Functions — The Basics
Notation · Domain & range · Types of functions · Mapping diagrams
What is a function?
A function is a rule that maps every input to exactly one output. Each x-value maps to one and only one y-value.
NOTATION
\(f(x) = 2x + 1\) — reads as "f of x"
\(f: x \mapsto 2x+1\) — mapping notation
\(f: \mathbb{R} \to \mathbb{R}\) — domain to codomain
\(f: x \mapsto 2x+1\) — mapping notation
\(f: \mathbb{R} \to \mathbb{R}\) — domain to codomain
Key rule: One input → one output. If any x maps to two outputs, it is NOT a function. Use the vertical line test on a graph.
Mapping diagram
Domain
The set of input values (x-values) for which the function is defined.
COMMON RESTRICTIONS
\(\dfrac{1}{x}\) → exclude \(x = 0\)
\(\sqrt{x}\) → require \(x \geq 0\)
\(\ln(x)\) → require \(x > 0\)
Given explicitly → just use it
\(\sqrt{x}\) → require \(x \geq 0\)
\(\ln(x)\) → require \(x > 0\)
Given explicitly → just use it
Exam tip: Always state domain restrictions. If not given, assume the largest possible set.
Range
The set of output values (y-values) that the function actually produces.
HOW TO FIND RANGE
1. Sketch the graph
2. Read off min/max y-values
3. State using inequalities or set notation
4. Watch for asymptotes
2. Read off min/max y-values
3. State using inequalities or set notation
4. Watch for asymptotes
Exam tip: Range depends on both the rule AND the domain. Change the domain → range changes.
Types of functions
ONE-TO-ONE
Each input → unique output. Has an inverse.
MANY-TO-ONE
Multiple inputs → same output. Still a function. No inverse.
ONE-TO-MANY ✗
One input → multiple outputs. NOT a function.
Vertical line test — is it a function?
Draw a vertical line anywhere across the graph.
Passes (✓ function) — vertical line crosses graph at most once everywhere.
Fails (✗ not a function) — vertical line crosses graph more than once anywhere.
Example: a circle fails because a vertical line through the centre crosses it twice.
Passes (✓ function) — vertical line crosses graph at most once everywhere.
Fails (✗ not a function) — vertical line crosses graph more than once anywhere.
Example: a circle fails because a vertical line through the centre crosses it twice.
Composite Functions
f(g(x)) · Order matters · Domain of composite
What is a composite function?
Apply one function, then feed the result into another. The output of g becomes the input of f.
\[fg(x) = f(g(x))\]
\[gf(x) = g(f(x))\]
Order matters: \(fg(x) \neq gf(x)\) in general. Do the right-most function first.
Flow diagram
Finding fg(x) — method
1
Write out both functions clearly
2
Substitute g(x) into f — replace every x in f with the entire expression for g(x)
3
Simplify
4
State the domain of fg — must satisfy domain of g AND g(x) must be in domain of f
Example — \(f(x)=x^2+1,\; g(x)=2x-3\)
fg
\(fg(x) = f(2x-3) = (2x-3)^2 + 1\)
\(= 4x^2 - 12x + 9 + 1 = 4x^2 - 12x + 10\)
\(= 4x^2 - 12x + 9 + 1 = 4x^2 - 12x + 10\)
gf
\(gf(x) = g(x^2+1) = 2(x^2+1) - 3\)
\(= 2x^2 + 2 - 3 = 2x^2 - 1\)
\(= 2x^2 + 2 - 3 = 2x^2 - 1\)
✓
\(fg(x) \neq gf(x)\) — order matters
Domain of a composite function
For \(fg(x)\), the domain must satisfy:
1. x is in the domain of g
2. g(x) is in the domain of f
The domain of fg is the set of x satisfying both conditions simultaneously.
1. x is in the domain of g
2. g(x) is in the domain of f
The domain of fg is the set of x satisfying both conditions simultaneously.
EXAMPLE
\(f(x) = \sqrt{x},\; x\geq 0\)
\(g(x) = x - 3,\; x \in \mathbb{R}\)
For \(fg(x)\): need \(g(x) \geq 0\)
\(\Rightarrow x - 3 \geq 0 \Rightarrow x \geq 3\)
Domain of \(fg\) is \(x \geq 3\)
\(g(x) = x - 3,\; x \in \mathbb{R}\)
For \(fg(x)\): need \(g(x) \geq 0\)
\(\Rightarrow x - 3 \geq 0 \Rightarrow x \geq 3\)
Domain of \(fg\) is \(x \geq 3\)
Inverse Functions
f⻹(x) · Finding the inverse · Graphical relationship
What is an inverse?
The inverse function undoes what f does. If f maps x to y, then f⻹ maps y back to x.
\[ff^{-1}(x) = f^{-1}f(x) = x\]
Only one-to-one functions have inverses. Many-to-one functions do not — unless you restrict the domain first.
Graphical relationship
The graph of \(f^{-1}(x)\) is the reflection of \(f(x)\) in the line \(y = x\).
Blue = f(x) · Green = f⻹(x) · Dashed = y = x
Finding f⻹(x) — method
1
Write \(y = f(x)\)
2
Rearrange to make x the subject
3
Swap x and y — replace x with \(f^{-1}(x)\)
4
State the domain of \(f^{-1}\) — it equals the range of f
Example — \(f(x) = 3x - 5\)
1
\(y = 3x - 5\)
2
\(y + 5 = 3x \Rightarrow x = \dfrac{y+5}{3}\)
3
\(f^{-1}(x) = \dfrac{x+5}{3}\)
4
Domain of \(f^{-1}\) = Range of \(f\) = \(\mathbb{R}\)
Self-inverse functions
A function is self-inverse if \(f^{-1}(x) = f(x)\) — the function is its own inverse.
This happens when the graph is symmetric about \(y = x\).
Classic examples: \(f(x) = x\), \(f(x) = \dfrac{1}{x}\), \(f(x) = -x + c\)
This happens when the graph is symmetric about \(y = x\).
Classic examples: \(f(x) = x\), \(f(x) = \dfrac{1}{x}\), \(f(x) = -x + c\)
TEST FOR SELF-INVERSE
Find \(f^{-1}(x)\) normally.
If \(f^{-1}(x) = f(x)\), it is self-inverse.
If \(f^{-1}(x) = f(x)\), it is self-inverse.
Modulus Functions
|f(x)| · −|f(x)| · Solving equations · Graph sketching
What is the modulus?
The modulus (absolute value) of x is its distance from zero — always non-negative.
\[|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}\]
y = |f(x)| — Draw f(x). Reflect any part below the x-axis upward. Parts above stay unchanged. No graph exists below the x-axis.
y = −|f(x)| — Entire graph flips below the x-axis. Vertex becomes a maximum. The equation −|f(x)| = k has no solutions when k > 0.
INTERACTIVE — Y = A·|X + B| + C (DRAG A NEGATIVE TO SEE −|X|)
a = 1 (neg = flip)
b = 0 (h. shift)
c = 0 (v. shift)
y = | x |
a > 0 — normal V
Opens upward. Vertex = minimum. Try a = 2.
a < 0 — inverted V
Opens downward. Vertex = maximum. Try a = −1.
Solving |f(x)| = k
1
Split into two cases: \(f(x) = k\) and \(f(x) = -k\)
2
Solve each equation separately
3
Check solutions satisfy the original equation
Note: If k < 0, there are no solutions — modulus is always ≥ 0.
Solving |f(x)| = g(x)
1
Case 1: \(f(x) = g(x)\) and check \(f(x) \geq 0\)
2
Case 2: \(-f(x) = g(x)\) and check \(f(x) < 0\)
3
Or: square both sides — \([f(x)]^2 = [g(x)]^2\) — then check all solutions
Squaring tip: Squaring both sides avoids cases but may introduce extraneous solutions — always check.
Worked example — solve |2x − 3| = 5
- Case 1: \(2x - 3 = 5\)
\(2x = 8 \Rightarrow x = 4\) - Case 2: \(2x - 3 = -5\)
\(2x = -2 \Rightarrow x = -1\) - Check: \(|2(4)-3|=5\) ✓ \(|2(-1)-3|=5\) ✓
- Answer: \(x = 4\) or \(x = -1\)
GRAPHICAL INTERPRETATION
Draw \(y=|2x-3|\) and \(y=5\). Intersections give the solutions. The horizontal line cuts the V twice → two solutions.
For \(-|f(x)| = k\) where k > 0 → no solutions since −|f(x)| ≤ 0 always.
For \(-|f(x)| = k\) where k > 0 → no solutions since −|f(x)| ≤ 0 always.
Graph Sketching
High mark topic — f(x) · |f(x)| · f(|x|) · −f(x) · transformations
The four key graph types from f(x)
y = f(x) — the original
Draw as given. This is your reference — all other transformations start here.
y = |f(x)| — reflect below x-axis up
Any part of f(x) below the x-axis gets reflected upward. Parts above stay the same. No graph below x-axis.
y = f(|x|) — reflect right half to left
Keep the right half (x ≥ 0) of f(x). Reflect it to the left. The result is always symmetric about the y-axis.
y = −f(x) — reflect in x-axis
Flip the entire graph upside down. Every y-value becomes −y. Maxima become minima and vice versa.
Interactive — see all four transformations live
y = f(x)
y = |f(x)| — reflect below up
y = f(|x|) — reflect right to left
y = −f(x) — flip in x-axis
Exam technique — how to sketch for marks
- Always sketch f(x) first — even if not asked
- Label all intercepts with coordinates — x-intercepts and y-intercept
- Label turning points / vertex if present
- Show asymptotes as dashed lines and label them
- For |f(x)|: mark where the reflection happens (where f(x) = 0)
- For f(|x|): show the y-axis as a line of symmetry
COMMON EXAM MISTAKES
✗ Forgetting to reflect — just copying f(x)
✗ Wrong reflection for f(|x|) — some reflect left to right instead
✗ Missing coordinates on axes
✗ Not showing cusp/corner at the reflection point for |f(x)|
✗ Wrong reflection for f(|x|) — some reflect left to right instead
✗ Missing coordinates on axes
✗ Not showing cusp/corner at the reflection point for |f(x)|
MARK ALLOCATION GUIDE
Correct shape — 1 mark
Correct position/translation — 1 mark
Key coordinates labelled — 1–2 marks
Asymptotes correct — 1 mark
Correct position/translation — 1 mark
Key coordinates labelled — 1–2 marks
Asymptotes correct — 1 mark
Practice Questions
3 Easy · 3 Medium · 3 Hard — Cambridge 0606 Functions
Easy 1
[2]
The function \(f(x) = 2x - 5\). Find \(f(3)\) and \(f^{-1}(x)\).
SOLUTION
f(3)
\(f(3)=2(3)-5=\boldsymbol{1}\)
Inverse
Let \(y=2x-5 \Rightarrow x=\frac{y+5}{2}\)
\(\boldsymbol{f^{-1}(x)=\dfrac{x+5}{2}}\)
\(\boldsymbol{f^{-1}(x)=\dfrac{x+5}{2}}\)
Easy 2
[2]
Given \(f:x\mapsto x^2+1\) for \(x\geq0\), state the range of \(f\).
SOLUTION
Minimum
When \(x=0\): \(f(0)=0+1=1\) (minimum value)
Range
\(x\geq0\) means \(x^2\geq0\), so \(f(x)\geq1\)
\(\boldsymbol{f(x)\geq1}\) or \(\boldsymbol{[1,\infty)}\)
\(\boldsymbol{f(x)\geq1}\) or \(\boldsymbol{[1,\infty)}\)
Easy 3
[2]
\(f(x)=3x+1\) and \(g(x)=x^2\). Find \(fg(x)\).
SOLUTION
Composite
\(fg(x)=f(g(x))=f(x^2)=3x^2+1\)
\(\boldsymbol{fg(x)=3x^2+1}\)
\(\boldsymbol{fg(x)=3x^2+1}\)
Remember: in \(fg(x)\), apply \(g\) FIRST, then \(f\). Order matters!
Medium 1
[4]
\(f(x)=2x^2-3\) for \(x\geq0\). Find \(f^{-1}(x)\) and state its domain.
STEP-BY-STEP SOLUTION
Step 1
Let \(y=2x^2-3\). Solve for \(x\):
\(x^2=\dfrac{y+3}{2} \Rightarrow x=\sqrt{\dfrac{y+3}{2}}\) (positive root since \(x\geq0\))
\(x^2=\dfrac{y+3}{2} \Rightarrow x=\sqrt{\dfrac{y+3}{2}}\) (positive root since \(x\geq0\))
Inverse
\(\boldsymbol{f^{-1}(x)=\sqrt{\dfrac{x+3}{2}}}\)
Domain
Domain of \(f^{-1}\) = range of \(f\).
When \(x\geq0\): \(f(x)=2x^2-3\geq-3\)
Domain of \(f^{-1}\): \(\boldsymbol{x\geq-3}\)
When \(x\geq0\): \(f(x)=2x^2-3\geq-3\)
Domain of \(f^{-1}\): \(\boldsymbol{x\geq-3}\)
Medium 2
[4]
\(f(x)=\dfrac{2x+1}{x-3}\). Find \(f^{-1}(x)\) and state the value of \(x\) excluded from the domain of \(f\).
STEP-BY-STEP SOLUTION
Domain of f
Denominator \(\neq0\): \(x-3\neq0 \Rightarrow \boldsymbol{x\neq3}\)
Find inverse
\(y=\dfrac{2x+1}{x-3} \Rightarrow y(x-3)=2x+1\)
\(xy-3y=2x+1 \Rightarrow xy-2x=3y+1\)
\(x(y-2)=3y+1 \Rightarrow x=\dfrac{3y+1}{y-2}\)
\(xy-3y=2x+1 \Rightarrow xy-2x=3y+1\)
\(x(y-2)=3y+1 \Rightarrow x=\dfrac{3y+1}{y-2}\)
Answer
\(\boldsymbol{f^{-1}(x)=\dfrac{3x+1}{x-2}}\); domain excludes \(x\neq2\)
Medium 3
[4]
\(g(x)=x^2-4x+7\) for \(x\geq2\). Express \(g\) in completed square form. Hence find \(g^{-1}(x)\).
STEP-BY-STEP SOLUTION
Complete square
\(g(x)=(x-2)^2+3\)
Find inverse
\(y=(x-2)^2+3 \Rightarrow (x-2)^2=y-3\)
\(x-2=\sqrt{y-3}\) (positive root since \(x\geq2\))
\(\boldsymbol{g^{-1}(x)=2+\sqrt{x-3}}\)
\(x-2=\sqrt{y-3}\) (positive root since \(x\geq2\))
\(\boldsymbol{g^{-1}(x)=2+\sqrt{x-3}}\)
Domain
Range of \(g\) when \(x\geq2\): min at \(x=2\), \(g(2)=3\)
Domain of \(g^{-1}\): \(\boldsymbol{x\geq3}\)
Domain of \(g^{-1}\): \(\boldsymbol{x\geq3}\)
Hard 1
[5]
\(f(x)=x^2-2x\) for \(x\geq1\) and \(g(x)=\sqrt{x+2}\) for \(x\geq-2\). Find \(gf(x)\) and state its domain and range.
STEP-BY-STEP SOLUTION
Composite
\(gf(x)=g(x^2-2x)=\sqrt{x^2-2x+2}=\sqrt{(x-1)^2+1}\)
Domain
Domain of \(f\) is \(x\geq1\). Check \(f(x)\geq-2\): \(x^2-2x\geq-2\), i.e. \((x-1)^2\geq-1\) always true.
Domain of \(gf\): \(\boldsymbol{x\geq1}\)
Domain of \(gf\): \(\boldsymbol{x\geq1}\)
Range
At \(x=1\): \((0)^2+1=1\), so min value under root is 1.
Range: \(\boldsymbol{gf(x)\geq1}\)
Range: \(\boldsymbol{gf(x)\geq1}\)
Hard 2
[5]
The function \(h(x) = \dfrac{ax+b}{x+c}\) is such that \(h(0)=2\), \(h(1)=3\), and \(h(h(x))=x\) for all valid \(x\). Find \(a\), \(b\), \(c\).
STEP-BY-STEP SOLUTION
h(0)=2
\(\dfrac{b}{c}=2 \Rightarrow b=2c\) ...[1]
h(1)=3
\(\dfrac{a+b}{1+c}=3 \Rightarrow a+b=3+3c\) ...[2]
Self-inverse
\(h(h(x))=x\) means \(h=h^{-1}\). For this: \(a+c=0 \Rightarrow a=-c\) ...[3]
Solve
From [1]: \(b=2c\). From [3]: \(a=-c\).
Sub into [2]: \(-c+2c=3+3c \Rightarrow c=-3\)
\(a=3,\; b=-6,\; c=-3\)
Sub into [2]: \(-c+2c=3+3c \Rightarrow c=-3\)
\(a=3,\; b=-6,\; c=-3\)
Answer
\(\boldsymbol{a=3,\; b=-6,\; c=-3}\)
Hard 3
[5]
\(f:x\mapsto 3x-k\) and \(g:x\mapsto \dfrac{x}{x-2}\) where \(x\neq2\). Given that \(gf(2)=4\), find \(k\). Hence find \(f^{-1}g^{-1}(x)\).
STEP-BY-STEP SOLUTION
gf(2)=4
\(f(2)=6-k\). Then \(g(6-k)=\dfrac{6-k}{6-k-2}=\dfrac{6-k}{4-k}=4\)
\(6-k=4(4-k) \Rightarrow 6-k=16-4k \Rightarrow 3k=10 \Rightarrow \boldsymbol{k=\frac{10}{3}}\)
\(6-k=4(4-k) \Rightarrow 6-k=16-4k \Rightarrow 3k=10 \Rightarrow \boldsymbol{k=\frac{10}{3}}\)
f⊃-1;
\(f^{-1}(x)=\dfrac{x+k}{3}=\dfrac{x+\frac{10}{3}}{3}=\dfrac{3x+10}{9}\)
g⊃-1;
\(y=\dfrac{x}{x-2}\Rightarrow yx-2y=x \Rightarrow x=\dfrac{2y}{y-1}\), so \(g^{-1}(x)=\dfrac{2x}{x-1}\)
f⊃-1;g⊃-1;(x)
\(f^{-1}g^{-1}(x)=f^{-1}\!\left(\dfrac{2x}{x-1}\right)=\dfrac{3\cdot\frac{2x}{x-1}+10}{9}=\boldsymbol{\dfrac{6x+10(x-1)}{9(x-1)}=\dfrac{16x-10}{9(x-1)}}\)
Past Year Paper Questions
Cambridge IGCSE 0606 style — Functions
0606 Style
Composite & Inverse
[11]
Functions \(f\) and \(g\) are defined for \(x\in\mathbb{R}\) by \(f(x)=2x+3\) and \(g(x)=x^2-1\).
(a) Find \(fg(x)\) and \(gf(x)\). [3]
(b) Solve \(fg(x)=gf(x)\). [3]
(c) Find \(f^{-1}(x)\). [2]
(d) Solve \(f^{-1}f(x^2-1)=3\). [3]
(a) Find \(fg(x)\) and \(gf(x)\). [3]
(b) Solve \(fg(x)=gf(x)\). [3]
(c) Find \(f^{-1}(x)\). [2]
(d) Solve \(f^{-1}f(x^2-1)=3\). [3]
FULL WORKED SOLUTION
(a)
\(fg(x)=2(x^2-1)+3=2x^2+1\)
\(gf(x)=(2x+3)^2-1=4x^2+12x+8\)
\(gf(x)=(2x+3)^2-1=4x^2+12x+8\)
(b)
\(2x^2+1=4x^2+12x+8\)
\(2x^2+12x+7=0\)
\(x=\dfrac{-12\pm\sqrt{144-56}}{4}=\dfrac{-12\pm\sqrt{88}}{4}=\dfrac{-6\pm\sqrt{22}}{2}\)
\(2x^2+12x+7=0\)
\(x=\dfrac{-12\pm\sqrt{144-56}}{4}=\dfrac{-12\pm\sqrt{88}}{4}=\dfrac{-6\pm\sqrt{22}}{2}\)
(c)
\(y=2x+3 \Rightarrow x=\dfrac{y-3}{2}\)
\(\boldsymbol{f^{-1}(x)=\dfrac{x-3}{2}}\)
\(\boldsymbol{f^{-1}(x)=\dfrac{x-3}{2}}\)
(d)
\(f^{-1}f(x^2-1)=x^2-1\) (they cancel)
\(x^2-1=3 \Rightarrow x^2=4 \Rightarrow \boldsymbol{x=\pm2}\)
\(x^2-1=3 \Rightarrow x^2=4 \Rightarrow \boldsymbol{x=\pm2}\)
0606 Style
Restricted Domain & Range
[9]
\(h(x)=3-2(x-1)^2\) for \(x\leq1\).
(a) State the range of \(h\). [2]
(b) Find \(h^{-1}(x)\) and state its domain. [4]
(c) On the same axes sketch \(y=h(x)\) and \(y=h^{-1}(x)\), showing the line of symmetry. [3]
(a) State the range of \(h\). [2]
(b) Find \(h^{-1}(x)\) and state its domain. [4]
(c) On the same axes sketch \(y=h(x)\) and \(y=h^{-1}(x)\), showing the line of symmetry. [3]
FULL WORKED SOLUTION
(a) Range
At \(x=1\): \(h(1)=3\) (maximum). As \(x\to-\infty\), \(h\to-\infty\).
Range: \(\boldsymbol{h\leq3}\)
Range: \(\boldsymbol{h\leq3}\)
(b) Inverse
\(y=3-2(x-1)^2\)
\(2(x-1)^2=3-y\)
\((x-1)^2=\dfrac{3-y}{2}\)
\(x-1=-\sqrt{\dfrac{3-y}{2}}\) (negative root since \(x\leq1\))
\(\boldsymbol{h^{-1}(x)=1-\sqrt{\dfrac{3-x}{2}}}\)
Domain: \(x\leq3\)
\(2(x-1)^2=3-y\)
\((x-1)^2=\dfrac{3-y}{2}\)
\(x-1=-\sqrt{\dfrac{3-y}{2}}\) (negative root since \(x\leq1\))
\(\boldsymbol{h^{-1}(x)=1-\sqrt{\dfrac{3-x}{2}}}\)
Domain: \(x\leq3\)
(c) Symmetry
\(y=h(x)\) and \(y=h^{-1}(x)\) are reflections in the line \(\boldsymbol{y=x}\).