Mathematics 9709 · Pure Mathematics 1

Definition of a function (domain and range)

A function is a machine. Put one number in. Get exactly one number out.
This topic asks two things. Which numbers can go in (the domain)? Which numbers come out (the range)?

The domain is the numbers you put in. The range is the numbers that come out. The domain sets the range. So write the range using $\mathrm{f}(x)$ , never $x$ .

The one idea behind the whole chapter

Every topic here is the same machine. Range asks what comes out. Composite puts the output of one machine into the input of the next. Inverse runs the machine backwards. Transformation moves the inputs and outputs on the graph.

This one idea explains the two facts students find strange. An inverse runs the machine backwards. So its inputs and outputs are the old ones swapped. That is why domain and range swap (§2.3, §2.4). Changing the input means asking: what $x$ do I put in to get the old value? So anything inside the bracket runs backwards too. That is why $\mathrm{f}(x+3)$ moves the graph left (§2.5). Both facts come from one thing: input and output keep swapping roles.

A machine with one input, one output

In $\mathrm{f}(x)=x+2$ , $x$ is the input and $x+2$ is the output. Put in $5$ , get $\mathrm{f}(5)=7$ .
One rule makes it a function. Each input gives one and only one output.
$\mathrm{f}:x\mapsto x+2$ (“f maps $x$ to $x+2$ ”) means the same machine.

Two words get the marks here:

the domain is the set of inputs you are allowed to put in (how far the graph goes left to right);
the range is the set of outputs that come out (how far the graph goes up and down).

You cannot read the range from the rule on its own.
Make the inputs smaller and the outputs get smaller too.
Find the domain first, then find the range.

One-one vs many-one

One-one: different inputs always give different outputs, like $\mathrm{f}(x)=x+2$ .
Many-one: two inputs can share an output, like $\mathrm{f}(x)=x^2$ , where $-3$ and $3$ both give $9$ .

Horizontal-line test: if no horizontal line cuts the graph more than once, it is one-one.
Only one-one functions have an inverse. So the word “one-one” in a question means an inverse is coming.

One-one = passes the horizontal-line test = has an inverse. These three all mean the same thing.

Reading the range off the graph

The range is how high and how low the graph goes.
Draw the curve over its domain.
Then read off the lowest and highest heights.

Here is $\mathrm{f}(x)=(x-3)^2+8$ on the domain $-1\le x\le 9$ . The bottom of the range is not an end point. It is the turning point $(3,8)$ , which sits inside the domain. The top is the far end point $x=9$ .

The range is the gold vertical line: lowest output 8 (the vertex) up to highest 44 (the end point x = 9).

The outputs go from $8$ up to $44$ :

Answer

8\le \mathrm{f}(x)\le 44

Why the lowest output is not always at an end point

On a straight line, the highest and lowest outputs are always at the two ends of the domain. So testing both ends is enough. A curve bends. So a turning point can go lower (or higher) than either end.

For $(x-3)^2+8$ , the squared bracket is never negative. So its smallest value is $0$ , at $x=3$ . And $3$ is inside $-1\le x\le 9$ . So that minimum really happens, giving $0+8=8$ . (You find that vertex by completing the square, the Quadratics §1.2 skill.) The top is the end point further from the vertex. Here that is $x=9$ , giving $(9-3)^2+8=44$ .

Say the domain was $5\le x\le 9$ instead. Then the vertex $x=3$ would be outside it. So you would just test the two ends.

Worked examples

Worked example

Textbook 2.1: state the range of $\mathrm{f}(x)=5-2x$ for $-4\le x\le 5$

A straight line has no turning point, so its biggest and smallest outputs are at the two ends of the domain, so test both ends.

Left end: $\mathrm{f}(-4)=5-2(-4)=5+8=13$ .
Right end: $\mathrm{f}(5)=5-2(5)=5-10=-5$ .
The line falls steadily (gradient $-2$ ) from $13$ down to $-5$ , so the range is everything in between, written in $\mathrm{f}(x)$ .

Answer

-5\le \mathrm{f}(x)\le 13

Worked example

Real exam: find the range of $\mathrm{f}(x)=x^2+\tfrac{16}{x}+2$ for $x>0$ (9709/12 Oct/Nov 2021 Q10)

This curve has no end points, because the domain has no top, so the range is set by the turning point, not an end, and the question gives you the stationary point at $x=2$ .

Find the lowest output by substituting the stationary value $x=2$ : $\mathrm{f}(2)=2^2+\tfrac{16}{2}+2=4+8+2=14$ .
Confirm it is a minimum: $\mathrm{f}''(x)=2+\tfrac{32}{x^3}$ , and $\mathrm{f}''(2)=6>0$ , so the curve turns upward there.
It is the only stationary point, so the curve never drops below $14$ . Write the range in $\mathrm{f}(x)$ .

Answer

\mathrm{f}(x)\ge 14

(9709/12 Oct/Nov 2021 Q10(c)) The mark scheme awards the second mark for writing it as $\mathrm{f}(x)\ge 14$ (or $y\ge 14$ ), and explicitly refuses $x\ge 14$ .

Where the marks go

Common mistake

Write a range using

\mathrm{f}(x)

(or

y

), like

\mathrm{f}(x)\ge 8

. Never write it using

x

. Writing the range as an

x

-interval is the most common mistake in this topic. The mark scheme says no marks for it. (9709/12 Oct/Nov 2021 Q10)

Common mistake

For a curve, do not just test the two ends. A turning point inside the domain gives the real lowest or highest value. So check first if that turning point is allowed. (9709/12 Oct/Nov 2021 Q10)

Now you try

The function $\mathrm{f}(x)=(x-2)^2+1$ is defined for $x\in\mathbb{R}$ . Find the range of $\mathrm{f}$ , and say whether it is one-one or many-one. (textbook Explore 2.2-style)

Your turn— tap to reveal the worked answer (textbook Explore 2.2-style)

The squared bracket is never negative, so its smallest value is $0$ , at the vertex $x=2$ , which makes the lowest output $0+1=1$ , and from there the curve goes up with no top.

Answer

\mathrm{f}(x)\ge 1

Over all of $\mathbb{R}$ , the two sides of the parabola give the same outputs. So it is many-one. That is why you would need a smaller domain before you could find an inverse.

PreviousHidden quadratics NextComposite functions

Definition of a function (domain and range)

A function is a machine. Put one number in. Get exactly one number out.
This topic asks two things. Which numbers can go in (the domain)? Which numbers come out (the range)?

The domain is the numbers you put in. The range is the numbers that come out. The domain sets the range. So write the range using $\mathrm{f}(x)$ , never $x$ .

The one idea behind the whole chapter

A machine with one input, one output

In $\mathrm{f}(x)=x+2$ , $x$ is the input and $x+2$ is the output. Put in $5$ , get $\mathrm{f}(5)=7$ .
One rule makes it a function. Each input gives one and only one output.
$\mathrm{f}:x\mapsto x+2$ (“f maps $x$ to $x+2$ ”) means the same machine.

Two words get the marks here:

the domain is the set of inputs you are allowed to put in (how far the graph goes left to right);
the range is the set of outputs that come out (how far the graph goes up and down).

You cannot read the range from the rule on its own.
Make the inputs smaller and the outputs get smaller too.
Find the domain first, then find the range.

One-one vs many-one

One-one: different inputs always give different outputs, like $\mathrm{f}(x)=x+2$ .
Many-one: two inputs can share an output, like $\mathrm{f}(x)=x^2$ , where $-3$ and $3$ both give $9$ .

Horizontal-line test: if no horizontal line cuts the graph more than once, it is one-one.
Only one-one functions have an inverse. So the word “one-one” in a question means an inverse is coming.

One-one = passes the horizontal-line test = has an inverse. These three all mean the same thing.

Reading the range off the graph

The range is how high and how low the graph goes.
Draw the curve over its domain.
Then read off the lowest and highest heights.

The range is the gold vertical line: lowest output 8 (the vertex) up to highest 44 (the end point x = 9).

The outputs go from $8$ up to $44$ :

Answer

8\le \mathrm{f}(x)\le 44

Why the lowest output is not always at an end point

Say the domain was $5\le x\le 9$ instead. Then the vertex $x=3$ would be outside it. So you would just test the two ends.

Worked examples

Worked example

Textbook 2.1: state the range of $\mathrm{f}(x)=5-2x$ for $-4\le x\le 5$

A straight line has no turning point, so its biggest and smallest outputs are at the two ends of the domain, so test both ends.

Left end: $\mathrm{f}(-4)=5-2(-4)=5+8=13$ .
Right end: $\mathrm{f}(5)=5-2(5)=5-10=-5$ .
The line falls steadily (gradient $-2$ ) from $13$ down to $-5$ , so the range is everything in between, written in $\mathrm{f}(x)$ .

Answer

-5\le \mathrm{f}(x)\le 13

Worked example

Real exam: find the range of $\mathrm{f}(x)=x^2+\tfrac{16}{x}+2$ for $x>0$ (9709/12 Oct/Nov 2021 Q10)

This curve has no end points, because the domain has no top, so the range is set by the turning point, not an end, and the question gives you the stationary point at $x=2$ .

Find the lowest output by substituting the stationary value $x=2$ : $\mathrm{f}(2)=2^2+\tfrac{16}{2}+2=4+8+2=14$ .
Confirm it is a minimum: $\mathrm{f}''(x)=2+\tfrac{32}{x^3}$ , and $\mathrm{f}''(2)=6>0$ , so the curve turns upward there.
It is the only stationary point, so the curve never drops below $14$ . Write the range in $\mathrm{f}(x)$ .

Answer

\mathrm{f}(x)\ge 14

(9709/12 Oct/Nov 2021 Q10(c)) The mark scheme awards the second mark for writing it as $\mathrm{f}(x)\ge 14$ (or $y\ge 14$ ), and explicitly refuses $x\ge 14$ .

Where the marks go

Common mistake

Write a range using

\mathrm{f}(x)

(or

y

), like

\mathrm{f}(x)\ge 8

. Never write it using

x

. Writing the range as an

x

-interval is the most common mistake in this topic. The mark scheme says no marks for it. (9709/12 Oct/Nov 2021 Q10)

Common mistake

For a curve, do not just test the two ends. A turning point inside the domain gives the real lowest or highest value. So check first if that turning point is allowed. (9709/12 Oct/Nov 2021 Q10)

Now you try

The function $\mathrm{f}(x)=(x-2)^2+1$ is defined for $x\in\mathbb{R}$ . Find the range of $\mathrm{f}$ , and say whether it is one-one or many-one. (textbook Explore 2.2-style)

Your turn— tap to reveal the worked answer (textbook Explore 2.2-style)

The squared bracket is never negative, so its smallest value is $0$ , at the vertex $x=2$ , which makes the lowest output $0+1=1$ , and from there the curve goes up with no top.

Answer

\mathrm{f}(x)\ge 1

Over all of $\mathbb{R}$ , the two sides of the parabola give the same outputs. So it is many-one. That is why you would need a smaller domain before you could find an inverse.