# Composite, Inverse Functions and Modulus Graphs

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u1-composite-inverse-functions-and-modulus/

This guide covers core function concepts tested in CIE IGCSE Additional Mathematics 0606, including composite function order rules, inverse function calculation and sketching, and modulus graph construction for common function types.

**Prerequisites:** Basic function notation and domain/range definitions; Sketching linear, quadratic, and trigonometric graphs

## Learning objectives

- Calculate composite functions $fg(x)$ and $gf(x)$ and identify order dependence
- Find the inverse of a one-one function using correct $f^{-1}$ notation
- Sketch a function and its inverse as reflections over the line $y=x$
- Plot and interpret $y = |f(x)|$ graphs for linear, quadratic, cubic, and trigonometric functions

## Composite Functions

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

**Composite Function** — A function formed by substituting the output of one function into the input of another function.

*Notation:* $fg(x) = f(g(x))$

*Example:* If $f(x) = 2x + 3$ and $g(x) = x^2$, then $fg(x) = f(x^2) = 2x^2 + 3$.

**Worked example:** Given $f(x) = 3x - 1$ and $g(x) = \frac{x + 2}{4}$, calculate (a) $fg(2)$ (b) $gf(x)$ 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 $gf(x) = g(f(x))$ 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: $\text{Domain of } gf \subseteq \text{Domain of } f$ and $\text{Range of } gf \subseteq \text{Range of } g$. Sometimes you must restrict the domain of $f$ so that $gf$ exists at all. For example, if $f(x) = x - 4$ and $g(x) = \sqrt{x}$ (which needs a non-negative input), then $gf(x) = \sqrt{x - 4}$ 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.

## Inverses of One-One Functions

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 $f^{-1}(x)$.

**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 $f(x) = 2x + 5$ 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 $f(x) = \frac{2x - 5}{3}$.

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 $ff^{-1}(x) = x$

   $$f\left(\frac{3x +5}{2}\right) = \frac{2\left(\frac{3x+5}{2}\right) -5}{3} = \frac{3x +5 -5}{3} = x$$

**Worked example:** Find the inverse of $f(x) = e^{2x}$, 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 $e^{2x} > 0$ 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.

## Sketching Functions and Their Inverses

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

1. First sketch $f(x) = 2x +1$: 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.

## Modulus Graphs $y = |f(x)|$

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.

**Modulus Function** — The non-negative value of $a$, such that $|a| = a$ if $a \geq 0$ and $|a| = -a$ if $a < 0$.

*Notation:* $|a|$

**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 $\pi$ instead of $2\pi$.

## Common pitfalls

- **Wrong:** Calculating $fg(x)$ as $f(x) \times g(x)$ instead of $f(g(x))$
  - Why it fails: Confusing composite function notation with multiplication notation
  - Correct: Remember that $fg(x)$ means function composition, apply the rightmost function first, use $f(x)g(x)$ explicitly if you mean multiplication
- **Wrong:** Finding an inverse for a function that is not one-one
  - Why it fails: Forgetting that inverse functions only exist for one-one functions where each output maps to exactly one input
  - Correct: Always confirm a function passes the horizontal line test before calculating its inverse, or use the restricted domain given in the question
- **Wrong:** Reflecting functions across the y-axis instead of $y=x$ when sketching inverses
  - Why it fails: Mixing up reflection axes for inverse functions
  - Correct: Always draw and label the line $y=x$ first when sketching function-inverse pairs, use it as your guide for reflection
- **Wrong:** Deleting parts of the function below the x-axis when drawing $y=|f(x)|$
  - Why it fails: Misunderstanding that modulus reflects negative parts, not removes them
  - Correct: Reflect all segments of $f(x)$ below the x-axis across the x-axis, leaving the rest of the graph unchanged
- **Wrong:** Swapping the order of composite functions, e.g. calculating $gf(x)$ when asked for $fg(x)$
  - Why it fails: Forgetting that function composition is not commutative
  - Correct: Always read the order from right to left: the rightmost function is applied first

## Cheatsheet

| Concept | Rule | Example |
| --- | --- | --- |
| Composite Function | $fg(x) = f(g(x))$, apply rightmost function first | $f(x)=x+2, g(x)=3x$: $fg(x)=3x+2$ |
| Inverse Function | Only exists for one-one functions, reflect over $y=x$ | $f(x)=2x-1$: $f^{-1}(x)=\frac{x+1}{2}$ |
| $y=\|f(x)\|$ 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 $x > 2$ |

## 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.

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