IB Maths AI HL Further Functions & Graphs Paper 1 & 2 Mappings, domain, range ~8 min read

Functions & Mappings

A function is a special kind of mapping — every input has exactly one output. The rest of the topic is bookkeeping: which set of inputs are allowed (the domain), which outputs come out (the range), and how to write it all down using f(x) notation.

📘 What you need to know

Mappings — the four types

A mapping sends each input to some output. The four types describe whether inputs can share outputs and whether one input can have several outputs. The first two are functions (one output per input); the last two are not.

The four mapping types — only the top row are functions ONE-TO-ONE FUNCTION ✓ 123 1827 e.g. f(x) = x³ MANY-TO-ONE FUNCTION ✓ −2−112 41 e.g. f(x) = x² ONE-TO-MANY NOT A FUNCTION ✗ 14 1−12−2 e.g. x → ±√x MANY-TO-MANY NOT A FUNCTION ✗ 246 123 e.g. x → factors of x
Top row: each input has exactly one output — functions. Bottom row: at least one input has multiple outputs — not functions.

Functions, notation, domain and range

If f is a function and x is the input, then f(x) is the output. So f(2) = 5 puts the point (2, 5) on the graph of y = f(x). The domain is the set of allowed inputs — written in terms of x — and the range is the set of outputs that come out — written in terms of f(x). The range depends on the domain: shrinking the domain shrinks the range.

If no domain is stated, assume the largest possible domain — usually all real numbers, restricted only by features that block specific inputs: √x needs x ≥ 0, 1/x needs x ≠ 0, and similar.

Sketching helps: when the domain is restricted, the cleanest way to find the range is to sketch y = f(x) over the given x-interval. Check the endpoints and any turning points inside the interval.
Function notation and number sets f(x) = output when input is xf(a) = b  ⇔  (a, b) on the graph ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ   (naturals, integers, rationals, reals)

Piecewise functions

A piecewise function uses different rules on different intervals of x. To evaluate f(x): pick the rule whose interval contains x, then substitute. The intervals must not overlap, but together they make up the function’s domain.

A piecewise function may or may not be continuous at a boundary. To check at x = a, evaluate both pieces at the boundary value: if they agree, the function is continuous there; if they disagree, it has a jump.

🧭 Recipe — finding the range from a given domain

  1. Write down the function and the domain interval.
  2. Check for turning points inside the domain (e.g. the vertex of a parabola).
  3. Evaluate f(x) at the endpoints of the domain and at any turning points inside it.
  4. Compare those values to find the smallest and largest outputs.
  5. State the range as an inequality in f(x), e.g. −3 ≤ f(x) ≤ 6.

Worked examples

WE 1

Classify the mappings

Classify each as one-to-one, many-to-one, one-to-many or many-to-many, and state whether it is a function.

(a) x → 2x + 1  ·  (b) xx2  ·  (c) x → ±√x

(a) distinct inputs give distinct outputs 2(1)+1 = 3, 2(2)+1 = 5, … all different (a) one-to-one — function ✓ (b) different inputs can share an output (−2)² = 2² = 4; one output, two inputs (b) many-to-one — function ✓ (c) a single input gives two outputs 4 → +2 and −2 (c) one-to-many — not a function ✗
WE 2

Function evaluation

For f(x) = 3x2 − 2x + 1, find f(−2).

substitute x = −2 f(−2) = 3(−2)² − 2(−2) + 1 = 3(4) + 4 + 1 = 12 + 4 + 1 f(−2) = 17 brackets matter: (−2)² = 4, not −4.
WE 3

State the domain and range

State the domain and range of f(x) = (x − 1)2 + 3.

no restrictions on input domain: x ∈ ℝ parabola, vertex at (1, 3), opens upward (x − 1)² ≥ 0 for all x, so (x − 1)² + 3 ≥ 3 range: f(x) ≥ 3 the minimum 3 is reached at x = 1; no upper limit.
WE 4

Range from a restricted domain

Find the range of f(x) = x2 − 4x + 1 for 0 ≤ x ≤ 5.

vertex at x = −b/(2a) = 4/2 = 2 (inside domain) f(2) = 4 − 8 + 1 = −3  (minimum) check endpoints f(0) = 1 f(5) = 25 − 20 + 1 = 6  (maximum) range: −3 ≤ f(x) ≤ 6 when the vertex is inside the domain, the minimum is at the vertex; the maximum is at the farther endpoint.
WE 5

Largest possible domain

State the largest possible domain of f(x) = √(2x − 6).

the argument of a square root must be ≥ 0 2x − 6 ≥ 0 2x ≥ 6 x ≥ 3 largest domain: x ≥ 3 other typical restrictions: 1/(…) needs … ≠ 0; log(…) needs … > 0.
WE 6

Piecewise function: evaluate and check continuity

For f(x) = {x2 if x ≤ 1;   3x − 2 if 1 < x ≤ 5;   −x + 8 if x > 5}: (a) find f(0), f(3), f(10). (b) Determine whether f is continuous at x = 1 and at x = 5.

(a) pick the correct piece for each input f(0): 0 ≤ 1 ⇒ f(0) = 0² = 0 f(3): 1 < 3 ≤ 5 ⇒ f(3) = 3(3) − 2 = 7 f(10): 10 > 5 ⇒ f(10) = −10 + 8 = −2 f(0) = 0 · f(3) = 7 · f(10) = −2 (b) compare both pieces at each boundary at x = 1: 1² = 1 and 3(1) − 2 = 1 ⇒ equal, continuous ✓ at x = 5: 3(5) − 2 = 13 and −5 + 8 = 3 ⇒ jump, not continuous ✗ continuous at x = 1; not continuous at x = 5

💡 Top tips

âš  Common mistakes

Next up: Graphing Functions & Their Key Features — sketching y = f(x) and identifying intercepts, turning points and asymptotes.

Need help with AI HL Functions?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →