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Lesson 2 — Functions: definition, domain, range

Function as a mathematical object: unique correspondence rule between two sets. Domain, codomain, range. Cartesian graph. Injective, surjective, bijective functions.

Used in: 1.º ano do EM (15 anos) · Math I japonês cap. 2 · Klasse 10 alemã

f : A \to B,\quad x \mapsto f(x)

A function from A to B is a rule that assigns each element of A to exactly one element of B. A is the domain, B the codomain, and the set of values actually attained is the range.

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous definition

"A function is a relation in which each input value produces exactly one output value." — OpenStax College Algebra 2e, §3.1

Each element of the domain points to exactly one element of the codomain. Note that $x_3$ can map to the same output as $f(x_1)$ — a function can send different inputs to the same destination.

Worked examples

Five examples of increasing difficulty — from the most direct (numerical evaluation and reading the domain) to real-world modeling (composition in a production pipeline). Each example cites its source: the original problem always comes from an open book.

Example— 1· Direct evaluation and domain (application)

Problem: For $f(x) = 3x^2 - 2x + 1$ , compute $f(-1)$ , $f(0)$ , and $f(2)$ . Determine the maximum domain.

Strategy: Evaluating a function at a point means substituting the variable with the given value and computing. The maximum domain of a polynomial is all of $\mathbb{R}$ , since polynomials are defined at every real number.

Solution:

$f(-1)$ : Substitute $x = -1$ : $f(-1) = 3(-1)^2 - 2(-1) + 1 = 3(1) + 2 + 1 = 6$ .
$f(0)$ : Substitute $x = 0$ : $f(0) = 3(0)^2 - 2(0) + 1 = 0 - 0 + 1 = 1$ .
$f(2)$ : Substitute $x = 2$ : $f(2) = 3(2)^2 - 2(2) + 1 = 3(4) - 4 + 1 = 12 - 4 + 1 = 9$ .
Domain: Since $f$ is a polynomial, there is no restriction. Domain = $\mathbb{R}$ .

Check: All three values are finite. To confirm consistency, note that $f(0) = 1$ is the constant term — this always holds for any polynomial.

Source. OpenStax — College Algebra 2e §3.1 — license CC-BY 4.0.

Example— 2· Maximum domain with combined restrictions (intermediate)

Problem: Determine the maximum domain of $f(x) = \dfrac{\sqrt{x + 2}}{x - 3}$ .

Strategy: There are two simultaneous restrictions: the radicand $x + 2$ must be non-negative (for the square root to exist in the reals), and the denominator $x - 3$ cannot be zero. The domain is the intersection of both conditions.

Solution:

Square root restriction: $x + 2 \geq 0 \Rightarrow x \geq -2$ , i.e., $x \in [-2, +\infty)$ .
Denominator restriction: $x - 3 \neq 0 \Rightarrow x \neq 3$ , i.e., $x \in \mathbb{R} \setminus \{3\}$ .
Intersection: $[-2, +\infty) \cap (\mathbb{R} \setminus \{3\}) = [-2, 3) \cup (3, +\infty)$ .

Check:

$x = -2$ : $\sqrt{0}/(-5) = 0$ . OK (included).
$x = 3$ : division by zero. Excluded. OK.
$x = -3$ : $\sqrt{-1}$ does not exist in the reals. Excluded. OK.
$x = 5$ : $\sqrt{7}/2$ . OK.

Source. OpenStax — College Algebra 2e §3.2 — license CC-BY 4.0.

Example— 3· Composition of functions (intermediate)

Problem: Let $f(x) = x^2 + 1$ and $g(x) = 2x - 3$ . Compute $(f \circ g)(x)$ and $(g \circ f)(x)$ . Show they are different at $x = 1$ .

Strategy: Composition $(f \circ g)(x)$ means "apply $g$ first, then $f$ ". That is, $f(g(x))$ . Swapping the order produces generally different functions — a classic warning.

Solution:

$(f \circ g)(x) = f(g(x))$ : substitute $g(x) = 2x - 3$ in place of $x$ in $f$ : $f(2x - 3) = (2x - 3)^2 + 1 = 4x^2 - 12x + 9 + 1 = 4x^2 - 12x + 10$ .
$(g \circ f)(x) = g(f(x))$ : substitute $f(x) = x^2 + 1$ in place of $x$ in $g$ : $g(x^2 + 1) = 2(x^2 + 1) - 3 = 2x^2 + 2 - 3 = 2x^2 - 1$ .
At $x = 1$ :
- $(f \circ g)(1) = 4 - 12 + 10 = 2$ .
- $(g \circ f)(1) = 2 - 1 = 1$ . Different — composition does not commute.

Check: Also try $x = 0$ : $(f \circ g)(0) = 10$ and $(g \circ f)(0) = -1$ . Different at all points except perhaps solutions of $4x^2 - 12x + 10 = 2x^2 - 1$ , i.e., $2x^2 - 12x + 11 = 0$ .

Source. Stitz–Zeager — Precalculus §5.1 — license CC-BY-NC-SA.

Example— 4· Inverse and domain restriction (advanced)

Problem: Let $f(x) = x^2 - 4x + 5$ defined on $\mathbb{R}$ . Show that $f$ is not injective on $\mathbb{R}$ . Restrict the domain to an interval where $f$ is injective and find $f^{-1}$ on that restriction.

Strategy: A quadratic function is not injective on $\mathbb{R}$ : points symmetric about the parabola's axis have the same output. The axis of symmetry is $x = 2$ (vertex). Restricting to $[2, +\infty)$ makes $f$ strictly increasing, hence injective; then the inverse can be computed.

Solution:

Non-injectivity. Complete the square: $f(x) = (x - 2)^2 + 1$ . Then $f(0) = 4 + 1 = 5$ and $f(4) = 4 + 1 = 5$ . Since $0 \neq 4$ but $f(0) = f(4)$ , $f$ is not injective.
Restriction: consider $f: [2, +\infty) \to [1, +\infty)$ . On this restriction $f$ is increasing (the right branch of the parabola), hence injective; the range is $[1, +\infty)$ , so it is also surjective — bijective.
Inverse. Set $y = (x-2)^2 + 1 \Rightarrow (x-2)^2 = y - 1 \Rightarrow x - 2 = \pm\sqrt{y - 1}$ . Since $x \geq 2$ , we choose the positive branch: $x = 2 + \sqrt{y - 1}$ . Therefore $f^{-1}(y) = 2 + \sqrt{y - 1}$ , with $y \geq 1$ .

Check: $f(f^{-1}(y)) = (2 + \sqrt{y-1} - 2)^2 + 1 = (y - 1) + 1 = y$ . ✓

Source. OpenStax — College Algebra 2e §3.7 — license CC-BY 4.0.

Example— 5· Real-world modeling: tiered electricity bill (real modeling)

Problem: A utility charges energy consumption $k$ (in kWh) in tiers: up to 100 kWh costs R$ 0.60 per kWh; between 100 and 200 kWh costs R$ 0.90 on the excess; above 200 kWh costs R$ 1.30 on the excess. Model the total bill $C(k)$ as a continuous piecewise function.

Strategy: Each tier adds a new "marginal price" applied only to consumption within that tier. The resulting function is piecewise affine and continuous at the boundary points (unlike a step-charge model).

Solution:

Tier 1 (up to 100 kWh): $C(k) = 0{,}60\, k$ .
Tier 2 (100 to 200 kWh): the first 100 kWh cost $0{,}60 \times 100 = 60$ reais; the excess $(k - 100)$ is charged at R$ 0.90: $C(k) = 60 + 0{,}90\,(k - 100) = 0{,}90\,k - 30$ .
Tier 3 (above 200 kWh): the first 200 kWh cost $60 + 0{,}90 \times 100 = 150$ reais; the excess $(k - 200)$ at R$ 1.30: $C(k) = 150 + 1{,}30\,(k - 200) = 1{,}30\,k - 110$ .
Piecewise function: $C(k) = \begin{cases} 0{,}60\,k & k \leq 100 \\ 0{,}90\,k - 30 & 100 < k \leq 200 \\ 1{,}30\,k - 110 & k > 200 \end{cases}$

Continuity check: at $k = 100$ : branch 1 gives 60, branch 2 gives $90 - 30 = 60$ . ✓ At $k = 200$ : branch 2 gives $180 - 30 = 150$ , branch 3 gives $260 - 110 = 150$ . ✓ The function is continuous — correct marginal billing model.

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 2 — free license.

Exercise list

42 exercises · 10 with worked solution (25%)

Application 26Understanding 7Modeling 4Challenge 5

Sources

Only books that directly informed the text and exercises. Full catalog at /livros.

OpenStax College Algebra 2e — Jay Abramson et al. · 2022 · EN · CC-BY 4.0 · §3.1–3.7. Primary source for blocks A, B, D.
Stitz–Zeager Precalculus — Carl Stitz, Jeff Zeager · 2013, v3 · EN · CC-BY-NC-SA · §1.4–1.6, §2.3, §5.1–5.2.
Yoshiwara — Modeling, Functions, and Graphs — Katherine Yoshiwara · 2020 · EN · free · chs. 1–2, 4, 6. Source for block E (modeling).
Active Calculus — Matt Boelkins · 2024, ed. 2.0 · EN · CC-BY-NC-SA · §1.1.
Hammack — Book of Proof — Richard Hammack · 2018 · EN · free · §12 (functions and bijections). Source for exercise 2.20.

Rigorous definition

Classification

Worked examples

Sources