What is a function in the context of Edexcel IGCSE Mathematics?

In Edexcel IGCSE Mathematics, a function is a relation between a set of inputs and a set of possible outputs where each input is related to exactly one output. Functions are often represented using equations, graphs, or tables. Understanding functions is crucial as they form the basis for analyzing relationships between variables in mathematics.

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 on its graph. If a vertical line intersects the graph at more than one point, the relation is not a function. Alternatively, check that each input value has only one corresponding output value. This ensures that the relation meets the definition of a function.

What are the different types of functions covered in the Edexcel IGCSE curriculum?

The Edexcel IGCSE curriculum covers various types of functions, including linear functions, quadratic functions, and exponential functions. Linear functions are represented by straight lines, quadratic functions form parabolas, and exponential functions show rapid growth or decay. Each type has distinct characteristics and applications in solving mathematical problems.

How do you find the inverse of a function?

To find the inverse of a function, swap the input (x) and output (y) variables and solve for the new output variable. This process essentially reverses the function's operations. Not all functions have inverses, and a function must be one-to-one (bijective) to have an inverse. Graphically, the inverse is a reflection of the original function across the line y = x.

Why are functions important in understanding sequences and graphs in mathematics?

Functions are fundamental in understanding sequences and graphs because they provide a systematic way to describe how one quantity changes with another. In sequences, functions can define the rule for generating terms, while in graphs, they illustrate the relationship between variables visually. Mastery of functions enables students to analyze and interpret mathematical models effectively.

Why is "Doing the wrong function first in composite" a common mistake in Functions?

Why it happens: Reading left-to-right as written. How to avoid it: fg(x) means f(g(x)) — apply g FIRST. The function CLOSEST to x is applied first. Read right-to-left. Source: 4MA1/1H May/Jun 2024 — examiner report Q16

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

Why it happens: Notation looks like reciprocal. How to avoid it: f^-1 is the INVERSE FUNCTION (undoes f), NOT the reciprocal. Use the swap method.

Why is "Substituting before composing" a common mistake in Functions?

Why it happens: Trying to evaluate immediately. How to avoid it: For fg(x), first FIND fg(x) as an expression, THEN substitute. Or evaluate inside-out for numbers.

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Detailed Study Notes

Detailed notes on Sequences, Functions and Graphs for Edexcel IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Functions Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

Function notation, evaluation, composite functions, and inverses. A* topic — guaranteed on every Higher Tier paper.

At a glance

$f(x)$ : rule mapping input to output.
Evaluate: substitute number for $x$ .
$fg(x) = f(g(x))$ : apply $g$ FIRST, then $f$ .
Inverse $f^{-1}(x)$ : undoes $f$ .
Method for inverse: swap $x, y$ and solve.
$f^{-1}$ is NOT the reciprocal $1/f$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.2 — Use function notation $f(x)$ .
3.2 — Evaluate functions for given values.
3.2 — Find composite functions $fg(x)$ .
3.2 — Find inverse functions $f^{-1}(x)$ .

Function notation and evaluation

$f(x)$ is a recipe. $f(5)$ means 'apply the recipe to $5$ '.

Function notation:

$f(x)$ reads 'f of x'.
$x$ is the INPUT.
$f(x)$ is the OUTPUT.

Examples:

A function takes an input, applies its rule, and produces one output.

If $f(x) = 2x + 3$ :

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

Substitution rule.

Wherever you see $x$ in the function definition, replace with the input. Use BRACKETS around the input.

For $f(x) = x^{2} + 3x$ :

$f(2) = (2)^{2} + 3(2) = 4 + 6 = 10$ .
$f(-2) = (-2)^{2} + 3(-2) = 4 - 6 = -2$ . (Note: brackets matter!)

Edexcel tip. Always show the substitution explicitly: ' $f(5) = 3(5) - 7 = 8$ '. Mark schemes credit each step.

$f(x)$ = function notation.
Input goes in; output comes out.
Use brackets when substituting.
Show every step.

Composite functions

$fg(x) = f(g(x))$ . Apply $g$ first, $f$ second.

Composite function: $fg(x) = f(g(x))$ .

Read right-to-left like onion layers — the inner one is applied first.

Method (algebraic):

Find $g(x)$ as an expression.
Substitute that into $f$ wherever $x$ appears.
Simplify.

Example. $f(x) = 2x + 3$ , $g(x) = x^{2} - 1$ .

(a) $fg(x) = f(g(x)) = f(x^{2} - 1) = 2(x^{2} - 1) + 3 = 2x^{2} + 1$ .

(b) $gf(x) = g(f(x)) = g(2x + 3) = (2x + 3)^{2} - 1 = 4x^{2} + 12x + 9 - 1 = 4x^{2} + 12x + 8$ .

Numerical composite. Easier — work inside-out.

$fg(4)$ with same $f, g$ :

$g(4) = 16 - 1 = 15$ .
$f(15) = 2(15) + 3 = 33$ .

In fg(x) the function nearest the bracket, g, is applied first; then f acts on its output.

Worked qualitative. Why does $fg(x) \ne gf(x)$ in general?

Order matters. 'Take square then add 1' is NOT the same as 'add 1 then square'.
Example: $f(x) = x^{2}, g(x) = x + 1$ .
$fg(x) = (x + 1)^{2} = x^{2} + 2x + 1$ .
$gf(x) = x^{2} + 1$ .
Different.

Edexcel tip. Mark schemes give M1 for correct order, M1 for substitution, A1 for simplified answer.

$fg(x) = f(g(x))$ .
$g$ first, then $f$ .
$fg \ne gf$ in general.
Inside-out for numerical inputs.

Inverse functions

$f^{-1}$ undoes $f$ . Method: swap $x, y$ , solve.

Inverse function $f^{-1}(x)$ undoes $f$ .

If $f(2) = 7$ , then $f^{-1}(7) = 2$ .

Property: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ .

The inverse f⁻¹ undoes f — it maps every output straight back to its input.

Method to find $f^{-1}$ :

Write $y = f(x)$ .
SWAP $x$ and $y$ .
SOLVE for $y$ .
The result is $f^{-1}(x)$ .

Example. $f(x) = \dfrac{2x - 5}{3}$ .

$y = \dfrac{2x - 5}{3}$ .
Swap: $x = \dfrac{2y - 5}{3}$ .
Solve for $y$ $y$ :
- $3x = 2y - 5$ .
- $2y = 3x + 5$ .
- $y = \dfrac{3x + 5}{2}$ .
$f^{-1}(x) = \dfrac{3x + 5}{2}$ .

Verify. $f(f^{-1}(x)) = f\left(\dfrac{3x + 5}{2}\right) = \dfrac{2 \cdot (3x + 5)/2 - 5}{3} = \dfrac{3x + 5 - 5}{3} = \dfrac{3x}{3} = x$ . ✓

Worked qualitative. Why is $f^{-1}(x) \ne 1/f(x)$ ?

$f^{-1}$ means INVERSE FUNCTION (undo).
$1/f(x)$ means RECIPROCAL.
The notation is overloaded but they're DIFFERENT.
$f(x) = 2x$ has inverse $f^{-1}(x) = x/2$ , but reciprocal $1/(2x) = 1/(2x)$ . Different.

Edexcel tip. When asked for $f^{-1}$ , ALWAYS use the swap method. Showing the swap step earns the M1.

$f^{-1}$ undoes $f$ .
Method: swap $x, y$ , solve.
$f^{-1} \ne 1/f$ — different.
Verify by composition.

Quick recap

$f(x)$ : function notation.
Evaluate by substitution.
$fg(x) = f(g(x))$ — $g$ first.
Inverse: swap $x, y$ , solve.
$f^{-1} \ne 1/f$ .

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$fg(x) = f(g(x))$ — composite.
Inverse method: swap, solve.
$f^{-1}$ undoes $f$ .
$f^{-1}$ is NOT $1/f$ .

How it’s examined

Functions on every Higher Tier paper (4-7 marks). Composite + inverse standard. Examiner reports flag (1) wrong order in composite, (2) interpreting $f^{-1}$ as reciprocal, (3) skipping swap step.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Take this whole topic with you

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
Foundation• function, evaluate
Question
$f(x) = 3x - 7$ . Find $f(5)$ and $f(-2)$ .
Step-by-step solution
1. Step 1
  $f(5) = 3(5) - 7 = 8$ .
2. Step 2
  $f(-2) = 3(-2) - 7 = -13$ .
Answer
$f(5) = 8$ , $f(-2) = -13$
2Composite functions $fg(x)$
Higher• Adapted from 4MA1/1H May/Jun 2024 Q16• composite
Question
$f(x) = 2x + 3$ and $g(x) = x^{2} - 1$ . Find (a) $fg(x)$ , (b) $gf(x)$ .
Step-by-step solution
1. Step 1
  (a) $fg(x) = f(g(x))$ . Apply $g$ FIRST, then $f$ .
2. Step 2
  $g(x) = x^{2} - 1$ . Substitute into $f$ : $f(g(x)) = 2(x^{2} - 1) + 3 = 2x^{2} + 1$ .
3. Step 3
  (b) $gf(x) = g(f(x))$ . Apply $f$ first.
4. Step 4
  $f(x) = 2x + 3$ . Substitute: $g(f(x)) = (2x + 3)^{2} - 1 = 4x^{2} + 12x + 9 - 1 = 4x^{2} + 12x + 8$ .
Answer
(a) $fg(x) = 2x^{2} + 1$ (b) $gf(x) = 4x^{2} + 12x + 8$
Examiner tip
$fg(x)$ means 'do $g$ first, then $f$ '. Read RIGHT to LEFT — like onion layers.
3Find an inverse function
Higher• Adapted from 4MA1/2H Jan 2024 Q13• inverse
Question
$f(x) = \dfrac{2x - 5}{3}$ . Find $f^{-1}(x)$ .
Step-by-step solution
1. Step 1
  Set $y = f(x) = \dfrac{2x - 5}{3}$ .
2. Step 2
  Swap $x$ and $y$ : $x = \dfrac{2y - 5}{3}$ .
3. Step 3
  Solve for $y$ : $3x = 2y - 5 \implies 2y = 3x + 5 \implies y = \dfrac{3x + 5}{2}$ .
4. Step 4
  So $f^{-1}(x) = \dfrac{3x + 5}{2}$ .
Answer
$f^{-1}(x) = \dfrac{3x + 5}{2}$
Examiner tip
Method: write $y = f(x)$ , swap $x$ and $y$ , solve for $y$ . The result is $f^{-1}(x)$ .
4Evaluate a composite
Higher• composite, evaluate
Question
$f(x) = 2x + 1$ and $g(x) = 3x - 2$ . Find $fg(4)$ .
Step-by-step solution
1. Step 1
  Compute $g(4) = 3(4) - 2 = 10$ .
2. Step 2
  Compute $f(10) = 2(10) + 1 = 21$ .
Answer
$fg(4) = 21$
Examiner tip
For numerical composite: evaluate inside-out. $g(4)$ first, then $f$ of that result.

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 Pearson Edexcel IGCSE 4MA1 sitting.

Function
Examiner keyword
A rule that assigns to each input EXACTLY ONE output. Notation: $f(x)$ reads 'f of x'.
Example
$f(x) = 2x + 3$ assigns to each $x$ the value $2x + 3$ .
Composite function
Examiner keyword
$fg(x)$ means 'do $g$ first, then apply $f$ '. $fg(x) = f(g(x))$ .
Example
If $f(x) = x + 1$ and $g(x) = 2x$ , then $fg(x) = 2x + 1$ .
Inverse function
Examiner keyword
Notation $f^{-1}(x)$ . The function that 'undoes' $f$ : $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ .
Example
If $f(x) = 3x + 1$ , then $f^{-1}(x) = (x - 1)/3$ .

Common Mistakes and Misconceptions — Functions

The traps other students keep falling into on functions questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Doing the wrong function first in composite
4MA1/1H May/Jun 2024 — examiner report Q16
Why it happens
Reading left-to-right as written.
How to avoid it
$fg(x)$ means $f(g(x))$ — apply $g$ FIRST. The function CLOSEST to $x$ is applied first. Read right-to-left.
✕Writing $f^{-1}(x) = 1/f(x)$
Why it happens
Notation looks like reciprocal.
How to avoid it
$f^{-1}$ is the INVERSE FUNCTION (undoes $f$ ), NOT the reciprocal. Use the swap method.
✕Substituting before composing
Why it happens
Trying to evaluate immediately.
How to avoid it
For $fg(x)$ , first FIND $fg(x)$ as an expression, THEN substitute. Or evaluate inside-out for numbers.

Functions — frequently asked questions

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

Edexcel IGCSE Mathematics (4MA1)

Functions

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Short Notes - Functions

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Detailed Study Notes

Detailed notes on Sequences, Functions and Graphs for Edexcel IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Functions Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

Function notation, evaluation, composite functions, and inverses. A* topic — guaranteed on every Higher Tier paper.

At a glance

$f(x)$ : rule mapping input to output.
Evaluate: substitute number for $x$ .
$fg(x) = f(g(x))$ : apply $g$ FIRST, then $f$ .
Inverse $f^{-1}(x)$ : undoes $f$ .
Method for inverse: swap $x, y$ and solve.
$f^{-1}$ is NOT the reciprocal $1/f$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.2 — Use function notation $f(x)$ .
3.2 — Evaluate functions for given values.
3.2 — Find composite functions $fg(x)$ .
3.2 — Find inverse functions $f^{-1}(x)$ .

Function notation and evaluation

$f(x)$ is a recipe. $f(5)$ means 'apply the recipe to $5$ '.

Function notation:

$f(x)$ reads 'f of x'.
$x$ is the INPUT.
$f(x)$ is the OUTPUT.

Examples:

A function takes an input, applies its rule, and produces one output.

If $f(x) = 2x + 3$ :

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

Substitution rule.

Wherever you see $x$ in the function definition, replace with the input. Use BRACKETS around the input.

For $f(x) = x^{2} + 3x$ :

$f(2) = (2)^{2} + 3(2) = 4 + 6 = 10$ .
$f(-2) = (-2)^{2} + 3(-2) = 4 - 6 = -2$ . (Note: brackets matter!)

Edexcel tip. Always show the substitution explicitly: ' $f(5) = 3(5) - 7 = 8$ '. Mark schemes credit each step.

$f(x)$ = function notation.
Input goes in; output comes out.
Use brackets when substituting.
Show every step.

Composite functions

$fg(x) = f(g(x))$ . Apply $g$ first, $f$ second.

Composite function: $fg(x) = f(g(x))$ .

Read right-to-left like onion layers — the inner one is applied first.

Method (algebraic):

Find $g(x)$ as an expression.
Substitute that into $f$ wherever $x$ appears.
Simplify.

Example. $f(x) = 2x + 3$ , $g(x) = x^{2} - 1$ .

(a) $fg(x) = f(g(x)) = f(x^{2} - 1) = 2(x^{2} - 1) + 3 = 2x^{2} + 1$ .

(b) $gf(x) = g(f(x)) = g(2x + 3) = (2x + 3)^{2} - 1 = 4x^{2} + 12x + 9 - 1 = 4x^{2} + 12x + 8$ .

Numerical composite. Easier — work inside-out.

$fg(4)$ with same $f, g$ :

$g(4) = 16 - 1 = 15$ .
$f(15) = 2(15) + 3 = 33$ .

In fg(x) the function nearest the bracket, g, is applied first; then f acts on its output.

Worked qualitative. Why does $fg(x) \ne gf(x)$ in general?

Order matters. 'Take square then add 1' is NOT the same as 'add 1 then square'.
Example: $f(x) = x^{2}, g(x) = x + 1$ .
$fg(x) = (x + 1)^{2} = x^{2} + 2x + 1$ .
$gf(x) = x^{2} + 1$ .
Different.

Edexcel tip. Mark schemes give M1 for correct order, M1 for substitution, A1 for simplified answer.

$fg(x) = f(g(x))$ .
$g$ first, then $f$ .
$fg \ne gf$ in general.
Inside-out for numerical inputs.

Inverse functions

$f^{-1}$ undoes $f$ . Method: swap $x, y$ , solve.

Inverse function $f^{-1}(x)$ undoes $f$ .

If $f(2) = 7$ , then $f^{-1}(7) = 2$ .

Property: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ .

The inverse f⁻¹ undoes f — it maps every output straight back to its input.

Method to find $f^{-1}$ :

Write $y = f(x)$ .
SWAP $x$ and $y$ .
SOLVE for $y$ .
The result is $f^{-1}(x)$ .

Example. $f(x) = \dfrac{2x - 5}{3}$ .

$y = \dfrac{2x - 5}{3}$ .
Swap: $x = \dfrac{2y - 5}{3}$ .
Solve for $y$ $y$ :
- $3x = 2y - 5$ .
- $2y = 3x + 5$ .
- $y = \dfrac{3x + 5}{2}$ .
$f^{-1}(x) = \dfrac{3x + 5}{2}$ .

Verify. $f(f^{-1}(x)) = f\left(\dfrac{3x + 5}{2}\right) = \dfrac{2 \cdot (3x + 5)/2 - 5}{3} = \dfrac{3x + 5 - 5}{3} = \dfrac{3x}{3} = x$ . ✓

Worked qualitative. Why is $f^{-1}(x) \ne 1/f(x)$ ?

$f^{-1}$ means INVERSE FUNCTION (undo).
$1/f(x)$ means RECIPROCAL.
The notation is overloaded but they're DIFFERENT.
$f(x) = 2x$ has inverse $f^{-1}(x) = x/2$ , but reciprocal $1/(2x) = 1/(2x)$ . Different.

Edexcel tip. When asked for $f^{-1}$ , ALWAYS use the swap method. Showing the swap step earns the M1.

$f^{-1}$ undoes $f$ .
Method: swap $x, y$ , solve.
$f^{-1} \ne 1/f$ — different.
Verify by composition.

Quick recap

$f(x)$ : function notation.
Evaluate by substitution.
$fg(x) = f(g(x))$ — $g$ first.
Inverse: swap $x, y$ , solve.
$f^{-1} \ne 1/f$ .

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$fg(x) = f(g(x))$ — composite.
Inverse method: swap, solve.
$f^{-1}$ undoes $f$ .
$f^{-1}$ is NOT $1/f$ .

How it’s examined

Functions on every Higher Tier paper (4-7 marks). Composite + inverse standard. Examiner reports flag (1) wrong order in composite, (2) interpreting $f^{-1}$ as reciprocal, (3) skipping swap step.

Take this whole topic with you

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
Foundation• function, evaluate
Question
$f(x) = 3x - 7$ . Find $f(5)$ and $f(-2)$ .
Step-by-step solution
1. Step 1
  $f(5) = 3(5) - 7 = 8$ .
2. Step 2
  $f(-2) = 3(-2) - 7 = -13$ .
Answer
$f(5) = 8$ , $f(-2) = -13$
2Composite functions $fg(x)$
Higher• Adapted from 4MA1/1H May/Jun 2024 Q16• composite
Question
$f(x) = 2x + 3$ and $g(x) = x^{2} - 1$ . Find (a) $fg(x)$ , (b) $gf(x)$ .
Step-by-step solution
1. Step 1
  (a) $fg(x) = f(g(x))$ . Apply $g$ FIRST, then $f$ .
2. Step 2
  $g(x) = x^{2} - 1$ . Substitute into $f$ : $f(g(x)) = 2(x^{2} - 1) + 3 = 2x^{2} + 1$ .
3. Step 3
  (b) $gf(x) = g(f(x))$ . Apply $f$ first.
4. Step 4
  $f(x) = 2x + 3$ . Substitute: $g(f(x)) = (2x + 3)^{2} - 1 = 4x^{2} + 12x + 9 - 1 = 4x^{2} + 12x + 8$ .
Answer
(a) $fg(x) = 2x^{2} + 1$ (b) $gf(x) = 4x^{2} + 12x + 8$
Examiner tip
$fg(x)$ means 'do $g$ first, then $f$ '. Read RIGHT to LEFT — like onion layers.
3Find an inverse function
Higher• Adapted from 4MA1/2H Jan 2024 Q13• inverse
Question
$f(x) = \dfrac{2x - 5}{3}$ . Find $f^{-1}(x)$ .
Step-by-step solution
1. Step 1
  Set $y = f(x) = \dfrac{2x - 5}{3}$ .
2. Step 2
  Swap $x$ and $y$ : $x = \dfrac{2y - 5}{3}$ .
3. Step 3
  Solve for $y$ : $3x = 2y - 5 \implies 2y = 3x + 5 \implies y = \dfrac{3x + 5}{2}$ .
4. Step 4
  So $f^{-1}(x) = \dfrac{3x + 5}{2}$ .
Answer
$f^{-1}(x) = \dfrac{3x + 5}{2}$
Examiner tip
Method: write $y = f(x)$ , swap $x$ and $y$ , solve for $y$ . The result is $f^{-1}(x)$ .
4Evaluate a composite
Higher• composite, evaluate
Question
$f(x) = 2x + 1$ and $g(x) = 3x - 2$ . Find $fg(4)$ .
Step-by-step solution
1. Step 1
  Compute $g(4) = 3(4) - 2 = 10$ .
2. Step 2
  Compute $f(10) = 2(10) + 1 = 21$ .
Answer
$fg(4) = 21$
Examiner tip
For numerical composite: evaluate inside-out. $g(4)$ first, then $f$ of that result.

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 Pearson Edexcel IGCSE 4MA1 sitting.

Function
Examiner keyword
A rule that assigns to each input EXACTLY ONE output. Notation: $f(x)$ reads 'f of x'.
Example
$f(x) = 2x + 3$ assigns to each $x$ the value $2x + 3$ .
Composite function
Examiner keyword
$fg(x)$ means 'do $g$ first, then apply $f$ '. $fg(x) = f(g(x))$ .
Example
If $f(x) = x + 1$ and $g(x) = 2x$ , then $fg(x) = 2x + 1$ .
Inverse function
Examiner keyword
Notation $f^{-1}(x)$ . The function that 'undoes' $f$ : $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ .
Example
If $f(x) = 3x + 1$ , then $f^{-1}(x) = (x - 1)/3$ .

Common Mistakes and Misconceptions — Functions

The traps other students keep falling into on functions questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Doing the wrong function first in composite
4MA1/1H May/Jun 2024 — examiner report Q16
Why it happens
Reading left-to-right as written.
How to avoid it
$fg(x)$ means $f(g(x))$ — apply $g$ FIRST. The function CLOSEST to $x$ is applied first. Read right-to-left.
✕Writing $f^{-1}(x) = 1/f(x)$
Why it happens
Notation looks like reciprocal.
How to avoid it
$f^{-1}$ is the INVERSE FUNCTION (undoes $f$ ), NOT the reciprocal. Use the swap method.
✕Substituting before composing
Why it happens
Trying to evaluate immediately.
How to avoid it
For $fg(x)$ , first FIND $fg(x)$ as an expression, THEN substitute. Or evaluate inside-out for numbers.

Functions — frequently asked questions

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

Functions

Short Notes - Functions

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Evaluate a function

2Composite functions fg(x)fg(x)fg(x)

3Find an inverse function

4Evaluate a composite

Function

Composite function

Inverse function

✕Doing the wrong function first in composite

✕Writing f−1(x)=1/f(x)f^{-1}(x) = 1/f(x)f−1(x)=1/f(x)

✕Substituting before composing

Functions — frequently asked questions

Functions

Short Notes - Functions

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Evaluate a function

2Composite functions fg(x)fg(x)fg(x)

3Find an inverse function

4Evaluate a composite

Function

Composite function

Inverse function

✕Doing the wrong function first in composite

✕Writing f−1(x)=1/f(x)f^{-1}(x) = 1/f(x)f−1(x)=1/f(x)

✕Substituting before composing

Functions — frequently asked questions

2Composite functions $fg(x)$

✕Writing $f^{-1}(x) = 1/f(x)$

2Composite functions $fg(x)$

✕Writing $f^{-1}(x) = 1/f(x)$