Functions
Welcome to this comprehensive guide on functions for IGCSE Mathematics (0580). This post covers everything you need to know about functions, including function notation, inverse functions, composite functions, and exponential functions - all essential topics for your exams.
1. Functions, Domain, Range, and Notation
Understanding the Basics
Function: A relation where every input (x-value) has exactly one output (y-value). Think of it as a machine: you put one number in, and get one specific number out.
Domain: The set of all possible input values (x-values) for the function.
Range: The set of all possible output values (y-values) that result from using the domain.
Notation: A function is usually written as f(x), read as "f of x". The 'f' is the function's name, and 'x' is the input variable. Other letters like g, h, p can also be used.
Example 1: Finding Outputs
Given f(x) = 3x − 5, find f(2) and f(−1).
Solution:
- f(2) = 3(2) − 5 = 6 − 5 = 1
- f(−1) = 3(−1) − 5 = −3 − 5 = −8
Example 2: Domain Considerations
Find the domain for:
a) f(x) = 3x − 5
b) p(x) =
Solution:
a) f(x) = 3x − 5 is a simple linear function. You can input any real number.
Domain: All real numbers, x ∈ ℝ.
b) p(x) =
Domain: x ∈ ℝ, x ≠ −2.
2. Inverse Functions f⁻¹(x)
Key Concepts
Inverse Function f⁻¹(x): A function that "reverses" the action of the original function f(x). If f(a) = b, then f⁻¹(b) = a.
How to find the inverse:
- Replace f(x) with y.
- Swap every x and y in the equation.
- Solve this new equation for y.
- Replace y with f⁻¹(x).
Example 1: Linear Function
Find the inverse of f(x) = 3x − 5.
Solution:
- y = 3x − 5
- Swap x and y: x = 3y − 5
- Solve for y:
x + 5 = 3y
y =x + 53 - f⁻¹(x) = x + 53
Example 2: Function with a Fraction
Find the inverse of f(x) =
Solution:
- y = 3x + 2
- Swap x and y: x = 3y + 2
- Solve for y:
x(y + 2) = 3
xy + 2x = 3
xy = 3 − 2x
y =3 − 2xx - f⁻¹(x) = 3 − 2xx, x ≠ 0
3. Composite Functions gf(x)
Understanding Composition
Composite Function: Combining two functions where the output of one function becomes the input of the other.
Notation: gf(x) means g(f(x)). You work from the inside out.
Important: Order matters! gf(x) is not usually the same as fg(x).
Example 1: Basic Composition
Given f(x) = 3x − 5 and g(x) =
Solution:
- First, find the inner function f(2): f(2) = 3(2) − 5 = 1
- Now, use this result as the input for g(x): gf(2) = g(f(2)) = g(1)
- g(1) = 3(1 + 4)5=3 × 55= 3
- So, gf(2) = 3
Example 2: Finding a General Expression
Given f(x) =
Solution:
- fg(x) = f(g(x)). So, the entire function g(x) becomes the input for f(x).
- In f(x), the input is x. We replace this input with g(x), which is (3x + 5)².
- fg(x) = f(g(x)) = 3(3x + 5)² + 2
- This is already a fraction in its simplest form.
4. Exponential Functions and Their Inverses
Working with Exponents
Exponential Functions: Functions where the variable is in the exponent, e.g., h(x) = 2ˣ, h(x) = 3ˣ
Key Properties:
- a⁰ = 1 for any a ≠ 0
- a¹ = a
- a⁻ⁿ = 1aⁿ
Example 1: Evaluating Exponential Functions
Given h(x) = 2ˣ, find:
a) h(3)
b) h(0)
c) h(−2)
Solution:
- h(3) = 2³ = 8
- h(0) = 2⁰ = 1
- h(−2) = 2⁻² = 12²=14
Example 2: Finding Inverse of Exponential Functions
Find the inverse of h(x) = 3ˣ.
Solution:
- y = 3ˣ
- Swap x and y: x = 3ʸ
- Solve for y: Since we cannot use logs at IGCSE level, we express this as:
If x = 3ʸ, then we can say y is the power we raise 3 to get x - h⁻¹(x) = the power such that 3 raised to that power equals x
5. Important Function Properties
Key Relationships
Function and Inverse Relationship:
- f(f⁻¹(x)) = x
- f⁻¹(f(x)) = x
Self-Inverse Functions: Functions where f⁻¹(x) = f(x)
Identity Property: h(h⁻¹(x)) = x for any function h
Example: Verifying Properties
Given f(x) = 2x + 3, verify that f(f⁻¹(x)) = x
Solution:
- First find f⁻¹(x):
1. y = 2x + 3
2. Swap: x = 2y + 3
3. Solve: x − 3 = 2y, so y =x − 32
4. f⁻¹(x) =x − 32 - Now find f(f⁻¹(x)):
f(f⁻¹(x)) = f(x − 32) = 2(x − 32) + 3 = (x − 3) + 3 = x ✓
6. Advanced Algebraic Manipulation with Functions
Complex Operations
Combining Functions: When adding/subtracting functions with different denominators, find a common denominator
Squaring Functions: (f(x))² means square the entire function, not just the x
Example 1: Writing as Single Fraction
Given f(x) =
Solution:
- f(x) + g(x) + 1 = 3x+2+ x + 1
- Common denominator: x+2
- = 3x+2+x(x+2)x+2+1(x+2)x+2
- = 3 + x(x+2) + (x+2)x+2
- = 3 + x² + 2x + x + 2x+2
- = x² + 3x + 5x+2
Example 2: Complex Composite Pattern
Given g(x) = x² + 1, find gg(x) in the form ax⁴ + bx² + c
Solution:
- gg(x) = g(g(x)) = g(x² + 1)
- = (x² + 1)² + 1
- = (x⁴ + 2x² + 1) + 1
- = x⁴ + 2x² + 2
- So a = 1, b = 2, c = 2
Practice Exercises
Exercise 1: Basic Function Notation
1. For h(x) = 2x² + 3, find h(0) and h(3).
Show Answer
h(0) = 2(0)² + 3 = 3
h(3) = 2(3)² + 3 = 2(9) + 3 = 21
2. For k(x) =
Show Answer
x = 4 is not in the domain because it makes the denominator zero.
Exercise 2: Inverse Functions
1. Find the inverse of g(x) =
Show Answer
1. y =
2. Swap: x =
3. Solve: 5x = 3(y + 4)
5x = 3y + 12
5x − 12 = 3y
y =
4. g⁻¹(x) =
Exercise 3: Composite Functions
1. Using f(x) = 3x − 5 and g(x) =
Show Answer
fg(x) = f(g(x)) = 3[g(x)] − 5 = 3[
=
Key Takeaways
- Always check domain restrictions, especially with fractions
- Remember the step-by-step process for finding inverse functions
- Work from the inside out when dealing with composite functions
- Practice identifying function properties and relationships
- Master exponential functions and their inverses for exam success
📐 Function rules – IGCSE Maths
Inverse function rules
f⁻¹(y) = x ⇔ f(x) = y💡 Idea: The inverse “undoes” the original function. If \(f(x)=y\), then \(f^{-1}(y)=x\).
\(f^{-1}(f(x)) = x\) | \(f(f^{-1}(y)) = y\)
Domain of \(f^{-1}\) = Range of \(f\)
\(g(x)=2x+3\)
\(g^{-1}(7)=?\) → \(2x+3=7\) → \(x=2\)
✅ \(g^{-1}(7)=2\)
Composite function rules
(f ∘ g)(x) = f(g(x))🔁 Order matters! First apply the inner function, then the outer one.
\((f \circ g)(x)=(x+1)^2\)
\((g \circ f)(x)=x^2+1\)
⚠️ Usually \(f \circ g \neq g \circ f\)
🔄 Inverse Functions – Quick Notes
The only rule you need
That’s it. No long steps. No solving for \( f^{-1}(x) \). Just turn inverse questions into normal function questions.
What it looks like
| If you see ... | It means ... | What to do |
|---|---|---|
| \( f^{-1}(5) = 2 \) | \( f(2) = 5 \) | Put 2 into \( f \) |
| \( f^{-1}(x) = 3 \) | \( f(3) = x \) | Find \( f(3) \) |
| \( f^{-1}(7) = x \) | \( f(x) = 7 \) | Solve \( f(x) = 7 \) |
Example with \( f(x) = 2x + 1 \)
Means: \( f(?) = 5 \)
\( 2x + 1 = 5 \) → \( x = 2 \)
✅ \( \boxed{2} \)
Means: \( f(3) = x \)
\( 2(3) + 1 = 7 \)
✅ \( \boxed{7} \)
Means: \( f(x) = 7 \)
\( 2x + 1 = 7 \) → \( x = 3 \)
✅ \( \boxed{3} \)
Quick summary
- \( f^{-1}(y) = x \) ⇔ \( f(x) = y \)
- To find \( x \) when \( f^{-1}(k) = x \): set \( f(x) = k \) and solve.
- To find \( f^{-1}(k) \): solve \( f(x) = k \) for \( x \).
- No need to find \( f^{-1}(x) \) — just use the meaning.
Even & odd functions
symmetry rules| Type | Rule | Graph symmetry | Example |
|---|---|---|---|
| Even | \(f(-x)=f(x)\) | mirror over y‑axis | \(x^2,\; \cos x\) |
| Odd | \(f(-x)=-f(x)\) | 180° rotation about origin | \(x^3,\; \sin x\) |
| Neither | doesn't satisfy either | no special symmetry | \(x^2+x\) |
Domain & Range
allowed inputs → possible outputs- Domain = set of all possible \(x\)-values (no division by zero, no square root of negative).
- Range = all possible \(y\)-values that the function produces.
- For inverses: \(\text{Domain}(f^{-1}) = \text{Range}(f)\) and \(\text{Range}(f^{-1}) = \text{Domain}(f)\).
Domain: \(x \neq 2\)
Range: \(y \ge 0\)
Graph transformations
shifts, stretches, flips| Change | Rule | Effect (visual) |
|---|---|---|
| Shift up | \(f(x)+a\) | moves graph upward |
| Shift down | \(f(x)-a\) | moves graph downward |
| Shift right | \(f(x-a)\) | slides right |
| Shift left | \(f(x+a)\) | slides left |
| Vertical stretch | \(a\cdot f(x),\; a>1\) | steeper/narrower |
| Reflect in x‑axis | \(-f(x)\) | flip upside down |
| Reflect in y‑axis | \(f(-x)\) | mirror left‑right |
Exponential & log rules
same base → add/subtract exponentsDifferentiation rules (power & more)
gradient of a function| Rule | Formula |
|---|---|
| Power rule | \(\frac{d}{dx}(x^n)=n x^{n-1}\) |
| Sum rule | \(\frac{d}{dx}(f+g)=f'+g'\) |
| Product rule | \(\frac{d}{dx}(fg)=f'g+fg'\) |
| Chain rule | \(\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\) |
Integration rules (reverse of derivative)
area under a curveTrigonometric rules
sine, cosine & friendsOne‑to‑one (injective) functions
must be one‑to‑one to have an inverseRule: \(f(a)=f(b) \implies a=b\) (different inputs → different outputs).
Horizontal line test: any horizontal line crosses the graph at most once.
🧠 ⚡ Law of indices (same base) – super important!
If \(a^m = a^n\) and \(a>0,\;a\neq 1\) ⇒ \(m = n\)
Example: \(3^{2x-1}=27\) → \(3^{2x-1}=3^3\) → \(2x-1=3\) → \(x=2\) ✅
📐 Function rules – Examples & exam practice
1. Inverse function rules
✨ 5 examples
➜ \(y=3x-5\) → swap \(x,y\): \(x=3y-5\) → \(3y=x+5\) → \(f^{-1}(x)=\frac{x+5}{3}\).
➜ \(2x+7=11\) → \(2x=4\) → \(x=2\). So \(g^{-1}(11)=2\).
➜ \(y=\frac{x}{4}-3\) → swap: \(x=\frac{y}{4}-3\) → \(x+3=\frac{y}{4}\) → \(y=4(x+3)\).
➜ \(5^x=125=5^3\) → \(x=3\). Hence \(f^{-1}(125)=3\).
➜ \(x^2=9\) → \(x=3\) (domain restriction).
📝 5 exam‑style questions
2. Composite functions
✨ 5 examples
📝 5 exam‑style questions
3. Even & odd functions
✨ 5 examples
📝 5 exam‑style questions
4. Domain & range
✨ 5 examples
📝 5 exam‑style questions
5. Exponential & log (same base)
✨ 5 examples
📝 5 exam‑style questions
6. Graph transformations
✨ 5 examples
📝 5 exam‑style questions
📐 Functions: Inverse & Composite
The only two rules you need
🧩 Composite: \( (f \circ g)(x) = f(g(x)) \) (do \( g \) first, then \( f \))
✅ No finding inverses. No inequality limits. Just apply the rules.
20 mixed examples
🔁 Inverse (1–10)
\( f(4)=11 \) ⇒ \( x=11 \)
\( f(3)=13 \) ⇒ \( x=13 \)
\( f(6)=36 \) ⇒ \( x=36 \)
\( f(4)=23 \) ⇒ \( x=23 \)
\( f(5)=32 \) ⇒ \( x=32 \)
\( f(4)=81 \) ⇒ \( x=81 \)
\( f(3)=66 \) ⇒ \( x=66 \)
\( f(-2)=-16 \) ⇒ \( x=-16 \)
\( f(5)=37 \) ⇒ \( x=37 \)
\( f(2)=25 \) ⇒ \( x=25 \)
🧩 Composite (11–20)
\( g(4)=7, f(7)=14 \)
\( f(3)=9, g(9)=7 \)
\( g(2)=4, f(4)=13 \)
\( f(1)=6, g(6)=12 \)
\( g(5)=7, f(7)=25 \)
\( g(3)=2, f(2)=4 \)
\( f(2)=5, g(5)=15 \)
\( g(4)=16, f(16)=27 \)
\( g(2)=9, f(9)=45 \)
\( g(2)=3, f(3)=9 \)
20 practice questions
🔁 Inverse (1–10)
🧩 Composite (11–20)
Answers
📐 Domain & Range – Learner Guide
What are domain and range?
All possible input values (\(x\) values).
👉 "What \(x\) are allowed?"
All possible output values (\(y\) values).
👉 "What \(y\) come out?"
Step‑by‑step method
📌 1. Identify restrictions
- Denominator ≠ 0 → exclude values that make denominator zero
- Square root: inside must be ≥ 0
- Logarithm: inside must be > 0
- If no restrictions → domain = all real numbers (\(\mathbb{R}\))
📌 2. Find the range
- Think about outputs: squares are ≥0, exponentials >0, etc.
- For quadratics: find vertex, check direction (opens up/down)
- For rational functions: sketch or solve \(y = f(x)\) for \(x\)
Common functions – domain & range at a glance
| Function | Domain | Range |
|---|---|---|
| \(f(x)=mx+c\) | \(\mathbb{R}\) | \(\mathbb{R}\) |
| \(f(x)=x^2\) | \(\mathbb{R}\) | \(y \ge 0\) |
| \(f(x)=x^2 + k\) | \(\mathbb{R}\) | \(y \ge k\) |
| \(f(x)= -x^2 + k\) | \(\mathbb{R}\) | \(y \le k\) |
| \(f(x)=\frac{1}{x}\) | \(x \neq 0\) | \(y \neq 0\) |
| \(f(x)=\frac{1}{x-a}\) | \(x \neq a\) | \(y \neq 0\) |
| \(f(x)=\sqrt{x}\) | \(x \ge 0\) | \(y \ge 0\) |
| \(f(x)=\sqrt{x-h}\) | \(x \ge h\) | \(y \ge 0\) |
| \(f(x)=a^x\) \((a>0)\) | \(\mathbb{R}\) | \(y > 0\) |
| \(f(x)=a^x + k\) | \(\mathbb{R}\) | \(y > k\) |
| \(f(x)=\log x\) | \(x > 0\) | \(\mathbb{R}\) |
Worked examples (step‑by‑step)
\(f(x)=3x-2\)
✅ Domain: all real numbers \(\mathbb{R}\) or \((-\infty,\infty)\)
✅ Range: all real numbers \(\mathbb{R}\) or \((-\infty,\infty)\)
\(f(x)=x^2-4x+5\)
▶ Domain: all real numbers
▶ Vertex: \(x=2\) → \(y=1\) (opens up)
✅ Range: \(y \ge 1\) or \([1,\infty)\)
\(f(x)=\frac{2}{x-1}\)
▶ Denominator zero when \(x=1\)
✅ Domain: \(x \neq 1\) → \((-\infty,1)\cup(1,\infty)\)
✅ Range: \(y \neq 0\) → \((-\infty,0)\cup(0,\infty)\)
\(f(x)=\sqrt{2x-6}\)
▶ Inside ≥ 0 → \(2x-6 \ge 0 \Rightarrow x \ge 3\)
✅ Domain: \([3,\infty)\)
✅ Range: \([0,\infty)\)
\(f(x)=3^x - 2\)
✅ Domain: all real numbers
▶ \(3^x > 0\) → subtract 2 → \(y > -2\)
✅ Range: \((-2,\infty)\)
Practice questions
📍 Domain? Range?
📍 Domain? Range?
📍 Domain? Range?
📍 Domain? Range?
📍 Domain? Range?
✅ Click to see answers
1. Domain: \(x \neq -2\) Range: \(y \neq 0\)
2. Domain: \(x \ge 4\) Range: \(y \ge 0\)
3. Domain: all reals Range: \(y > 5\)
4. Domain: all reals Range: \(y \ge 1\) (vertex at \(x=3, y=1\))
5. Domain: \(x \neq 0\) Range: \(y > 0\)
Notation & quick checklist
| Notation | Meaning | Example |
|---|---|---|
| \((a,b)\) | between \(a\) and \(b\), not including \(a,b\) | \((2,5)\) |
| \([a,b]\) | between \(a\) and \(b\), including \(a,b\) | \([2,5]\) |
| \((a,\infty)\) | greater than \(a\) | \((2,\infty)\) |
| \((-\infty,a)\) | less than \(a\) | \((-\infty,5)\) |
| \(\mathbb{R}\) | all real numbers | \(\mathbb{R}\) |
| \(x \neq a\) | all real numbers except \(a\) | \(x \neq 3\) |
☐ Denominator ≠ 0 ☐ Square root: inside ≥ 0 ☐ Logarithm: inside > 0
☐ If none → all real numbers
☐ Sketch or think about possible outputs ☐ Quadratics: vertex + direction
☐ Exponentials: base>0 → y>0 (plus shift) ☐ Reciprocals: y ≠ 0