- Identify the function’s full domain and codomain, including whether zero or negatives are permitted.
- Substitute x=0,y=0,x=y and y=−x where those inputs are legal.
- Swap x and y and compare the two equations when they both apply.
- Try a simple candidate, then verify it for arbitrary inputs; do not confuse existence with uniqueness.
- Check any continuity, monotonicity, boundedness or other regularity assumptions before classifying all solutions.
Independent practice
Try the following: prove f(0)=0 when f is additive on the reals; derive f(−x)=−f(x); if f:ℤ→ℝ is additive with f(1)=7, find f(−3); give two different functions satisfying f(x+2)=f(x); and show that f(x)=x² solves f(x+y)=f(x)+f(y)+2xy.
The answers are obtained by substituting zero, using x+(−x)=0, finding f(−3)=−21, choosing for example f(x)=0 and f(x)=sin(πx), and expanding (x+y)². Every step should explicitly mention which domain or property makes it legal.
Frequently asked questions
Are functional equations a separately named 2027 H3 syllabus topic?
The 2027 H3 Mathematics syllabus 9820 includes mathematical proof, non-routine problems and unfamiliar texts. This functional-equations guide is enrichment illustrating those skills, not a claim that functional equations form a separately named syllabus unit.
Does finding one function solve the question?
Only if the task asks for an example or existence. If it asks for every solution, you must prove the classification and address any exceptional cases or regularity conditions.
Should JC1 learners first master H2 functions?
Yes. Domains, inverse functions, transformations and algebraic identities form the foundation. These exercises should enrich rather than replace current H2 study.
Official resources and related Mathematics learning
Use the SEAB 2027 A-Level syllabus directory, definitions and theorems and invented operations for related reasoning.
The unchanged eduKateSG three-student Mathematics reference describes the tutorial approach near Sixth Avenue MRT. It does not establish current JC H3 specialist availability. Enquire directly about tutor expertise, timetables, fees and suitable classes.
Did you know? The first good substitution often reveals more about a functional equation than the first clever guess. A clear domain check tells you which substitutions you are actually allowed to make.
Further substitutions, branches and regularity conditions
Extended example 1: Zero and the exponential functional equation
Let f:ℝ→ℝ satisfy f(x+y)=f(x)f(y). Set y=0: f(x)=f(x)f(0). If some f(x) is nonzero, then f(0)=1. Otherwise f is identically zero. The constant zero and constant one functions both satisfy the equation. Separating these branches is required before dividing by f(0) or claiming that a solution is exponential.
Extended example 2: Negatives in the exponential equation
If f(x+y)=f(x)f(y) and f is not the zero function, then f(0)=1. Setting y=−x gives f(x)f(−x)=1 for every real x. Thus no function value can equal zero, and f(−x)=1/f(x). This conclusion follows from the equation; it is not a guess about what the function graph should look like.
Extended example 3: A familiar exponential candidate
The function f(x)=2ˣ satisfies f(x+y)=f(x)f(y) because 2ˣ⁺ʸ=2ˣ2ʸ. The same holds for f(x)=eˣ and f(x)=1. These demonstrate multiple solutions. If a problem asks for all solutions, extra assumptions and a classification proof are needed. An example, however convincing, is not a complete general solution.
Extended example 4: Continuity narrows the exponential family
For a positive-valued function f:ℝ→(0,∞) satisfying f(x+y)=f(x)f(y), one may take logarithms to define g(x)=ln f(x). Then g is additive. If f is continuous, so is g, making g(x)=cx and f(x)=eᶜˣ. The positivity and continuity assumptions are indispensable to this classification; they are not decorations.
Extended example 5: A multiplicative equation forces a one-value condition
For f(xy)=f(x)f(y) on real numbers, take y=1. If f is not identically zero, some f(x) is nonzero, so f(1)=1. For a nonzero input x, f(x)f(1/x)=f(1)=1, so f(x) cannot vanish. This is a strong structural consequence from only two simple substitutions.
Extended example 6: Negative one under a multiplicative equation
If nonzero f satisfies f(xy)=f(x)f(y) for all real x,y, then f(−1)²=f(1)=1, so f(−1)=1 or f(−1)=−1. For any x, f(−x)=f(−1)f(x). Thus the behaviour under changing sign is governed by one of two cases. The original domain must include −1 for this step.
Extended example 7: An odd part can be isolated
Suppose f(x)=f(−x)+2x for every real x. Set h(x)=f(x)−x. Then h(−x)=f(−x)+x=f(x)−x=h(x), so h is even. Thus all such solutions are f(x)=x+h(x), where h is any even function. A general characterisation needs this freedom, not only the example f(x)=x.
Extended example 8: A fixed reflected sum
If f(x)+f(−x)=4, setting x=0 yields 2f(0)=4, so f(0)=2. Write h(x)=f(x)−2; then h(x)+h(−x)=0, meaning h is odd. Every function f(x)=2+h(x) with odd h satisfies the relation. The equation determines the even part without fixing every output.
Extended example 9: An impossible reflected equation
Suppose f(x)+f(1−x)=x for every real x. At x=0, f(0)+f(1)=0. At x=1, the same left-hand sum equals one, a contradiction. Therefore no such function exists on all reals. This is a model example of swapping related inputs before attempting to guess a formula.
Extended example 10: A constant function forced by two free variables
Suppose f(x+y)=f(x−y) for every pair of real x,y. Given arbitrary reals u,v, choose x=(u+v)/2 and y=(u−v)/2. Then x+y=u and x−y=v, so f(u)=f(v). As u,v were arbitrary, f is constant. This proof depends on being able to choose those real input pairs.
Extended example 11: A shifted quadratic normalisation
If f(x+y)=f(x)+f(y)+xy over real inputs, set g(x)=f(x)−x²/2. The identity (x+y)²/2=x²/2+y²/2+xy yields g(x+y)=g(x)+g(y). This converts the problem to an additive equation. With appropriate continuity, g(x)=cx, giving f(x)=x²/2+cx.
Extended example 12: A composition equation requires the full domain
If f(f(x))=x for all x in a set D and f maps D into itself, then f is injective: f(a)=f(b) implies a=f(f(a))=f(f(b))=b. It is also onto D because every y equals f(f(y)). Thus f is a bijection whose inverse is itself. The closure f(D)⊆D is necessary for the composition to make sense.
Extended example 13: An equation with f(0) but no zero in the domain
Suppose f:(0,∞)→ℝ satisfies a functional equation for positive x,y. Choosing x=0 is not legal, even if it would simplify the relation dramatically. A valid solution must use allowed positive inputs, perhaps x=y=1 or y=1/x. The first step of solving is to read the domain, not to reach for the most convenient number.
Extended example 14: Find an input fixed by an involution
For f(x)=1−x, solving f(x)=x gives 1−x=x, or x=1/2. The other values are paired under the involution: f(0)=1 and f(1)=0. The equation f(f(x))=x holds, but most inputs are not fixed. Distinguishing fixed points from paired points prevents an incorrect claim that f(x)=x everywhere.
Extended example 15: A recurrence on reals with periodic freedom
The functional relation f(x+1)=f(x)+1 admits f(x)=x+sin(2πx) and f(x)=x as two distinct continuous real solutions. In both cases the sine term repeats after one unit. Thus even continuity does not establish uniqueness for this shift-only equation. A complete solution needs additional values or constraints within one period.
Extended example 16: Check the proposed answer for every variable
For f(x)=x², verify f(x+y)=(x+y)²=x²+y²+2xy=f(x)+f(y)+2xy. This is a valid substitution proof on all real x,y. A calculation at x=1,y=2 merely checks one case. When a functional equation claims a law for arbitrary inputs, the final verification must be general.
Extended example 17: The role of a finite initial segment
For a recurrence f(n+1)=2f(n) defined on integers n≥0, the initial value f(0)=3 uniquely determines f(n)=3×2ⁿ. By contrast, when the same shift is defined on all real x, values on different fractional parts may be independent. The number system determines what counts as sufficient initial information.
Extended example 18: A solution can be a family rather than a formula
The relation f(x)+f(−x)=2x² admits f(x)=x²+x, f(x)=x²−x and countless other choices. The general form is f(x)=x²+h(x) with h odd. The equation fixes only the symmetric component. An answer giving one function is complete only when the question asks for an example.
Extended example 19: Why two substitutions are often enough to contradict a claim
In an unfamiliar functional equation, evaluate first at one legal pair (x,y), then at a strategically related pair such as (y,x) or (−x,−y). If the left sides must agree but the right sides differ, there is no function satisfying the claimed universal relation. This approach is stronger than searching numerical candidate formulas.
Extended example 20: A rigorous final-check routine
After finding a candidate function, substitute it into the original equation for general permitted inputs, simplify both sides, and check any base conditions. Then ask whether the argument proved uniqueness or merely existence. Finally inspect the regularity assumptions: continuity, boundedness or monotonicity may be exactly what makes an “all functions” classification possible.
A practical parent and tutor follow-up
A useful tutorial begins with a functional equation whose domain is the integers; ask the learner to find f(0), f(1), f(−1) and then a general formula. Next switch the domain to the real numbers and show how a shift rule alone can leave periodic freedom. The point is to make assumptions visible, not to introduce difficulty for its own sake.
A week later, offer an unlabelled invented equation with two variables and ask the learner which inputs they will try and why. Improvement is visible when they check the domain first, compare related substitutions, and verify a proposed function for arbitrary arguments rather than a few numerical samples.
JC1 H3 Mathematics Tutor Bukit Timah | Functional Equations and Substitution Strategies. Did you know? A mathematical claim can become much easier once its domain, assumptions and essential structure are understood. This guide explains functional equations with concrete examples, careful reasoning and an emphasis on what students can independently verify.
This is a preparation resource for JC1 learners in Bukit Timah, Beauty World and Sixth Avenue who are strengthening H2 Mathematics before any possible later H3 pathway. It is not a claim that a JC1 H3 specialist or class is available. H3 subject selection is determined by the student’s junior college.
The official 2027 H3 Mathematics 9820 syllabus includes mathematical reasoning, proof and non-routine problem solving, while assuming H2 Mathematics. The worked topic below is educational enrichment, not a claim that its exact title is a separately listed compulsory syllabus subtopic.
What the learner should be able to explain
A strong answer identifies the permitted inputs, states the method, performs exact calculations and checks the conclusion against the original conditions. It distinguishes a successful example from a general theorem, and a bound from an attainable optimum whenever those questions arise.
Reference standard and learning progression
The unchanged eduKateSG small-group Mathematics reference describes the established three-student tutorial approach near Sixth Avenue MRT. Continue from current H2 school topics towards proof-based extensions at a pace suited to the learner.
Additional reading and local enquiries
Use the SEAB 2027 A-Level directory for examination-year verification, and contact eduKate Singapore about tutor expertise, group size, schedules and fees. No tuition outcome or H3 place is guaranteed.
Functional equation toolkit: substitutions before guesses
Worked example 1: What a functional equation asks
A functional equation is an equation whose unknown is a function rather than a single number. For example, f(x+y)=f(x)+f(y) relates several outputs for all permitted x,y.
The task is to find or characterise functions satisfying every specified input relation. A few numerical matches are only examples; a complete proof must check the proposed formula for arbitrary allowed arguments.
Worked example 2: Substitute zero first
For f(x+y)=f(x)+f(y) over the reals, set y=0. We get f(x)=f(x)+f(0), hence f(0)=0.
The input zero is often the simplest legal substitution, but this reasoning depends on zero belonging to the domain. If a problem defines f only on positive reals, the substitution y=0 may be illegal.
Worked example 3: Find f(-x) by making the sum zero
With the same additive equation, choose y=−x. Then f(0)=f(x)+f(−x)=0, so f(−x)=−f(x).
This proves oddness of an additive function on the reals. The result follows from the functional equation, not from an unsupported assumption that every function looks symmetric around the origin.
Worked example 4: Find f(2x)
Setting y=x in f(x+y)=f(x)+f(y) gives f(2x)=2f(x). Applying it repeatedly shows f(nx)=nf(x) for positive integers n.
These consequences are valid for any additive function on an additive domain. They do not by themselves identify its value at every real x; additional regularity may be needed.
Worked example 5: Additivity with an integer domain
Let f:ℤ→ℝ satisfy f(m+n)=f(m)+f(n) for all integers m,n, and suppose f(1)=3. Repeated addition gives f(n)=3n for positive n.
Since f(0)=0 and f(−n)=−f(n), the same formula holds for zero and negative integers. On this domain, the given value f(1) determines the entire function.
Worked example 6: Additivity over the rational numbers
Suppose an additive function f:ℚ→ℝ has f(1)=2. Then f(p)=2p for any integer p. Because qf(p/q)=f(p), we obtain f(p/q)=2p/q for nonzero integer q.
Thus f(x)=2x for every rational x. The reasoning uses only additivity and integer division in the rational domain, not an assumption about continuity on the reals.
Worked example 7: A warning about the real Cauchy equation
For f:ℝ→ℝ satisfying f(x+y)=f(x)+f(y), the condition f(1)=2 does not by itself justify f(x)=2x for every real x.
The equation determines values on rational inputs, but without regularity there exist nonstandard additive functions on the real numbers. A complete solution must check whether continuity, measurability or another condition was supplied.
Worked example 8: Continuity can make an additive function linear
Suppose an additive real function is continuous at zero and f(1)=2. Its rational values are f(q)=2q. For any real x, choose rational numbers q approaching x.
Additivity plus continuity transfers the rational formula to x, proving f(x)=2x. The hypothesis of continuity is doing essential work; omitting it would overstate what additivity alone establishes.
Worked example 9: A simple affine relation
Let f:ℝ→ℝ satisfy f(x+y)=f(x)+f(y)+1. Taking y=0 gives f(0)=−1. Define g(x)=f(x)+1.
Then g(x+y)=g(x)+g(y), so g is additive. With continuity and a known value g(1), g becomes linear; without such conditions, state the general additive structure rather than assuming one particular line.
Worked example 10: A functional equation known at integer steps
Suppose f:ℤ→ℤ satisfies f(n+1)=f(n)+4 with f(0)=3. The positive terms are 3,7,11,15, and reversing the relation gives values for negative integers.
Hence f(n)=4n+3 for every integer n. The initial value and recurrence determine the function on integers, unlike a real-domain recurrence that leaves the values on an entire interval unspecified.
Worked example 11: The same shift rule on all real numbers
Suppose f:ℝ→ℝ satisfies f(x+1)=f(x)+4. The function f(x)=4x+3 is one solution, but the functional equation does not force its exact intercept or remove periodic variation.
For any one-periodic function h, f(x)=4x+h(x) also works. Values on one representative interval are needed to determine the solution more precisely. Do not infer uniqueness from a shift relation alone.
Worked example 12: An equation symmetric around one-half
If f(x)=f(1−x) for all real x, the function is symmetric under reflection about x=1/2. One example is f(x)=(x−1/2)².
The equation does not force f to be constant. It pairs inputs x and 1−x, and a student should distinguish symmetry from a claim that all outputs are equal.
Worked example 13: The odd-function relation
If f(−x)=−f(x) for every real x, then substituting x=0 gives f(0)=−f(0), so f(0)=0. Examples include f(x)=x and f(x)=x³.
The equation states a symmetry property, not one unique formula. A proof that proposes a candidate must verify the property, while a uniqueness claim would require additional information.
Worked example 14: The even-function relation
If f(−x)=f(x), the graph has symmetry about the vertical axis. Functions f(x)=x² and f(x)=cos x satisfy the condition.
The function need not be constant, and the condition alone does not determine f(1). It constrains paired input values rather than prescribing every individual output.
Worked example 15: An involution equation
If f(f(x))=x for every permitted x, the function is called an involution. The examples f(x)=x, f(x)=−x and f(x)=1−x on the reals all work.
The equation implies f is one-to-one: if f(a)=f(b), applying f gives a=b. The key is verifying compositions and checking that the domain supports both applications.
Worked example 16: A square-output equation has many branches
If f(x)²=x² for every real x, then for each x the value f(x) can equal x or −x. The functions f(x)=x and f(x)=−x are two examples.
Without continuity or other restrictions, different branches can be chosen for different inputs. The equation is a constraint on outputs but does not automatically identify one unique function.
Worked example 17: A sum of reflected values
Consider f(x)+f(1−x)=1 for every real x. Setting x=1/2 gives 2f(1/2)=1, so f(1/2)=1/2.
The simple function f(x)=x satisfies the relation, but so do many other functions. The midpoint value is forced; the entire function is not determined by that single reflected-sum condition.
Worked example 18: A fixed shift of period two
Suppose f(x+2)=f(x) for every real x. The function is periodic with period two. Examples include f(x)=sin(πx) and a constant function.
Knowing f(0)=0 does not determine f everywhere; the function may have arbitrary allowed behaviour elsewhere in one period. A periodicity condition repeats information rather than fixing all initial values.
Worked example 19: A multiplicative functional equation at zero
Let f:ℝ→ℝ satisfy f(xy)=f(x)f(y) for every real x,y. Setting x=0 gives f(0)=f(0)f(y) for all y.
If f(0)≠0, dividing shows f(y)=1 for every y, the constant-one solution. Otherwise f(0)=0. The special input zero divides the problem into distinct cases.
Worked example 20: A multiplicative equation at one
For f(xy)=f(x)f(y), set y=1. The equality becomes f(x)=f(x)f(1). If f is not identically zero, there is some x with f(x)≠0, forcing f(1)=1.
The identically zero function is another valid solution. This is why a functional-equation proof must handle exceptional branches before cancelling or dividing by a function value.
Worked example 21: A familiar power is one candidate
The function f(x)=x² satisfies f(xy)=f(x)f(y) for all real x,y, because (xy)²=x²y². The function f(x)=x³ also satisfies the same relation.
These examples prove existence but not uniqueness. Simply guessing one power function does not characterise every possible solution without further hypotheses.
Worked example 22: A function determined by a composition
Suppose f(x)=2x+1 and g(x)=3x−4. Then f(g(x))=6x−7 while g(f(x))=6x−1.
The two compositions differ even though both functions are linear. If a functional equation relates f(g(x)) and g(f(x)), the order must be preserved throughout each substitution.
Worked example 23: A swap of two variables
If a functional equation holds for all real x,y, it also holds with x and y interchanged. Comparing the two resulting equations may eliminate common terms or reveal a symmetry.
For example, f(x+y)=f(x)+f(y) is symmetric automatically. But if the given right side includes x alone, swapping can yield genuinely new information. The method is valid because both input orders are permitted.
Worked example 24: A reflection at minus x can expose a constant
Consider f(x)+f(−x)=2x² for all real x. The function f(x)=x²+x is one solution because the linear terms cancel.
But f(x)=x²−x also works. The equation fixes the even part of f but leaves freedom in its odd part. A student who announces one candidate as unique has not solved the full functional equation.
Worked example 25: A useful normalisation with a quadratic term
If f(x+y)=f(x)+f(y)+2xy, define g(x)=f(x)−x². Since (x+y)²=x²+y²+2xy, substitution gives g(x+y)=g(x)+g(y).
The original equation has been transformed into the additive Cauchy equation. The substitution is justified by an exact identity; any further claim that g is linear needs suitable domain and regularity hypotheses.
Worked example 26: A simple quadratic solution
The function f(x)=x² satisfies f(x+y)=f(x)+f(y)+2xy, as direct expansion shows. If f is required continuous and f(1)=1, the normalised additive function has g(1)=0.
Continuity then forces g(x)=0 for all real x, giving f(x)=x². Without continuity, a broader family of additive perturbations can satisfy the same functional equation.
Worked example 27: An operation with one fixed input
Suppose f(x+1)=f(x)+1 for integer x and f(0)=5. The formula f(n)=n+5 is forced for all integers n.
On real inputs the same shift relation admits f(x)=x+h(x) where h is one-periodic. The domain decides whether the recurrence and starting value determine the entire function.
Worked example 28: A false attempt to divide by f(x)
From f(x)f(y)=0, a learner might divide by f(x) and conclude f(y)=0. That step requires f(x)≠0, which may not have been given.
When a functional equation contains products of unknown function values, explicitly separate the zero and nonzero cases. Cancelling a function value without checking it can remove an entire family of solutions.
Worked example 29: A functional equation and a missing domain
The relation f(xy)=f(x)f(y) has different consequences on positive reals, all reals and integers. Input zero and negative values can supply additional constraints.
When the problem says f:(0,∞)→(0,∞), substituting zero or −1 is illegal. A reader should write down the domain before choosing useful substitutions or declaring a general solution.
Worked example 30: An output bound changes what is possible
Suppose f:ℝ→ℝ is additive and bounded on a nontrivial interval. Such a regularity condition rules out pathological real additive functions and forces f(x)=cx for some constant c.
This is more information than additivity alone supplies. Advanced Mathematical reasoning often hinges on the hypothesis that allows a classification; do not erase it in the final theorem.
A checklist for new functional-equation texts
- Identify the function’s full domain and codomain, including whether zero or negatives are permitted.
- Substitute x=0,y=0,x=y and y=−x where those inputs are legal.
- Swap x and y and compare the two equations when they both apply.
- Try a simple candidate, then verify it for arbitrary inputs; do not confuse existence with uniqueness.
- Check any continuity, monotonicity, boundedness or other regularity assumptions before classifying all solutions.
Independent practice
Try the following: prove f(0)=0 when f is additive on the reals; derive f(−x)=−f(x); if f:ℤ→ℝ is additive with f(1)=7, find f(−3); give two different functions satisfying f(x+2)=f(x); and show that f(x)=x² solves f(x+y)=f(x)+f(y)+2xy.
The answers are obtained by substituting zero, using x+(−x)=0, finding f(−3)=−21, choosing for example f(x)=0 and f(x)=sin(πx), and expanding (x+y)². Every step should explicitly mention which domain or property makes it legal.
Frequently asked questions
Are functional equations a separately named 2027 H3 syllabus topic?
The 2027 H3 Mathematics syllabus 9820 includes mathematical proof, non-routine problems and unfamiliar texts. This functional-equations guide is enrichment illustrating those skills, not a claim that functional equations form a separately named syllabus unit.
Does finding one function solve the question?
Only if the task asks for an example or existence. If it asks for every solution, you must prove the classification and address any exceptional cases or regularity conditions.
Should JC1 learners first master H2 functions?
Yes. Domains, inverse functions, transformations and algebraic identities form the foundation. These exercises should enrich rather than replace current H2 study.
Official resources and related Mathematics learning
Use the SEAB 2027 A-Level syllabus directory, definitions and theorems and invented operations for related reasoning.
The unchanged eduKateSG three-student Mathematics reference describes the tutorial approach near Sixth Avenue MRT. It does not establish current JC H3 specialist availability. Enquire directly about tutor expertise, timetables, fees and suitable classes.
Did you know? The first good substitution often reveals more about a functional equation than the first clever guess. A clear domain check tells you which substitutions you are actually allowed to make.
Further substitutions, branches and regularity conditions
Extended example 1: Zero and the exponential functional equation
Let f:ℝ→ℝ satisfy f(x+y)=f(x)f(y). Set y=0: f(x)=f(x)f(0). If some f(x) is nonzero, then f(0)=1. Otherwise f is identically zero. The constant zero and constant one functions both satisfy the equation. Separating these branches is required before dividing by f(0) or claiming that a solution is exponential.
Extended example 2: Negatives in the exponential equation
If f(x+y)=f(x)f(y) and f is not the zero function, then f(0)=1. Setting y=−x gives f(x)f(−x)=1 for every real x. Thus no function value can equal zero, and f(−x)=1/f(x). This conclusion follows from the equation; it is not a guess about what the function graph should look like.
Extended example 3: A familiar exponential candidate
The function f(x)=2ˣ satisfies f(x+y)=f(x)f(y) because 2ˣ⁺ʸ=2ˣ2ʸ. The same holds for f(x)=eˣ and f(x)=1. These demonstrate multiple solutions. If a problem asks for all solutions, extra assumptions and a classification proof are needed. An example, however convincing, is not a complete general solution.
Extended example 4: Continuity narrows the exponential family
For a positive-valued function f:ℝ→(0,∞) satisfying f(x+y)=f(x)f(y), one may take logarithms to define g(x)=ln f(x). Then g is additive. If f is continuous, so is g, making g(x)=cx and f(x)=eᶜˣ. The positivity and continuity assumptions are indispensable to this classification; they are not decorations.
Extended example 5: A multiplicative equation forces a one-value condition
For f(xy)=f(x)f(y) on real numbers, take y=1. If f is not identically zero, some f(x) is nonzero, so f(1)=1. For a nonzero input x, f(x)f(1/x)=f(1)=1, so f(x) cannot vanish. This is a strong structural consequence from only two simple substitutions.
Extended example 6: Negative one under a multiplicative equation
If nonzero f satisfies f(xy)=f(x)f(y) for all real x,y, then f(−1)²=f(1)=1, so f(−1)=1 or f(−1)=−1. For any x, f(−x)=f(−1)f(x). Thus the behaviour under changing sign is governed by one of two cases. The original domain must include −1 for this step.
Extended example 7: An odd part can be isolated
Suppose f(x)=f(−x)+2x for every real x. Set h(x)=f(x)−x. Then h(−x)=f(−x)+x=f(x)−x=h(x), so h is even. Thus all such solutions are f(x)=x+h(x), where h is any even function. A general characterisation needs this freedom, not only the example f(x)=x.
Extended example 8: A fixed reflected sum
If f(x)+f(−x)=4, setting x=0 yields 2f(0)=4, so f(0)=2. Write h(x)=f(x)−2; then h(x)+h(−x)=0, meaning h is odd. Every function f(x)=2+h(x) with odd h satisfies the relation. The equation determines the even part without fixing every output.
Extended example 9: An impossible reflected equation
Suppose f(x)+f(1−x)=x for every real x. At x=0, f(0)+f(1)=0. At x=1, the same left-hand sum equals one, a contradiction. Therefore no such function exists on all reals. This is a model example of swapping related inputs before attempting to guess a formula.
Extended example 10: A constant function forced by two free variables
Suppose f(x+y)=f(x−y) for every pair of real x,y. Given arbitrary reals u,v, choose x=(u+v)/2 and y=(u−v)/2. Then x+y=u and x−y=v, so f(u)=f(v). As u,v were arbitrary, f is constant. This proof depends on being able to choose those real input pairs.
Extended example 11: A shifted quadratic normalisation
If f(x+y)=f(x)+f(y)+xy over real inputs, set g(x)=f(x)−x²/2. The identity (x+y)²/2=x²/2+y²/2+xy yields g(x+y)=g(x)+g(y). This converts the problem to an additive equation. With appropriate continuity, g(x)=cx, giving f(x)=x²/2+cx.
Extended example 12: A composition equation requires the full domain
If f(f(x))=x for all x in a set D and f maps D into itself, then f is injective: f(a)=f(b) implies a=f(f(a))=f(f(b))=b. It is also onto D because every y equals f(f(y)). Thus f is a bijection whose inverse is itself. The closure f(D)⊆D is necessary for the composition to make sense.
Extended example 13: An equation with f(0) but no zero in the domain
Suppose f:(0,∞)→ℝ satisfies a functional equation for positive x,y. Choosing x=0 is not legal, even if it would simplify the relation dramatically. A valid solution must use allowed positive inputs, perhaps x=y=1 or y=1/x. The first step of solving is to read the domain, not to reach for the most convenient number.
Extended example 14: Find an input fixed by an involution
For f(x)=1−x, solving f(x)=x gives 1−x=x, or x=1/2. The other values are paired under the involution: f(0)=1 and f(1)=0. The equation f(f(x))=x holds, but most inputs are not fixed. Distinguishing fixed points from paired points prevents an incorrect claim that f(x)=x everywhere.
Extended example 15: A recurrence on reals with periodic freedom
The functional relation f(x+1)=f(x)+1 admits f(x)=x+sin(2πx) and f(x)=x as two distinct continuous real solutions. In both cases the sine term repeats after one unit. Thus even continuity does not establish uniqueness for this shift-only equation. A complete solution needs additional values or constraints within one period.
Extended example 16: Check the proposed answer for every variable
For f(x)=x², verify f(x+y)=(x+y)²=x²+y²+2xy=f(x)+f(y)+2xy. This is a valid substitution proof on all real x,y. A calculation at x=1,y=2 merely checks one case. When a functional equation claims a law for arbitrary inputs, the final verification must be general.
Extended example 17: The role of a finite initial segment
For a recurrence f(n+1)=2f(n) defined on integers n≥0, the initial value f(0)=3 uniquely determines f(n)=3×2ⁿ. By contrast, when the same shift is defined on all real x, values on different fractional parts may be independent. The number system determines what counts as sufficient initial information.
Extended example 18: A solution can be a family rather than a formula
The relation f(x)+f(−x)=2x² admits f(x)=x²+x, f(x)=x²−x and countless other choices. The general form is f(x)=x²+h(x) with h odd. The equation fixes only the symmetric component. An answer giving one function is complete only when the question asks for an example.
Extended example 19: Why two substitutions are often enough to contradict a claim
In an unfamiliar functional equation, evaluate first at one legal pair (x,y), then at a strategically related pair such as (y,x) or (−x,−y). If the left sides must agree but the right sides differ, there is no function satisfying the claimed universal relation. This approach is stronger than searching numerical candidate formulas.
Extended example 20: A rigorous final-check routine
After finding a candidate function, substitute it into the original equation for general permitted inputs, simplify both sides, and check any base conditions. Then ask whether the argument proved uniqueness or merely existence. Finally inspect the regularity assumptions: continuity, boundedness or monotonicity may be exactly what makes an “all functions” classification possible.
A practical parent and tutor follow-up
A useful tutorial begins with a functional equation whose domain is the integers; ask the learner to find f(0), f(1), f(−1) and then a general formula. Next switch the domain to the real numbers and show how a shift rule alone can leave periodic freedom. The point is to make assumptions visible, not to introduce difficulty for its own sake.
A week later, offer an unlabelled invented equation with two variables and ask the learner which inputs they will try and why. Improvement is visible when they check the domain first, compare related substitutions, and verify a proposed function for arbitrary arguments rather than a few numerical samples.