What is a function in the context of IB DP Mathematics?

In IB DP Mathematics, a function is a relation between a set of inputs and a set of possible outputs where each input is related to exactly one output. Functions are often represented as f(x), where x is the input variable, and f(x) is the output. Understanding functions is crucial as they form the basis for many mathematical concepts and real-world applications.

How do you determine if a relation is a function?

To determine if a relation is a function, you can use the vertical line test on its graph. If any vertical line intersects the graph at more than one point, the relation is not a function. This is because a function can only have one output for each input. Alternatively, you can check the set of ordered pairs to ensure that each input corresponds to only one output.

What are the different types of functions studied in IB DP Mathematics?

In IB DP Mathematics, students study various types of functions including linear, quadratic, polynomial, rational, exponential, logarithmic, and trigonometric functions. Each type has unique properties and applications, and understanding these differences is essential for solving complex mathematical problems.

How do you find the inverse of a function?

To find the inverse of a function, you need to swap the roles of the input and output variables and solve for the new output variable. For a function f(x), its inverse is denoted as f^(-1)(x). Not all functions have inverses; a function must be bijective (both injective and surjective) to have an inverse. Graphically, the inverse is a reflection of the function across the line y = x.

What is the significance of domain and range in functions?

The domain of a function is the set of all possible input values, while the range is the set of all possible output values. Understanding the domain and range is crucial because they define the limits within which the function operates. In IB DP Mathematics, analyzing the domain and range helps in graphing functions and solving equations accurately.

Why is "Computing $(f \circ g)(x)$ as $g(f(x))$." a common mistake in Functions?

Why it happens: Reading left-to-right. How to avoid it: Use the bracket: f(g(x)) — inner first.

Why is "Finding $f^{-1}$ without stating its domain." a common mistake in Functions?

Why it happens: Treating the formula as the whole answer. How to avoid it: Always state 'so f^-1(x) = for x '.

Why is "Writing $(f \circ g)^{-1} = f^{-1} \circ g^{-1}$." a common mistake in Functions?

Why it happens: Default assumption that order is preserved. How to avoid it: Order REVERSES: (f g)^-1 = g^-1 f^-1.

IB DP Mathematics (AA)

Functions

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Short Notes - Functions

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Detailed Study Notes

Detailed notes on Functions for IB DP Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Functions — IB Maths AA HL: domain, range, composition, inverse and self-inverse functions

Function foundations for AA HL. This note covers function notation, domain and range, composition, finding inverses (with restricted domains), and the HL-relevant self-inverse and one-to-one concepts.

At a glance

FUNCTION: each input → exactly one output.
DOMAIN: allowed inputs; RANGE: outputs actually produced.
COMPOSITE: (f ∘ g)(x) = f(g(x)) — apply g FIRST.
INVERSE $f^{-1}$ : defined iff $f$ is ONE-TO-ONE (passes horizontal line test).
FINDING INVERSE: swap $x$ and $y$ , then solve for $y$ ; state new domain.
INVERSE GRAPH: reflection of $f$ in $y = x$ .
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ — REVERSE order.

What you’ll learn

Mapped to the IB DP Maths AA HL subject guide (2021 onwards (applies to 2026 exams)).

AO1 — Use function notation and state domain/range.
AO2 — Compute composite and inverse functions.
AO2 — Restrict the domain to force invertibility.
AO3 — Justify domain restrictions explicitly.

Notation, domain and range

Set-builder or interval notation.

Function $f : X \to Y$ assigns to each $x \in X$ exactly one $y \in Y$ .

Domain $D_f$ : allowed inputs. Range $R_f$ : actually produced outputs.

Common domain restrictions:

$\div 0$ : denominator must be nonzero.
$\sqrt{\text{...}}$ : argument $\ge 0$ over reals.
$\ln(\text{...})$ : argument $> 0$ .

Worked example. Find domain and range of $f(x) = \dfrac{1}{\sqrt{4 - x^2}}$ .

Need $4 - x^2 > 0$ (denominator nonzero AND argument of $\sqrt{}$ positive): $-2 < x < 2$ .

So $D_f = (-2, 2)$ .

Range: $\sqrt{4 - x^2}$ ranges over $(0, 2]$ , so $1/\sqrt{4 - x^2}$ ranges over $[1/2, \infty)$ .

So $R_f = [1/2, \infty)$ .

Vertical line test. A graph in the $xy$ -plane represents a function iff every vertical line crosses it at most once.

Domain forbidden: $\div 0$ , $\sqrt{\text{negative}}$ , $\ln(\le 0)$ .
Range usually requires sketching the function.
Vertical line test for functions.

Composite functions

$(f \circ g)(x) = f(g(x))$ — apply $g$ first.

Composition $(f \circ g)(x) = f(g(x))$ .

Domain of $f \circ g$ : $\{x \in D_g : g(x) \in D_f\}$ . Both restrictions apply.

Worked example. $f(x) = \sqrt{x}$ , $g(x) = x - 3$ . Find $(f \circ g)(x)$ and its domain.

$(f \circ g)(x) = \sqrt{x - 3}$ . Domain: $x - 3 \ge 0$ , so $x \ge 3$ .

Worked example (evaluate). $f(x) = 2x + 1$ , $g(x) = x^2 - 5$ . Find $(g \circ f)(3)$ .

$f(3) = 7$ . $g(7) = 49 - 5 = 44$ .

Worked example (solve composite equation). Given $f(x) = x + 4$ and $g(x) = x^2$ , solve $(f \circ g)(x) = 13$ .

$g(x) = x^2$ , $f(g(x)) = x^2 + 4 = 13 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$ .

State BOTH.

Order matters. $f \circ g$ and $g \circ f$ are generally different functions.

Inner function applied first.
Domain restricted by BOTH the inner's domain and the requirement that inner output be in outer's domain.
Solve composite equations as you would any equation — state all solutions.

Inverse functions

Swap, solve, restrict the domain.

$f^{-1}$ exists iff $f$ is one-to-one (no two different inputs give the same output). Horizontal line test is the visual check.

Method.

Write $y = f(x)$ .
Swap $x$ and $y$ .
Solve for $y$ — this is $f^{-1}(x)$ .
State the new domain (= range of $f$ ).

Worked example. Find $f^{-1}$ for $f(x) = \dfrac{2x + 1}{x - 3}$ .

$y = \dfrac{2x + 1}{x - 3}$ . Swap: $x = \dfrac{2y + 1}{y - 3}$ .

$x(y - 3) = 2y + 1 \Rightarrow y(x - 2) = 3x + 1 \Rightarrow y = \dfrac{3x + 1}{x - 2}$ .

$f^{-1}(x) = \dfrac{3x + 1}{x - 2}$ , domain $\mathbb{R} \setminus \{2\}$ .

Verification. $f \circ f^{-1} = \text{id}$ and $f^{-1} \circ f = \text{id}$ .

Restricting domain to invert. $f(x) = x^2$ is not invertible over $\mathbb{R}$ . Restrict to $D_f = [0, \infty)$ : inverse is $f^{-1}(x) = \sqrt{x}$ .

Graph property. $y = f^{-1}(x)$ is the REFLECTION of $y = f(x)$ in $y = x$ .

Composition of inverses (HL). $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ — order REVERSES.

Worked example. Verify $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ with $f(x) = x + 5$ and $g(x) = 3x$ .

$(f \circ g)(x) = 3x + 5$ . Inverse: swap $y = 3x + 5 \to x = 3y + 5 \to y = (x - 5)/3$ . So $(f \circ g)^{-1}(x) = (x - 5)/3$ .

$f^{-1}(x) = x - 5$ , $g^{-1}(x) = x/3$ . $(g^{-1} \circ f^{-1})(x) = g^{-1}(x - 5) = (x - 5)/3$ ✓.

Inverse exists iff one-to-one.
Swap-and-solve method.
Always state new domain (range of $f$ ).
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ .

Quick recap

Function: one output per input.
Domain and range: state explicitly with set or interval notation.
$(f \circ g)(x) = f(g(x))$ .
Inverse: swap and solve; state restricted domain.
Inverse graph reflects in $y = x$ .
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ — order reverses.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

$(f \circ g)(x) = f(g(x))$
$f(f^{-1}(x)) = x$ , $f^{-1}(f(x)) = x$
Vertical line test: function. Horizontal: invertible.
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$

How it’s examined

Paper 1: composition, inverses, domain restrictions. Paper 2: applied problems. Paper 3: deeper structural questions. Examiner reports stress the explicit domain statement when defining $f^{-1}$ .

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-31.

Take this whole topic with you

Step-by-step worked examples — Functions

Step-by-step solutions to past-paper-style questions on functions, written exactly the way a tutor would explain them at the board.

1Find domain and range
Getting started• domain, range
Question
Find $D_f$ and $R_f$ for $f(x) = 2 - \sqrt{x + 1}$ .
Step-by-step solution
1. Step 1
  $x + 1 \ge 0 \Rightarrow x \ge -1$ . So $D_f = [-1, \infty)$ .
2. Step 2
  $\sqrt{x+1} \ge 0$ , so $2 - \sqrt{x+1} \le 2$ . Range $(-\infty, 2]$ .
Answer
$D_f = [-1, \infty)$ , $R_f = (-\infty, 2]$ .
2Find inverse of rational function
Building confidence• inverse
Question
Find $f^{-1}(x)$ for $f(x) = \dfrac{3x - 1}{x + 4}$ .
Step-by-step solution
1. Step 1
  $y = \dfrac{3x - 1}{x + 4}$ . Swap: $x = \dfrac{3y - 1}{y + 4}$ .
2. Step 2
  $x(y + 4) = 3y - 1 \Rightarrow y(x - 3) = -1 - 4x$ .
3. Step 3
  $y = \dfrac{-1 - 4x}{x - 3} = \dfrac{4x + 1}{3 - x}$ .
Answer
$f^{-1}(x) = \dfrac{4x + 1}{3 - x}$ , $x \ne 3$ .
3Restrict domain to make $f$ invertible
Stretch• inverse
Question
$f(x) = x^2 - 6x + 5$ . Find the largest domain containing $x = 0$ on which $f$ is invertible, and find $f^{-1}$ .
Step-by-step solution
1. Step 1
  Vertex at $x = 3$ . Parabola opens up, so $f$ is decreasing on $(-\infty, 3]$ and increasing on $[3, \infty)$ . Since $x = 0$ is in the first piece, restrict to $(-\infty, 3]$ .
2. Step 2
  Vertex form: $f(x) = (x - 3)^2 - 4$ .
3. Step 3
  Inverse: $y = (x-3)^2 - 4 \Rightarrow x = (y - 3)^2 - 4 \Rightarrow (y - 3)^2 = x + 4$ .
4. Step 4
  Since $y \le 3$ : $y - 3 = -\sqrt{x + 4}$ , so $y = 3 - \sqrt{x + 4}$ . Domain $[-4, \infty)$ .
Answer
Restrict to $(-\infty, 3]$ ; $f^{-1}(x) = 3 - \sqrt{x + 4}$ , $x \ge -4$ .
4Composite with domain restriction
Stretch• composite
Question
$f(x) = \ln x$ , $g(x) = 4 - x^2$ . Find $(f \circ g)(x)$ and state its domain.
Step-by-step solution
1. Step 1
  $(f \circ g)(x) = \ln(4 - x^2)$ .
2. Step 2
  Domain requires $4 - x^2 > 0$ , so $|x| < 2$ , i.e. $-2 < x < 2$ .
Answer
$\ln(4 - x^2)$ , domain $(-2, 2)$ .

Key Definitions and Keywords — Functions

Definitions to memorise and the exact keywords mark schemes credit for functions answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA HL sitting.

One-to-one (injective)
Examiner keyword
$f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ . Required for invertibility.
Inverse $f^{-1}$
Examiner keyword
Function satisfying $f(f^{-1}(x)) = x$ for $x$ in range of $f$ , and $f^{-1}(f(x)) = x$ for $x$ in domain of $f$ .
Inverse of composite
Examiner keyword
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ — order reverses.

Common Mistakes and Misconceptions — Functions

The traps other students keep falling into on functions questions — taken from recent IB DP Maths AA HL examiner reports and mark schemes — and how to avoid them.

✕Computing $(f \circ g)(x)$ as $g(f(x))$ .
Why it happens
Reading left-to-right.
How to avoid it
Use the bracket: $f(g(x))$ — inner first.
✕Finding $f^{-1}$ without stating its domain.
Why it happens
Treating the formula as the whole answer.
How to avoid it
Always state 'so $f^{-1}(x) = \ldots$ for $x \in \ldots$ '.
✕Writing $(f \circ g)^{-1} = f^{-1} \circ g^{-1}$ .
Why it happens
Default assumption that order is preserved.
How to avoid it
Order REVERSES: $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ .

Functions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.