Functions: Domain, Inverses, Composites & Transformations

Q: What's the difference between domain and range?

Domain is the set of valid inputs (found by checking what makes the formula undefined); range is the set of actual outputs, read from the graph rather than assumed.

Q: How do you know if a function has an inverse?

It must be one-to-one — every output comes from exactly one input. Prove this algebraically by assuming $f(x_1)=f(x_2)$ and showing $x_1=x_2$ must follow.

Q: When is a rational function self-inverse?

For $f(x)=\dfrac{ax+b}{cx+d}$, it's self-inverse exactly when $a+d=0$. Always check this condition rather than assuming it from the function's shape.

Q: Why does the order matter in $y=af(x)+k$?

Because $a$ multiplies $f(x)$ before $k$ is added — applying $k$ first gives a completely different (and wrong) transformed graph.

Q: How do I find the domain of a composite function?

Take the domain of $g$, then remove any $x$-values where $g(x)$ falls outside the domain of $f$ — both conditions must hold together.

Q: Do I need the formula for $g$ to transform its graph?

No — if you're given key points, use the coordinate mapping $(x,y)\to(x/b+h,\ ay+k)$ to transform them directly, without ever writing out $g$'s equation.

The IB AI HL toolkit for functions — asymptotes, self-inverse shortcuts and the one equation that covers every transformation.

Graph showing a function, its inverse reflected in y=x, and asymptotes labelled

Subject

Maths AI

Curriculum

IB Diploma Programme

Grade

Topic

Functions

Reading

8 min

Difficulty

Advanced

Quick facts

Difficulty

★★★★☆

Exam weight

~15-20% of AI HL (Topic 2)

Prerequisites

Algebraic manipulation, graphing basics

You'll learn

Domain/range, inverses, composites, asymptotes, transformations

Revision time

45-60 min

Functions is one of the most exam-dense topics in IB Maths AI HL — it rarely appears alone and instead fuses with algebra or drives a full Paper 2 modelling question. At its core you need five skills: reading domain and range correctly (not assuming), proving one-to-one algebraically, finding inverse functions and stating their domain properly, composing functions in the right order, and applying the single transformation equation $y = a f (b (x - h)) + k$ to any graph. Rational functions add a layer — vertical and horizontal asymptotes control both domain and range, and a neat shortcut tells you instantly when a function is self-inverse. This teaser pulls out the five ideas examiners test most, with the traps students fall into every year. For the full worked examples, proofs and practice set, the complete revision note has you covered.

What you’ll be able to do

✓Determine domain and range from a function's formula and graph

✓Prove a function is one-to-one using algebra, not just a graph

✓Find $f^{-1}(x)$ and correctly state its domain as the range of $f$

✓Evaluate and find the domain of composite functions $(f\circ g)(x)$

✓Identify vertical and horizontal asymptotes of a rational function

✓Recognise the self-inverse condition $a+d=0$

✓Apply $y=af(b(x-h))+k$ to identify any combination of transformations

✓Transform key coordinate points without needing $f$'s formula

Domain and Range: Read the Graph, Don't Assume

Domain is what makes the formula undefined — check denominators equal to zero, negatives under square roots, or a context restriction like $t \geq 0$ . Range must come from the actual graph, not a guess: a rational function's range typically excludes its horizontal asymptote value. At HL, examiners expect you to justify these with working, especially near asymptotes and endpoints.

Rational function graph with vertical and horizontal asymptotes labelled with domain and range shaded

Check	What to look for
Denominator	Set equal to zero, exclude that $x$
Square root	Set the inside $\geq 0$
Context	Real-world restriction, e.g. $t \geq 0$
Range	Read from the graph near asymptotes/endpoints

Exam tip

For 'state the asymptotes' questions worth 2 marks, write full equations like $x = 3$ and $y = 2$ — bare numbers can cost you a mark.

Common mistake

Assuming domain is all reals without checking denominators or roots first.

Mini summary

Domain = what's undefined; range = what the graph actually produces.

One-to-One Functions and Finding $f^{-1}$

A function needs to be one-to-one before an inverse exists without restricting the domain. Prove it algebraically: assume $f (x_{1}) = f (x_{2})$ and show this forces $x_{1} = x_{2}$ — examiners specifically check that the algebra steps (like two terms cancelling) are shown, not skipped. To find $f^{- 1}$ : swap $x$ and $y$ , solve for $y$ , and state its domain as the range of $f$ — this last step is the one everyone forgets under time pressure.

A function and its inverse reflected across the line y=x

Exam tip

When a question gives points on $y = f (x)$ and asks where $f (x) = f^{- 1} (x)$ , solve the faster equation $f (x) = x$ instead of solving from scratch.

Common mistake

Isolating a restricted expression (like a square root) correctly but forgetting to state its sign restriction before squaring.

Mini summary

Prove one-to-one with algebra; find $f^{- 1}$ by swapping variables, and always state its domain.

Composite Functions: Order Always Matters

$(f \circ g) (x) = f (g (x))$ means apply $g$ first, then $f$ — swapping the order generally gives a different function entirely. The domain of $f \circ g$ is restricted to $x$ -values in the domain of $g$ for which $g (x)$ also lands inside the domain of $f$ . This domain-chaining step is exactly where HL questions like to catch students out.

Diagram showing input passing through function g then function f

Common mistake

Forgetting to check that outputs of $g$ actually lie in the domain of $f$ before accepting the composite's domain.

Mini summary

Composite order matters: $g$ acts first, then $f$ ; domain must satisfy both functions.

Rational Functions: Asymptotes and the Self-Inverse Shortcut

For $y = \frac{a x + b}{c x + d}$ , the vertical asymptote is $x = - d / c$ and the horizontal asymptote is $y = a / c$ — these control both domain and range. A powerful shortcut: this type of function is self-inverse exactly when $a + d = 0$ . Don't assume self-inverse just because the shape looks familiar — always check the condition before claiming $f = f^{- 1}$ .

Rational function graph with asymptotes and a highlighted self-inverse condition

Common mistake

Assuming a rational function is self-inverse purely from its shape without checking $a + d = 0$ .

Mini summary

Asymptotes: $x = - d / c$ , $y = a / c$ ; self-inverse only when $a + d = 0$ .

Transformations: One Equation Covers Everything

Every translation, stretch and reflection you'll be asked about is one instance of $y = a f (b (x - h)) + k$ . Outside changes ( $a$ , $k$ ) act on $y$ as expected — vertical stretch and shift. Inside changes ( $b$ , $h$ ) act on $x$ and behave opposite to intuition — always factorise first to correctly identify $h$ before describing any horizontal shift or stretch.

Original graph and its transformed image with labelled points showing the effect of a, b, h, k

Parameter	Effect
$k$	Vertical translation
$a$	Vertical stretch factor $∣ a ∣$ ; reflection in $x$ -axis if $a < 0$
$h$	Horizontal translation (from factorised $(x - h)$ )
$b$	Horizontal stretch factor $1/∣ b ∣$ ; reflection in $y$ -axis if $b < 0$

Exam tip

Given key points but no formula for $g$ , transform coordinates directly using $(x, y) \to (x / b + h, a y + k)$ .

Common mistake

Reading $y = f (2 x - 6)$ as 'shift 6 right, then stretch' using the raw number instead of factorising to $f (2 (x - 3))$ first.

Mini summary

Identify $a, b, h, k$ from the factorised form, then apply the coordinate mapping rule.

Quick formula sheet

y = \frac{a x + b}{c x + d} ⟹ vertical asymptote x = - \frac{d}{c}, horizontal asymptote y = \frac{a}{c}

Asymptotes of a rational function directly from its coefficients. — Vertical asymptote: set denominator to zero. Horizontal: ratio of leading coefficients.

f self-inverse ⟺ a + d = 0 for f (x) = \frac{a x + b}{c x + d}

Quick test for whether a rational function equals its own inverse. — 'a plus d, add to zero' means the function mirrors itself.

(f \circ g) (x) = f (g (x))

Composite function: apply

g

first, then

f

. — Read right to left: g acts first, closest to x.

domain (f^{- 1}) = range (f), range (f^{- 1}) = domain (f)

The domain and range swap between a function and its inverse. — Inverse swaps x and y, so domain and range swap too.

y = a f (b (x - h)) + k

General transformation equation covering every translation, stretch and reflection. — Outside (a,k) acts on y as expected; inside (b,h) acts on x, opposite to intuition.

(x, y) \to (\frac{x}{b} + h, a y + k)

Maps a key point on

y = f (x)

to its image on

y = a f (b (x - h)) + k

. — Divide x-part by b then add h; multiply y-part by a then add k.

Practice questions

Easy

State the domain of $f (x) = \frac{1}{x - 5}$ .
Write down the vertical and horizontal asymptotes of $y = \frac{3 x + 2}{x - 1}$ .
If $f (x) = 2 x + 3$ , find $f^{- 1} (x)$ .

Medium

Prove that $f (x) = \frac{2 x - 1}{x + 3}$ is one-to-one for $x \neq = - 3$ .
Given $f (x) = x^{2} + 1$ for $x \geq 0$ and $g (x) = x - 3$ , find $(f \circ g) (x)$ and its domain.
Describe fully the transformations that map $y = f (x)$ to $y = 3 f (2 x - 6) + 1$ .

Challenge

Show that $f (x) = \frac{2 x + 1}{x - 2}$ is self-inverse, and verify by finding $f^{- 1} (x)$ directly.
For $f (x) = x + 4 + 1$ , $x \geq - 4$ , find $f^{- 1} (x)$ stating its domain, then find where $y = f (x)$ meets $y = f^{- 1} (x)$ .
A curve $y = g (x)$ passes through $(2, 4)$ . Find the image of this point under $y = 2 g (3 (x + 1)) - 4$ .

Frequently asked questions

What's the difference between domain and range?+

Domain is the set of valid inputs (found by checking what makes the formula undefined); range is the set of actual outputs, read from the graph rather than assumed.

How do you know if a function has an inverse?+

It must be one-to-one — every output comes from exactly one input. Prove this algebraically by assuming $f (x_{1}) = f (x_{2})$ and showing $x_{1} = x_{2}$ must follow.

When is a rational function self-inverse?+

For $f (x) = \frac{a x + b}{c x + d}$ , it's self-inverse exactly when $a + d = 0$ . Always check this condition rather than assuming it from the function's shape.

Why does the order matter in $y=af(x)+k$?+

Because $a$ multiplies $f (x)$ before $k$ is added — applying $k$ first gives a completely different (and wrong) transformed graph.

How do I find the domain of a composite function?+

Take the domain of $g$ , then remove any $x$ -values where $g (x)$ falls outside the domain of $f$ — both conditions must hold together.

Do I need the formula for $g$ to transform its graph?+

No — if you're given key points, use the coordinate mapping $(x, y) \to (x / b + h, a y + k)$ to transform them directly, without ever writing out $g$ 's equation.

Master Functions with the Full IB Maths AI Revision Notes

Full worked examples for inverses, composites and self-inverse proofs Step-by-step transformation walkthroughs with coordinate mapping Complete formula sheet plus examiner traps and mark-scheme insights Mock exam-style questions to test every skill in this topic

Get the Functions notes on RevisionPrep →

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