Functions and Graph Transformations

A-Level Maths · Pure Mathematics

Functions

A function maps each input to exactly one output. Key terms:

  • Domain = the set of allowed inputs (x-values); range = the set of possible outputs (y-values).
  • A function must be one-to-one (or restricted to be) to have an inverse.

Composite functions

fg(x) means "do g first, then f": fg(x) = f(g(x)). Order matters — fg is generally not the same as gf.

  • Example: f(x) = 2x, g(x) = x + 3 → fg(x) = 2(x + 3) = 2x + 6; gf(x) = 2x + 3.

Inverse functions

The inverse f⁻¹(x) reverses f: if f(a) = b then f⁻¹(b) = a.

  • Method: write y = f(x), swap x and y, rearrange to make y the subject.
  • The graph of f⁻¹ is the reflection of f in the line y = x.
  • The domain of f⁻¹ = the range of f (and vice versa).

The modulus function

|x| gives the positive value (distance from zero). To solve |f(x)| = a, solve f(x) = a and f(x) = −a. The graph of y = |f(x)| reflects any part below the x-axis above it.

Graph transformations

For y = f(x):

TransformationEffect
f(x) + atranslate up by a
f(x + a)translate left by a (opposite sign)
af(x)vertical stretch, scale factor a
f(ax)horizontal stretch, scale factor 1/a
−f(x)reflect in the x-axis
f(−x)reflect in the y-axis

Inside the bracket → horizontal (and "backwards"); outside → vertical (as expected).

Worked example

Given f(x) = 3x − 1, find f⁻¹(x).

  • y = 3x − 1 → swap: x = 3y − 1 → x + 1 = 3y → y = (x + 1)/3.
  • f⁻¹(x) = (x + 1)/3. ✓

Common mistakes

  • Doing fg as "f then g" — it's g first, then f.
  • Getting horizontal translations the wrong way: f(x + a) moves left.
  • Forgetting a function must be one-to-one for an inverse to exist.

Exam tips

  • State domain and range, especially for inverses (they swap).
  • Apply transformations in the right order; inside the bracket affects x (horizontally, reversed).
  • For modulus equations, consider both the positive and negative cases.

Key facts to remember

  • Composite fg(x) = f(g(x)) (g first); inverse found by swapping x and y — reflection in y = x; domain/range swap.
  • |x| = positive value; solve |f(x)| = a via f(x) = ±a.
  • Transformations: outside bracket = vertical (as written); inside bracket = horizontal and reversed.
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