# Functions

> CIE A-Level Mathematics · 9709
> Source: https://www.owlsprep.com/study/cie-9709-u1-functions/

This module covers core function concepts for CIE 9709 AS Pure 1, including definitions, domain, range, composite functions and inverse functions, with exam-focused examples to master common question types.

**Prerequisites:** [Basic algebra and coordinate geometry](https://www.owlsprep.com/study/cie-9709-u1-algebra-basics/)

## Learning objectives

- Define and identify functions, using the vertical line test
- Find the domain and range of common functions
- Form composite functions and determine their domain
- Find inverse functions for one-to-one and restricted domain functions
- Describe the four graph transformations and simple combinations

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P1 past papers (2016–2025): **Functions** appears **100 times** in the last 10 years — **14.5%** of all P1 questions (100 of 692).

Most-tested forms: Inverse functions (61), Composite functions (48), Domain and range (33), Graph transformations (33).

## Functions, Domain and Range

**Function** — A mapping from a set of inputs (domain) to a set of outputs where each input maps to exactly one output.

*Notation:* f(x)

*Example:* $f(x) = 2x+1$ is a function; $x^2 + y^2 = 1$ is not a function for all $x$.

The **vertical line test** checks if a graph represents a function: any vertical line will intersect the graph at most once if it is a function.

The *domain* of a function is the set of all valid input values $x$. The *range* is the set of all possible output values $f(x)$.

**Worked example:** Find the domain and range of $f(x) = \frac{1}{x - 3}$ for $x \in \mathbb{R}$.

1. First, find any $x$ that makes the function undefined. Denominators cannot equal zero:
2. $$x - 3 = 0 \implies x = 3$$
3. All other real values of $x$ are valid, so the domain is:
4. $$\text{Domain: } x \in \mathbb{R}, x \neq 3$$
5. To find the range, note that $\frac{1}{x-3}$ can never equal zero (the numerator is a non-zero constant 1). All other non-zero real values are possible:
6. $$\text{Range: } f(x) \in \mathbb{R}, f(x) \neq 0$$

> **Exam tip:** Always check for two common domain restrictions: denominators cannot be zero, and expressions under square roots cannot be negative.

## Composite Functions

**Composite Function** — A combination of two functions where the output of the inner function $g$ becomes the input of the outer function $f$. This means *apply $g$ first, then $f$*.

*Notation:* f \circ g = f(g(x))

*Example:* If $f(x) = x^2$ and $g(x) = x+1$, $f(g(x)) = (x+1)^2$.

> **info**
>
> A composite $fg(x)=f(g(x))$ can only be formed when the **range of $g$ lies within the domain of $f$** — otherwise some outputs of $g$ have no valid input into $f$. For example, $f(x)=\sqrt{x}$ only accepts $x\ge 0$: if $g(x)=x^2+1$ (range $\ge 1$, all non-negative) then $fg$ is fine; if $g(x)=x-5$ can output negatives, $fg$ breaks there.

To find the domain of a composite function $f(g(x))$, two conditions must always be satisfied: 1. $x$ is in the domain of $g$, 2. $g(x)$ is in the domain of $f$.

**Worked example:** Given $f(x) = \frac{1}{x}$, $x \neq 0$, and $g(x) = x - 4$, $x \in \mathbb{R}$, find the domain of $f(g(x))$.

1. First, confirm that for $f(g(x))$ to exist, $g(x)$ must be a valid input to $f$:
2. $f$ is only defined for non-zero inputs, so we require $g(x) \neq 0$:
3. $$x - 4 = 0 \implies x = 4$$
4. Next, check the domain of $g$: $g$ is defined for all real $x$, so the only restriction is $x \neq 4$:
5. $$\text{Domain: } x \in \mathbb{R}, x \neq 4$$

> **Exam tip:** Order matters for composite functions: $f(g(x))$ is not the same as $g(f(x))$ in most cases. Always apply the inner function first.

## Inverse Functions

**Inverse Function** — A function that reverses the mapping of an original one-to-one (injective) function. If $f$ maps $x$ to $y$, $f^{-1}$ maps $y$ back to $x$.

*Notation:* f^{-1}(x)

*Example:* If $f(x) = 2x + 3$, $f^{-1}(x) = \frac{x-3}{2}$.

A function is one-to-one if different inputs always give different outputs — equivalently, each output value comes from just one input. (Every function already sends each input to a single output; one-to-one is the extra condition on the *outputs*.) Use the *horizontal line test* to check: any horizontal line intersects the graph at most once for a one-to-one function. A key property: $\text{domain of } f^{-1} = \text{range of } f$, $\text{range of } f^{-1} = \text{domain of } f$.

**Worked example:** Find the inverse of $f(x) = 2x - 5$, $x \in \mathbb{R}$.

1. Step 1: Let $y = f(x)$:
2. $$y = 2x - 5$$
3. Step 2: Rearrange the equation to make $x$ the subject:
4. $$y + 5 = 2x \implies x = \frac{y + 5}{2}$$
5. Step 3: Swap $x$ and $y$ to get $f^{-1}(x)$:
6. $$f^{-1}(x) = \frac{x + 5}{2}$$
7. Verify: $f(f^{-1}(x)) = 2\left(\frac{x+5}{2}\right) - 5 = x$, which confirms the inverse is correct.

> **info**
>
> The graph of $f^{-1}(x)$ is always the reflection of the graph of $f(x)$ in the line $y = x$.

## Inverses with Restricted Domains

Many common functions like quadratics are not one-to-one over their entire domain. To get an inverse, we first restrict the domain to make the function one-to-one. This is one of the most frequently tested concepts in CIE 9709.

**Worked example:** Given $f(x) = x^2$, $x \geq 0$, find $f^{-1}(x)$ and state its domain and range.

1. Step 1: Confirm $f(x)$ is one-to-one on the restricted domain. For $x \geq 0$, it passes the horizontal line test, so an inverse exists.
2. Step 2: Let $y = x^2$, rearrange for $x$:
3. $$y = x^2 \implies x = \pm \sqrt{y}$$
4. Step 3: The original domain is $x \geq 0$, so we only take the positive root:
5. $$x = \sqrt{y}$$
6. Step 4: Swap $x$ and $y$ to get the inverse:
7. $$f^{-1}(x) = \sqrt{x}$$
8. Step 5: State domain and range, using the inverse property:
9. $$\text{Domain of } f^{-1} = \text{Range of } f = x \geq 0 \\ \text{Range of } f^{-1} = \text{Domain of } f = f^{-1}(x) \geq 0$$

**Worked example:** For the same $f(x) = x^2$, $x \geq 0$ (so $f^{-1}(x) = \sqrt{x}$), sketch $f$ and $f^{-1}$ on the same axes and show how they are related.

1. Sketch $f(x) = x^2$ for $x \geq 0$: the right-hand half of the parabola, starting at the origin and rising. Two easy points to plot are $(1, 1)$ and $(2, 4)$.
2. Sketch $f^{-1}(x) = \sqrt{x}$ for $x \geq 0$ by swapping the coordinates of every point on $f$. So $(1, 1)$ stays put and $(2, 4)$ moves to $(4, 2)$:
3. $$(2,\, 4) \;\longmapsto\; (4,\, 2)$$
4. Draw the line $y = x$. The graph of $f^{-1}$ is the mirror image of $f$ in this line: each point of $f$ reflects onto the matching point of $f^{-1}$, and points where the curves meet lie on $y = x$ (here $(0,0)$ and $(1,1)$).
5. In CIE, a sketch of a function and its inverse should always include an indication of the mirror line $y = x$ — draw it in and label it.

> **Exam tip:** Always pick the correct root (positive or negative) based on the original restricted domain. This is the most commonly missed mark in inverse function questions.

## Transformations of Graphs

**Transformation of y = f(x)** — A change to the equation y = f(x) that moves or stretches its graph in a predictable way, without redrawing it point by point.

*Notation:* y = f(x)+a,\quad f(x+a),\quad af(x),\quad f(ax)

*Example:* If $f(x)=x^2$, then $y=f(x)+3$ is the same parabola raised 3 units, and $y=f(x-2)$ is that parabola shifted 2 units to the right.

One idea unlocks every transformation in this topic:

**A change on the *outside* of $f$ acts *vertically* and does exactly what you expect. A change on the *inside* of $f$ (applied to $x$ before $f$ acts) works *horizontally* and does the *opposite* of what you expect.**

That is why $f(x)+a$ moves the graph *up* (outside, intuitive) but $f(x+a)$ moves it *left* (inside, reversed).

**Translations (shifts).**

- $y = f(x) + a$: shift **up** by $a$ (down if $a<0$). Outside $f$, so vertical.
- $y = f(x + a)$: shift **left** by $a$ (right if $a<0$). Inside $f$, so horizontal and reversed.

**Stretches.**

- $y = a\,f(x)$: **vertical** stretch, scale factor $a$. Every $y$-value is multiplied by $a$; points on the $x$-axis stay fixed.
- $y = f(ax)$: **horizontal** stretch, scale factor $\frac{1}{a}$. Every $x$-value is divided by $a$; points on the $y$-axis stay fixed.

**Reflections** are just stretches with a negative factor: $y = -f(x)$ reflects in the $x$-axis, and $y = f(-x)$ reflects in the $y$-axis.

> **info**
>
> Master rule: **outside = vertical = intuitive; inside = horizontal = reversed.** Hold on to this one sentence and you never have to memorise the four cases separately.

**Worked example:** The function $f$ is defined by $f(x) = x^2$. Describe fully the single transformation that maps $y = f(x)$ onto each of (a) $y = f(x) + 4$, (b) $y = f(x - 3)$, (c) $y = 2f(x)$, (d) $y = f(2x)$.

1. (a) The $+4$ is *outside* $f$, so it acts vertically and intuitively:
2. $$y = f(x) + 4 = x^2 + 4 \quad\Rightarrow\quad \text{translation 4 units up}$$
3. (b) The $-3$ is *inside* $f$, so it acts horizontally and in reverse — subtracting moves the graph to the *right*:
4. $$y = f(x-3) = (x-3)^2 \quad\Rightarrow\quad \text{translation 3 units right}$$
5. (c) The $2$ multiplies the whole function (*outside*), so it is a vertical stretch:
6. $$y = 2f(x) = 2x^2 \quad\Rightarrow\quad \text{vertical stretch, scale factor 2}$$
7. (d) The $2$ multiplies $x$ *inside* $f$, so it is a horizontal stretch with the reciprocal factor:
8. $$y = f(2x) = (2x)^2 = 4x^2 \quad\Rightarrow\quad \text{horizontal stretch, scale factor } \frac{1}{2}$$

**Worked example:** The curve $y = f(x)$ passes through the point $(4, 3)$. It is transformed to $y = f(2x) + 1$. Describe the two transformations and find the image of $(4, 3)$.

1. Read off each change. $f(2x)$ is *inside* $f$ → horizontal stretch, scale factor $\frac{1}{2}$. The $+1$ is *outside* → translation 1 unit up. The horizontal change acts on the $x$-coordinate, the vertical change on the $y$-coordinate.
2. Horizontal stretch, factor $\frac{1}{2}$, multiplies the $x$-coordinate by $\frac{1}{2}$:
3. $$x:\quad 4 \times \frac{1}{2} = 2$$
4. The $+1$ raises the $y$-coordinate by 1:
5. $$y:\quad 3 + 1 = 4$$
6. So the image of the point is:
7. $$(4,\, 3) \;\longmapsto\; (2,\, 4)$$
8. Sanity check: a factor of 2 *inside* $f$ gives a stretch of $\frac{1}{2}$, pulling points towards the $y$-axis ($x=4 \to x=2$) — never a stretch of 2. That is 'inside = reversed' in action.

> **info**
>
> **Does the order matter?** When you combine **two vertical** transformations (or **two horizontal** ones), the order you apply them in *can* change the result. When you combine **one horizontal with one vertical** — like $f(2x)+1$ above — the order never matters, because they act on different coordinates. Vertical steps follow the normal order of operations; horizontal steps follow the *reverse* order.

**Worked example:** The curve $y = f(x)$ passes through $(3, 5)$. Describe the transformations in $y = 2f(x) - 3$, find the image of $(3, 5)$, and explain why the order matters here.

1. Both changes are *outside* $f$, so both are vertical: $2f(x)$ is a vertical stretch of factor 2, and the $-3$ is a translation 3 units down. Two vertical transformations follow the normal order of operations — multiply first, then subtract.
2. The $x$-coordinate is unchanged (nothing acts horizontally). Apply the stretch $\times 2$ to the $y$-coordinate first, then subtract 3:
3. $$y:\quad 5 \times 2 - 3 = 7$$
4. So the image of the point is:
5. $$(3,\, 5) \;\longmapsto\; (3,\, 7)$$
6. Order matters because both steps act on the same ($y$) coordinate: subtracting 3 *first* and then stretching would give $2(5 - 3) = 4$, not $7$. This is unlike $f(2x)+1$, where the horizontal and vertical steps act on different coordinates and can be done in either order.

> **Exam tip:** Two traps examiners rely on: $f(x+a)$ moves the graph *opposite* to the sign of $a$ (so $+a$ is a shift left), and $f(ax)$ stretches by $\frac{1}{a}$, not by $a$. When unsure, track one easy point through the change and see where it lands.

## Common pitfalls

- **Wrong:** Claiming $x^2 + y^2 = 1$ is a function of $x$
  - Why it fails: A single $x$ value maps to two different $y$ values, so it fails the vertical line test
  - Correct: Recognize this is a relation, not a function, because each $x$ between $-1$ and $1$ has two outputs
- **Wrong:** Forgetting to check the domain of the inner function for composites
  - Why it fails: Only checking the restriction on the outer function misses existing restrictions on the inner function
  - Correct: Always check two conditions: $x$ is in the domain of the inner function, and the inner output is valid for the outer function
- **Wrong:** Swapping order of composite functions, writing $g(f(x))$ for $f(g(x))$
  - Why it fails: Misremembering that the notation $f(g(x))$ means the inner function $g$ is applied first
  - Correct: Always remember: inner function first, outer function second for composite functions
- **Wrong:** Keeping both roots when finding the inverse of a restricted quadratic
  - Why it fails: The original domain restricts the sign of $x$, so only one root is valid for the inverse
  - Correct: Check the original domain to select the correct positive or negative root for the inverse
- **Wrong:** Reading $y = f(x + 3)$ as a translation 3 units to the right
  - Why it fails: The $+3$ is inside $f$, applied to $x$ before the function acts, so it works horizontally and in reverse — the graph moves 3 units to the *left*
  - Correct: Use 'inside = horizontal = reversed', or set the bracket to zero ($x+3=0 \Rightarrow x=-3$) to see the graph shifts to $x=-3$, i.e. 3 units left

## Cheatsheet

| Concept | Key Rule |
| --- | --- |
| Function | Each input maps to exactly one output (passes vertical line test) |
| Domain | All valid $x$; exclude values that make denominators zero / square roots negative |
| Composite $f(g(x))$ | Apply $g$ first, then $f$; domain: $x \in \text{dom}(g), g(x) \in \text{dom}(f)$ |
| Inverse $f^{-1}(x)$ | Only exists for one-to-one functions; $\text{dom}(f^{-1}) = \text{ran}(f)$; graph reflected in $y=x$ |
| Finding inverse | 1. Let $y=f(x)$ 2. Rearrange for $x$ 3. Swap $x$ and $y$ |
| Transformations | $f(x)+a$ up; $f(x+a)$ left; $af(x)$ vertical $\times a$; $f(ax)$ horizontal $\times \frac{1}{a}$ |

## What's next

Functions are a foundational topic for almost all other areas of pure mathematics, including calculus, coordinate geometry, and trigonometry. Understanding domain and range helps you correctly interpret limits, derivatives, and integrals, while inverse functions are essential for working with logarithmic and inverse trigonometric functions. The graph transformations covered in the final section of this guide are a frequent CIE 9709 Paper 1 topic in their own right and often combine with domain, range and inverse questions, so building a solid understanding here will help you tackle more complex questions later in the course.

- [Quadratics](https://www.owlsprep.com/study/cie-9709-u1-quadratics/)
- [Differentiation](https://www.owlsprep.com/study/cie-9709-u1-differentiation/)
- [Coordinate geometry](https://www.owlsprep.com/study/cie-9709-u1-coordinate-geometry/)

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