6.1 Function Definitions, Relations, Domain, Range & Notation

6.1 Function Definitions, Relations, Domain, Range & Notation

1. Set-Theoretic Foundations: Relations vs. Functions

In secondary mathematics, algebraic reasoning begins with the formal distinction between arbitrary mathematical relations and well-defined functions. Given two non-empty sets $X$ (the set of inputs) and $Y$ (the set of outputs), a relation $R$ from $X$ to $Y$ is formally defined as any subset of the Cartesian product $X \times Y$:

$R \subseteq X \times Y = \{(x, y) \mid x \in X, \; y \in Y\}$

A relation places elements of $X$ in correspondence with elements of $Y$ without restriction. An input may associate with zero, one, or multiple outputs.

A function $f: X \to Y$ is a specialized relation that satisfies a strict uniqueness condition: every element $x$ in the domain $X$ must be paired with exactly one element $y$ in the codomain $Y$. In formal first-order logic:

$\forall x \in X, \; \exists ! y \in Y \text{ such that } (x, y) \in f$

Equivalently, if $(x, y_1) \in f$ and $(x, y_2) \in f$, then $y_1 = y_2$. The set $X$ is called the domain of $f$ (denoted $\operatorname{dom}(f)$), representing all permissible inputs. The set of all realized outputs in $Y$ is the range (or image) of $f$:

$\operatorname{ran}(f) = \{y \in Y \mid \exists x \in X \text{ such that } f(x) = y\}$

While every function is a relation, not every relation is a function. Secondary educators must train students to distinguish relations from functions across four standard mathematical representations:

1. Arrow Mapping Diagrams: In a function mapping, exactly one directed arrow originates from each element of the domain set. If multiple arrows originate from a single input, the relation fails to be a function. If an element in the designated domain has no outgoing arrow, the relation is not fully defined on that domain.
2. Sets of Ordered Pairs: A finite set of coordinates represents a function if and only if no two distinct ordered pairs share the same first coordinate. For instance, ${(1, 4), (2, 7), (3, 7)}$ is a function (outputs may repeat), whereas ${(1, 4), (1, 5), (2, 7)}$ is merely a relation because the input $1$ maps to both $4$ and $5$.
3. Input-Output Tables: A table represents a function if each unique row in the independent variable column is associated with a single, unambiguous entry in the dependent variable column.
4. Cartesian Graphs and the Vertical Line Test (VLT): A geometric curve in the Cartesian coordinate plane represents a single-valued function $y = f(x)$ if and only if every vertical line $x = c$ intersects the graph at no more than one point. If any vertical line intersects the curve at two or more points, the single input $x = c$ produces multiple distinct outputs $y$, violating the function definition. Standard non-functions include circles ($x^2 + y^2 = r^2$), ellipses, and sideways parabolas ($x = y^2$).

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2. Implied Algebraic Domain Restrictions

When a function is defined by an algebraic expression $y = f(x)$ without an explicitly stated domain, conventions dictate that the domain is the natural (implied) domain: the maximal subset of real numbers $\mathbb{R}$ for which the expression produces a well-defined real output. Secondary mathematics focuses on three primary domain hazards:

Odd-indexed radicals, such as $\sqrt[3]{g(x)}$ or $\sqrt[5]{g(x)}$, impose no domain restrictions on their radicands because negative real numbers possess unique real odd roots (e.g., $\sqrt[3]{-8} = -2$). Thus, the domain of $\sqrt[3]{g(x)}$ is simply the domain of $g(x)$.

Comprehensive Multi-Constraint Worked Exemplar

Find the natural domain of the function:

$f(x) = \frac{\sqrt{2x + 10}}{\ln(6 - 2x)} - \frac{1}{x^2 - 4}$

To determine the implied domain, establish and solve each independent algebraic constraint:

1. Radicand of the even root: The numerator contains $\sqrt{2x + 10}$. The radicand must be non-negative: $2x + 10 \ge 0 \implies 2x \ge -10 \implies x \ge -5$
2. Argument of the natural logarithm: The denominator contains $\ln(6 - 2x)$. The logarithmic argument must be strictly positive: $6 - 2x > 0 \implies 6 > 2x \implies x < 3$
3. Logarithmic denominator non-zero: The logarithm appears in the denominator, so it cannot equal zero. Since $\ln(u) = 0 \iff u = 1$: $6 - 2x \neq 1 \implies -2x \neq -5 \implies x \neq \frac{5}{2}$
4. Rational denominator non-zero: The second term has denominator $x^2 - 4$. Setting $x^2 - 4 \neq 0$: $(x - 2)(x + 2) \neq 0 \implies x \neq 2 \quad \text{and} \quad x \neq -2$

Now intersect all four solution sets on the real number line:

• From constraints 1 and 2: $x \in [-5, 3)$.
• Exclude internal boundary points from constraints 3 and 4: $x \neq -2$, $x \neq 2$, and $x \neq \frac{5}{2} = 2.5$. All three excluded points lie strictly within $[-5, 3)$.

Expressing the intersection in standard union-of-intervals notation:

$\operatorname{dom}(f) = [-5, -2) \cup (-2, 2) \cup \left(2, \frac{5}{2}\right) \cup \left(\frac{5}{2}, 3\right)$

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3. Function Evaluation and the Difference Quotient

Evaluating functions requires substituting numeric values or variable expressions into every instance of the independent variable $x$. A ubiquitous misconception among secondary algebra students is the distributive misinterpretation of function notation, where students treat $f(x + h)$ as multiplication: $f \cdot (x + h) = f(x) + f(h)$ or $f(x) + h$.

The foundational algebraic construct connecting function evaluation to differential calculus is the difference quotient:

$\frac{\Delta y}{\Delta x} = \frac{f(x + h) - f(x)}{h}, \quad h \neq 0$

Geometrically, the difference quotient computes the slope of the secant line passing through the points $(x, f(x))$ and $(x + h, f(x + h))$ on the graph of $f$. As $h \to 0$, this secant slope approaches the instantaneous rate of change (derivative) at $x$.

Rigorous Algebraic Simplification of the Difference Quotient

Consider the quadratic function $f(x) = 3x^2 - 4x + 7$. Compute and completely simplify the difference quotient.

Step 1: Evaluate $f(x + h)$ using precise algebraic expansion. $f(x + h) = 3(x + h)^2 - 4(x + h) + 7$ Expand the squared binomial and distribute all scalar coefficients: $f(x + h) = 3(x^2 + 2xh + h^2) - 4x - 4h + 7 = 3x^2 + 6xh + 3h^2 - 4x - 4h + 7$

Step 2: Formulate the numerator difference $f(x + h) - f(x)$. Subtract the entire original function, enclosing $f(x)$ in grouping parentheses to ensure proper distribution of the negative sign: $f(x + h) - f(x) = (3x^2 + 6xh + 3h^2 - 4x - 4h + 7) - (3x^2 - 4x + 7)$ Eliminate identical terms: $3x^2 - 3x^2 = 0$, $-4x - (-4x) = 0$, and $7 - 7 = 0$. This leaves: $f(x + h) - f(x) = 6xh + 3h^2 - 4h$ Notice that every remaining term contains $h$ as a factor. If non-$h$ terms remain at this stage, an algebraic sign or expansion error has occurred.

Step 3: Divide by $h$ and cancel the non-zero common factor. $\frac{f(x + h) - f(x)}{h} = \frac{h(6x + 3h - 4)}{h} = 6x - 4 + 3h$ This linear expression represents the exact secant slope for any step size $h \neq 0$.

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4. Piecewise-Defined & Step Functions

Many real-world relationships cannot be modeled by a single algebraic formula across their entire domain. A piecewise-defined function applies different algebraic rules over mutually disjoint subsets of its domain:

$f(x) = \begin{cases} -2x + 1, & x < -1 \\ x^2 - 3, & -1 \le x \le 2 \\ 5, & x > 2 \end{cases}$

When evaluating piecewise functions, examine the input $x$ against the inequality conditions first to determine which sub-rule governs that input. For example:

• To evaluate $f(-1)$, look at the domain conditions: $x = -1$ satisfies $-1 \le x \le 2$, so evaluate using the middle rule: $f(-1) = (-1)^2 - 3 = 1 - 3 = -2$. A student evaluating the top rule would arrive at an invalid result because $-1 < -1$ is false.
• At the transition point $x = 2$, $f(2) = (2)^2 - 3 = 1$. Just to the right of $2$ (e.g., $x = 2.001$), $f(x) = 5$. This mismatch produces a jump discontinuity on the graph.

A prominent subcategory of piecewise functions tested on the FTCE is the step function, most notably the greatest integer function (or floor function), denoted $f(x) = \lfloor x \rfloor$ or $[x]$. The floor function assigns to any real number $x$ the greatest integer less than or equal to $x$:

$\lfloor x \rfloor = \max \{n \in \mathbb{Z} \mid n \le x\}$

• For integers: $\lfloor 4 \rfloor = 4$, $\lfloor -3 \rfloor = -3$.
• For non-integer positives: $\lfloor 3.8 \rfloor = 3$, $\lfloor 0.25 \rfloor = 0$.
• For non-integer negatives: $\lfloor -2.3 \rfloor = -3$ (not $-2$, since $-3 \le -2.3 < -2$).

The companion ceiling function $\lceil x \rceil$ rounds upward to the least integer greater than or equal to $x$ (e.g., $\lceil -2.3 \rceil = -2$, $\lceil 3.1 \rceil = 4$). The domain of $f(x) = \lfloor x \rfloor$ is $\mathbb{R}$, while its range is the discrete set of integers $\mathbb{Z}$. The graph consists of horizontal half-open line segments $[k, k+1)$ with a closed circle at the left endpoint $(k, k)$ and an open circle at the right endpoint $(k+1, k)$.