Functions

Syllabus coverage

Understand the terms function, domain, range, one-one function, inverse function and composition of functions
Identify the range of a given function in simple cases, and find the composition of two given functions, including the condition that a composite function $g f$ can only be formed when the range of $f$ is within the domain of $g$
Determine whether or not a given function is one-one, and find the inverse of a one-one function in simple cases
Illustrate in graphical terms the relation between a one-one function and its inverse, with sketches showing the mirror line $y = x$
Understand and use the transformations of the graph of $y = f (x)$ given by $y = f (x) + a$ , $y = f (x + a)$ , $y = a f (x)$ , $y = f (a x)$ and simple combinations of these
Use the terms translation, reflection and stretch when describing a transformation

Part of the AS & A Level Maths syllabus 9709.

A function as a machine:

f (x) = 3 x + 2

takes one input and gives exactly one output, so

f (4) = 14

and

f (- 1) = - 1

. The inverse machine undoes the steps in reverse - subtract 2, then divide by 3.

A function is a rule plus a domain

A function takes each value from a set of allowed inputs and gives back exactly one output. The set of allowed inputs is the domain, and the set of outputs that actually occur is the range. A function is not properly defined until both are known, which is why a question always writes something like $f (x) = x^{2} + 3$ for $x \geq 0$ : the condition after the rule is part of the function, not a decoration.

The rule 'one output for each input' is what separates a function from a relation. $y^{2} = x$ is not a function of $x$ , because $x = 4$ gives both $y = 2$ and $y = - 2$ . On a graph, a function has a single y-value for every x in its domain, so no vertical line through the domain can meet the graph twice. That is the vertical line test.

Notation matters as well. $f (x)$ is the output when the input is $x$ , and $f (3)$ means 'put 3 in'. If $f (x) = 3 x + 2$ then $f (3) = 11$ . The letter $x$ is only a placeholder, so $f (t)$ is the same function written with a different input letter.

Term	Meaning
Function	a rule that gives exactly one output for each input
Domain	the set of inputs the function is allowed to take
Range	the set of outputs the function actually produces
One-one	no two different inputs give the same output
Inverse	the function that reverses the rule; it exists only when the function is one-one

Read the domain condition on the same line as the rule before you start any question in this topic. Almost every mark lost here comes from ignoring a condition such as

x \geq 1

Finding the range

The range is the set of outputs the function really produces. For a straight line it is quick: $f (x) = 9 - 2 x$ for $x \geq 1$ decreases as $x$ increases, so the largest output is at the smallest input, $f (1) = 7$ , and the range is $f (x) \leq 7$ . A decreasing function swaps the ends round: the left end of the domain gives the biggest output.

For a quadratic you need the turning point, and completing the square finds it. $g (x) = x^{2} + 6 x + 5 = (x + 3)^{2} - 4$ has least value $- 4$ at $x = - 3$ , so on the whole real line the range is $g (x) \geq - 4$ . If the domain is restricted to $x \geq - 3$ the range does not change, because $- 3$ is still allowed. If the domain were $x \geq 0$ the turning point would lie outside it, the function would be increasing throughout the domain, and the range would be $g (x) \geq 5$ .

Some functions have a restricted range for a structural reason rather than because of a turning point. $\sqrt{x - 2}$ can never be negative, so its range is $f (x) \geq 0$ however generous the domain is. The reciprocal $\frac{1}{x}$ never reaches zero, so its range is every real number except 0.

Restricting a domain can only leave the range the same or make it smaller. If your range for a restricted domain is wider than the range on the whole real line, something has gone wrong.

One-one functions

A function is one-one if no two different inputs produce the same output. $f (x) = 3 x + 2$ is one-one: a larger input always gives a larger output, so two different inputs can never collide. $f (x) = x^{2}$ on the whole real line is not one-one, because $f (2)$ and $f (- 2)$ are both 4.

On a graph, a one-one function passes the horizontal line test: no horizontal line meets the graph more than once. The parabola $y = x^{2}$ fails it, while the right-hand branch, $y = x^{2}$ for $x \geq 0$ , passes it comfortably.

One-one matters because it is exactly the condition for an inverse function to exist. To reverse a rule, every output must have come from a single input - otherwise 'the input that produced this output' would not name one number. That is why a question that wants the inverse of a quadratic restricts the domain first, and why you must say which branch you are working on.

$f (x) = 2 x - 7$ for all real $x$ : one-one
$f (x) = x^{2}$ for all real $x$ : not one-one
$f (x) = x^{2}$ for $x \geq 0$ : one-one
$f (x) = \frac{1}{x}$ for $x \neq 0$ : one-one

Inverse functions

The inverse $f^{- 1}$ undoes $f$ : if $f (a) = b$ then $f^{- 1} (b) = a$ . To find it, write $y = f (x)$ , make $x$ the subject, and then relabel $y$ as $x$ . One fact about the domain carries most of the marks: the domain of $f^{- 1}$ is the range of $f$ , and the range of $f^{- 1}$ is the domain of $f$ .

Worked through: $f (x) = x^{2} - 2 x$ for $x \geq 1$ . Completing the square gives $y = (x - 1)^{2} - 1$ , so $(x - 1)^{2} = y + 1$ . Because $x \geq 1$ , the quantity $x - 1$ is not negative, so the positive root is the one that applies: $x = \sqrt{y + 1} + 1$ . Relabelling gives $f^{- 1} (x) = \sqrt{x + 1} + 1$ , and since the range of $f$ is $f (x) \geq - 1$ , the domain of $f^{- 1}$ is $x \geq - 1$ .

The two functions undo each other exactly: $f^{- 1} (f (x)) = x$ for every $x$ in the domain of $f$ , and $f (f^{- 1} (x)) = x$ for every $x$ in the domain of $f^{- 1}$ . Never read $f^{- 1} (x)$ as $\frac{1}{f (x)}$ : the $- 1$ is a label meaning 'inverse', not a power.

The square root step is where the domain condition earns its marks. Without

x \geq 1

you could not choose between

\sqrt{y + 1} + 1

and

- \sqrt{y + 1} + 1

, and each would be a correct answer to a different question.

Composite functions

A composite function applies one function and then another. In $g f (x)$ , read from the inside outwards: apply $f$ first, then apply $g$ to the result. So $g f (x) = g (f (x))$ , and the function written closest to the $x$ is the one that acts first.

With $f (x) = 3 x - 1$ and $g (x) = x^{2} + 2$ : $g f (x) = g (3 x - 1) = (3 x - 1)^{2} + 2 = 9 x^{2} - 6 x + 3$ , while $f g (x) = f (x^{2} + 2) = 3 (x^{2} + 2) - 1 = 3 x^{2} + 5$ . Two different answers from the same pair of functions, so the order is part of the question rather than a detail.

$g f$ can only be formed when the range of $f$ fits inside the domain of $g$ , because the outputs of $f$ have to be legal inputs for $g$ . If $g$ is defined for $x \geq 5$ and the range of $f$ is $f (x) \geq 1$ , then outputs of $f$ between 1 and 5 have nowhere to go, and $g f$ does not exist. Checking the two ranges before substituting is worth a mark on its own.

Composite	Means	Order of work
$f g (x)$	$f (g (x))$	$g$ first, then $f$
$g f (x)$	$g (f (x))$	$f$ first, then $g$
$f f (x)$	$f (f (x))$	$f$ applied twice

Composite functions are usually not commutative:

f g (x)

and

g f (x)

are different functions unless the question is built to make them the same. Never assume the two are equal.

The graph of a function and its inverse

The graph of the inverse of a one-one function is the reflection of the graph of the function in the line $y = x$ . The reason is arithmetic rather than geometric: if $f$ sends the input $a$ to the output $b$ , then the point $(a, b)$ lies on the graph of $f$ , and $f^{- 1}$ sends $b$ back to $a$ , so $(b, a)$ lies on the graph of $f^{- 1}$ . Swapping the two coordinates is exactly what reflection in $y = x$ does.

Two results follow immediately. A point on one graph has a mirror-image partner on the other, so a question that gives you a point on $f$ and asks about $f^{- 1}$ is a one-line question. And the domain of $f^{- 1}$ is the range of $f$ , which is why the inverse graph lives where you would expect: a range $f (x) \geq - 1$ becomes a domain $x \geq - 1$ .

If the graph of a function meets its own inverse, the meeting point lies on the line $y = x$ , because a point that survives the swap of coordinates must have equal coordinates. The whole inverse graph is found by reflecting the curve, so a vertical feature of $f$ becomes a horizontal feature of $f^{- 1}$ : a value that $f$ never reaches becomes a value that $f^{- 1}$ is never given.

When a question asks you to sketch

f^{- 1}

alongside

f

, draw the line

y = x

first and then reflect the key points - intercepts, turning points and any end points of the domain. Cambridge mark schemes specifically want the mirror line shown.

Transformations of the graph of y = f(x)

Four transformations cover everything the syllabus asks for, and they are easiest to remember by asking what each one does to the coordinates. Changes made outside the function, $y = f (x) + a$ and $y = a f (x)$ , act on the output, so they move or stretch the graph vertically. Changes made inside the function, $y = f (x + a)$ and $y = f (a x)$ , act on the input, so they move or stretch the graph horizontally.

The counter-intuitive one is $y = f (x + a)$ , which slides the graph $a$ units to the left rather than to the right. Test it with $a = 2$ : the curve $y = f (x + 2)$ takes the value $f (3)$ when $x = 1$ , so whatever used to happen at $x = 3$ now happens at $x = 1$ . Inside the bracket everything happens earlier, and the shift is the opposite of the sign you can see.

Stretches work the same way round. $y = f (2 x)$ squashes the graph horizontally: the value that $f$ used to reach at $x = 4$ is now reached at $x = 2$ , so every x-coordinate is halved. A negative $a$ adds a reflection: $y = - f (x)$ is a reflection in the x-axis, and $y = f (- x)$ is a reflection in the y-axis. Combinations are read in the order the notation suggests, so in $y = f (x - 1) + 5$ the graph moves 1 unit to the right and then 5 units up.

Transformation	What it does to the graph	What happens to the coordinates
$y = f (x) + a$	translation of $a$ units in the positive y-direction	every y-coordinate increases by $a$
$y = f (x + a)$	translation of $a$ units in the negative x-direction	every x-coordinate decreases by $a$
$y = a f (x)$	stretch parallel to the y-axis, scale factor $a$	every y-coordinate is multiplied by $a$ ; x-intercepts do not move
$y = f (a x)$	stretch parallel to the x-axis, scale factor $\frac{1}{a}$	every x-coordinate is divided by $a$ ; the y-intercept does not move

Name the transformation the way the mark scheme wants it: 'translation of 4 units in the positive y-direction' or 'stretch parallel to the y-axis with scale factor 3'. 'It moves up' and 'it gets taller' will not score the description mark.

Key terms you must be able to define

Function: A rule that gives exactly one output for each input in its domain.

Domain: The set of values the input of a function is allowed to take.

Range: The set of values the output of a function actually takes.

Image: The output that a particular input is mapped to; the image of $a$ under $f$ is $f (a)$ .

One-one function: A function in which different inputs always give different outputs, so no horizontal line meets its graph twice.

Inverse function: The function $f^{- 1}$ that reverses $f$ , so that $f^{- 1} (f (x)) = x$ . It exists only when $f$ is one-one.

Composite function: A function formed by applying one function to the output of another, written $g f (x) = g (f (x))$ .

Composite condition: $g f$ can be formed only when the range of $f$ lies inside the domain of $g$ .

Mapping: The rule that sends each element of the domain to its image. A function is a one-one or many-to-one mapping, never one-to-many.

Vertical line test: A graph is the graph of a function if no vertical line meets it more than once.

Horizontal line test: A graph is the graph of a one-one function if no horizontal line meets it more than once.

Translation: A transformation that slides a graph without turning or stretching it. $y = f (x) + a$ and $y = f (x + a)$ are translations.

Reflection: A transformation that mirrors a graph in a line. $y = - f (x)$ reflects in the x-axis and $y = f (- x)$ reflects in the y-axis.

Stretch: A transformation that scales a graph in one direction only. $y = a f (x)$ stretches parallel to the y-axis and $y = f (a x)$ stretches parallel to the x-axis.

Mirror line: The line $y = x$ , in which the graph of a function and the graph of its inverse are reflections of each other.

Restricted domain: A domain that is deliberately limited, such as $x \geq 1$ , usually so that the function becomes one-one and can be inverted.

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

Range on a restricted domain

The function $f$ is defined by $f (x) = 4 x - 3$ for $0 \leq x \leq 5$ . The rule increases, so the smallest output is at the left end of the domain and the largest at the right end: $f (0) = - 3$ and $f (5) = 17$ , so the range is $- 3 \leq f (x) \leq 17$ .

Inverse of a linear function

$f (x) = 5 - 2 x$ for all real $x$ . Write $y = 5 - 2 x$ , so $2 x = 5 - y$ and $x = \frac{5 - y}{2}$ . Relabelling gives $f^{- 1} (x) = \frac{5 - x}{2}$ . Check it: $f (3) = - 1$ , and $f^{- 1} (- 1) = 3$ , so the inverse really does undo the function.

A composite in one direction only

$f (x) = x^{2}$ for $x \geq 0$ , and $g (x) = x + 4$ for all real $x$ . Then $g f (x) = g (x^{2}) = x^{2} + 4$ , which is fine because every output of $f$ is at least 0. But $f g (x) = f (x + 4) = (x + 4)^{2}$ is defined only for $x \geq - 4$ , because $f$ will not accept an input below 0.

Transforming a known point

The curve $y = f (x)$ has a minimum at $(3, - 2)$ . For $y = 2 f (x)$ every y-coordinate doubles, so the minimum becomes $(3, - 4)$ . For $y = f (2 x)$ every x-coordinate halves, so it becomes $(1.5, - 2)$ . For $y = f (x - 1) + 5$ it becomes $(4, 3)$ .

Reflecting to get the inverse

$f (x) = 2 x + 1$ passes through $(0, 1)$ and $(2, 5)$ . Its inverse is $f^{- 1} (x) = \frac{x - 1}{2}$ , which passes through $(1, 0)$ and $(5, 2)$ - the same two points with the coordinates swapped, which is what reflection in $y = x$ does.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What are the domain and the range of a function?

The domain is the set of allowed inputs; the range is the set of outputs that actually occur.

Q2How do you decide whether a graph is the graph of a one-one function?

Use the horizontal line test: if any horizontal line meets the graph more than once, the function is not one-one.

Q3Why does an inverse function only exist for a one-one function?

Reversing the rule needs each output to have come from one input. If two inputs share an output, the reverse rule would not give a single answer.

Q4What is the domain of

f^{- 1}

The range of

f

. The range of

f^{- 1}

is the domain of

f

Q5Find the inverse of

f (x) = 5 - 2 x

f^{- 1} (x) = \frac{5 - x}{2}

. Write

y = 5 - 2 x

, rearrange to

x = \frac{5 - y}{2}

, then relabel.

Q6What does

g f (x)

mean?

g (f (x))

: apply

f

first, then apply

g

to the result.

Q7When can the composite

g f

be formed?

Only when the range of

f

lies inside the domain of

g

, so that every output of

f

is a legal input for

g

Q8Is

f g (x)

the same as

g f (x)

Usually not. Composite functions are not commutative unless the question is built that way.

Q9How is the graph of

f^{- 1}

related to the graph of

f

It is the reflection of the graph of

f

in the line

y = x

. Each point

(a, b)

becomes

(b, a)

Q10Describe the transformation

y = f (x) + a

A translation of

a

units in the positive y-direction. Every y-coordinate increases by

a

Q11Describe the transformation

y = f (x + a)

A translation of

a

units in the negative x-direction. Inside the bracket, the shift is the opposite of the sign.

Q12Describe the transformation

y = a f (x)

A stretch parallel to the y-axis with scale factor

a

. Every y-coordinate is multiplied by

a

, and x-intercepts stay put.

Q13Describe the transformation

y = f (a x)

A stretch parallel to the x-axis with scale factor

\frac{1}{a}

. Every x-coordinate is divided by

a

, and the y-intercept stays put.

Q14What is the range of

g (x) = x^{2} + 6 x + 5

g (x) \geq - 4

, because

g (x) = (x + 3)^{2} - 4

and a square is never negative.

Q15A curve

y = f (x)

has a minimum at

(3, - 2)

. Where is the minimum of

y = f (x - 1) + 5

(4, 3)

: the graph moves 1 unit to the right and 5 units up.

0 of 15 marked as known

Exam technique for this topic

Write the domain down first and refer to it at every step. A question that says 'for $x \geq 1$ ' is telling you which square root to take, what the range of the inverse is, and often where a composite stops being valid.
For a range question, decide first whether the function is increasing, decreasing or turning. Straight lines are decided by their gradient, quadratics by completing the square, and a restricted domain may cut the turning point out altogether.
Answer an inverse question in three moves: write $y = f (x)$ , make $x$ the subject, relabel. State the domain of $f^{- 1}$ even when the question does not ask for it - it is often the next part of the question, and it is the part students lose.
Before substituting into a composite, check that the range of the inner function sits inside the domain of the outer one. When the question asks you to explain why a composite cannot be formed, the marks are for naming both the range of the inner function and the domain of the outer one.
Describe a transformation with its type, its direction and its size: 'translation of 3 units in the negative x-direction' or 'stretch parallel to the x-axis with scale factor $\frac{1}{2}$ '. Test a point from the original graph to decide the direction of a horizontal shift, because the sign inside the bracket is the opposite of the shift you see.

Why students lose marks here

A student wrote

f (x) = x^{2}

for all real

x

, so

f^{- 1} (x) = \sqrt{x}

Why it loses marksOn the whole real line

f

is not one-one, so it has no inverse at all. Restricting the domain to

x \geq 0

is what makes the inverse exist, and that restriction must be stated.

The answer that scoresFor

f (x) = x^{2}

with domain

x \geq 0

f^{- 1} (x) = \sqrt{x}

with domain

x \geq 0

A student wrote

g f (x)

means

f (g (x))

Why it loses marksThe function nearest the

x

acts first.

g f (x)

applies

f

first and then

g

, which is the opposite of what the letters look like in order.

The answer that scoresWith

f (x) = 3 x - 1

and

g (x) = x^{2} + 2

g f (x) = g (3 x - 1) = 9 x^{2} - 6 x + 3

, while

f g (x) = f (x^{2} + 2) = 3 x^{2} + 5

A student wrote

f (x) = 9 - 2 x

for

x \geq 1

, so the range is

f (x) \geq 7

Why it loses marksThe rule decreases as

x

increases, so the smallest input gives the largest output. The range runs downwards from 7, not upwards.

The answer that scores

f (1) = 7

is the largest output, so the range is

f (x) \leq 7

A student wrote

y = f (x + 3)

moves the graph 3 units to the right.

Why it loses marksA change inside the bracket acts on the input and does the opposite of what the sign suggests. The value

f (4)

is now reached at

x = 1

, so the graph has moved to the left.

The answer that scores

y = f (x + 3)

is a translation of 3 units in the negative x-direction.

A student wrote

y = f (2 x)

doubles every x-coordinate.

Why it loses marksIt is the other way round:

f (2 x)

reaches a given height twice as quickly, so each x-coordinate is halved.

y = 2 f (x)

is the one that doubles, and it doubles y-coordinates.

The answer that scores

y = f (2 x)

is a stretch parallel to the x-axis with scale factor

\frac{1}{2}

Syllabus coverage

A function is a rule plus a domain

Finding the range

One-one functions

Inverse functions

Composite functions

The graph of a function and its inverse

Transformations of the graph of y = f(x)

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here