find $f(2)$ and $f(-3)$.

(b) Find and simplify $f(a + 1)$.

(c) Solve $f(x) = 4$.

Solution

(a) $f(2) = 10 - 4 = 6$. $\qquad f(-3) = -15 - 9 = -24$.

(b) Replace every $x$ by $(a + 1)$:

\[ f(a + 1) = 5(a + 1) - (a + 1)^2 = 5a + 5 - a^2 - 2a - 1 = -a^2 + 3a + 4 . \]

(c) $5x - x^2 = 4 \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow (x - 1)(x - 4) = 0$, so $x = 1$ or $x = 4$.

Graphical representation of functions

The graph of a function $f$ is the set of all points $(x, y)$ with $y = f(x)$, where $x$ is in the domain.

For example, let $f(x) = (x - 1)^2 + 2$ with domain $0 \leq x \leq 3$ (Figure 2).

Figure 2: The domain is on the $x$-axis and the range is on the $y$-axis.

The graph helps us in three ways.

It shows where the function starts and where it ends: at $(0, 3)$ and $(3, 6)$.
It shows the domain on the $x$-axis and the range on the $y$-axis. Here the domain is $\{x : 0 \leq x \leq 3\}$ and the range is $\{y : 2 \leq y \leq 6\}$. The lowest point is the vertex $(1, 2)$, not an end point.
It shows that each $x$ gives exactly one $y$. For example $f(0) = 3$ and $f(1) = 2$.

Testing functions: the vertical line test

Not every curve is the graph of a function. The circle $x^2 + y^2 = 25$ passes through $(3, 4)$ and $(3, -4)$, so the input $x = 3$ has two outputs. A circle is not a function.

Look at Figure 3.

Left: $y = x^3 - 3x + 1$, $\;-2 \leq x \leq 2.5$. Every vertical line $x = a$ (with $a$ in the domain) meets the curve once. For example $x = -1$ meets it only at $(-1, 3)$, and $x = 1$ only at $(1, -1)$. This is a function.
Right: $y^2 = x + 2$. The line $x = 2$ meets the curve at $(2, 2)$ and $(2, -2)$. This is not a function.

Vertical line test Not in the formula booklet

A graph represents a function if and only if every vertical line $x = a$, with $a$ in the domain, meets the graph exactly once.

Example 3 Use the graphs in Figure 4 to decide whether each relation is a function. Give a reason.

(a) $y = (x - 1)^2,\; x \in \mathbb{R}$

(b) $y = -\sqrt{x},\; x \geq 0$

(c) $y^2 = 4x,\; x \geq 0$

Solution

(a) Every vertical line meets the parabola exactly once. It is a function.

(b) For each $x \geq 0$ the vertical line meets the curve exactly once, at $\big(x, -\sqrt{x}\big)$. It is a function.

(c) $y^2 = 4x$ means $y = \pm 2\sqrt{x}$. For $x > 0$ the vertical line meets the curve twice, for example at $(1, 2)$ and $(1, -2)$. It is not a function.

Notice: the symbol $\sqrt{x}$ always means the non-negative square root. So $y = \sqrt{x}$ and $y = -\sqrt{x}$ are functions, but $y = \pm\sqrt{x}$ is not.

A function as a mathematical model

Functions are used to model real situations. When we use a function as a model, we must also think about the domain that makes sense in the real situation.

Example 4 A stone is thrown upwards. Its height in metres after $t$ seconds is modelled by

\[ h(t) = 20t - 5t^2 . \]

(a) Find $h(1)$ and explain what it means.

(b) Find the time when the stone lands, and hence state a sensible domain for $h$.

(c) Find the range of $h$ for this domain.

Solution

(a) $h(1) = 20 - 5 = 15$. After $1$ second the stone is $15$ m above the ground.

(b) $h(t) = 0 \Rightarrow 5t(4 - t) = 0 \Rightarrow t = 0$ or $t = 4$. The stone lands after $4$ s. A sensible domain is $0 \leq t \leq 4$ (negative times, or times after landing, have no meaning).

(c) $h(t) = -5(t^2 - 4t) = -5(t - 2)^2 + 20$. The maximum height is $20$ m at $t = 2$, and the least height is $0$. The range is $0 \leq h(t) \leq 20$.

Common errors

✗ Wrong	✓ Right	Why
“Two inputs give the same output, so it is not a function.”	It can still be a function.	Only one input with two outputs breaks the rule.
$f(a + 1) = 5a + 1 - a^2 + 1$	$f(a + 1) = 5(a + 1) - (a + 1)^2$	Replace every $x$ by the whole bracket $(a + 1)$.
Range of Figure 2 is $3 \leq y \leq 6$	$2 \leq y \leq 6$	The least value is at the vertex, not at an end point.
Using a horizontal line to test for a function	Use vertical lines	A vertical line fixes one input $x$.
$y = \pm\sqrt{x}$ is a function	It is not	Each $x > 0$ gives two values of $y$.

Exercises

1 The relation $R$ maps $\{0, 1, 2, 3\}$ to $\{1, 2, 3, 4, 5\}$ by $0 \mapsto 1$, $1 \mapsto 3$, $2 \mapsto 5$, $3 \mapsto 3$. Is $R$ a function? If so, state its domain and range.

Solution

Every input has exactly one output, so $R$ is a function. Domain $\{0, 1, 2, 3\}$; range $\{1, 3, 5\}$.

2 A function $g$ has domain $\{-1, 0, 1, 2\}$ and $g(x) = 2x^2 - 1$. List the range of $g$.

Solution

$g(-1) = 1$, $g(0) = -1$, $g(1) = 1$, $g(2) = 7$. The range is $\{-1, 1, 7\}$.

3 Let $f(x) = 3 - 2x$. Find

(a) $f(4)$

(b) $f\!\left(-\dfrac{1}{2}\right)$

(c) the value of $x$ for which $f(x) = 11$.

Solution

(a) $3 - 8 = -5$.

(b) $3 + 1 = 4$.

(c) $3 - 2x = 11 \Rightarrow x = -4$.

4 Let $f(x) = x^2 - 4x$. Find and simplify $f(x + 2)$.

Solution

$f(x + 2) = (x + 2)^2 - 4(x + 2) = x^2 + 4x + 4 - 4x - 8 = x^2 - 4$.

5 Let $g(x) = \dfrac{6}{x + 1}$, $x \neq -1$. Solve $g(x) = 3x - 1$.

Solution

$6 = (3x - 1)(x + 1) = 3x^2 + 2x - 1 \Rightarrow 3x^2 + 2x - 7 = 0$.

\[ x = \frac{-2 \pm \sqrt{4 + 84}}{6} = \frac{-2 \pm \sqrt{88}}{6} = \frac{-1 \pm \sqrt{22}}{3} . \]

Neither value is $-1$, so both are solutions.

6 Which of these are functions of $x$? Give a reason.

(a) $y = x^3 - x$

(b) $x^2 + y^2 = 9$

(c) $x = 4$

(d) $y = 4$

Solution

(a) Yes: each $x$ gives one value $x^3 - x$.

(b) No: for example $x = 0$ gives $y = 3$ and $y = -3$.

(c) No: this is a vertical line, so the input $4$ has infinitely many outputs.

(d) Yes: every $x$ gives the one value $4$ (a constant function).

7 The function $f(x) = 2x + 1$ has domain $-1 \leq x \leq 3$. Sketch the graph and state the range.

Solution

The graph is a straight line segment from $(-1, -1)$ to $(3, 7)$. The function is increasing, so the range is $-1 \leq f(x) \leq 7$.

8 The function $h(x) = x^2 - 2x$ has domain $-1 \leq x \leq 4$. Find the range of $h$.

Solution

$h(x) = (x - 1)^2 - 1$, with vertex $(1, -1)$ inside the domain. End points: $h(-1) = 3$, $h(4) = 8$.

The least value is $-1$ and the greatest is $8$. Range: $-1 \leq h(x) \leq 8$.

9 A taxi company charges a fixed amount of $4 plus $1.50 per kilometre.

(a) Write down a function $C(d)$ for the cost of a journey of $d$ km.

(b) Find $C(12)$.

(c) A journey cost $31. How long was it?

Solution

(a) $C(d) = 4 + 1.5d$, $\;d \geq 0$.

(b) $C(12) = 4 + 18 = \$22$.

(c) $4 + 1.5d = 31 \Rightarrow 1.5d = 27 \Rightarrow d = 18$ km.

10 A ball is kicked upwards. Its height in metres after $t$ seconds is $h(t) = 1 + 14t - 7t^2$.

(a) Find the maximum height.

(b) Find, to 3 significant figures, the time when it hits the ground.

(c) State a sensible domain for $h$.

Solution

(a) $h(t) = -7(t^2 - 2t) + 1 = -7(t - 1)^2 + 8$. The maximum height is $8$ m (at $t = 1$).

(b) $-7(t - 1)^2 + 8 = 0 \Rightarrow (t - 1)^2 = \dfrac{8}{7} \Rightarrow t = 1 + \sqrt{\dfrac{8}{7}} = 2.07$ s (the other root is negative).

(c) $0 \leq t \leq 2.07$ (approximately).