Functions

How to Use This Guide

Functions is Topic 2 of the IB Math AA HL syllabus and is one of the highest-yield topics for Paper 1 (non-calculator). The concepts here underpin all of calculus, statistics, and algebra — you cannot afford gaps. This guide covers subtopics 2.1–2.5 in syllabus order plus the AHL extension: language of functions, composite and inverse functions, transformations, rational functions, the modulus function, and polynomial division with the factor and remainder theorems. Every section includes worked examples drawn from past IB papers, exam alert boxes flagging the Paper 1 traps that cost marks most often, and MCQ practice at the end of each section. Use the jump links below to navigate directly to the section you need.

Jump links: Language of Functions (2.1) · Composite & Inverse (2.2) · Transformations (2.3) · Rational Functions (2.4) · Modulus & Inequalities (2.5) · Polynomial Division & Theorems (2.6 AHL) · MCQ Practice

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Section 1: Language of Functions (2.1)

A function is a rule that assigns to each element of a set (the domain) exactly one element of another set (the codomain). The set of all output values actually produced is called the range (or image). Note: the range is a subset of the codomain, but they are not necessarily equal.

$f: X \to Y$

where $X$ is the domain, $Y$ is the codomain, and the range is $\{f(x) : x \in X\} \subseteq Y$.

1.1 Key Vocabulary

Term	Definition	Notation
Domain	Set of permitted input values	$\text{dom}(f)$ or $X$
Codomain	Set within which outputs must fall	$Y$
Range / Image	Set of actual output values produced	$\text{range}(f)$
Image of a point	The specific output for a given input	$f(a)$
Pre-image	Input(s) that map to a given output	$f^{-1}(\{b\})$

Arrow diagram notation: In an arrow diagram, each arrow begins at a domain element and ends at the corresponding codomain element. A mapping is a function if and only if every domain element has exactly one outgoing arrow.

1.2 Types of Function

One-to-one (injective): Every element of the range is the image of exactly one domain element. No two different inputs produce the same output. Equivalently: $f(a) = f(b) \Rightarrow a = b$.

Many-to-one: At least one output value is the image of more than one input. Example: $f(x) = x^2$ maps both $x = 2$ and $x = -2$ to $4$.

Onto (surjective): The range equals the codomain — every element of the codomain is the image of at least one domain element.

Bijective: Both injective and surjective. Only bijective functions have a well-defined inverse over their full domain.

1.3 Finding the Domain and Range

Natural domain: the largest subset of $\mathbb{R}$ for which $f(x)$ is defined. Restrictions arise from:

• Square roots: the expression under the radical must be $\geq 0$.
• Denominators: the denominator must be $\neq 0$.
• Logarithms: the argument must be $> 0$.

Quick Recall — Section 1

Try to answer without scrolling up:

1. What is the difference between range and codomain?
2. State the horizontal line test.
3. What restrictions on the domain arise from a square root?

Reveal answers

1. The codomain is the declared target set; the range is the set of outputs actually produced. Range $\subseteq$ codomain, but they need not be equal.
2. A function is one-to-one if and only if every horizontal line intersects its graph at most once.
3. The expression under the radical must be $\geq 0$.

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Section 2: Composite and Inverse Functions (2.2)

2.1 Composite Functions

The composite function $f \circ g$ (read “f composed with g”) is defined by:

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

Think of $g$ as the inner function (applied first) and $f$ as the outer function (applied second). Order matters: $f \circ g \neq g \circ f$ in general.

Domain of $f \circ g$: The domain is the set of all $x$ in $\text{dom}(g)$ such that $g(x) \in \text{dom}(f)$.

$\text{dom}(f \circ g) = \{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}$

2.2 Inverse Functions

If $f$ is one-to-one (injective), then the inverse function $f^{-1}$ is defined so that:

$f^{-1}(f(x)) = x \quad \text{for all } x \in \text{dom}(f)$ $f(f^{-1}(x)) = x \quad \text{for all } x \in \text{dom}(f^{-1})$

Geometrically, the graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$.

Key relationships:

• $\text{dom}(f^{-1}) = \text{range}(f)$
• $\text{range}(f^{-1}) = \text{dom}(f)$

Algorithm for finding $f^{-1}(x)$:

1. Write $y = f(x)$.
2. Rearrange to express $x$ in terms of $y$.
3. Swap $x$ and $y$ (replace every $y$ with $x$ and every $x$ with $y$).
4. The result is $y = f^{-1}(x)$.
5. State the domain of $f^{-1}$ (which equals the range of $f$).

2.3 Restricting the Domain for Invertibility

A function that is not one-to-one does not have an inverse over its full domain. You can, however, restrict the domain to a maximal interval on which the function is one-to-one, then find the inverse on that restricted domain.

Quick Recall — Section 2

Try to answer without scrolling up:

1. What is $\text{dom}(f \circ g)$ in terms of the individual domains?
2. State the two properties $f^{-1}$ must satisfy.
3. What does “restrict the domain” mean for a non-injective function, and why is it necessary?

Reveal answers

1. $\text{dom}(f \circ g) = \{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}$.
2. $f^{-1}(f(x)) = x$ for all $x \in \text{dom}(f)$, and $f(f^{-1}(x)) = x$ for all $x \in \text{dom}(f^{-1})$.
3. Choosing a sub-interval of the domain on which the function is one-to-one, so that an inverse exists on that interval.

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Section 3: Transformations of Functions (2.3)

A transformation maps a parent function $y = f(x)$ to a new function by shifting, reflecting, or stretching its graph. The IB AA HL syllabus requires you to identify and apply six transformation types fluently.

3.1 Transformation Summary Table

Transformation	Function form	Effect on graph
Horizontal translation right by $a$	$f(x - a)$	Every point moves $a$ units right
Horizontal translation left by $a$	$f(x + a)$	Every point moves $a$ units left
Vertical translation up by $b$	$f(x) + b$	Every point moves $b$ units up
Vertical translation down by $b$	$f(x) - b$	Every point moves $b$ units down
Reflection in the $y$-axis	$f(-x)$	$x$-coordinates change sign
Reflection in the $x$-axis	$-f(x)$	$y$-coordinates change sign
Vertical stretch by factor $a$	$a\,f(x)$	$y$-coordinates multiplied by $a$
Horizontal stretch by factor $\frac{1}{b}$	$f(bx)$	$x$-coordinates divided by $b$

3.2 Combining Transformations

When multiple transformations are combined, apply them in the correct order:

1. Horizontal stretch/compression (changes to $x$ inside brackets)
2. Horizontal translation
3. Vertical stretch/compression
4. Vertical translation

For $y = a\,f(b(x - h)) + k$:

• $|b| > 1$: horizontal compression (narrowing)
• $0 < |b| < 1$: horizontal stretch (widening)
• $|a| > 1$: vertical stretch (tallening)
• $0 < |a| < 1$: vertical compression (flattening)
• $h$: horizontal shift right by $h$ (left if $h < 0$)
• $k$: vertical shift up by $k$ (down if $k < 0$)

3.3 Transformation Diagram

The diagram below shows how the transformations $y = x^2 \to y = (x-2)^2 \to y = (x-2)^2 + 1$ shift the parabola.

3.4 Invariant Points

A point is invariant under a transformation if it maps to itself. For reflections in the $x$-axis, invariant points satisfy $f(x) = 0$ (points on the $x$-axis). For reflections in the line $y = x$ (which corresponds to finding $f^{-1}$), invariant points satisfy $f(x) = x$.

Quick Recall — Section 3

Try to answer without scrolling up:

1. Does $f(x - 3)$ shift the graph left or right? By how much?
2. Write the combined transformation $y = a\,f(b(x-h)) + k$ — which parameter controls horizontal stretch?
3. How do you find invariant points under the reflection in $y = x$?

Reveal answers

1. Right, by 3 units. Changes inside the brackets act in the opposite direction to their sign.
2. Parameter $b$: the graph is stretched horizontally by factor $\frac{1}{b}$ (compressed if $|b| > 1$, stretched if $0 < |b| < 1$).
3. Solve $f(x) = x$.

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Section 4: Rational Functions (2.4)

A rational function has the form $f(x) = \dfrac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials and $q(x) \not\equiv 0$. The key features are asymptotes and holes.

4.1 Vertical Asymptotes

A vertical asymptote at $x = a$ occurs where the denominator is zero and the numerator is non-zero. The function grows without bound as $x \to a$.

Finding vertical asymptotes:

1. Factor the denominator fully.
2. Find zeros of the denominator.
3. Check whether the numerator is also zero at those points.
   • If numerator $\neq 0$: vertical asymptote at $x = a$.
   • If numerator $= 0$: there is a hole (removable discontinuity) at $x = a$, not an asymptote.

4.2 Horizontal Asymptotes

A horizontal asymptote at $y = L$ occurs when $f(x) \to L$ as $x \to \pm\infty$.

Let $\deg p = m$ and $\deg q = n$:

$y = \begin{cases} 0 & m < n \\[4pt] \dfrac{\text{leading coeff of } p}{\text{leading coeff of } q} & m = n \\[8pt] \text{no horizontal asymptote} & m > n \end{cases}$

When $m > n$, divide $p$ by $q$ using polynomial long division to find an oblique (slant) asymptote of the form $y = ax + b$.

4.3 Oblique Asymptotes HL

When $\deg p = \deg q + 1$, divide to obtain:

$f(x) = ax + b + \frac{r(x)}{q(x)}$

where $r$ is the remainder. As $x \to \pm\infty$, the remainder term $\to 0$, so $y = ax + b$ is the oblique asymptote.

4.4 Sketching Rational Functions

Systematic approach for $f(x) = \dfrac{p(x)}{q(x)}$:

1. Find all vertical asymptotes (zeros of $q$ where $p \neq 0$).
2. Find all holes (zeros of $q$ where $p = 0$ too — cancel the common factor first).
3. Find the horizontal or oblique asymptote using degree comparison / long division.
4. Find $x$-intercepts: set $p(x) = 0$ (after cancellation).
5. Find $y$-intercept: compute $f(0)$.
6. Determine sign of $f$ in each interval between vertical asymptotes using a sign table.
7. Sketch, ensuring the curve approaches each asymptote from the correct side.

4.5 Rational Function Diagram

Quick Recall — Section 4

Try to answer without scrolling up:

1. How do you find vertical asymptotes of a rational function?
2. State the horizontal asymptote rule when $\deg p = \deg q$.
3. When does a rational function have an oblique asymptote, and how do you find it?

Reveal answers

1. Set the denominator equal to zero. Check the numerator at each root — if the numerator is non-zero there, it is a vertical asymptote; if the numerator is also zero, it is a removable hole.
2. $y = \dfrac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$.
3. When the degree of the numerator is exactly one more than the degree of the denominator. Find it by polynomial long division; the quotient (excluding the remainder term) is the oblique asymptote.

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Section 5: Modulus Function and Inequalities (2.5)

The modulus function (absolute value) is defined as:

$|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}$

Geometrically, $|x|$ is the distance of $x$ from the origin on the number line.

5.1 Solving Equations Involving Modulus

Basic form $|f(x)| = k$ (where $k > 0$):

$|f(x)| = k \implies f(x) = k \quad \text{or} \quad f(x) = -k$

Always check both solutions by substituting back into the original equation, especially when the original problem places restrictions on the domain.

5.2 Modulus Inequalities

Form $|f(x)| < k$ (strict; $k > 0$):

$|f(x)| < k \iff -k < f(x) < k$

Form $|f(x)| > k$ (strict; $k > 0$):

$|f(x)| > k \iff f(x) > k \quad \text{or} \quad f(x) < -k$

5.3 Graphs of $|f(x)|$ and $f(|x|)$

These two operations produce different graphs and are a common source of confusion.

$y = |f(x)|$: Take the graph of $y = f(x)$, then reflect all parts below the $x$-axis upward. Parts above the axis remain unchanged.

$y = f(|x|)$: Take the right half of $y = f(x)$ (for $x \geq 0$) and reflect it in the $y$-axis to produce the left half. The original left half of the graph is discarded.

Quick Recall — Section 5

Try to answer without scrolling up:

1. Write the solution set for $|f(x)| < k$.
2. Write the solution set for $|f(x)| > k$.
3. Describe the graph of $y = f(|x|)$ in terms of the graph of $y = f(x)$.

Reveal answers

1. $-k < f(x) < k$, a single bounded interval.
2. $f(x) > k$ or $f(x) < -k$, two unbounded intervals.
3. Take the right half of the graph of $y = f(x)$ ($x \geq 0$) and reflect it in the $y$-axis. The left half of the original graph is replaced.

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Section 6: Polynomial Division, Factor & Remainder Theorems HL

This section covers AHL content for Topic 2. The factor and remainder theorems give you a systematic toolkit for factorising and evaluating polynomials — skills that appear in Paper 1 (factor theorem, quick remainder evaluation) and Paper 2 (full long division to find quotients).

6.1 Polynomial Long Division

When you divide a polynomial $p(x)$ by a divisor $d(x)$, you obtain a quotient $q(x)$ and a remainder $r(x)$, where the degree of $r(x)$ is strictly less than the degree of $d(x)$:

$p(x) = d(x) \cdot q(x) + r(x)$

For a linear divisor $(x - a)$, the remainder $r$ is a constant.

Method — long division by $(x - a)$:

1. Write the dividend in descending powers, inserting $0$ coefficients for any missing terms.
2. Divide the leading term of the current dividend by the leading term of the divisor. Write the result above the line as the next term of the quotient.
3. Multiply the divisor by that term and subtract.
4. Bring down the next term and repeat until the degree of the remainder is less than the degree of the divisor.

6.2 The Factor Theorem

The factor theorem gives you a fast way to test whether a linear expression divides evenly into a polynomial, without performing long division. Once one factor is confirmed, long division (or synthetic division) yields the remaining quotient.

Rational Root Theorem (useful for finding candidates): If $p(x) = a_n x^n + \cdots + a_0$ has integer coefficients, any rational root has the form $\pm \dfrac{\text{factor of } a_0}{\text{factor of } a_n}$.

6.3 The Remainder Theorem

The remainder theorem lets you find the remainder of a polynomial division without performing the full algorithm — just evaluate the polynomial at the appropriate value.

6.4 Combining the Theorems — Solving Cubic Equations

The standard strategy for solving a cubic $p(x) = 0$:

1. Find one root by testing $\pm 1, \pm 2, \ldots$ (rational root candidates) using the factor theorem.
2. Extract the linear factor $(x - a)$, then divide $p(x)$ by $(x - a)$ to obtain a quadratic quotient.
3. Solve the quadratic by factorisation, completing the square, or the quadratic formula.

If $p(a) = 0$, then $(x - a)$ divides $p(x)$ exactly, and the quotient has degree one lower than $p(x)$.

Quick Recall — Section 6

Try to answer without scrolling up:

1. State the Factor Theorem.
2. State the Remainder Theorem.
3. When dividing $p(x)$ by $(2x - 3)$, what value do you substitute to find the remainder using the Remainder Theorem?

Reveal answers

1. $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$.
2. When $p(x)$ is divided by $(x - a)$, the remainder equals $p(a)$.
3. Set $2x - 3 = 0$, so $x = \dfrac{3}{2}$. The remainder is $p\!\left(\dfrac{3}{2}\right)$.

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MCQ Practice

These questions are styled after IB AA HL Paper 1. Work each one before revealing the answer.

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Section 1 MCQs — Language of Functions

Q1. The function $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = x^2 - 4$. Which of the following statements is correct?

A. $f$ is one-to-one and its range equals its codomain.

B. $f$ is many-to-one and its range equals its codomain.

C. $f$ is one-to-one and its range is $[-4, \infty)$.

D. $f$ is many-to-one and its range is $[-4, \infty)$.

Reveal answer and explanation

Answer: D

$f(x) = x^2 - 4$ is many-to-one because, for example, $f(2) = f(-2) = 0$. The minimum value of $x^2$ is $0$ (at $x = 0$), so the minimum of $f$ is $-4$. The range is $[-4, \infty)$, which is a proper subset of the codomain $\mathbb{R}$. Options A and C are wrong because $f$ is not one-to-one. Option B is wrong because the range does not equal $\mathbb{R}$.

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Q2. The natural domain of $g(x) = \ln(2x - 6)$ is:

A. $x > 0$

B. $x > 3$

C. $x \geq 3$

D. $x \in \mathbb{R}$

Reveal answer and explanation

Answer: B

For the logarithm to be defined, the argument must be strictly positive: $2x - 6 > 0 \Rightarrow x > 3$. The domain is the open interval $(3, \infty)$, which matches option B. Note the strict inequality — $x = 3$ gives $\ln(0)$, which is undefined, so C is wrong.

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Section 2 MCQs — Composite and Inverse Functions

Q3. Let $f(x) = 2x + 1$ and $g(x) = x^2 - 3$. Find $(g \circ f)(x)$.

A. $2x^2 - 5$

B. $4x^2 + 4x - 2$

C. $4x^2 + 4x + 1 - 3$, simplified to $4x^2 + 4x - 2$

D. $2x^2 - 5$ (same as A)

Reveal answer and explanation

Answer: B (equivalently C)

$(g \circ f)(x) = g(f(x)) = g(2x+1) = (2x+1)^2 - 3 = 4x^2 + 4x + 1 - 3 = 4x^2 + 4x - 2$.

Note that A is wrong — a common error is to compute $f(g(x)) = 2(x^2-3)+1 = 2x^2-5$ instead of $g(f(x))$. Always apply the inner function first.

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Q4. The function $f(x) = \dfrac{3x - 2}{x + 1}$, $x \neq -1$. Find $f^{-1}(2)$.

A. $x = 4$

B. $x = -4$

C. $x = 4$ (confirmed)

D. $x = \dfrac{1}{2}$

Reveal answer and explanation

Answer: A

Method 1 (direct): $f^{-1}(2)$ asks “which $x$ satisfies $f(x) = 2$?” Set $\dfrac{3x-2}{x+1} = 2$, so $3x - 2 = 2x + 2$, giving $x = 4$.

Method 2 (full inverse): Let $y = \dfrac{3x-2}{x+1}$. Then $y(x+1) = 3x-2$, so $yx + y = 3x - 2$, giving $x(y-3) = -2 - y$, so $x = \dfrac{-2-y}{y-3} = \dfrac{2+y}{3-y}$. Thus $f^{-1}(x) = \dfrac{x+2}{3-x}$ and $f^{-1}(2) = \dfrac{4}{1} = 4$.

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Section 3 MCQs — Transformations

Q5. The graph of $y = f(x)$ is translated by $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$. The equation of the resulting graph is:

A. $y = f(x - 3) + 2$

B. $y = f(x + 3) - 2$

C. $y = f(x + 3) + 2$

D. $y = f(x - 3) - 2$

Reveal answer and explanation

Answer: C

A translation of $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$ means 3 units left and 2 units up. Moving left by 3 means replacing $x$ with $x + 3$ inside the function. Moving up by 2 means adding 2 outside. Result: $y = f(x + 3) + 2$. The trap here is option A: $x - 3$ inside the brackets means a rightward shift, not leftward.

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Q6. The point $(4, -6)$ lies on the graph of $y = f(x)$. What point lies on the graph of $y = -f(2x)$?

A. $(2, 6)$

B. $(-4, 6)$

C. $(2, -6)$

D. $(8, 6)$

Reveal answer and explanation

Answer: A

If $(4, -6)$ is on $y = f(x)$, then $f(4) = -6$.

Under $y = -f(2x)$: the horizontal compression by factor $\frac{1}{2}$ maps $x = 4$ to $x = 2$; the reflection $-f$ negates the $y$-value, so $-f(2 \cdot 2) = -f(4) = -(-6) = 6$.

The image point is $(2, 6)$.

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Section 4 MCQs — Rational Functions

Q7. The function $h(x) = \dfrac{x^2 - 1}{x^2 - x - 2}$. Which of the following correctly describes its behaviour?

A. One vertical asymptote at $x = 2$ and a hole at $x = -1$.

B. Vertical asymptotes at $x = 2$ and $x = -1$.

C. One vertical asymptote at $x = -1$ and a hole at $x = 2$.

D. No vertical asymptotes; a hole at $x = 1$.

Reveal answer and explanation

Answer: A

Factor: numerator $= (x-1)(x+1)$; denominator $= (x-2)(x+1)$.

Cancel the common factor $(x+1)$: the simplified form is $\dfrac{x-1}{x-2}$, valid for $x \neq -1$.

At $x = -1$: both numerator and denominator were zero — hole (removable discontinuity) at $x = -1$.

At $x = 2$: denominator $= 0$, numerator $\neq 0$ — vertical asymptote at $x = 2$.

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Q8. As $x \to \infty$, the function $f(x) = \dfrac{3x^2 - 2x + 1}{6x^2 + x - 5}$ approaches:

A. $3$

B. $\dfrac{1}{2}$

C. $\infty$

D. $0$

Reveal answer and explanation

Answer: B

Both numerator and denominator have degree $2$. The horizontal asymptote is the ratio of leading coefficients: $\dfrac{3}{6} = \dfrac{1}{2}$.

Option A is wrong (confusing the leading coefficient ratio with just the numerator coefficient). Option C is wrong because the degrees are equal. Option D would only apply if the numerator’s degree were smaller.

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Section 5 MCQs — Modulus and Inequalities

Q9. Solve $|5 - 2x| < 3$.

A. $x < 1$ or $x > 4$

B. $1 < x < 4$

C. $x < -4$ or $x > 1$

D. $-4 < x < 1$

Reveal answer and explanation

Answer: B

$|5 - 2x| < 3$ gives $-3 < 5 - 2x < 3$.

Subtract 5: $-8 < -2x < -2$.

Divide by $-2$ and reverse both inequality signs: $4 > x > 1$, i.e. $1 < x < 4$.

The trap is forgetting to reverse the inequalities when dividing by a negative number.

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Q10. Which of the following is the graph of $y = |x^2 - 4|$?

A. The parabola $y = x^2 - 4$ with the portion between $x = -2$ and $x = 2$ reflected above the $x$-axis.

B. The parabola $y = x^2 - 4$ with the portions outside $[-2, 2]$ reflected above the $x$-axis.

C. A parabola with vertex at $(0, 4)$ opening downward.

D. The graph of $y = x^2 - 4$ unchanged.

Reveal answer and explanation

Answer: A

$y = x^2 - 4$ is negative between its roots $x = -2$ and $x = 2$ (since the parabola dips below the $x$-axis there). Applying $|\cdot|$ reflects that negative portion upward. Outside $[-2, 2]$, the parabola is already non-negative, so it is unchanged. The result is a “W-shape” with the dip between $-2$ and $2$ folded upward, touching the $x$-axis at $x = \pm 2$.

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Section 6 MCQs — Polynomial Division, Factor & Remainder Theorems

Q11. Find the remainder when $p(x) = x^3 + 2x^2 - 3x + 4$ is divided by $(x + 1)$.

A. $0$

B. $4$

C. $8$

D. $-4$

Reveal answer and explanation

Answer: C

By the Remainder Theorem, the remainder equals $p(-1)$:

$p(-1) = (-1)^3 + 2(-1)^2 - 3(-1) + 4 = -1 + 2 + 3 + 4 = 8$

Option A would mean $(x + 1)$ is a factor — but $p(-1) = 8 \neq 0$, so it is not. Option D is a sign error from mishandling $-3(-1)$.

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Q12. If $(x - 2)$ is a factor of $p(x) = x^3 - kx^2 + 3x - 2$, find the value of $k$.

A. $k = 1$

B. $k = 2$

C. $k = 3$

D. $k = 4$

Reveal answer and explanation

Answer: C

By the Factor Theorem, $(x - 2)$ is a factor if and only if $p(2) = 0$:

$p(2) = (2)^3 - k(2)^2 + 3(2) - 2 = 8 - 4k + 6 - 2 = 12 - 4k$

Set equal to zero: $12 - 4k = 0 \Rightarrow k = 3$.

A common error is substituting $x = -2$ instead of $x = 2$ (confusing the factor $(x - 2)$ with the root $x = -2$). Always set the factor equal to zero to find the substitution value.

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