Functions, domain & range — notes, question types & practice · Functions, Graphs & Logarithms

01

A function is a machine

Think of a currency-exchange counter. You hand over a number, the machine applies one fixed rule, and hands back exactly one number. That is a function.

We write $f(x) = 2x + 3$ . Here $f$ is the rule, $x$ is the input, and $f(x)$ is the output.

Feed in $x = 4$ : $f(4) = 2(4) + 3 = 11$ .

Rule: One input must give exactly one output. If some input gives two different outputs, it is not a function.

02

Checking a relation is a function

Take the pairs $(1, 2)$ , $(2, 5)$ , $(1, 7)$ . The input $1$ gives two outputs, $2$ and $7$ . Not a function.

Now $(1, 2)$ , $(2, 5)$ , $(3, 5)$ . Inputs $2$ and $3$ share the output $5$ — that is allowed. Two inputs may share one output; one input may not split into two outputs.

On a graph this becomes the vertical line test: a vertical line must cut the curve at most once.

03

Domain: which inputs are allowed

The domain is the set of inputs the machine accepts. Two things break a machine:

Division by zero. In $f(x) = \dfrac{1}{x-3}$ , the input $x = 3$ gives $\dfrac{1}{0}$ . Reject it. Domain: all real numbers except $3$ .
Square roots of negatives. In $f(x) = \sqrt{x-5}$ , the input must keep $x - 5 \ge 0$ , so $x \ge 5$ .

Polynomials like $x^2 + 3x$ carry no such traps. Every real number is allowed.

Watch: For $\dfrac{1}{\sqrt{x-3}}$ both traps act together. The root needs $x - 3 \ge 0$ and the bottom cannot be zero, so $x > 3$ strictly.

Worked: domain of $\dfrac{x+1}{x^2 - 5x + 6}$ . Factor the bottom: $(x-2)(x-3)$ . Reject $x = 2$ and $x = 3$ .

04

Range: which outputs can come out

The range is the set of outputs the machine can actually produce.

A straight line like $f(x) = 2x + 5$ climbs forever in both directions. It can output every real number.

A parabola has a floor or a ceiling. Complete the square to find it:

f(x) = x^2 - 4x + 7 = (x-2)^2 + 3

A square is never negative, so the smallest output is $3$ , reached at $x = 2$ . Range: $f(x) \ge 3$ .

If the $x^2$ term carries a minus sign, the parabola opens down. Then the vertex value is the maximum.

Tip: For $a(x-h)^2 + k$ , the range is $[k, \infty)$ when $a > 0$ and $(-\infty, k]$ when $a < 0$ .

05

Even and odd functions

Replace $x$ by $-x$ and simplify.

Same result: $f(-x) = f(x)$ — even. The graph mirrors in the $y$ -axis. Example: $x^4 + x^2$ stays itself.
Sign flips: $f(-x) = -f(x)$ — odd. The graph looks the same after a half turn. Example: $x^3 + x$ flips sign cleanly.
Neither: $f(x) = x + 1$ gives $-x + 1$ , which is neither $f(x)$ nor $-f(x)$ .

Note: Test the rule, not the powers. $x^2 + 1$ is even though $1$ is an odd number; $x^3 + 1$ is neither.

06

Reading the question

Exams ask value, domain, range and symmetry as four separate question types.

"For which values is $f$ defined?" — hunt denominators and roots first.
"The minimum value of $f(x)$ " — complete the square.
"Find $k$ given the minimum" — vertex form, then compare.
"Even, odd or neither" — compute $f(-x)$ fully before deciding.

Write the restriction first, then answer. That order catches almost every domain error.

07

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Find the value of a function

How to spot it:

A rule like $f(x) = 3x - 4$ is given and $f(5)$ or $f(a) + f(b)$ is asked.

f(a) = \text{(rule with } x = a)

Method

Write the rule.
Replace every $x$ by the given number, in brackets.
Simplify once, carefully.
For $f(a) + f(b)$ , evaluate each part separately.

Why it works:

A function value is just the output of the rule at that input, so substitution is the whole method.

Try this

If $f(x) = x^2 + 2x$ , find $f(3)$ .

Show solution

$f(3) = 3^2 + 2(3)$ .
$= 9 + 6 = 15$ .

Answer

15

Type 2very common2 practice Q

Domain of a rational function

How to spot it:

The rule is a fraction. 'For what values of $x$ is $f$ defined?' is asked.

\text{denominator} \ne 0

Method

Set the bottom equal to zero.
Solve for the rejected values.
State the domain as all reals except those values.
For a quadratic bottom, factor or use the formula.

Why it works:

Division by zero is undefined, so exactly the zeros of the bottom drop out of the domain.

Try this

Find the domain of $f(x) = \dfrac{x+1}{x^2 - 5x + 6}$ .

Show solution

$x^2 - 5x + 6 = (x-2)(x-3)$ .
Reject $x = 2$ and $x = 3$ .

Answer

All real numbers except $x = 2$ and $x = 3$

Type 3common3 practice Q

Domain with a square root

How to spot it:

The rule contains $\sqrt{\;}$ of an expression in $x$ , alone or under a fraction.

\text{(inside of root)} \ge 0

Method

Keep the inside of the root $\ge 0$ .
Solve the inequality for $x$ .
If the root sits in a denominator, make it strict: $> 0$ .
Combine both conditions when both appear.

Why it works:

Square roots of negative numbers are not real, so the inside must stay on the non-negative side.

Try this

Find the domain of $f(x) = \sqrt{2x - 8}$ .

Show solution

$2x - 8 \ge 0$ .
$x \ge 4$ .

Answer

x ≥ 4

Type 4common2 practice Q

Range of a linear function

How to spot it:

$f(x) = ax + b$ with $a \ne 0$ , and the set of possible outputs is asked.

a \ne 0 \Rightarrow \text{range} = \text{all real numbers}

Method

Check the slope $a$ is not zero.
A non-horizontal line climbs or falls forever.
Conclude the range is all real numbers.

Why it works:

A slanted line keeps rising or falling without end, so no output value is missed.

Try this

Find the range of $f(x) = 2x + 5$ .

Show solution

Slope $2 \ne 0$ .
As $x$ runs over all reals, $2x + 5$ also runs over all reals.

Answer

All real numbers

Type 5very common2 practice Q

Range or minimum of a quadratic

How to spot it:

$ax^2 + bx + c$ is given and its minimum (or maximum) value, or the full range, is asked.

a(x-h)^2 + k \Rightarrow \text{min/max} = k

Method

Complete the square, or use $x = -b/2a$ .
Note the sign of $a$ : $a > 0$ gives a minimum.
Substitute the vertex input back into $f$ .
Write the range from the vertex value.

Why it works:

The parabola turns around at its vertex, so the vertex value bounds the range on one side.

Try this

Find the range of $f(x) = x^2 - 4x + 7$ .

Show solution

$f(x) = (x-2)^2 + 3$ .
Smallest value $3$ at $x = 2$ .

Answer

f(x) ≥ 3

Type 6common2 practice Q

Even, odd or neither

How to spot it:

' $f(x) = \ldots$ is:' with options even / odd / neither. Symmetry is asked without drawing.

f(-x) = f(x) \text{ even}; \quad f(-x) = -f(x) \text{ odd}

Method

Replace $x$ by $-x$ everywhere.
Simplify the new rule.
Compare with $f(x)$ and $-f(x)$ .
If neither matches, answer 'neither'.

Why it works:

Even means the rule is blind to the sign of $x$ ; odd means the output flips sign with the input.

Try this

Classify $f(x) = x^4 + x^2$ .

Show solution

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

Answer

Even

08

Formula sheet

Domain: denominator

x \ne \text{(zeros of the denominator)}

Reject every input that makes the bottom zero.

Domain: square root

\sqrt{A} \text{ needs } A \ge 0

The inside of a root stays on the non-negative side.

Range of a quadratic

a(x-h)^2 + k \Rightarrow \text{range} = [k, \infty) \text{ if } a > 0

Complete the square; $k$ is the floor or the ceiling.

Vertex input

x = -\frac{b}{2a}

Substitute back to get the minimum or maximum value.

Even / odd test

f(-x) = f(x) \text{ (even)}; \quad f(-x) = -f(x) \text{ (odd)}

Replace $x$ by $-x$ and simplify fully before deciding.

09

Shortcuts that save time

⚡ Hunt the two troublemakers

For a domain, only two things can go wrong: a zero bottom or a negative root. Scan the rule for $\div$ and $\sqrt{}$ , write the condition, solve.

Example

Find the domain of $f(x) = \dfrac{1}{x - 7}$ .

Show solution

Bottom is $x - 7$ ; it must not be zero.
$x \ne 7$ ; everything else is fine.

Answer

All real numbers except $x = 7$

⚡ Complete the square for range

Write $ax^2 + bx + c$ as $a(x - h)^2 + k$ . The square cannot go below $0$ , so $k$ is the smallest (or largest) output.

Example

Find the minimum value of $f(x) = x^2 - 4x + 7$ .

Show solution

$f(x) = (x-2)^2 + 3$ .
$(x-2)^2 \ge 0$ , so the least value is $3$ .

Answer

3

⚡ Test $f(-1)$ for even / odd

Compute $f(-1)$ and compare with $f(1)$ and $-f(1)$ . Equal means even, opposite means odd. Always confirm with the full rule $f(-x)$ .

Example

Is $f(x) = x^3 - x$ even, odd or neither?

Show solution

$f(-x) = -x^3 + x = -(x^3 - x)$ .
This equals $-f(x)$ .

Answer

Odd

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Treating $x = 0$ as always forbidden — only a denominator equal to zero is forbidden, not the input zero itself.

Mistake 02

Giving the domain of $\sqrt{x-4}$ as all real numbers — the inside must stay $\ge 0$ , so $x \ge 4$ .

Mistake 03

Swapping domain and range — domain is the set of inputs, range is the set of outputs.

Mistake 04

For $\dfrac{1}{\sqrt{x-2}}$ , writing $x \ge 2$ — the bottom cannot be zero, so $x > 2$ strictly.

Mistake 05

Testing even/odd at a single value and concluding without checking the rule $f(-x)$ for all $x$ .

Mistake 06

Judging even/odd from the powers alone — $x^2 + 1$ is even and $x^3 + 1$ is neither; only the test with $f(-x)$ decides.

11

Quick revision

Read this the night before the exam.

Function: one input gives exactly one output.
Domain traps: denominator $\ne 0$ ; inside of $\sqrt{}$ $\ge 0$ .
$1/\sqrt{x-a}$ needs $x > a$ strictly.
Line $ax + b$ , $a \ne 0$ : range is all real numbers.
Quadratic $a(x-h)^2 + k$ : least or greatest value is $k$ .
Even: $f(-x) = f(x)$ . Odd: $f(-x) = -f(x)$ .

12

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.