Do I need to label the line when sketching functions and their inverses?

Yes, examiners require you to draw and label the line explicitly to demonstrate you understand the reflection relationship between and .

Can I delete parts of below the x-axis when drawing ?

No. Reflect all segments of that lie below the x-axis (where ) across the x-axis, leaving all parts above the x-axis unchanged. No sections of the domain are removed.

Composite, Inverse Functions and Modulus Graphs

CIE IGCSE Additional Mathematics· 25 min read

1. Composite Functions★★☆☆☆⏱ 6 min

Composite functions combine two functions by applying one function to the output of another. The order of composition is critical: $f g (x)$ means apply $g$ first, then apply $f$ to the result of $g (x)$ , while $g f (x)$ means apply $f$ first then $g$ .

📘 Definition

Composite Function

$f g (x) = f (g (x))$

A function formed by substituting the output of one function into the input of another function.

Example:

If $f (x) = 2 x + 3$ and $g (x) = x^{2}$ , then $f g (x) = f (x^{2}) = 2 x^{2} + 3$ .

📐 Worked Example

Given $f (x) = 3 x - 1$ and $g (x) = \frac{x + 2}{4}$ , calculate (a) $f g (2)$ (b) $g f (x)$ in its simplest form.

1
For part (a), first calculate $g (2)$
$g(2) = \frac{2 + 2}{4} = 1$
2
Substitute $g (2) = 1$ into $f (x)$
$fg(2) = f(1) = 3(1) - 1 = 2$
3
For part (b), substitute $f (x)$ into $g (x)$
$gf(x) = g(3x - 1) = \frac{(3x -1) + 2}{4}$
4
Simplify the expression
$gf(x) = \frac{3x + 1}{4}$

For a composite such as $g f (x) = g (f (x))$ to be defined, every output of the inner function $f$ must be a valid input of the outer function $g$ . This gives two containment rules: $Domain of g f \subseteq Domain of f$ and $Range of g f \subseteq Range of g$ . Sometimes you must restrict the domain of $f$ so that $g f$ exists at all. For example, if $f (x) = x - 4$ and $g (x) = x$ (which needs a non-negative input), then $g f (x) = x - 4$ is only defined for $x - 4 \geq 0$ , so the domain of $f$ must be restricted to $x \geq 4$ .

Exam tip:

Always write intermediate steps for composite function calculations to earn method marks even if your final answer is incorrect.

2. Inverses of One-One Functions★★★☆☆⏱ 7 min

An inverse function reverses the operation of the original function, and only exists for one-one (injective) functions, where each output maps to exactly one input. The inverse of function $f$ is written as $f^{- 1} (x)$ .

📘 Definition

One-One Function

A function where no two distinct inputs produce the same output, satisfying the horizontal line test when sketched.

Example:

The linear function $f (x) = 2 x + 5$ is one-one, while the quadratic $f (x) = x^{2}$ is not one-one over all real numbers.

📐 Worked Example

Find the inverse of the one-one function $f (x) = \frac{2 x - 5}{3}$ .

1
Set $y = f (x)$
$y = \frac{2x - 5}{3}$
2
Rearrange to make $x$ the subject of the formula
$3y = 2x -5 \\ 2x = 3y + 5 \\ x = \frac{3y + 5}{2}$
3
Swap $x$ and $y$ , then replace $y$ with $f^{- 1} (x)$
$f^{-1}(x) = \frac{3x + 5}{2}$
4
Verify by checking $f f^{- 1} (x) = x$
$f\left(\frac{3x +5}{2}\right) = \frac{2\left(\frac{3x+5}{2}\right) -5}{3} = \frac{3x +5 -5}{3} = x$

📐 Worked Example

Find the inverse of $f (x) = e^{2 x}$ , defined for all real $x$ , and state its domain.

1
Set $y = f (x)$ :
$y = e^{2x}$
2
Take natural logarithms of both sides ( $ln$ undoes $e$ ):
$\ln y = 2x$
3
Make $x$ the subject, then swap $x$ and $y$ :
$x = \tfrac{1}{2}\ln y \implies f^{-1}(x) = \tfrac{1}{2}\ln x$
4
The domain of $f^{- 1}$ equals the range of $f$ . Since $e^{2 x} > 0$ for every $x$ , the range of $f$ is $f (x) > 0$ , so $f^{- 1}$ has domain $x > 0$ (and range all real values). This exponential-logarithm pair is the standard example from syllabus 1.6; the exponential and log functions themselves are studied in Unit 6.
$\text{Domain of } f^{-1}: x > 0$

Exam tip:

Always confirm your inverse function is correct by checking that composing it with the original function gives $x$ , this catches most algebraic errors.

3. Sketching Functions and Their Inverses★★★☆☆⏱ 6 min

The graph of an inverse function $f^{- 1} (x)$ is the reflection of the graph of $f (x)$ across the line $y = x$ . Key features of the original function map directly to the inverse: intercepts swap axes, and the domain of $f$ becomes the range of $f^{- 1}$ , and vice versa.

📐 Worked Example

Sketch the function $f (x) = 2 x + 1$ and its inverse $f^{- 1} (x)$ on the same axes, labeling all key features.

1
First sketch $f (x) = 2 x + 1$ : it has a y-intercept at (0, 1) and x-intercept at (-0.5, 0), slope 2.
2
Sketch and label the line $y = x$ as your reflection axis.
3
Reflect the intercepts of $f (x)$ across $y = x$ to get intercepts of $f^{- 1} (x)$ : (1, 0) and (0, -0.5).
4
Draw the line for $f^{- 1} (x)$ passing through these points, label both functions clearly.

Exam tip:

Examiners require explicit labeling of $y = x$ , $f (x)$ , and $f^{- 1} (x)$ to award full marks for sketch questions, even if your reflection is accurate.

4. Modulus Graphs $y = |f(x)|$★★★★☆⏱ 6 min

The modulus (absolute value) function returns the non-negative value of its input. For $y = ∣ f (x) ∣$ , all parts of the graph of $f (x)$ that lie below the x-axis (where $f (x) < 0$ ) are reflected across the x-axis, while parts above the x-axis remain unchanged.

📘 Definition

Modulus Function

$∣ a ∣$

The non-negative value of $a$ , such that $∣ a ∣ = a$ if $a \geq 0$ and $∣ a ∣ = - a$ if $a < 0$ .

📐 Worked Example

Sketch the graph of $y = ∣ x^{2} - 4∣$ , labeling all intercepts and turning points.

1
First sketch the unmodulated quadratic $f (x) = x^{2} - 4$ : it crosses the x-axis at (-2, 0) and (2, 0), y-intercept at (0, -4), minimum turning point at (0, -4).
2
Reflect the part of $f (x)$ below the x-axis (the segment between x=-2 and x=2) across the x-axis.
3
The resulting graph has intercepts at (-2, 0), (2, 0) and (0, 4), with a local maximum turning point at (0, 4), and the same upward opening arms outside $x = \pm 2$ as the original quadratic.

📐 Worked Example

Sketch the graph of $y = ∣ x^{3} - x ∣$ , showing where the curve meets the $x$ -axis.

1
First sketch the cubic $f (x) = x^{3} - x$ . Factorise to find where it crosses the $x$ -axis:
$x^3 - x = x(x - 1)(x + 1)$
2
The roots are $x = - 1, 0, 1$ . Checking the sign of $f (x)$ : it lies below the $x$ -axis on $x < - 1$ and on $0 < x < 1$ , and above it on $- 1 < x < 0$ and on $x > 1$ .
3
For $y = ∣ f (x) ∣$ , reflect only the sections that lie below the $x$ -axis (the parts on $x < - 1$ and on $0 < x < 1$ ) up across the $x$ -axis. The sections already above the axis are left unchanged.
4
The graph still touches the $x$ -axis at $x = - 1, 0, 1$ , but each of these crossing points becomes a sharp corner (a cusp) where the curve is folded upward. Every $y$ -value is now non-negative.

Exam tip:

For trigonometric modulus graphs like $y = ∣ sin x ∣$ , reflect all negative troughs above the x-axis, resulting in a graph with range $0 \leq y \leq 1$ and period $π$ instead of $2 π$ .

5. Common Pitfalls

Wrong move:

Calculating $f g (x)$ as $f (x) \times g (x)$ instead of $f (g (x))$

Why:

Confusing composite function notation with multiplication notation

Correct move:

Remember that $f g (x)$ means function composition, apply the rightmost function first, use $f (x) g (x)$ explicitly if you mean multiplication

Wrong move:

Finding an inverse for a function that is not one-one

Why:

Forgetting that inverse functions only exist for one-one functions where each output maps to exactly one input

Correct move:

Always confirm a function passes the horizontal line test before calculating its inverse, or use the restricted domain given in the question

Wrong move:

Reflecting functions across the y-axis instead of $y = x$ when sketching inverses

Why:

Mixing up reflection axes for inverse functions

Correct move:

Always draw and label the line $y = x$ first when sketching function-inverse pairs, use it as your guide for reflection

Wrong move:

Deleting parts of the function below the x-axis when drawing $y = ∣ f (x) ∣$

Why:

Misunderstanding that modulus reflects negative parts, not removes them

Correct move:

Reflect all segments of $f (x)$ below the x-axis across the x-axis, leaving the rest of the graph unchanged

Wrong move:

Swapping the order of composite functions, e.g. calculating $g f (x)$ when asked for $f g (x)$

Why:

Forgetting that function composition is not commutative

Correct move:

Always read the order from right to left: the rightmost function is applied first

6. Quick Reference Cheatsheet

Concept	Rule	Example
Composite Function	$f g (x) = f (g (x))$ , apply rightmost function first	$f (x) = x + 2, g (x) = 3 x$ : $f g (x) = 3 x + 2$
Inverse Function	Only exists for one-one functions, reflect over $y = x$	$f (x) = 2 x - 1$ : $f^{- 1} (x) = \frac{x + 1}{2}$
$y = ∣ f (x) ∣$ Graph	Reflect all parts of $f (x)$ below x-axis upwards	$y = ∣ x - 3∣$ is V-shaped, vertex at (3,0)
Inverse Domain/Range	Domain of $f$ = Range of $f^{- 1}$ , Range of $f$ = Domain of $f^{- 1}$	If $f (x)$ has range $y > 2$ , $f^{- 1} (x)$ has domain $x > 2$

7. Frequently Asked

Do I need to label the $y = x$ line when sketching functions and their inverses?

Yes, examiners require you to draw and label the line $y = x$ explicitly to demonstrate you understand the reflection relationship between $f (x)$ and $f^{- 1} (x)$ .

Can I delete parts of $f (x)$ below the x-axis when drawing $y = ∣ f (x) ∣$ ?

No. Reflect all segments of $f (x)$ that lie below the x-axis (where $f (x) < 0$ ) across the x-axis, leaving all parts above the x-axis unchanged. No sections of the domain are removed.

Going deeper

syllabusCIE 0606 2025-2027 SyllabusRefer to section 1.4, 1.6-1.8 for official topic guidance

What's Next

Now that you have mastered composite, inverse functions, and modulus graphs, you are ready to apply these skills to more advanced function problems in CIE IGCSE Additional Mathematics 0606. These concepts form the foundation for topics including function transformations, solving modulus equations, and even calculus applications later in the syllabus. You will frequently encounter these questions on both Paper 1 (non-calculator) and Paper 2 (calculator) of the exam, so regular practice of sketching and algebraic manipulation is critical to maximize your score.