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A Level Maths 9709 · Paper 1 Pure Mathematics 1 · 1.2 · ~11 min read

Functions and transformations

Domain and range, one-one functions, inverse functions and their graphs, composite functions and when they exist, and the four graph transformations described in the exact words examiners credit.

Domain & rangeOne-one functionsInverse functionsComposite functionsTranslationsStretchesReflectionsCombined transformations
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A function sends every input in its domain to exactly one output. The set of outputs is the range.
  • Only a one-one function has an inverse. The graph of f⁻¹ is the reflection of the graph of f in y = x, and the domain and range swap over.
  • fg(x) means f(g(x)): do g first. It only exists when every output of g is an allowed input for f.
  • Transformations: changes outside f act on y in the obvious way; changes inside f act on x in the opposite way. Always name the type ("translation", "stretch", "reflection") with its vector, factor or line.

By the end you can

  • Find the range of a function, including quadratics on a restricted domain.
  • Decide whether a function is one-one and find its inverse, with the correct domain.
  • Form composite functions, decide whether they exist, and find their ranges.
  • Describe and apply translations, stretches and reflections, including combinations.

1Domain, range and one-one functions

Key words

Domain — the set of input values (x-values) the function is defined for. It is part of the definition: f(x) = x² for x ≥ 0 is a different function from f(x) = x² for x ∈ ℝ.

Range — the set of output values (y-values) the function actually produces.

One-one function — every output comes from exactly one input. Test: every horizontal line meets the graph at most once.

Many-one function — some output comes from two or more inputs, e.g. f(x) = x² on ℝ, since f(2) = f(−2) = 4.

  • Finding a range: sketch the graph over the given domain. For quadratics, complete the square to find the vertex, then check the endpoints of the domain.
  • Notation: write the range in terms of f(x) or y, e.g. f(x) ≥ 2, not x ≥ 2. The June 2024 report flagged incomplete range inequalities.
Worked example

f(x) = x² − 6x + 11 for x ≥ 3. Find the range of f.

  1. x² − 6x + 11 = (x − 3)² + 2 M1
  2. the vertex (3, 2) is at the left end of the domain and the graph rises from there
  3. range: f(x) ≥ 2 A1
Worked example

f(x) = x² + 4x for x ≤ k. Find the largest value of k for which f is one-one.

  1. x² + 4x = (x + 2)² − 4, so the vertex is at x = −2 M1
  2. f is one-one on one side of the vertex only
  3. largest k: k = −2 A1

2Inverse functions

Finding f⁻¹

  • 1 · Write y = f(x). 2 · Rearrange to make x the subject. 3 · Swap x and y (or replace y by x) to get f⁻¹(x). 4 · State the domain of f⁻¹: it is the range of f.
domain of f⁻¹ = range of f, range of f⁻¹ = domain of f LEARN IT
and f⁻¹f(x) = ff⁻¹(x) = x
Worked example

f(x) = x² − 6x + 11 for x ≥ 3. Find f⁻¹(x) and state its domain.

  1. y = (x − 3)² + 2 ⇒ (x − 3)² = y − 2 M1
  2. x − 3 = +√(y − 2): take the positive root because x ≥ 3 M1
  3. f⁻¹(x) = 3 + √(x − 2), for x ≥ 2 A1 B1
  4. check: f(4) = 3 and f⁻¹(3) = 3 + 1 = 4 ✓
Worked example

g(x) = (2x + 3)/(x − 1) for x > 1. Find g⁻¹(x).

  1. y(x − 1) = 2x + 3 ⇒ xy − y = 2x + 3 M1
  2. xy − 2x = y + 3 ⇒ x(y − 2) = y + 3 ⇒ x = (y + 3)/(y − 2) M1
  3. g⁻¹(x) = (x + 3)/(x − 2), x > 2 A1 (the range of g is g(x) > 2)
Common mistakeChoosing the wrong sign of the square root. For x ≥ 3, f⁻¹(x) = 3 − √(x − 2) is wrong: it gives outputs below 3. The 2024 report saw f⁻¹(x) = −√(x + 4)-type answers that ignored the range of the inverse.

The graph of f⁻¹

xy2468102460y = f(x) = x² + 2, x ≥ 0y = f⁻¹(x) = √(x − 2)y = x(1, 3)(3, 1)Domain of f⁻¹ = range of f (x ≥ 2). Range of f⁻¹ = domain of f (y ≥ 0). Mirror line y = x.
f and f⁻¹ are reflections of each other in y = x: the point (1, 3) on f becomes (3, 1) on f⁻¹.
  • Draw the line y = x on any sketch of f and f⁻¹, and use the same scale on both axes. Examiners deducted marks when the mirror line was missing.
  • Where do f and f⁻¹ meet? For an increasing function, any intersection lies on y = x, so you can solve f(x) = x instead of f(x) = f⁻¹(x).
Exam tip"Explain why f has no inverse" needs one sentence: "f is not one-one" (for example, f(1) = f(5)). That is the accepted reason.

3Composite functions

fg(x) = f(g(x))

  • Do the inside function first. fg(x): apply g, then apply f to the result. gf is usually different.
  • fg exists only if the range of g lies inside the domain of f.
  • Range of fg: find the range of g, then find what f does to that set. Don't just simplify the formula and guess.
Worked example

f(x) = 2x + 1 and g(x) = x² − 3, both for x ∈ ℝ. Find fg(x) and gf(x), and solve gf(x) = 6.

  1. fg(x) = f(x² − 3) = 2(x² − 3) + 1 = 2x² − 5 B1
  2. gf(x) = g(2x + 1) = (2x + 1)² − 3 B1
  3. (2x + 1)² − 3 = 6 ⇒ (2x + 1)² = 9 ⇒ 2x + 1 = ±3 M1
  4. x = 1 or x = −2 A1
Worked example

f(x) = 1/(x − 2) for x > 2 and g(x) = x² + 1 for x ∈ ℝ. Explain why fg does not exist, and find the range of gf.

  1. range of g is g(x) ≥ 1, but f needs inputs greater than 2 (e.g. g(0) = 1 is not in the domain of f), so fg does not exist B1
  2. range of f is f(x) > 0, which is inside the domain of g, so gf exists: gf(x) = 1/(x − 2)² + 1 M1
  3. 1/(x − 2)² takes every positive value, so gf(x) > 1 A1
Common mistakeWriting fg(x) as f(x) × g(x). The notation means "f of g of x", not a product.

4Transformations of graphs

The four basic transformations

EquationTransformation (use these words)Effect on a point (x, y)
y = f(x) + atranslation by the vector 0a(x, y + a)
y = f(x + a)translation by the vector −a0(x − a, y)
y = af(x)stretch parallel to the y-axis, scale factor a(x, ay)
y = f(ax)stretch parallel to the x-axis, scale factor 1/a(x/a, y)
y = −f(x)reflection in the x-axis(x, −y)
y = f(−x)reflection in the y-axis(−x, y)
xyy = f(x) + 1.5 (translate up)xyy = f(x + 2) (translate left)xyy = 2f(x) (vertical stretch ×2)xyy = f(2x) (horizontal stretch ×½)Grey dashed curve = the original y = f(x)
Outside the brackets: acts on y as written. Inside the brackets: acts on x the opposite way (f(x + 2) moves left; f(2x) squashes by ½).
Common mistakeVague descriptions. "Moves down 2" or "shifts" lose marks. Write "translation by the vector 0−2". For a stretch, give the direction and the factor: "stretch parallel to the x-axis, scale factor ½".

Combinations: order matters

  • Vertical and horizontal changes are independent, so their order doesn't matter: y = f(2x) − 3 is a stretch parallel to the x-axis (factor ½) and a translation by 0−3, in either order.
  • Two changes in the same direction do depend on order. y = 2f(x) + 3 is "stretch ×2 then translate up 3". y = 2[f(x) + 3] is "translate up 3 then stretch ×2" (the curve ends up 6 higher).
  • Work from the equation outwards: for y = af(x) + b, deal with the multiplication before the addition, as you would when calculating.
Worked example

Describe a sequence of transformations that maps y = x² onto y = 3(x + 1)² − 4.

  1. translation by −10 gives y = (x + 1)² B1
  2. stretch parallel to the y-axis, scale factor 3, gives y = 3(x + 1)² B1
  3. translation by 0−4 gives y = 3(x + 1)² − 4 B1
  4. (the stretch must come before the downward translation)
Worked example

The curve y = x² − 4x is reflected in the x-axis and then translated by 30. Find the equation of the new curve.

  1. reflection: y = −(x² − 4x) B1
  2. translation by 30: replace x by x − 3 M1
  3. y = −[(x − 3)² − 4(x − 3)] = −(x² − 10x + 21)
  4. y = −x² + 10x − 21 A1
Common mistakeReversing the order or replacing x by x/2 instead of 2x. The June 2024 report listed both as common errors. Test one point: under y = f(2x), the point (4, 1) on y = f(x) moves to (2, 1).

Trigonometric functions

  • The same rules apply to trig graphs: y = 3 − 2 sin x is y = sin x stretched by factor 2 parallel to the y-axis, reflected in the x-axis, then translated by 03.
  • Range of a sin or cos expression: sin x lies between −1 and 1, so 3 − 2 sin x lies between 1 and 5.
Linking questionThe completed square from Quadratics gives ranges and inverses; trig transformations come back in Trigonometry; and inverse functions are why eˣ and ln x are reflections of each other in Logarithms and exponential functions.

Test yourself — cover the answers first

Quick check

  1. What condition must a function satisfy to have an inverse?
  2. f(x) = 3x − 2. Find f⁻¹(x).
  3. What is the domain of f⁻¹ in terms of f?
  4. Describe the transformation y = f(x) → y = f(x − 5).
  5. f(x) = x + 4, g(x) = 2x. Find gf(1) and fg(1).

Exam-style

  1. f(x) = 4 − (x − 2)² for x ≥ 2. (a) State the range of f. (b) Find f⁻¹(x) and its domain. (c) Sketch y = f(x) and y = f⁻¹(x) on one diagram.
  2. Describe fully two transformations that map y = cos x onto y = 2 cos(x/3).
  3. f(x) = 1/(x + 1) for x > −1 and g(x) = 2x − 1 for x > a. Find the least value of a for which fg exists.
Show answers
  1. it must be one-one
  2. (x + 2)/3
  3. the range of f
  4. translation by the vector (5, 0)
  5. gf(1) = 10; fg(1) = 6
  6. (a) f(x) ≤ 4
    (b) f⁻¹(x) = 2 + √(4 − x), x ≤ 4
    (c) reflections of each other in y = x, meeting on that line
  7. stretch parallel to the y-axis, scale factor 2; stretch parallel to the x-axis, scale factor 3
  8. need 2x − 1 > −1 for every input, so x > 0: a = 0
Practise this chapter
A Level practice papers
Next chapterCoordinate geometry: lines and circles →
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.