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Mathematics I · Week 2 — Functions and domains
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Mathematics I · Week 2 — Functions and domains
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10 min read
2026-07-19
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Week 2 — functions and domains
Start with the input set. A correct rule applied to a forbidden input is still not a valid function value. This chapter is about keeping the mapping, the allowed inputs, and the outputs actually attained separate.
Before any calculation
Write the domain. Then ask whether each allowed input gets exactly one output. Only after that should you talk about range, composition, inverse, or graph shape.
Learning objectives
By the end of this week, you should be able to:
- Distinguish a function’s domain, codomain, and range without treating them as interchangeable.
- Decide whether a relation is a function, injective, surjective, or bijective on its stated sets.
- Derive the real-valued domain of a formula before simplifying it.
- Check the intermediate permission condition in a composition.
- Test whether a proposed inverse is genuinely a function with the right domain.
Prerequisites
- Set notation, especially membership (
x ∈ A) and subset (A ⊆ B) language from Week 1. - Ordered pairs and coordinate reading; a function’s graph is a set of permitted input-output pairs.
- Basic algebraic rearrangement, factoring, square roots, and logarithm notation.
Core content
What makes a function?
A function assigns exactly one output to every input in its domain. The codomain is the declared target set; the range is the set of outputs actually reached. The distinction is useful: a function can have codomain
ℝ while its range is only the non-negative real numbers.A relation is not a function if an allowed input has two distinct outputs. Different inputs are allowed to share an output; that simply means the function is not one-to-one.
| Question | Test | Non-example to avoid |
|---|---|---|
| Is it a function? | Each domain input has one output. | One input mapped to two outputs. |
| Is it injective? | Different inputs never share an output. | Calling x² injective on all reals. |
| Is it onto? | Every codomain value is hit. | Confusing codomain with range. |
| Is an inverse function possible? | It must be injective on the stated domain. | Swapping x and y without checking. |
Build the mapping before naming it
Use the mapper as a short diagnostic, not a game of making every status badge green. First build a relation that fails to be a function by sending one input to two outputs. Then clear it and build a function which is not one-to-one. Finally, make a bijection. After every version, say which exact definition changed the verdict.
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Domain, codomain, range—three different jobs
The domain is the set you must account for. The codomain is the set you promised to aim at. The range is the outputs you actually reached. In the mapper, leave one target unused to see a function that is not onto.
Domain first
For a formula, find restrictions before simplification:
- Denominators cannot be zero.
- Even roots over the real numbers need a non-negative radicand.
- Logarithms need a positive argument.
- A stated context may impose further restrictions.
For
f(x) = (x² - 9)/(x - 3), factoring produces x + 3 only for x ≠ 3. The original function has a hole at 3; cancellation does not repair it. This is a non-example of treating an algebraically similar expression as the same function on every real input.Original worked example — the rule is not the whole function
Let
g(x) = √(x-2). Over the real numbers, x-2 must be non-negative, so the domain is x ≥ 2. Square roots output non-negative values, so the range is y ≥ 0. The codomain can be declared as ℝ, but that does not make negative outputs appear. Write each statement separately: domain tells you what may enter; range tells you what actually leaves.A restriction decision table
When a formula contains more than one sensitive operation, collect every condition and keep their intersection. “Or” is usually wrong here: an input must survive every operation in the expression.
| Feature in the rule | Local condition | Example | What to record |
|---|---|---|---|
| Denominator | denominator ≠ 0 | 1/(x-4) | x ≠ 4 |
| Even root | radicand ≥ 0 | √(5-x) | x ≤ 5 |
| Logarithm | argument > 0 | ln(x+1) | x > -1 |
| Composition | inner output belongs to outer domain | √(1/(x-2)) | first x ≠ 2, then 1/(x-2) ≥ 0 |
For the final row,
1/(x-2) is non-negative exactly when x > 2; it can never equal zero. So the composite rule has domain (2,∞). The important part is the order: identify the inner expression, then ask whether the outer operation accepts its output.A domain restriction is not a visual preference
A graph can look smooth near a forbidden input if the screen resolution hides a hole. The original rule, not the apparent drawing, decides whether the input is allowed.
Function taxonomy
- Injective / one-to-one: no two distinct domain inputs share an output.
- Surjective / onto: every codomain value is reached.
- Bijective: both; an inverse function exists between the stated sets.
- Composite:
(f ∘ g)(x) = f(g(x)); outputs ofgmust be valid inputs forf.
An inverse relation swaps input and output. It is an inverse function only when the original mapping is injective over its domain. Restricting a domain can sometimes make this possible.
For example,
f(x)=x² on all real numbers is not injective because f(2)=f(-2). If its domain is restricted to x ≥ 0, it becomes injective and has inverse √x on its range. The rule did not change; the declared domain changed the mapping’s properties.Graphs and transformations
A graph represents permitted input-output pairs. The vertical-line test diagnoses whether a graph is a function. The horizontal-line test is a graphical clue for injectivity, but an algebraic domain statement still matters.
For transformations, keep track of what changes first:
f(x-a) shifts an input before the rule acts, while f(x) + a changes the output afterwards. A small table of inputs and outputs prevents sign-based guessing.Composition: a two-stage permission check
(f ∘ g)(x) means first apply g, then feed the result into f. It exists only if the output of g is a permitted input for f. Do not settle for “both formulas exist”; check the middle value.- Start with the domain of
g. - Keep only inputs whose
g(x)output is in the domain off. - Apply the two rules in order.
Original worked example — when a composition loses inputs
Let
f(u)=√u and g(x)=x-5. Both rules are meaningful by themselves, but (f∘g)(x)=√(x-5) needs g(x)≥0. Thus the composition is defined only for x≥5.Now reverse the order:
(g∘f)(x)=√x-5. This time the square root acts first, so the only restriction is x≥0. Composition is therefore not commutative: the two expressions use the same rules but admit different input sets and usually produce different outputs.Make the middle stage explicit in a small table before simplifying:
| x | g(x)=x-5 | Can f accept it? | (f∘g)(x) |
|---|---|---|---|
| 3 | -2 | no: a real square root needs non-negative input | undefined over ℝ |
| 5 | 0 | yes | 0 |
| 9 | 4 | yes | 2 |
The word undefined here does not mean “the answer is zero” or “skip the question.” It means the chosen rule and domain do not produce a real output for that input.
Inverses: reverse the mapping, then verify it
An inverse is best treated as a two-way check, not a memorised recipe. If
f:A→B is bijective, its inverse f⁻¹:B→A satisfies both f⁻¹(f(x))=x for allowed inputs in A and f(f⁻¹(y))=y for allowed outputs in B.For
h(x)=3x-7 on ℝ, solve y=3x-7 for x: x=(y+7)/3. Therefore h⁻¹(y)=(y+7)/3. Check a value in both directions: h(5)=8 and h⁻¹(8)=5; also h(h⁻¹(8))=8. A check catches sign slips that symbolic rearrangement can hide.For
q(x)=x² on all reals, solving y=x² gives x=±√y. That ± is evidence that the inverse relation gives two possible outputs, so it is not an inverse function. Restricting q to x≥0 removes the negative branch and makes q⁻¹(y)=√y a function with domain y≥0.A fast inverse audit
Ask three questions: Was the original function one-to-one on its stated domain? Does the proposed inverse accept precisely the original range? Does composing in both permitted directions return the input? If any answer is no, do not call it an inverse function yet.
Key formulas and translation sheet
| Idea | Compact form | Read it as |
|---|---|---|
| Function rule | f : A → B | Every input in A has exactly one output in B. |
| Range | f(A) | The outputs actually attained; it may be smaller than B. |
| Composition | (f ∘ g)(x) = f(g(x)) | Apply g first; its output must be allowed by f. |
| Inverse check | f⁻¹(f(x)) = x | Reverse the mapping on the original domain. |
| Other inverse check | f(f⁻¹(y)) = y | Return every permitted output to itself. |
| Rational restriction | denominator ≠ 0 | Record exclusions before cancelling factors. |
| Root / logarithm restriction | radicand ≥ 0; log argument > 0 | Keep only real inputs accepted by the operation. |
Common pitfalls
Codomain is not automatically the range
The codomain is a declared target set. The range is evidence: list or reason about the outputs that actually occur. A function into
ℝ may still never produce a negative number.Cancellation does not refill a hole
Simplifying
(x² - 9)/(x - 3) gives a matching rule away from x = 3; it does not add 3 back to the original domain. State the exclusion first, then simplify.An inverse needs one output per input
Solving an equation is not enough. If one output traces back to two possible inputs, the inverse relation is not an inverse function until the original domain is restricted.
Composition has a middle gate
Both individual formulas can be valid while the composition rejects some inputs. Track
g(x) explicitly, then ask whether that value belongs to the domain of f.A graph cannot override a labelled domain
A tiny hole can disappear on a rough graph. Trust the original rule and stated domain over the visual approximation.
Retrieval ladder
- Explain the difference between codomain and range using a function whose outputs are never negative.
- State the domain of a rational expression and name the exact restriction causing it.
- Give one reason
x²needs a restricted domain before it has an inverse function. - Write a two-step composition and identify the intermediate permission check.
- Build a mapping with every domain element used exactly once but one codomain element unused. Name every property it has and every property it lacks.
Five-minute exit check
You are ready when you can say: a function is a rule plus a domain; range is attained outputs, not merely the declared target; and cancellation never restores a forbidden input.
Retrieval practice
Define a function with a denominator and state three things separately: its domain, one output that occurs, and one input that is excluded. Then explain, in one sentence, why a relation with two outputs for the same input fails the function rule.