IB DP Maths AA · Topic 2 Functions · SL 2.1–2.5 · AHL 2.14 · ~6 min read

Functions, inverses and composites

Straight lines and their equations, what a function is, domain and range, key features of graphs, composite functions, inverse functions and their graphs, and (HL) odd and even functions, self-inverse functions and restricting domains.

Straight linesDomain & rangeKey features of graphsComposite functionsInverse functionsOdd & even functions (HL)Self-inverse (HL)

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No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

A function maps every input in its domain to exactly one output; the set of outputs is the range.
A composite (f ∘ g)(x) = f(g(x)) applies g first, then f. Order matters.
An inverse f⁻¹ undoes f. It exists only if f is one-to-one; its graph is the reflection of f in y = x, and domain and range swap.
Lines: gradient m = Δy/Δx; parallel lines have equal gradients, perpendicular lines have m₁m₂ = −1.
HL: even functions (f(−x) = f(x)), odd functions (f(−x) = −f(x)), self-inverse functions (f = f⁻¹) and restricting domains so an inverse exists.

By the end you can

Find gradients and equations of lines, including parallel and perpendicular lines.
Find the domain and range of a function and identify key features of its graph.
Form and evaluate composite functions.
Find inverse functions algebraically and sketch them by reflection in y = x.
Identify odd, even and self-inverse functions and restrict a domain to make an inverse exist. HL

1Straight lines

m = y₂ − y₁x₂ − x₁ FORMULA BOOKLET

gradient between two points

y = mx + c, ax + by + d = 0, y − y₁ = m(x − x₁) FORMULA BOOKLET

three forms of a line

Worked example

Find the line through (2, 3) perpendicular to y = 2x + 1, in the form ax + by + d = 0.

perpendicular gradient = −½ A1
y − 3 = −½(x − 2) M1
x + 2y − 8 = 0 A1

2Functions, domain and range

One output per input: a function. Two inputs to one output is fine; one input to two outputs is not.

Domain problems: you cannot divide by zero, take the square root of a negative, or take the log of zero or a negative.

f(x)	Domain	Range
√(x − 2)	x ≥ 2	f(x) ≥ 0
1x − 3	x ≠ 3	f(x) ≠ 0
ln(x + 1)	x > −1	f(x) ∈ ℝ
x² + 4	x ∈ ℝ	f(x) ≥ 4

Key features (SL 2.3–2.4): intercepts, maximum and minimum points, symmetry, vertical and horizontal asymptotes. Sketch them, label them with coordinates or equations.

GDC (calculator) methodPaper 2: graph, then use zero, maximum, minimum and intersect. Set the window to show every key feature, and copy the coordinates to 3 s.f.

3Composite functions

(f ∘ g)(x) = f(g(x))

do g first, then f

Worked example

f(x) = 2x + 1 and g(x) = x². Find (f ∘ g)(x), (g ∘ f)(x) and (f ∘ g)(3).

(f ∘ g)(x) = f(x²) = 2x² + 1 A1
(g ∘ f)(x) = (2x + 1)² A1
(f ∘ g)(3) = 2(9) + 1 = 19 A1

Common mistakeThinking f ∘ g means "f first". The function written closest to x acts first: in f(g(x)), g acts on x.

4Inverse functions

Method: write y = f(x), swap x and y (or rearrange for x), make y the subject, write f⁻¹(x) = …

Worked example

Find f⁻¹(x) for f(x) = 2x + 3x − 1, x ≠ 1.

y(x − 1) = 2x + 3 ⇒ xy − 2x = y + 3 M1
x = y + 3y − 2 A1
f⁻¹(x) = x + 3x − 2, x ≠ 2 A1

Reflect every point (a, b) to (b, a): that is reflection in y = x.

f(f⁻¹(x)) = f⁻¹(f(x)) = x. The domain of f⁻¹ is the range of f, and its range is the domain of f.

Exam tip"Solve f(x) = f⁻¹(x)" is quickest by solving f(x) = x, when the graphs meet on the line y = x (true for increasing functions).

5Odd, even and self-inverse functions HL

Even functions have mirror symmetry in the y-axis; odd functions have half-turn symmetry about the origin.

Worked example

Is f(x) = x³ − 4x odd, even or neither?

f(−x) = −x³ + 4x = −(x³ − 4x) = −f(x) M1
odd A1

Self-inverse: f(f(x)) = x, so the graph is symmetric in y = x. Examples: f(x) = 1/x, f(x) = −x + c, f(x) = x + 3x − 1.
Restricting the domain: f(x) = x² has no inverse on ℝ, but on x ≥ 0 it is one-to-one and f⁻¹(x) = √x.

Worked example

f(x) = (x − 2)² + 1. Find the largest domain of the form x ≥ a for which f⁻¹ exists, and find f⁻¹.

the vertex is at x = 2, so x ≥ 2 A1
y − 1 = (x − 2)² ⇒ x = 2 + √(y − 1) (positive root because x ≥ 2) M1
f⁻¹(x) = 2 + √(x − 1), x ≥ 1 A1

Linking questionInverse functions explain why ln undoes e^x (Exponents and logarithms) and why arcsin needs a restricted range (Trig identities). Composite functions are where the chain rule comes from (Differentiation).

Idea for your IA / explorationModel a real process as a composite (currency conversion then tax, temperature conversions) and explore when the inverse makes sense in context.

Test yourself — cover the answers first

Quick check

Find the gradient of the line through (1, 5) and (4, −1).
Give the domain of f(x) = 1/(x + 4).
If f(x) = x + 3 and g(x) = 2x, find g(f(1)).
Find f⁻¹(x) for f(x) = 5x − 2.
Which line is f⁻¹ a reflection in?
HL Is cos x odd or even?

Exam-style

f(x) = 3x − 1 and g(x) = x + 13. Show that g = f⁻¹ and find (f ∘ g)(x).
HL f(x) = e^2x − 1. Find f⁻¹(x) and state its domain.

Show answers

−2
x ≠ −4
8
(x + 2)/5
y = x
even
f(g(x)) = 3(x + 13) − 1 = x
and g(f(x)) = x, so g is the inverse
(f ∘ g)(x) = x
y + 1 = e^2x ⇒ x = ½ ln(y + 1)
f⁻¹(x) = ½ ln(x + 1)
domain x > −1 (the range of f)

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IB Maths AA practice papers

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