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Cambridge IGCSE Mathematics revision notes

Functions Revision Notes

Cambridge IGCSE revision notes for Functions.

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Functions

CambridgeIGCSEMathematicsExtended Algebra and graphs

Complete revision guide

Functions: Domain, Range, Inverse and Composite Functions

This note explains the concept of functions, including domain and range, how to use function notation, find inverse functions, and form composite functions. It includes worked examples and exam-focused advice, emphasising clear reasoning and simplification of answers.

Syllabus source

Section 3, Topic E2.13, page 42, objective E2.13.1

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What you should be able to do

  • ✓Understand what a function is and identify its domain and range.
  • ✓Use function notation correctly to express functions.
  • ✓Calculate inverse functions by interchanging variables and solving.
  • ✓Form composite functions by substituting one function into another.
  • ✓Simplify answers, especially fractions, to their simplest form.
1

Understanding Functions, Domain and Range

A function is a rule that assigns exactly one output value for each input value. The set of all possible inputs is called the domain, and the set of all possible outputs is called the range.

For example, consider the function f defined by f(x) = 3x - 5. Here, x is the input variable, and f(x) is the output. If the domain is all real numbers, then the range is also all real numbers because 3x - 5 can take any real value depending on x.

Functions can be represented in various ways, including equations, tables, or mapping diagrams. Mapping diagrams visually show how each input maps to exactly one output.

  • Function: a rule assigning one output per input.
  • Domain: all possible input values.
  • Range: all possible output values.
  • Function notation: f(x) means output of function f for input x.
2

Using Function Notation

Function notation is a concise way to write functions. For example, f(x) = 3x - 5 means the function f takes an input x and outputs 3 times x minus 5.

Other examples include g(x) = \frac{5}{2}, which is a constant function always outputting 2.5, and h(x) = 2x + 3, which doubles the input and adds 3.

When evaluating a function for a specific input, substitute the input value into the expression. For example, f(4) = 3(4) - 5 = 12 - 5 = 7.

  • f(x) = 3x - 5 means output = 3 × input - 5.
  • g(x) = 5/2 means output is always 2.5.
  • h(x) = 2x + 3 means output = 2 × input + 3.
  • Evaluate by substituting input values.
3

Finding Inverse Functions

An inverse function reverses the effect of the original function. If f maps x to y, then its inverse f^{-1} maps y back to x.

To find the inverse of a function given by y = f(x), swap x and y, then solve for y.

For example, if f(x) = 3x + 5, then y = 3x + 5. Swap to x = 3y + 5, then solve for y: x - 5 = 3y, so y = \frac{x - 5}{3}. Thus, f^{-1}(x) = \frac{x - 5}{3}.

  • Inverse function reverses input-output pairs.
  • Swap x and y in the equation y = f(x).
  • Solve for y to get f^{-1}(x).
  • Example: f(x) = 3x + 5 inverse is f^{-1}(x) = (x - 5)/3.
4

Forming Composite Functions

Composite functions combine two functions by applying one function to the result of another.

If f(x) and g(x) are functions, then the composite function fg(x) means f(g(x)) — apply g first, then f.

For example, if f(x) = 3x + 5 and g(x) = x + 2, then fg(x) = f(g(x)) = f(x + 2) = 3(x + 2) + 5 = 3x + 6 + 5 = 3x + 11. Similarly, gf(x) = g(f(x)) = g(3x + 5) = (3x + 5) + 2 = 3x + 7.

  • Composite fg(x) = f(g(x)) means apply g, then f.
  • Substitute g(x) into f(x).
  • Simplify the expression fully.
  • Example: f(x) = 3x + 5, g(x) = x + 2, then fg(x) = 3x + 11, gf(x) = 3x + 7.
5

Simplifying Answers and Expressing Fractions

When working with functions, especially inverse and composite functions, answers often involve fractions.

Always simplify fractions to their lowest terms for clarity and accuracy.

For example, if an inverse function is f^{-1}(x) = \frac{2x}{4}, simplify to f^{-1}(x) = \frac{x}{2}. This ensures answers are neat and exam-ready.

  • Simplify fractions fully.
  • Check for common factors in numerator and denominator.
  • Express answers clearly and neatly.
  • Example: \frac{6x + 4}{2} simplifies to 3x + 2.

At-a-glance summary

Comparison table

StepDetailExampleExam Use
Identify function and notationRecognise function rule and notation f(x)f(x) = 3x - 5Use correct notation to avoid confusion
Find inverseSwap x and y, solve for yf(x) = 3x + 5 → f^{-1}(x) = (x - 5)/3Show all algebraic steps clearly
Form compositeSubstitute one function into anotherfg(x) = f(g(x)) = 3(x + 2) + 5Write composite function explicitly
Simplify answerReduce fractions and expressions\frac{6x}{4} = \frac{3x}{2}Simplify to gain full marks

Active recall

Quick check

1. If f(x) = 2x + 3, what is f(4)?

f(4) = 2 × 4 + 3 = 8 + 3 = 11.

2. Find the inverse of f(x) = 5x - 7.

Swap x and y: x = 5y - 7 → 5y = x + 7 → y = (x + 7)/5, so f^{-1}(x) = (x + 7)/5.

3. Given f(x) = x^2 and g(x) = x + 1, find fg(2).

fg(2) = f(g(2)) = f(2 + 1) = f(3) = 3^2 = 9.

4. Simplify the composite function fg(x) if f(x) = 3x - 4 and g(x) = \frac{x}{2}.

fg(x) = f(g(x)) = f(x/2) = 3(x/2) - 4 = (3x/2) - 4.

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