H2 Maths Functions Formula Sheet | Domain, Range & Inverses
H2 Maths Functions Formula Sheet | Domain, Range & Inverses
Study guide/
H2 Maths functions formula sheet: domain and range rules, the one-one test, conditions for inverses to exist, domain and range swap under inversion, the reflection property of i...
Q: What does H2 Maths Notes (JC 1-2): 1.1) Functions cover? A: Domains, ranges, inverse checks, and composite workflows for Section A Topic 1.1 of the 2026 H2 Maths syllabus.
Before you revise Keep a running list of domain restrictions (denominators, even roots, logarithms). Test every claim with quick substitutions or GC trace-functions questions are marked on precision.
A function needs valid inputs and outputs: State the domain.
Inverses need one-to-one behaviour: Restrict the domain before finding the inverse.
Composite functions carry restrictions through each layer: Check the inner function's range against the outer function's domain.
Status: SEAB's current H2 Mathematics (9758) syllabus PDF is labelled for 2026. Topic 1.1 expectations are within Section A Pure Mathematics, which is assessed in Paper 1 (100 marks) and Paper 2 Section A (40 marks).
Formulas at a glance
Every key result, condition, and rule from this topic, on one screen. None of these definitions appear in MF27, so commit them to memory. Worked examples for each appear in the sections below.
Definitions and domain rules
Concept
Rule
Function
f:A→B assigns each x∈Df to exactly one output f(x).
Set of all outputs actually reached on Df. Use graph shape, monotonicity, or completed-square form.
One-one (injectivity) test
Method
Condition
What to write
Algebraic
Assume f(x1)=f(x2) and deduce x1=x2.
State the conclusion explicitly.
Derivative
f′(x)>0 or f′(x)<0 throughout Df
Horizontal-line test
No horizontal line meets the graph more than once on Df.
Use as support; pair with an algebraic or derivative argument.
Inverse functions
Concept
Rule
Existence condition
f−1 exists if and only if f is one-one on Df.
Domain swap
Df−1=Rf (over the restricted domain).
Range swap
Rf−1=Df (over the restricted domain).
Graph relationship
The graph of y=f−1(x) is the reflection of y=f(x) in the line y=x
Verification
f−1(f(x))=x for all x∈Df
Composite functions
Concept
Rule
Existence condition
fg exists if and only if Rg⊆Df.
Order of composition
(f∘g)(x)=f(g(x)): apply g first, then f.
Domain of fg
Dfg=Dg restricted so that g(x)∈Df
Inverse-order rule
(f∘g)−1=g−1∘f−1
Function question route map
Use the first cue in the question to choose the workflow before doing algebra. Most errors in functions questions come from finding a formula first and checking the domain too late.
Question cue
What to decide first
Working route
"State the domain"
Which inputs are illegal?
Check denominators, even roots, logarithms, then combine the restrictions.
"Find the range"
Which outputs can the function actually reach?
Use graph shape, monotonicity, completed square form, or a boundary-value table.
"Find the inverse"
Is the given domain one-to-one?
Restrict or verify the domain, rearrange, then swap the domain and range.
"Find a composite"
Does the inner output fit the outer input?
Find the inner function's range first, then apply the outer function only where valid.
"Solve an equation involving functions"
Which layer can be undone first?
Let the inner expression be a new variable, solve the outer equation, then back-substitute.
Trap check: If a question gives a restricted domain, use that restricted domain for the range and inverse. Do not silently switch back to the natural domain halfway through the solution.
Core Concepts
A function f:A→B assigns each x∈A to one output f(x)∈B; A is the domain, f(A) is the range.
Surjective (onto) functions cover the whole codomain; bijective functions are both injective and surjective and therefore invertible.
Composition (f∘g)(x)=f(g(x)) is only defined when g(x) lies inside the domain of f
Inverses reverse the mapping: f−1(f(x))=x for all x in the restricted domain.
Domain and Range Workflow
Determine the natural domain by eliminating forbidden inputs:
Even roots: require the radicand ≥0.
Logarithms: argument strictly >0.
Rational functions: denominator =0.
Composite functions: propagate restrictions through every stage.
State the range by analysing turning points, asymptotes, or monotonic intervals.
Restrict the domain if necessary to make f one-to-one before finding the inverse.
Example -- Square root domain
For f(x)=3x−5, require 3x−5≥0 so x≥35. The range is [0,∞).
Domain-range transfer checkpoint
For inverse questions, write the four sets in this order before rearranging the formula: original domain, original range, inverse domain, inverse range. This prevents a common mistake where the natural domain is used after the question has already restricted the function.
Item to state
Comes from
What to check
Domain of f
The question or the natural-domain restrictions
Denominators, even roots, logarithms, and any given interval.
Range of f
Outputs reached only on that domain
Boundary values, turning points, asymptotes, and monotonic branch chosen.
Domain of f−1
Range of the restricted f
Do not use the range from the unrestricted graph.
Range of f−1
Domain of the restricted f
Carry over the same endpoint inclusions.
Worked check: suppose f(x)=x2−4x+1 and the question restricts x≥2. Completing the square gives f(x)=(x−2)2−3, so the range of the restricted function is y≥−3. Therefore the domain of f−1 is x≥−3, and the range of f−1 is y≥2.
Misconception check: "swap x and y" is not the first step. First decide which branch of the original function is being inverted, then swap the restricted domain and range.
Injectivity proof checkpoint
Before finding an inverse, decide how you will justify that the function is one-to-one on the stated domain. Choose the shortest proof that matches the information given.
Given information
Fast proof route
What to write
A simple algebraic formula
Start from f(a)=f(b).
Show the equation forces a=b, so the function is one-to-one.
A differentiable function on an interval
Check the sign of f′(x).
If f′(x)>0 throughout the interval, f is strictly increasing; if f′(x)<0
A quadratic or modulus graph
Restrict to one branch first.
State the branch, then use monotonicity or a completed-square form to prove one-to-one behaviour.
A graph is supplied
Use the horizontal-line test as support.
Say no horizontal line cuts the restricted graph more than once, then state the domain used.
Worked check: for f(x)=x2−6x+5, completing the square gives f(x)=(x−3)2−4. On x≥3, the branch is increasing, so the function is one-to-one and an inverse can be found on that restricted domain. On all real x, it is not one-to-one because values on either side of x=3 repeat.
Misconception check: "the graph has a turning point" does not automatically mean the inverse is impossible. It means you must restrict to one side of the turning point before inverting.
Inverse Functions
Use algebraic manipulation to isolate x in y=f(x); then interchange x and y to express f−1(x).
Confirm f−1(f(x))=x on the chosen domain and state both domain and range for the inverse explicitly.
For quadratics, complete the square to reveal monotonic branches.
Example -- Quadratic inverse
Consider f(x)=2x2−3x−1.
Stationary point from f′(x)=4x−3=0 gives x=43.
Restrict domain to x≥43 so f is strictly increasing.
Complete the square: y=2(x−43)2−817
Swap variables: x=2(y−43)2−817
The inverse domain is x≥−817 and range y≥43
Composite Functions
Record intermediate ranges meticulously: if g:A→B and f:C→D, composition requires g(A)⊆C.
When solving (f∘g)(x)=k, set g(x) equal to an auxiliary variable first, solve f(u)=k, then back-substitute.
Use the identity (f∘g)−1=g−1∘f−1 only after verifying both inverses exist on the relevant domains.
Composite-domain checkpoint
For a composite function, the inner function is checked first, but the final domain must satisfy both the inner function and the outer function.
Task
First check
Then check
Common trap
Find (f∘g)(x)
Inputs must be valid for g.
Outputs from g must be valid inputs for f.
Only using the natural domain of the simplified final expression.
Find (g∘f)(x)
Inputs must be valid for f.
Outputs from f must be valid inputs for g.
Solve (f∘g)(x)=k
Solve f(u)=k for possible inner values u.
Use (f∘g)−1
Check that each inverse exists on the stated domain.
Reverse the order only after the restrictions are carried through.
Applying the inverse-order rule without one-to-one checks.
Worked check: let f(x)=x−2 and g(x)=x2−1. For (f∘g)(x), first g accepts all real x, then f requires g(x)≥2. So x2−1≥2, giving x≤−3 or x≥3. The expression x2−3 alone does not replace the domain check; it confirms the same restriction.
Misconception check: simplifying the algebra can hide the route taken by the composite. Always record the inner-output restriction before finalising the domain.
Example -- Composition with logarithms
Let f(x)=ln(x−1) and g(x)=x2+4.
Natural domain of g is all reals, range [4,∞).
f requires input >1, so g(x)>1 holds for all x.
Composition (f∘g)(x)=ln(x2+3).
Inverse sequence: f−1(x)=1+ex, g−1(x)=±x−4
Modelling with Functions
When modelling real data, specify domain in context (e.g. weeks, temperatures).
Want weekly guided practice on Functions? Our H2 Maths tuition programme builds fluency in this topic through structured problem sets and exam-style drills.
Common exam mistakes
Giving the range of f as the domain of f−1 without checking: The domain of f−1 equals the range of f, but only over the restricted domain. If you have restricted f to make it one-to-one, compute the range of the restricted f before stating the domain of f−1.
Attempting composition fg when the range of g is not a subset of the domain of f: The composition f∘g only exists when every output of g is a valid input for f
Restricting the domain to make f injective without verifying injectivity: Simply stating "restrict to x≥0" is insufficient if f is still not one-to-one on that interval. Always verify the horizontal-line test or use the derivative to confirm monotonicity.
Swapping x and y without keeping track of variable meaning: When deriving f−1, many students swap x↔y mid-working and lose track of which variable is the input. Write y=f(x)
Omitting the domain and range when stating the inverse function: A complete answer for an inverse function must state both f−1(x)=… and explicitly give the domain of f−1 (which equals the range of f). Omitting either loses a mark.
Frequently asked questions
Is there a formula sheet for H2 Maths functions? Yes - the "Formulas at a glance" section near the top of this page collects every key result you need: the definition of a function, natural domain rules, the one-one test (algebraic, derivative, and horizontal-line methods), the existence condition for inverses, the domain-range swap under inversion, the reflection property of inverse graphs, the composite-existence condition Rg⊆Df, the order of composition, and the domain of a composite. None of these appear in MF27, so commit them to memory.
Is Topic 1.1 (Functions) in Paper 1 or Paper 2? Topic 1.1 is Pure Mathematics and can appear in Paper 1 (100 marks) or Paper 2 Section A (40 marks). Functions questions are often in the early parts of structured questions and test precision with domain notation.
Do I need to prove injectivity algebraically, or is a GC sketch sufficient? The GC can support your reasoning but is not a substitute for an algebraic or derivative-based proof. To prove f is one-to-one, either show f(x1)=f(x2)⇒x1=x2 algebraically, or show f′(x)>0 (or <0) on the domain to establish strict monotonicity.
What is the difference between "domain" and "natural domain"? The natural domain is the largest set of real numbers for which f(x) is defined (no division by zero, no negative square roots, etc.). A restricted domain is a subset chosen to make f injective (for finding an inverse). Always check whether the question specifies a domain or asks you to state the natural domain.
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