Function Properties & Inverses — AHL 2.14 Odd and even functions; periodicity; inverse function with restriction (HL) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

The concepts of symmetry and periodicity in functions did not appear overnight — they grew out of centuries of mathematical exploration. Ancient Greek mathematicians studied the symmetry of geometric shapes, but it was not until the rise of analytic geometry in the 17th century that mathematicians began classifying functions by their algebraic and graphical properties. As calculus matured, the need to reverse operations — to "undo" a function — led to the formal theory of inverse functions. Understanding when a function can be reversed, and when its domain must be restricted to make reversal possible, became essential to fields ranging from engineering to cryptography.

~300 BC • Greek Symmetry

Euclid and other Greek geometers studied reflective symmetry in shapes, laying the groundwork for later ideas about even and odd functions.

1637 • Descartes & Analytic Geometry

René Descartes introduced the coordinate plane, enabling mathematicians to study functions visually and classify their symmetry properties algebraically.

1748 • Euler's Introductio

Leonhard Euler formalized the concept of a function and explored periodicity extensively through trigonometric and exponential functions.

1822 • Fourier's Theorem

Joseph Fourier showed that any periodic function can be decomposed into sums of sines and cosines, making odd/even decomposition a powerful analytical tool.

20th C • Modern Applications

Inverse functions and domain restrictions became central to computer science, signal processing, and the IB Mathematics curriculum as tools for solving real-world problems.

The central question this topic addresses is: How can we classify functions by their symmetry and repetition, and under what conditions can we reverse them? Answering this will give you powerful tools for the IB HL exam and for understanding mathematics at a deeper level.

SECTION 2

Core Principles & Definitions

This topic rests on three interconnected ideas: the symmetry classification of functions as even or odd, the concept of periodicity (repeating behaviour), and the construction of inverse functions — sometimes requiring a domain restriction to make the inverse valid. These ideas connect algebra, graphing, and problem-solving in a way the IB exam frequently tests.

Even Functions

A function f is even if f(−x) = f(x) for all x in its domain. Graphically, this means the curve is symmetric about the y-axis. Classic examples include f(x) = x² and f(x) = cos x.

Odd Functions

A function f is odd if f(−x) = −f(x) for all x in its domain. This gives the graph 180° rotational symmetry about the origin. Examples include f(x) = x³ and f(x) = sin x.

Periodic Functions

A function f is periodic with period p if f(x + p) = f(x) for all x, where p > 0 is the smallest such value. The graph repeats every p units horizontally. Sine and cosine have period 2π.

Inverse Functions

If f maps x to y, the inverse function f⁻¹ maps y back to x. It exists only when f is one-to-one (injective). The graph of f⁻¹ is the reflection of f in the line y = x.

Domain Restriction

When a function is not one-to-one over its entire domain (like x²), we restrict the domain to an interval where it is. This makes the inverse well-defined. For x², restricting to x ≥ 0 gives the inverse √x.

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KEY TAKEAWAY

Think of a function like a one-way street. An inverse function is driving that street in reverse. But if the street has a fork (the function is not one-to-one), you cannot retrace your route uniquely — you have to block off one fork (restrict the domain) so there is only one path backwards. Even functions are like roads with a perfect mirror image on both sides; odd functions look the same if you flip the whole road upside-down and left-to-right.

SECTION 3

Visual Explanation — Symmetry & Periodicity

The diagram below shows three function graphs side by side, illustrating the visual signatures of even, odd, and periodic functions. Recognising these shapes on sight is crucial for the IB exam, because you can often classify a function from its graph before doing any algebra.

Left: the parabola y = x² is symmetric about the y-axis (even). Centre: the cubic y = x³ has rotational symmetry about the origin (odd). Right: the sine curve repeats every 2π (periodic with period 2π). Note that sin x is also odd — a function can be both odd and periodic.

When you look at these three panels, notice the key visual tests. For an even function, fold the graph along the y-axis — the two halves should match perfectly. For an odd function, rotate the entire graph 180° about the origin — it should look the same. For a periodic function, slide the graph horizontally by one period — the shape repeats exactly. These visual checks are fast and reliable, especially under exam conditions.

SECTION 4

Mathematical Framework

Let us formalise the definitions you met in Section 2, then develop the algebra of inverse functions. Each equation below is one you should be able to state, verify, and apply on the IB exam.

EVEN FUNCTION TEST

f(−x) = f(x) for all x ∈ domain(f)

Replace every x with −x. If the output is identical to f(x), the function is even. Example: f(x) = x⁴ − 2x² → f(−x) = (−x)⁴ − 2(−x)² = x⁴ − 2x² = f(x). ✓

ODD FUNCTION TEST

f(−x) = −f(x) for all x ∈ domain(f)

Replace x with −x. If the result equals the negative of f(x), the function is odd. Example: f(x) = x³ − x → f(−x) = −x³ + x = −(x³ − x) = −f(x). ✓

PERIODICITY

f(x + p) = f(x) for all x, where p > 0 is the period

p is the smallest positive number for which this identity holds. For sin(bx), the period is 2π / |b|. For tan(bx), the period is π / |b|.

INVERSE FUNCTION CONDITION

f(f⁻¹(x)) = x and f⁻¹(f(x)) = x

f⁻¹ exists only when f is one-to-one (passes the horizontal line test). The domain of f becomes the range of f⁻¹, and vice versa.

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Finding f⁻¹ Algebraically

Step 1: Write y = f(x). Step 2: Swap x and y. Step 3: Solve for y — this gives y = f⁻¹(x). Step 4: State the domain and range of f⁻¹ (these are the range and domain of f, respectively). If you cannot solve uniquely for y, you need a domain restriction.

An important connection: if f is both odd and periodic with period p, then its graph has a rich symmetry structure. For instance, the function sin x is odd (symmetric about the origin) and periodic with period 2π. These properties together let you reconstruct the entire graph from just the interval [0, π/2]. This kind of efficiency is what makes these classifications so useful.

SECTION 5

Inverse Functions & Domain Restriction

Many common functions are not one-to-one over their natural domain. The parabola y = x², for example, fails the horizontal line test because both x = 2 and x = −2 give y = 4. To create an inverse, we must restrict the domain to a region where the function is strictly increasing (or strictly decreasing). The diagram below shows how this works for y = x² and introduces the standard restriction used for the IB.

Left panel: y = x² on all of ℝ fails the horizontal line test (the red dashed line hits the parabola twice). Right panel: when we restrict to x ≥ 0, the function is one-to-one, and the inverse f⁻¹(x) = √x exists. Notice that f and f⁻¹ are mirror images across the line y = x. Each point (a, b) on f corresponds to (b, a) on f⁻¹.

The same idea applies to trigonometric functions. The function sin x is periodic and therefore not one-to-one on ℝ. The standard restriction is to the interval [−π/2, π/2], where sin x is strictly increasing. This gives the inverse function arcsin x (also written sin⁻¹ x), with domain [−1, 1] and range [−π/2, π/2]. Similarly, cos x is restricted to [0, π] for arccos x, and tan x to (−π/2, π/2) for arctan x.

Function	Standard Restriction	Inverse	Domain of Inverse	Range of Inverse
sin x	[−π/2, π/2]	arcsin x	[−1, 1]	[−π/2, π/2]
cos x	[0, π]	arccos x	[−1, 1]	[0, π]
tan x	(−π/2, π/2)	arctan x	ℝ	(−π/2, π/2)
x²	[0, ∞)	√x	[0, ∞)	[0, ∞)

Standard domain restrictions for common inverse functions

SECTION 6

Worked Example

Let us work through a comprehensive example that tests all three ideas from AHL 2.14. This is the kind of multi-part question the IB loves.

Classifying symmetry, finding an inverse, and stating a restricted domain

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Step 1 — State the function

Consider f(x) = x³ − x, defined for x ∈ ℝ. We are asked to: (a) determine whether f is even, odd, or neither; (b) explain why f does not have an inverse on ℝ; (c) suggest a suitable domain restriction; (d) find the inverse on that restricted domain.
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Step 2 — Test for even/odd

Compute f(−x) = (−x)³ − (−x) = −x³ + x = −(x³ − x) = −f(x). Since f(−x) = −f(x) for all x, the function is odd. Its graph has 180° rotational symmetry about the origin.

f is odd
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Step 3 — Horizontal line test

f(x) = x³ − x = x(x² − 1) = x(x − 1)(x + 1) has roots at x = −1, 0, 1. The function has a local maximum at x = −1/√3 and a local minimum at x = 1/√3 (found by setting f′(x) = 3x² − 1 = 0). Because f increases, decreases, then increases again, a horizontal line can cross the graph up to three times. Therefore f is not one-to-one on ℝ, and no inverse exists on the full domain.

f fails the horizontal line test on ℝ
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Step 4 — Choose a domain restriction

We need an interval where f is strictly monotonic. Since f′(x) = 3x² − 1 > 0 when x > 1/√3 ≈ 0.577, the function is strictly increasing on [1/√3, ∞). A simpler choice that the IB often accepts is x ≥ 1, where f is clearly increasing. On x ≥ 1 the range is f(1) = 0 onwards, so range = [0, ∞).

Restrict to x ≥ 1, giving range [0, ∞)
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Step 5 — Find the inverse (on a simpler restriction for illustration)

For a cleaner algebra example, consider the restriction x ≥ 1 with a slight simplification. In practice, the inverse of x³ − x cannot be expressed in a neat closed form, so the IB might ask you to set up the equation and describe the process rather than solve it fully. You would write y = x³ − x, swap to get x = y³ − y, and note that x = f⁻¹(y) is defined implicitly. For the IB, stating the correct domain and range of f⁻¹ — domain [0, ∞), range [1, ∞) — and showing the reflection across y = x is typically sufficient.

f⁻¹: [0, ∞) → [1, ∞), defined implicitly by y³ − y = x

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Exam Tip

The IB often asks you to sketch f and f⁻¹ on the same axes. Always include the line y = x and label at least one pair of reflected points (a, b) and (b, a). This earns easy marks.

SECTION 7

Comparing Even, Odd, and Neither Functions

Most functions are neither even nor odd — the two categories are special cases. The table below helps you quickly classify a function and understand the implications for its graph, inverse, and integration.

Property	Even	Odd	Neither
Algebraic test	f(−x) = f(x)	f(−x) = −f(x)	Neither identity holds
Graphical symmetry	Reflection in y-axis	180° rotation about origin	No special symmetry
f(0) requirement	No restriction (f(0) can be any value)	f(0) = 0 (if 0 is in the domain)	No restriction
Examples	x², \|x\|, cos x, x⁴ + 1	x³, sin x, tan x, x	eˣ, x² + x, ln x
Definite integral ∫ from −a to a	= 2 × ∫ from 0 to a	= 0	Must compute fully

Quick classification guide for even, odd, and neither functions

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KEY TAKEAWAY

Here's a useful fact that links these topics: every function can be written as the sum of an even part and an odd part. Specifically, f(x) = ½[f(x) + f(−x)] + ½[f(x) − f(−x)]. The first bracket is always even, the second is always odd. This decomposition is used extensively in Fourier analysis and signal processing.

SECTION 8

Connection to Advanced Theory

The ideas you've learned in AHL 2.14 are not isolated — they connect forward to several advanced topics in the IB HL syllabus and beyond. Understanding these connections will deepen your mastery and help you see why the IB emphasises this material.

AHL 2.14 Concept	Advanced Connection	Where You'll See It
Even/odd classification	Fourier series — decomposing functions into sine (odd) and cosine (even) components	University maths, engineering, physics
Periodicity	Trigonometric identities, modelling oscillations (SHM), wave functions	AHL 3.9–3.12, Physics SL/HL
Inverse functions	Logarithms as inverses of exponentials, inverse trig in integration	AHL 2.9, AHL 5.16
Domain restriction	Defining branch cuts in complex analysis, principal values	University complex analysis

How AHL 2.14 connects to later mathematics

In particular, when you study calculus, the fact that the integral of an odd function over a symmetric interval [−a, a] equals zero will save you significant time. And when you encounter differential equations, the periodicity of solutions will be directly tied to the period of the forcing function. These are not abstract curiosities — they are computational shortcuts and conceptual insights that working mathematicians use every day.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A function f satisfies f(0) = 3. Can f be an odd function? Explain your reasoning.

Show answer

No, f cannot be odd. By the definition of an odd function, f(−x) = −f(x) for all x in the domain. Setting x = 0 gives f(0) = −f(0), which means 2f(0) = 0, so f(0) = 0. Since f(0) = 3 ≠ 0, the function cannot be odd. This is a quick check you can always apply: if a function passes through the origin's y-axis at a non-zero value, it immediately fails the odd function test.

Question 2: BASIC CALCULATION

Determine whether g(x) = 3x⁴ − 2x² + 7 is even, odd, or neither. Show your working.

Show answer

Compute g(−x) = 3(−x)⁴ − 2(−x)² + 7 = 3x⁴ − 2x² + 7 = g(x). Since g(−x) = g(x) for all x, the function g is even. Notice that every term has an even power of x (x⁴, x², and the constant 7 = 7x⁰), which is a reliable pattern: a polynomial with only even-powered terms is always an even function.

Question 3: INTERMEDIATE

The function h(x) = 2sin(3x) + 1 is defined for x ∈ ℝ. (a) Find the period of h. (b) Determine whether h is even, odd, or neither. (c) Explain why h does not have an inverse on ℝ.

Show answer

(a) The period of sin(3x) is 2π/3, and adding 1 does not change the period, so the period of h is 2π/3. (b) Compute h(−x) = 2sin(−3x) + 1 = −2sin(3x) + 1. Compare: h(x) = 2sin(3x) + 1. We see h(−x) ≠ h(x) (not even) and h(−x) ≠ −h(x) = −2sin(3x) − 1 (not odd). Therefore h is neither even nor odd. The vertical shift of +1 breaks the symmetry. (c) Since h is periodic, it repeats its values every 2π/3 units, so a horizontal line at any y-value in its range [−1, 3] will intersect the graph infinitely many times. This violates the horizontal line test, so h is not one-to-one on ℝ and has no inverse.

Question 4: APPLIED

A temperature sensor records T(t) = 20 − 8cos(πt/12), where T is in °C and t is hours after midnight. (a) State the period and explain what it means physically. (b) Find the inverse function T⁻¹ on a suitable restricted domain, and interpret T⁻¹(16) in context.

Show answer

(a) The period is 2π ÷ (π/12) = 24 hours, meaning the temperature cycle repeats every 24 hours — one full day, as expected. (b) The function T(t) = 20 − 8cos(πt/12) has range [12, 28]. The cosine part cos(πt/12) completes one period over 0 ≤ t < 24. On [0, 12], cos(πt/12) decreases from 1 to −1, so T increases from 12 to 28 — making T one-to-one on this interval. To invert: set y = 20 − 8cos(πt/12), so cos(πt/12) = (20 − y)/8, giving πt/12 = arccos((20 − y)/8), hence t = (12/π)arccos((20 − y)/8). Therefore T⁻¹(y) = (12/π)arccos((20 − y)/8). Evaluating: T⁻¹(16) = (12/π)arccos((20 − 16)/8) = (12/π)arccos(0.5) = (12/π)(π/3) = 4 hours. This means the temperature first reaches 16°C at 4:00 AM.

Question 5: CRITICAL THINKING

Prove that if f is both even and odd (simultaneously), then f(x) = 0 for all x in its domain. Then give an example showing that the sum of an even function and an odd function is generally neither even nor odd.

Show answer

Proof: If f is even, then f(−x) = f(x) for all x. If f is also odd, then f(−x) = −f(x) for all x. Setting these equal: f(x) = −f(x), which gives 2f(x) = 0, so f(x) = 0 for all x. The only function that is both even and odd is the zero function. For the second part: let e(x) = x² (even) and o(x) = x (odd). Then s(x) = x² + x. Test: s(−x) = x² − x. Compare: s(x) = x² + x ≠ x² − x (not even) and −s(x) = −x² − x ≠ x² − x (not odd). So x² + x is neither even nor odd, confirming that adding an even and an odd function typically destroys both symmetries.

SUMMARY

Lesson Summary

Summary

In this lesson you learned three powerful ways to classify functions. An even function satisfies f(−x) = f(x) and is symmetric about the y-axis, while an odd function satisfies f(−x) = −f(x) and has 180° rotational symmetry about the origin. A periodic function repeats its values every p units, where p is the period. The algebraic tests for these properties are straightforward: substitute −x and compare with f(x) for symmetry, or check f(x + p) = f(x) for periodicity.

An inverse function f⁻¹ reverses the mapping of f, and it exists only when f is one-to-one (passes the horizontal line test). When a function is not one-to-one — as with x², sin x, or cos x — you must apply a domain restriction to an interval where it is monotonic before the inverse can be defined. The graph of f⁻¹ is always the reflection of f in the line y = x, and the domain and range swap between f and f⁻¹. Mastering these ideas prepares you for IB exam questions on function analysis, trigonometric inverses, and modelling with periodic functions.