Domain and Range of Functions in Multiple Representations

Foundations of Domain, Range, and Relations

Mathematics Grade 9, Term 1, Week 3 (Part 2) under the revised K to 10 curriculum focuses on determining the domain and the range of a function expressed across four primary representations: ordered pairs, algebraic equations, graphs, and tables of values.

A relation is defined as a set of ordered pairs $(x, y)$. Within any relation or ordered pair, the input corresponds to the $x$-value and the output corresponds to the $y$-value.

The domain of a function is defined as the set of all allowed input values, or $x$-values, for which the function is defined. The range of a function is defined as the set of all possible output values, or $y$-values, that the function can produce. Consequently, the input is the domain and the output is the range.

Expressing Domain and Range Using Set-Builder Notation

Set-builder notation is the standard mathematical method used to express the domain and range of a function. In set-builder notation, the variable $x$ is utilized when describing the domain, while the variable $y$ is utilized when describing the range.

A set-builder expression is enclosed in curly brackets and begins with the variable followed by a vertical bar signifying "such that". For domain notation, the expression $\{x \mid 2 \le x \le 4\}$ is read aloud starting from the central variable as "the set of all $x$ such that $x$ is greater than or equal to two and less than or equal to four."

For range notation, replacing the domain variable $x$ with $y$ yields expressions such as $\{y \mid -2 \le y \le 5\}$, which is read as "the set of all $y$ such that $y$ is greater than or equal to negative two and less than or equal to five."

Determining Domain and Range from Ordered Pairs

When a function is presented as a set of ordered pairs, its domain and range are finite and measurable. The domain is determined by listing all distinct $x$-coordinates, and the range is determined by listing all distinct $y$-coordinates.

For the set of ordered pairs $\{(2, 5), (3, 5), (4, 9)\}$, the domain values obtained via the listing method are $2$, $3$, and $4$. The range values are $5$, $5$, and $9$. To express the domain in set-builder notation, identify the smallest value $2$ and the largest value $4$, yielding $\{x \mid 2 \le x \le 4\}$. For the range, identifying the smallest value $5$ and the largest value $9$ gives the set-builder notation $\{y \mid 5 \le y \le 9\}$.

For the set of ordered pairs $\{(-6, -2), (0, 5), (1, -8)\}$, the extracted $x$-values are $-6$, $0$, and $1$. In set-builder notation, because these values are listed in increasing order, the domain is written as $\{x \mid -6 \le x \le 1\}$. The extracted $y$-values are $-2$, $5$, and $-8$. When writing set-builder notation, values must first be arranged in increasing order from smallest to largest: $-8$, $-2$, and $5$. Thus, the range in set-builder notation is $\{y \mid -8 \le y \le 5\}$.

For the set of ordered pairs $\{(-3, -1), (0, 5), (2, -7)\}$, the extracted $x$-values are $-3$, $0$, and $2$, producing the domain $\{x \mid -3 \le x \le 2\}$. The extracted $y$-values are $-1$, $5$, and $-7$. Reordering these values from smallest to largest identifies $-7$ as the smallest value and $5$ as the largest value, producing the range set-builder notation $\{y \mid -7 \le y \le 5\}$.

Calculating Domain and Range from Algebraic Equations

Finding the domain and range of a function from its algebraic equation requires identifying mathematical restrictions such as division by zero or negative radicands under square roots.

For simple linear equations such as $y = x + 5$, or general linear forms such as $y = 2x + 5$, $y = 3x + 5$, $y = 10x$, or $y = 11x - 5$, there are no mathematical restrictions on the input or output values. The domain is the set of all real numbers, expressed in set-builder notation as $\{x \mid x \in \mathbb{R}\}$. Similarly, the range is the set of all real numbers, expressed as $\{y \mid y \in \mathbb{R}\}$.

For the quadratic equation $x^2 + y - 6 = 0$, determine the domain by rewriting the equation in terms of $y$: $y = 6 - x^2$ Because any real number can be substituted for $x$ and squared without restriction, the domain is the set of all real numbers, $\{x \mid x \in \mathbb{R}\}$. To determine the range, rewrite the equation in terms of $x$: $x^2 = 6 - y$ $x = \sqrt{6 - y}$ Because a negative radicand under a square root is not allowed, establish the restriction boundary by setting the radicand equal to zero: $6 - y = 0$ $y = 6$ Testing values demonstrates that if $y = 6$, then $\sqrt{6 - 6} = \sqrt{0} = 0$. If $y = 5$, then $\sqrt{6 - 5} = \sqrt{1} = 1$. However, if $y = 7$, then $\sqrt{6 - 7} = \sqrt{-1}$, which is undefined in real numbers. Thus, $y$ must be less than or equal to six, giving the range in set-builder notation as $\{y \mid y \le 6\}$.

For the radical equation $y = \sqrt{x + 2}$, the domain restriction dictates that the radicand cannot be negative. Setting the radicand to zero determines the boundary value: $x + 2 = 0$ $x = -2$ Testing values shows that substituting $x = -2$ gives $\sqrt{0} = 0$, and substituting $x = -1$ gives $\sqrt{1} = 1$. Substituting $x = -3$ yields $\sqrt{-1}$, which is not permitted. Thus, the domain is $\{x \mid x \ge -2\}$. Evaluating the range using the boundary value $x = -2$ produces $y = 0$, and larger values of $x$ produce positive outputs extending to positive infinity. Therefore, the range is $\{y \mid y \ge 0\}$.

For the absolute value equation $y = |x|$, inputting any real number yields a positive result or zero ($|2| = 2$, $|-3| = 3$). Since any real number can be evaluated, the domain is all real numbers, $\{x \mid x \in \mathbb{R}\}$. Because the absolute value of any real number is always non-negative, the range extends from zero to positive infinity, written as $\{y \mid y \ge 0\}$.

For the rational equation $y = \frac{2}{x - 5}$, a denominator of zero is not allowed because division by zero is undefined. To find the domain restriction, set the denominator equal to zero: $x - 5 = 0$ $x = 5$ The domain is the set of all real numbers except five, expressed as $\{x \mid x \in \mathbb{R}, x \neq 5\}$. To find the range, express the equation in terms of $x$ using cross-multiplication: $y(x - 5) = 2$ $xy - 5y = 2$ $xy = 5y + 2$ $x = \frac{5y + 2}{y}$ Setting the denominator equal to zero reveals that $y \neq 0$. Thus, the range is the set of all real numbers except zero, expressed as $\{y \mid y \in \mathbb{R}, y \neq 0\}$.

For the rational equation $y = \frac{3x + 1}{x - 2}$, find the domain restriction by setting the denominator to zero: $x - 2 = 0$ $x = 2$ The domain is all real numbers except two, written as $\{x \mid x \in \mathbb{R}, x \neq 2\}$. To determine the range, rewrite the equation in terms of $x$ using cross-multiplication and factoring: $y(x - 2) = 3x + 1$ $xy - 2y = 3x + 1$ $xy - 3x = 2y + 1$ $x(y - 3) = 2y + 1$ $x = \frac{2y + 1}{y - 3}$ Setting the denominator $y - 3 = 0$ gives $y = 3$. Thus, the range is the set of all real numbers except three, written as $\{y \mid y \in \mathbb{R}, y \neq 3\}$.

Evaluating Domain and Range from Graphs

Determining the domain and range from a graph requires analyzing the horizontal boundaries along the $x$-axis for the domain and the vertical boundaries along the $y$-axis for the range. Solid or shaded endpoints indicate inclusive bounds ($\le$), whereas open circle endpoints indicate strict inequalities ($<$).

For a line segment graph with shaded endpoints extending horizontally from $x = -4$ to $x = 3$ and vertically from $y = 1$ to $y = 2$, both boundaries are inclusive. The domain is $\{x \mid -4 \le x \le 3\}$ and the range is $\{y \mid 1 \le y \le 2\}$.

For a line segment graph where the left endpoint at $x = -3$ is shaded and the right endpoint at $x = 5$ is an open circle, the domain spans horizontally with an inclusive left bound and strict right bound, written as $\{x \mid -3 \le x < 5\}$. Vertically, if the open circle rests at $y = -2$ and the shaded endpoint reaches $y = 3$, the range is written as $\{y \mid -2 < y \le 3\}$.

For a parabolic curve graph with shaded endpoints spanning horizontally from $x = -1$ to $x = 5$, the domain is $\{x \mid -1 \le x \le 5\}$. If the lowest point on the curve along the $y$-axis is $-2$ and the highest point is $2$, both inclusive, the range is $\{y \mid -2 \le y \le 2\}$.

For a graph with an open circle endpoint on the left at $x = -4$ and a shaded endpoint on the right at $x = 5$, the domain is $\{x \mid -4 < x \le 5\}$. If the vertical span extends from a shaded endpoint at $y = 1$ to an open circle endpoint at $y = 3$, the range is $\{y \mid 1 \le y < 3\}$.

Determining Domain and Range from Tables of Values

A table of values lists corresponding $x$ and $y$ coordinates. To find the domain and range from a table of values, identify the smallest and largest values for $x$ and $y$.

For a table containing $x$-values $-2$, $-1$, $0$, and $5$, the domain extends from the minimum $x$-value $-2$ to the maximum $x$-value $5$. In set-builder notation, the domain is written as $\{x \mid -2 \le x \le 5\}$.

For the corresponding $y$-values in the table, assessing the smallest value $-3$ and the largest value $4$ yields the range. In set-builder notation, the range is written as $\{y \mid -3 \le y \le 4\}$.

Interactive Practice Problems and Solutions

In Activity A, relations presented as sets of ordered pairs are analyzed to determine if they represent functions, alongside their domain and range: Item 1 represents a function. Item 2 represents a function. Item 3 is not a function because the input $x = 3$ repeats with different outputs. Item 4 is not a function because the input $x = 5$ repeats. Item 5 represents a function.

In Activity B, domain and range are calculated for four algebraic equations: Item 1 yields domain $\{x \mid x \in \mathbb{R}\}$ and range $\{y \mid y \in \mathbb{R}\}$. Item 2 yields domain $\{x \mid x \in \mathbb{R}\}$ and range $\{y \mid y \le 6\}$. Item 3 yields domain $\{x \mid x \ge -2\}$ and range $\{y \mid y \ge 0\}$. Item 4 yields domain $\{x \mid x \in \mathbb{R}, x \neq 5\}$ and range $\{y \mid y \in \mathbb{R}, y \neq 0\}$.

In Activity C, domain and range are evaluated from a graph: The domain is $\{x \mid -4 \le x \le 3\}$ and the range is $\{y \mid 1 \le y \le 2\}$.

In Activity D, domain and range are evaluated from a table of values: The domain is $\{x \mid -2 \le x \le 5\}$ and the range is $\{y \mid -3 \le y \le 4\}$.

Self-Assessment Framework and Formative Assessment

A self-assessment scale using the acronym MATH allows learners to gauge their level of understanding: M stands for Mastered: fully understanding the content and able to solve problems confidently. A stands for Almost There: understanding most parts but still needing some guidance. T stands for Trying Hard: putting forth effort but getting confused by certain ideas. H stands for Haven't Got It: not understanding the material yet and unable to solve problems.

Formative assessment requires evaluating the domain and range for three specific representations: Item 1 provides a set of ordered pairs. Item 2 provides an algebraic equation. Item 3 provides a graph.