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Functions & Function Notation –

GCSEMathsAI Team·8 min read·18 March 2026

Functions and function notation are Higher-tier topics that underpin several areas of GCSE Maths — from graph transformations to inverse and composite functions. Understanding what f(x) means, how to evaluate a function at a given value, and how to set up and solve equations involving functions are all skills that AQA, Edexcel and OCR expect at Grade 7 and above. This guide breaks the topic down with clear definitions, step-by-step methods and fully worked examples.

What Is a Function?

A function is a rule that takes an input (usually x) and produces exactly one output. Functions are written using function notation:

f(x) = 2x + 5

This means "the function f takes x and gives back 2x + 5". The letter f is just a label — you may also see g(x), h(x) or other letters.

Key vocabulary

  • Input: the value you substitute into the function (the x-value).
  • Output: the value the function produces (the y-value or f(x) value).
  • Domain: the set of all possible input values.
  • Range: the set of all possible output values.

Evaluating a function

To find f(a), substitute x = a into the expression for f(x).

For example, if f(x) = 3x² − 1, then:

f(4) = 3(4)² − 1 = 3(16) − 1 = 48 − 1 = 47

f(−2) = 3(−2)² − 1 = 3(4) − 1 = 12 − 1 = 11


Step-by-Step Method

How to evaluate a function

  1. Write down the function definition: f(x) = ...
  2. Replace every x in the expression with the given input value.
  3. Calculate using the correct order of operations (BIDMAS).

How to solve f(x) = k

  1. Set the function expression equal to k: e.g., 2x + 5 = 13.
  2. Solve the equation for x using standard algebraic methods.
  3. Check by substituting your answer back into f(x).

How to find the input from an output

  1. Set up the equation: if f(a) = 20, write the expression with x = a and set it equal to 20.
  2. Solve for a.

Worked Example 1 — Evaluating Functions

Question: f(x) = 5x − 3. Find: (a) f(4), (b) f(−2), (c) the value of x when f(x) = 22.

(a) f(4) = 5(4) − 3 = 20 − 3 = 17

(b) f(−2) = 5(−2) − 3 = −10 − 3 = −13

(c) Set 5x − 3 = 22:

5x = 25

x = 5

Check: f(5) = 5(5) − 3 = 25 − 3 = 22 ✓


Worked Example 2 — Functions with Quadratics

Question: g(x) = x² + 4x − 7. Find: (a) g(3), (b) the values of x when g(x) = 0.

(a) g(3) = (3)² + 4(3) − 7 = 9 + 12 − 7 = 14

(b) Set x² + 4x − 7 = 0.

This does not factorise easily, so use the quadratic formula:

x = (−b ± √(b² − 4ac)) / 2a

a = 1, b = 4, c = −7

x = (−4 ± √(16 + 28)) / 2 = (−4 ± √44) / 2 = (−4 ± 2√11) / 2

x = −2 ± √11

x = −2 + √11 ≈ 1.32 or x = −2 − √11 ≈ −5.32 (to 2 d.p.)


Common Mistakes

  • Confusing f(x) with f × x. f(x) does not mean f multiplied by x. It means "the function f applied to x". Treat the brackets as instruction brackets, not multiplication brackets.
  • Squaring negatives incorrectly. When evaluating f(−3) for f(x) = x², remember (−3)² = 9, not −9. The brackets mean you square the whole number, including the negative.
  • Forgetting to substitute everywhere. If f(x) = 2x² + x, then f(3) = 2(3)² + (3) = 18 + 3 = 21. Do not forget the second x.
  • Not checking solutions. Always substitute your answer back into the original function to verify it gives the correct output.

Exam Tips

  1. Write the substitution out in full. Show f(3) = 2(3)² + (3) before simplifying. This earns method marks even if you make an arithmetic error.
  2. When asked "find x such that f(x) = k", set up and solve an equation. Do not try to work backwards informally — show clear algebraic working.
  3. On AQA papers, function questions often appear alongside composite or inverse functions. Knowing basic evaluation fluently saves time for the harder parts.
  4. If the function is quadratic, expect two solutions when solving f(x) = k. Do not stop after finding only one.

Practice Questions

Question 1: f(x) = 7 − 2x. Find f(5) and f(−3).

Answer: f(5) = 7 − 2(5) = 7 − 10 = −3. f(−3) = 7 − 2(−3) = 7 + 6 = 13.

Question 2: h(x) = x² − 6x + 10. Find the value of h(1) and solve h(x) = 2.

Answer: h(1) = 1 − 6 + 10 = 5. For h(x) = 2: x² − 6x + 10 = 2 → x² − 6x + 8 = 0 → (x − 2)(x − 4) = 0 → x = 2 or x = 4.

Question 3: f(x) = 4x + 1. Show that f(3) + f(5) = 2f(4).

Answer: f(3) = 13, f(5) = 21, so f(3) + f(5) = 34. f(4) = 17, so 2f(4) = 34. Therefore f(3) + f(5) = 2f(4). ✓

Sharpen your function skills with unlimited practice on GCSEMathsAI. Every question comes with a full worked solution, and the difficulty adapts to your level.



Summary

  • A function is a rule that maps each input to exactly one output, written as f(x).
  • To evaluate f(a), substitute a for x in the expression and calculate.
  • The domain is the set of allowed inputs; the range is the set of possible outputs.
  • To solve f(x) = k, set the function expression equal to k and solve the resulting equation.
  • Always show your substitution step in full for method marks.
  • Quadratic functions may give two solutions when you solve f(x) = k.
  • Function notation is the foundation for inverse functions, composite functions and graph transformations.