Functions, domain and representations

Review: from a rule to ordered pairs

A relation associates elements. A function is a relation where each allowed input has exactly one output. Consider the rule “associate each integer from 1 to 4 with its square”. The pairs are (1, 1), (2, 4), (3, 9), (4, 16). The list allows output uniqueness to be checked for every input of this finite domain.

In f(x), x is the input and f(x) the result. If f(x) = 3x − 2, then f(5) = 13. By contrast, f(x) = 13 is an equation whose solution is x = 5. The letter f names the function; we could call it g without changing the idea if the definition stayed the same. The input variable can also be called t, especially when it represents time.

Reviewing substitution is essential: if g(x) = x² + 2x, then g(−3) = (−3)² + 2(−3) = 9 − 6 = 3. Parentheses keep the whole negative input under the square. We can also have g(1) = 3. This does not violate the definition: two inputs can give the same output.

Domain, codomain and range

Writing f: D → C indicates that D is the domain and C the declared codomain. Every x in D must receive a value f(x) belonging to C. The range, Im(f), includes only values actually reached. Thus Im(f) is contained in C but need not equal C.

For example, define f: {−2, −1, 0, 1, 2} → ℝ by f(x) = x². The domain contains five inputs; the range is {0, 1, 4}; the codomain is all ℝ. The number 3 belongs to the codomain but not the range. If we declared codomain {0, 1, 4}, the same rule on the same inputs would reach the entire codomain.

For q: ℝ → ℝ, q(x) = x², the range is [0, +∞[. Every y ≥ 0 is reached by x = √y, whereas no negative y can be a real square. A function reaching its entire declared codomain is called surjective. A function where different inputs produce different outputs is injective; q is not injective on ℝ because q(−2) = q(2).

How to find a formula's domain

When a formula is given without explicit domain and we work in ℝ, we seek its natural domain: real values for which every operation makes sense. Polynomials such as x³ − 2x + 1 accept all reals. A denominator cannot be zero. An even-index root requires a nonnegative radicand; an odd-index root also permits negative numbers.

For f(x) = √(x − 2)/(x − 5), we need x − 2 ≥ 0 and x − 5 ≠ 0 simultaneously. The first condition gives x ≥ 2; the second excludes 5. The domain is [2, 5[ ∪ ]5, +∞[. The symbol ∪ means union: inputs from one interval or the other. “Simultaneously” corresponds to intersecting conditions, not choosing only one.

Simplification does not eliminate original restrictions. The expression (x² − 9)/(x − 3) matches x + 3 only for x ≠ 3. The first forbids input 3; the second, if defined on all ℝ, does not. Simplified formulas may describe the same values where both exist, but functions with different domains are not the same function.

Words, table, diagram and graph

A verbal description explains quantity meanings and assumptions; a table shows concrete pairs; an expression allows values to be calculated; a graph reveals trends and intersections. No representation automatically replaces all the others. A finite table alone does not determine a unique formula for all reals: many rules can agree at the listed points and differ between them.

In an arrow diagram, each domain element must have exactly one outgoing arrow. Several arrows may reach the same destination. On a Cartesian graph, a vertical line x = a cannot meet two points with different y-coordinates if the relation is a function of x. The vertical-line test geometrically expresses output uniqueness.

A complete circle is not the graph of a single function y of x: several x-coordinates have both positive and negative y-coordinates. The upper semicircle can be a function, for example y = √(r² − x²) on [−r, r], with r > 0. Restricting the relation selects a single output, but we need to declare that choice.

Read growth and average rates

A function is increasing on an interval when larger inputs give larger outputs; it is decreasing when larger inputs give smaller outputs. Always compare from left to right along the input axis. For f(x) = (x − 3)², the function decreases until x = 3 and increases afterwards. Positive outputs do not imply growth: between 0 and 2, values 9, 4 and 1 are positive and decrease.

The average rate of change between distinct inputs a and b is [f(b) − f(a)]/(b − a). The numerator measures output change; the denominator input change. If output is volume in litres and input time in minutes, the rate is litres per minute. An average rate does not guarantee that change occurred uniformly.

For V(t) = 120 − 6t, between t = 2 and t = 7, V changes from 108 to 78. Average rate is (78 − 108)/(7 − 2) = −30/5 = −6 L/min. The negative sign indicates decreasing volume. For this linear function, all average rates coincide; that is no longer true for a quadratic function.

Contextual domain and piecewise functions

A model can have a more restricted domain than its expression. If V(t) = 120 − 6t represents a tank emptying at constant rate, time starts at 0 and ends when V = 0, at t = 20 min. Using t = 30 would give volume −60 L, allowed by the calculation but incompatible with this physical model.

Some situations use different rules in different regions. A hypothetical service may charge C(n) = 100n for 0 ≤ n ≤ 5 and C(n) = 500 + 80(n − 5) for n > 5, where n is a whole number of units and C is in Kz. The second branch retains the cost of the first five and adds 80 per additional unit; it does not reapply the price to all units.

To evaluate C(7), first select the right branch: C(7) = 500 + 80 × 2 = 660 Kz. Merely seeing a letter and substituting into any expression is not enough. Branch endpoints need clear rules: if two branches include the same input with different outputs, the definition fails; if a domain input has no branch, it also fails.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

Range and codomain are always equal.

Range contains only outputs actually reached; codomain is the declared destination set.

The formula determines everything about a function.

Domain and codomain are part of the definition.

A repeated output prevents a function.

One input with two different outputs violates the definition.

A positive function must be increasing.

Positivity compares outputs with zero; growth compares outputs with each other as input increases.