What is a function in the context of IB MYP Mathematics?

In IB MYP Mathematics, a function is a special relationship between two sets, typically called the domain and the range, where each element in the domain is related to exactly one element in the range. Functions are often expressed as equations like f(x) = x + 2, where 'x' is the input and 'f(x)' is the output.

How do you determine if a relation is a function?

To determine if a relation is a function, you can use the vertical line test. If a vertical line intersects the graph of the relation at more than one point, then the relation is not a function. This is because a function can only have one output for each input.

What are the different types of functions studied in the IB MYP curriculum?

In the IB MYP curriculum, students study various types of functions including linear functions, quadratic functions, exponential functions, and sometimes more complex types like trigonometric functions. Each type has unique characteristics and equations that describe their behavior.

How do you find the domain and range of a function?

The domain of a function is the set of all possible input values (x-values), while the range is the set of all possible output values (f(x)-values). To find the domain, consider the values of x for which the function is defined. To find the range, determine the possible values of f(x) as x varies over the domain.

What is the importance of functions in Algebra and real-world applications?

Functions are fundamental in Algebra as they provide a way to model relationships between quantities. In real-world applications, functions are used to describe phenomena such as the growth of populations, the decay of radioactive materials, and financial calculations like interest rates. Understanding functions allows students to analyze and predict behaviors in various contexts.

Why is "Computing $(f \circ g)$ as $f \cdot g$." a common mistake in Functions?

Why it happens: Confusing composition with multiplication. How to avoid it: Composition means APPLY ONE AFTER THE OTHER, not multiply. (f g)(x) = f(g(x)).

Why is "Writing $f^{-1}(x) = \dfrac{1}{f(x)}$." a common mistake in Functions?

Why it happens: Misreading the -1 exponent. How to avoid it: f^-1 is the INVERSE FUNCTION, not the reciprocal. Find it by swap-and-solve.

Why is "Substituting a negative without brackets." a common mistake in Functions?

Why it happens: Forgetting. How to avoid it: Always use brackets: f(-3) = 2(-3)^2 + , not f(-3) = 2 · -3^2.

AlgebraIB MYP Maths Extended Level

Functions

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Functions — IB MYP Mathematics (Extended): mapping inputs to outputs

A function is a rule that takes an input and produces exactly one output. This MYP Mathematics Extended note covers function notation, evaluation, composition, and the vertical-line test.

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Use function notation; evaluate $f(x)$ .
MYP Mathematics A — Find inverse functions.
MYP Mathematics A — Compose two functions.
MYP Mathematics C — Use the vertical-line test on a graph.

What is a function?

A rule where each input has exactly one output.

A function is a rule that takes each input from a set (called the DOMAIN) and produces EXACTLY ONE output (in a set called the RANGE).

Notation: $f(x) = 2x + 3$ .

$f$ is the NAME of the function.
$x$ is the INPUT (or argument).
$f(x) = 2x + 3$ is the RULE (what to do with the input).

To EVALUATE $f$ at a specific value, substitute:

$f(0) = 2(0) + 3 = 3$ .
$f(5) = 2(5) + 3 = 13$ .
$f(-2) = 2(-2) + 3 = -1$ .

Important: 'exactly one' output.

A rule like 'plug in $x$ , get out $\pm\sqrt{x}$ ' is NOT a function — each positive input gives TWO outputs. (The square-root FUNCTION conventionally takes the positive root only.)

Vertical-line test. Given a graph in the $(x, y)$ plane, the graph represents a function iff every VERTICAL LINE intersects it at AT MOST ONE POINT.

The vertical-line test: a graph is a function iff every vertical line crosses it at most once. A circle fails.

Each input → exactly ONE output.
$f(x) =$ rule. $f(\text{input})$ = output.
Vertical-line test for graphs.

Inverse functions

The function that 'undoes' $f$ .

The inverse function $f^{-1}$ undoes whatever $f$ does. If $f$ takes 3 to 7, then $f^{-1}$ takes 7 back to 3.

To find $f^{-1}$ :

Write $y = f(x)$ .
SWAP $x$ and $y$ .
Solve for $y$ . This is $f^{-1}(x)$ .

Worked example. Find $f^{-1}$ for $f(x) = 2x + 3$ .

$y = 2x + 3$ .
Swap: $x = 2y + 3$ .
Solve: $y = \dfrac{x - 3}{2}$ .
So $f^{-1}(x) = \dfrac{x - 3}{2}$ .

Verify: $f(f^{-1}(x)) = f\left(\dfrac{x-3}{2}\right) = 2 \cdot \dfrac{x-3}{2} + 3 = (x-3) + 3 = x$ . ✓

Important: not all functions have inverses! A function has an inverse iff it is one-to-one — every output comes from EXACTLY ONE input. The HORIZONTAL-LINE TEST checks this.

For example $f(x) = x^2$ does NOT have a global inverse on all reals — both $2$ and $-2$ map to $4$ .

Graphical meaning. The graph of $f^{-1}$ is the REFLECTION of the graph of $f$ in the line $y = x$ .

Inverse: swap $x$ and $y$ , then solve.
$f(f^{-1}(x)) = x$ for all $x$ .
Horizontal-line test: function is invertible iff every horizontal line crosses it at most once.

Composing functions

Apply $g$ , then apply $f$ to the result.

Composition of two functions: $(f \circ g)(x) = f(g(x))$ . Apply $g$ first, then $f$ to the result.

Worked example. $f(x) = x^2$ and $g(x) = x + 1$ .

$(f \circ g)(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 2x + 1$ .
$(g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 1$ .

Crucial fact: $(f \circ g)$ is generally NOT the same as $(g \circ f)$ — order matters!

Worked example with specific input. $f(x) = 2x - 1$ , $g(x) = 3x$ . Find $(f \circ g)(5)$ .

Method 1: $g(5) = 15$ . Then $f(15) = 29$ .
Method 2: $(f \circ g)(x) = 2(3x) - 1 = 6x - 1$ . Then $(f \circ g)(5) = 29$ .

Either method works; choose whichever is easier.

Why composition matters. Real-world processes often combine: a CONVERSION (function $g$ ) followed by a COMPUTATION (function $f$ ). For example, $g$ converts km to miles; $f$ computes the cost of fuel for $g$ -many miles. The full process is $(f \circ g)$ .

$(f \circ g)(x) = f(g(x))$ .
Apply $g$ first, then $f$ .
Order matters: $(f \circ g) \ne (g \circ f)$ generally.

How it’s examined

Criterion A: evaluate, compose, invert functions. Criterion C: clearly show working with function notation. Criterion D: model real situations as functions.

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Functions, ready to print or save as PDF.

Step-by-step worked examples — Functions

Step-by-step solutions to past-paper-style questions on functions, written exactly the way a tutor would explain them at the board.

1Evaluate a function
Getting started• evaluation
Question
Given $f(x) = 3x^2 - 5$ , find $f(-2)$ .
Step-by-step solution
1. Step 1
  Substitute $x = -2$ with brackets: $f(-2) = 3(-2)^2 - 5$ .
2. Step 2
  $(-2)^2 = 4$ . So $f(-2) = 3(4) - 5 = 12 - 5 = 7$ .
Answer
$f(-2) = 7$
2Find an inverse
Building confidence• inverse
Question
Find $f^{-1}(x)$ for $f(x) = \dfrac{4x - 1}{3}$ .
Step-by-step solution
1. Step 1
  Write $y = \dfrac{4x - 1}{3}$ .
2. Step 2
  Swap: $x = \dfrac{4y - 1}{3}$ .
3. Step 3
  Multiply both sides by 3: $3x = 4y - 1$ .
4. Step 4
  Add 1: $3x + 1 = 4y$ . Divide: $y = \dfrac{3x + 1}{4}$ .
5. Step 5
  $f^{-1}(x) = \dfrac{3x + 1}{4}$ .
Answer
$f^{-1}(x) = \dfrac{3x + 1}{4}$
3Compose functions
Building confidence• composition
Question
Given $f(x) = x^2 + 1$ and $g(x) = 2x - 3$ , find $(g \circ f)(2)$ .
Step-by-step solution
1. Step 1
  Apply $f$ first: $f(2) = 4 + 1 = 5$ .
2. Step 2
  Then apply $g$ to the result: $g(5) = 2(5) - 3 = 7$ .
Answer
$(g \circ f)(2) = 7$
4Use the vertical-line test
Stretch• graph
Question
Is $y = \pm\sqrt{x}$ (both branches of a sideways parabola) a function?
Step-by-step solution
1. Step 1
  Test: for $x = 4$ , the relation gives $y = 2$ AND $y = -2$ . Two outputs for one input.
2. Step 2
  A vertical line at $x = 4$ would cross the graph TWICE. The vertical-line test FAILS.
3. Step 3
  Therefore $y = \pm\sqrt{x}$ is NOT a function. (The single-branch $y = \sqrt{x}$ , taking only the positive root, IS a function.)
Answer
NOT a function (vertical line crosses twice).

Key Definitions and Keywords — Functions

Definitions to memorise and the exact keywords mark schemes credit for functions answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Function
Examiner keyword
A rule that takes each input from the domain to EXACTLY ONE output in the range.
Domain
The set of all valid inputs for the function.
Range
The set of all possible outputs.
Inverse function $f^{-1}$
Examiner keyword
The function that undoes $f$ : $f(f^{-1}(x)) = x$ .
Function composition $f \circ g$
Examiner keyword
$(f \circ g)(x) = f(g(x))$ . Apply $g$ first, then $f$ .

Common Mistakes and Misconceptions — Functions

The traps other students keep falling into on functions questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕Computing $(f \circ g)$ as $f \cdot g$ .
Why it happens
Confusing composition with multiplication.
How to avoid it
Composition $\circ$ means APPLY ONE AFTER THE OTHER, not multiply. $(f \circ g)(x) = f(g(x))$ .
✕Writing $f^{-1}(x) = \dfrac{1}{f(x)}$ .
Why it happens
Misreading the $-1$ exponent.
How to avoid it
$f^{-1}$ is the INVERSE FUNCTION, not the reciprocal. Find it by swap-and-solve.
✕Substituting a negative without brackets.
Why it happens
Forgetting.
How to avoid it
Always use brackets: $f(-3) = 2(-3)^2 + \ldots$ , not $f(-3) = 2 \cdot -3^2$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Functions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

AlgebraIB MYP Maths Extended Level

Functions

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next.

Previous Next

Learn this Topic

Start with the slides for the quick version, then go deeper with the full study notes.

Short Study Notes in the form of Slides

Read the notes first. If the method in a worked example clicks, you're ready for the questions.

Short Study Notes — Functions

Start with these resources to cover the key concepts, then work through the practice questions.

Page 1 / 0

Detailed Notes

Full prose, callouts and a recap — built for A* mastery, not just a quick scan.

Take these study notes with you

Download a branded PDF — full prose, callouts, recap and memorise list for Functions, ready to print or save offline.

Functions — IB MYP Mathematics (Extended): mapping inputs to outputs

A function is a rule that takes an input and produces exactly one output. This MYP Mathematics Extended note covers function notation, evaluation, composition, and the vertical-line test.

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Use function notation; evaluate $f(x)$ .
MYP Mathematics A — Find inverse functions.
MYP Mathematics A — Compose two functions.
MYP Mathematics C — Use the vertical-line test on a graph.

What is a function?

A rule where each input has exactly one output.

A function is a rule that takes each input from a set (called the DOMAIN) and produces EXACTLY ONE output (in a set called the RANGE).

Notation: $f(x) = 2x + 3$ .

$f$ is the NAME of the function.
$x$ is the INPUT (or argument).
$f(x) = 2x + 3$ is the RULE (what to do with the input).

To EVALUATE $f$ at a specific value, substitute:

$f(0) = 2(0) + 3 = 3$ .
$f(5) = 2(5) + 3 = 13$ .
$f(-2) = 2(-2) + 3 = -1$ .

Important: 'exactly one' output.

A rule like 'plug in $x$ , get out $\pm\sqrt{x}$ ' is NOT a function — each positive input gives TWO outputs. (The square-root FUNCTION conventionally takes the positive root only.)

Vertical-line test. Given a graph in the $(x, y)$ plane, the graph represents a function iff every VERTICAL LINE intersects it at AT MOST ONE POINT.

The vertical-line test: a graph is a function iff every vertical line crosses it at most once. A circle fails.

Each input → exactly ONE output.
$f(x) =$ rule. $f(\text{input})$ = output.
Vertical-line test for graphs.

Inverse functions

The function that 'undoes' $f$ .

The inverse function $f^{-1}$ undoes whatever $f$ does. If $f$ takes 3 to 7, then $f^{-1}$ takes 7 back to 3.

To find $f^{-1}$ :

Write $y = f(x)$ .
SWAP $x$ and $y$ .
Solve for $y$ . This is $f^{-1}(x)$ .

Worked example. Find $f^{-1}$ for $f(x) = 2x + 3$ .

$y = 2x + 3$ .
Swap: $x = 2y + 3$ .
Solve: $y = \dfrac{x - 3}{2}$ .
So $f^{-1}(x) = \dfrac{x - 3}{2}$ .

Verify: $f(f^{-1}(x)) = f\left(\dfrac{x-3}{2}\right) = 2 \cdot \dfrac{x-3}{2} + 3 = (x-3) + 3 = x$ . ✓

Important: not all functions have inverses! A function has an inverse iff it is one-to-one — every output comes from EXACTLY ONE input. The HORIZONTAL-LINE TEST checks this.

For example $f(x) = x^2$ does NOT have a global inverse on all reals — both $2$ and $-2$ map to $4$ .

Graphical meaning. The graph of $f^{-1}$ is the REFLECTION of the graph of $f$ in the line $y = x$ .

Inverse: swap $x$ and $y$ , then solve.
$f(f^{-1}(x)) = x$ for all $x$ .
Horizontal-line test: function is invertible iff every horizontal line crosses it at most once.

Composing functions

Apply $g$ , then apply $f$ to the result.

Composition of two functions: $(f \circ g)(x) = f(g(x))$ . Apply $g$ first, then $f$ to the result.

Worked example. $f(x) = x^2$ and $g(x) = x + 1$ .

$(f \circ g)(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 2x + 1$ .
$(g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 1$ .

Crucial fact: $(f \circ g)$ is generally NOT the same as $(g \circ f)$ — order matters!

Worked example with specific input. $f(x) = 2x - 1$ , $g(x) = 3x$ . Find $(f \circ g)(5)$ .

Method 1: $g(5) = 15$ . Then $f(15) = 29$ .
Method 2: $(f \circ g)(x) = 2(3x) - 1 = 6x - 1$ . Then $(f \circ g)(5) = 29$ .

Either method works; choose whichever is easier.

$(f \circ g)(x) = f(g(x))$ .
Apply $g$ first, then $f$ .
Order matters: $(f \circ g) \ne (g \circ f)$ generally.

How it’s examined

Criterion A: evaluate, compose, invert functions. Criterion C: clearly show working with function notation. Criterion D: model real situations as functions.

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Functions, ready to print or save as PDF.

Step-by-step worked examples — Functions

Step-by-step solutions to past-paper-style questions on functions, written exactly the way a tutor would explain them at the board.

1Evaluate a function
Getting started• evaluation
Question
Given $f(x) = 3x^2 - 5$ , find $f(-2)$ .
Step-by-step solution
1. Step 1
  Substitute $x = -2$ with brackets: $f(-2) = 3(-2)^2 - 5$ .
2. Step 2
  $(-2)^2 = 4$ . So $f(-2) = 3(4) - 5 = 12 - 5 = 7$ .
Answer
$f(-2) = 7$
2Find an inverse
Building confidence• inverse
Question
Find $f^{-1}(x)$ for $f(x) = \dfrac{4x - 1}{3}$ .
Step-by-step solution
1. Step 1
  Write $y = \dfrac{4x - 1}{3}$ .
2. Step 2
  Swap: $x = \dfrac{4y - 1}{3}$ .
3. Step 3
  Multiply both sides by 3: $3x = 4y - 1$ .
4. Step 4
  Add 1: $3x + 1 = 4y$ . Divide: $y = \dfrac{3x + 1}{4}$ .
5. Step 5
  $f^{-1}(x) = \dfrac{3x + 1}{4}$ .
Answer
$f^{-1}(x) = \dfrac{3x + 1}{4}$
3Compose functions
Building confidence• composition
Question
Given $f(x) = x^2 + 1$ and $g(x) = 2x - 3$ , find $(g \circ f)(2)$ .
Step-by-step solution
1. Step 1
  Apply $f$ first: $f(2) = 4 + 1 = 5$ .
2. Step 2
  Then apply $g$ to the result: $g(5) = 2(5) - 3 = 7$ .
Answer
$(g \circ f)(2) = 7$
4Use the vertical-line test
Stretch• graph
Question
Is $y = \pm\sqrt{x}$ (both branches of a sideways parabola) a function?
Step-by-step solution
1. Step 1
  Test: for $x = 4$ , the relation gives $y = 2$ AND $y = -2$ . Two outputs for one input.
2. Step 2
  A vertical line at $x = 4$ would cross the graph TWICE. The vertical-line test FAILS.
3. Step 3
  Therefore $y = \pm\sqrt{x}$ is NOT a function. (The single-branch $y = \sqrt{x}$ , taking only the positive root, IS a function.)
Answer
NOT a function (vertical line crosses twice).

Key Definitions and Keywords — Functions

Definitions to memorise and the exact keywords mark schemes credit for functions answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Function
Examiner keyword
A rule that takes each input from the domain to EXACTLY ONE output in the range.
Domain
The set of all valid inputs for the function.
Range
The set of all possible outputs.
Inverse function $f^{-1}$
Examiner keyword
The function that undoes $f$ : $f(f^{-1}(x)) = x$ .
Function composition $f \circ g$
Examiner keyword
$(f \circ g)(x) = f(g(x))$ . Apply $g$ first, then $f$ .

Common Mistakes and Misconceptions — Functions

The traps other students keep falling into on functions questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕Computing $(f \circ g)$ as $f \cdot g$ .
Why it happens
Confusing composition with multiplication.
How to avoid it
Composition $\circ$ means APPLY ONE AFTER THE OTHER, not multiply. $(f \circ g)(x) = f(g(x))$ .
✕Writing $f^{-1}(x) = \dfrac{1}{f(x)}$ .
Why it happens
Misreading the $-1$ exponent.
How to avoid it
$f^{-1}$ is the INVERSE FUNCTION, not the reciprocal. Find it by swap-and-solve.
✕Substituting a negative without brackets.
Why it happens
Forgetting.
How to avoid it
Always use brackets: $f(-3) = 2(-3)^2 + \ldots$ , not $f(-3) = 2 \cdot -3^2$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Functions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Functions

Short Study Notes in the form of Slides

Short Study Notes — Functions

Detailed Notes

What you’ll learn

How it’s examined

1Evaluate a function

2Find an inverse

3Compose functions

4Use the vertical-line test

Function

Domain

Range

Inverse function f−1f^{-1}f−1

Function composition f∘gf \circ gf∘g

✕Computing (f∘g)(f \circ g)(f∘g) as f⋅gf \cdot gf⋅g.

✕Writing f−1(x)=1f(x)f^{-1}(x) = \dfrac{1}{f(x)}f−1(x)=f(x)1​.

✕Substituting a negative without brackets.

Practice questions

Past paper style quiz

Assess your understanding

Functions — frequently asked questions

Functions

Short Study Notes in the form of Slides

Short Study Notes — Functions

Detailed Notes

What you’ll learn

How it’s examined

1Evaluate a function

2Find an inverse

3Compose functions

4Use the vertical-line test

Function

Domain

Range

Inverse function f−1f^{-1}f−1

Function composition f∘gf \circ gf∘g

✕Computing (f∘g)(f \circ g)(f∘g) as f⋅gf \cdot gf⋅g.

✕Writing f−1(x)=1f(x)f^{-1}(x) = \dfrac{1}{f(x)}f−1(x)=f(x)1​.

✕Substituting a negative without brackets.

Practice questions

Past paper style quiz

Assess your understanding

Functions — frequently asked questions

Inverse function $f^{-1}$

Function composition $f \circ g$

✕Computing $(f \circ g)$ as $f \cdot g$ .

✕Writing $f^{-1}(x) = \dfrac{1}{f(x)}$ .

Inverse function $f^{-1}$

Function composition $f \circ g$

✕Computing $(f \circ g)$ as $f \cdot g$ .

✕Writing $f^{-1}(x) = \dfrac{1}{f(x)}$ .