Functions

Core Concepts

Representation of a Function A function is a mathematical relation that maps every element from a set of inputs (the domain) to exactly one element in a set of outputs (the codomain). If a function $f$ maps elements from set $A$ to set $B$, we write $f: A \to B$.

A function can be represented in four primary ways:

Verbally: Using words to describe the input-output relationship.
Numerically: Using tables of values or a set of ordered pairs $(x, y)$.
Visually: Using mapping diagrams (arrows pointing from inputs to outputs) or plotting the coordinates on a Cartesian graph.
Algebraically: Using an explicit equation or formula, such as $f(x) = x^2 + 3$.

A highly testable algebraic representation in the NECTA syllabus is the piecewise function. This is a function defined by multiple sub-equations, where each sub-equation applies to a different interval of the domain. It is written using curly braces: $$f(x) = \begin{cases} g(x) & \text{if } x < a \\ h(x) & \text{if } x \ge a \end{cases}$$

Domain and Range of a Function

Domain: The set of all valid input values (independent variable, usually $x$) for which the function produces a real output. To find the domain algebraically, look for restrictions:

For rational functions like $f(x) = \frac{1}{x-a}$, the denominator cannot be zero ($x \neq a$).
For square root functions like $f(x) = \sqrt{x-a}$, the radicand cannot be negative ($x \ge a$).
For standard polynomials, the domain is all real numbers, denoted as $\{x \in \mathbb{R}\}$.

Range: The set of all possible output values (dependent variable, usually $y$ or $f(x)$) produced by the function. To find the range algebraically, let $y = f(x)$, solve the equation for $x$ in terms of $y$, and determine the restrictions on $y$.

The Inverse of a Function The inverse of a function $f(x)$, denoted as $f^{-1}(x)$, reverses the action of $f$. If $f(a) = b$, then $f^{-1}(b) = a$. A fundamental property connecting a function and its inverse is the swapping of input and output sets:

Domain of $f^{-1}$ = Range of $f$
Range of $f^{-1}$ = Domain of $f$

Graph of a Function The graph of a function is the set of all points $(x, f(x))$ plotted on a coordinate plane.

Vertical Line Test: A curve on a graph represents a function if and only if no vertical line intersects the curve more than once.
Graphing Piecewise Functions: Since piecewise functions use different rules for different intervals, their graphs consist of broken segments or separate rays. When graphing boundary points, use a closed circle ($\bullet$) to show that a point is included ($\le$ or $\ge$) and an open circle ($\circ$) to show that a point is excluded ($<$ or $>$).

Worked Examples

Example 1: Evaluating a Piecewise Function A function $f$ is defined as: $$f(x) = \begin{cases} -2 & \text{if } 0 < x \le 5 \\ x + 1 & \text{if } -6 \le x < 0 \end{cases}$$ Find $f(4)$ and $f(-5)$.

Solution: To find $f(4)$, check which interval contains the input $x=4$. Since $0 < 4 \le 5$, we apply the first rule: $f(x) = -2$. $$f(4) = -2$$ To find $f(-5)$, check which interval contains $x=-5$. Since $-6 \le -5 < 0$, we apply the second rule: $f(x) = x + 1$. $$f(-5) = -5 + 1 = -4$$

Example 2: Domain and Range of a Rational Function Given the function $f(x) = \frac{1}{x-2}$, find its domain and range.

Solution: Domain: The function is undefined when the denominator is equal to zero. $$x - 2 = 0 \implies x = 2$$ Therefore, the domain is all real numbers except $2$. Domain = $\{x \in \mathbb{R} : x \neq 2\}$

Range: To find the range, set $y = f(x)$ and solve for $x$: $$y = \frac{1}{x-2}$$ Multiply both sides by $(x-2)$: $$y(x - 2) = 1 \implies xy - 2y = 1$$ Rearrange to isolate $x$: $$xy = 1 + 2y \implies x = \frac{1 + 2y}{y}$$ This resulting expression for $x$ is undefined when the denominator is zero ($y = 0$). Range = $\{y \in \mathbb{R} : y \neq 0\}$

Example 3: Graphing a Piecewise Function and Finding Domain/Range The function $f$ is defined by: $$f(x) = \begin{cases} -2 & \text{if } x < -1 \\ 0 & \text{if } x = -1 \\ x + 2 & \text{if } x > -1 \end{cases}$$ (Note: In the original 2018 NECTA exam, the third condition was printed as $x \ge -1$, creating a mathematical contradiction at $x = -1$. We use $x > -1$ here for validity.) Sketch the graph of $f$ and determine its domain and range.

Solution: Graphing:

For $x < -1$, draw a horizontal line at $y = -2$. The line ends at $(-1, -2)$ with an open circle.
For $x = -1$, plot a single solid point at $(-1, 0)$.
For $x > -1$, sketch the straight line $y = x + 2$. At the boundary $x = -1$, the value is $y = 1$, so draw an open circle at $(-1, 1)$. The line continues upwards through $(0, 2)$ and $(1, 3)$.

Domain and Range:

Looking horizontally (x-axis), the function covers values less than $-1$, exactly $-1$, and greater than $-1$.

Domain = $\{x \in \mathbb{R}\}$

Looking vertically (y-axis), the outputs are exactly $-2$, exactly $0$, and strictly greater than $1$.

Range = $\{-2, 0\} \cup \{y \in \mathbb{R} : y > 1\}$

Example 4: Evaluating an Inverse Function Given the function $f(x) = \frac{1}{x-2}$, find $f^{-1}\left(\frac{1}{3}\right)$.

Solution: Method: Find the inverse function equation algebraically, then substitute $x = \frac{1}{3}$. Let $y = \frac{1}{x-2}$. Swap $x$ and $y$ to begin finding the inverse: $$x = \frac{1}{y-2}$$ Solve for the new $y$: $$x(y - 2) = 1 \implies xy - 2x = 1$$ $$xy = 1 + 2x \implies y = \frac{1 + 2x}{x}$$ So, $f^{-1}(x) = \frac{1 + 2x}{x}$. Now, substitute $x = \frac{1}{3}$: $$f^{-1}\left(\frac{1}{3}\right) = \frac{1 + 2\left(\frac{1}{3}\right)}{\frac{1}{3}} = \frac{1 + \frac{2}{3}}{\frac{1}{3}} = \frac{\frac{5}{3}}{\frac{1}{3}} = \frac{5}{3} \times \frac{3}{1} = 5$$ (Alternative logic: Set the original function equal to $\frac{1}{3}$ and solve for $x$: $\frac{1}{x-2} = \frac{1}{3} \implies x-2 = 3 \implies x=5$.)

Example 5: Domain and Range of an Inverse Function Suppose a function $f$ is defined by $f(x) = (x + 2)^2$ for $x \ge -2$. Find the domain and range of the inverse of the function $f$.

Solution: Instead of calculating the algebraic inverse, we use the property that the Domain of $f^{-1}$ is the Range of $f$, and the Range of $f^{-1}$ is the Domain of $f$.

First, find the domain and range of the original function $f(x)$:

The domain is explicitly given as: Domain of $f$ = $\{x \in \mathbb{R} : x \ge -2\}$.
For the range, note that $(x+2)^2$ is a perfect square, which means its lowest possible value is $0$ (occurring at $x = -2$). As $x$ increases, $(x+2)^2$ grows infinitely large. Thus, the Range of $f$ = $\{y \in \mathbb{R} : y \ge 0\}$.

By swapping these sets: Domain of $f^{-1}$ = $\{x \in \mathbb{R} : x \ge 0\}$ Range of $f^{-1}$ = $\{y \in \mathbb{R} : y \ge -2\}$

Example 6: Real-World Application of Functions During an inter-school quiz competition, each correct answer $x$ was awarded a number of points $y$ related by the function $y = 3x^2$. How many points would a competitor collect by answering 8 questions correctly?

Solution: The function mapping correct answers to points is $f(x) = 3x^2$. We must evaluate the function at $x = 8$. $$y = 3(8)^2$$ $$y = 3(64) = 192$$ The competitor would collect 192 points.

Common Pitfalls & Misconceptions

Confusing "Inverse" with "Reciprocal":

A widespread error is interpreting the inverse function notation $f^{-1}(x)$ as a negative exponent, mistakenly writing $f^{-1}(x) = \frac{1}{f(x)}$. The notation $f^{-1}(x)$ means the compositional inverse (the process of undoing the function's operations to return to the original input), not algebraic division.

Ignoring Sub-intervals in Piecewise Functions:

When asked to evaluate a piecewise function, students sometimes substitute the given $x$-value into all equations provided and sum the results. The input must only be substituted into the single sub-equation whose domain interval explicitly contains that $x$-value.

Open vs. Closed Circles in Graphs:

When graphing piecewise functions, students frequently lose marks for terminating lines improperly. If an interval uses strict inequalities ($<$ or $>$), the line must end with an open circle ($\circ$). If it includes "or equal to" ($\le$ or $\ge$), it must end with a solid circle ($\bullet$).

Struggling with the Range of Rational Functions:

While most students successfully find the domain by setting the denominator equal to zero, they often freeze on finding the range. The most reliable method is algebraic manipulation: rewrite $y=f(x)$ to express $x$ entirely in terms of $y$, and then find what values make the new denominator zero.

NECTA Exam Focus

Based on the 2018–2025 NECTA CSEE past papers, the topic of Functions is highly predictable and primarily tests algebraic manipulation and visual interpretation.

Piecewise Functions are a NECTA Favorite: They frequently appear in Paper 1. You must be comfortable sketching them accurately, interpreting their graphs, reading domain and range from the visual plot, and evaluating them algebraically at specific integer values.
Rational Functions: You are routinely tested on finding the domain and range of equations like $f(x) = \frac{1}{x-2}$. This leads seamlessly into finding the explicit formula for $f^{-1}(x)$ and evaluating it.
Conceptual Inversion: There is a recurring trend of asking for the domain and range of an inverse function without demanding that you formulate the inverse equation (e.g., given $y = (x+2)^2$ or $y = 3x^2$). This tests your understanding of the principle that swapping the domains and ranges between $f$ and $f^{-1}$ is mathematically sufficient.
Word Problems: NECTA occasionally frames standard algebraic function evaluations inside real-world scenarios, such as quiz points modeled by polynomial functions.

(Note: In Section A of the exam, basic numerical fraction ordering—e.g., sorting $\frac{1}{2}, \frac{2}{9}, 0.\dot{6}$—is sometimes mixed alongside functions questions, though it is fundamentally an arithmetic skill.)

Practice Problems

Category 1: Evaluating and Graphing Piecewise Functions

A function $f$ is defined as $f(x) = \begin{cases} -2 & \text{if } 0 < x \le 5 \\ x + 1 & \text{if } -6 \le x < 0 \end{cases}$. Find $f(4)$ and $f(-5)$. (NECTA 2023)
State the domain and range of the function given in Question 1. (NECTA 2023)
The function $f$ is defined by $f(x) = \begin{cases} -2 & \text{if } x < -1 \\ 0 & \text{if } x = -1 \\ x + 2 & \text{if } x > -1 \end{cases}$. Sketch the graph of $f$. (Adapted from NECTA 2018)
Use your sketched graph from Question 3 to precisely determine the domain and range of $f$. (NECTA 2018)

Category 2: Rational Functions and Inverse Formulation

Given the function $f(x) = \frac{1}{x-2}$, find its domain and range. (NECTA 2022)
Using the rational function from Question 5, find $f^{-1}\left(\frac{1}{3}\right)$. (NECTA 2022)
Find the inverse function $g^{-1}(x)$ if $g(x) = \frac{2x+3}{x-1}$, and explicitly state the domain of $g^{-1}(x)$.

Category 3: Conceptual Domain/Range Swapping and Applications

Suppose a function $f$ is defined by $f(x) = (x + 2)^2$ restricted to the domain $x \ge -2$. Find the domain and range of the inverse of the function $f$. (NECTA 2020)
During an inter-school quiz competition, each correct answer $x$ was awarded a number of points $y$ related by the function $y = 3x^2$. Find the domain and range of the inverse of the given function, assuming $x \ge 0$. (NECTA 2024)
Using the scoring system from Question 9, how many points would a competitor collect by answering 12 questions correctly?

Subtopics

Representation of a function
Domain and range of a function
Graph of a function

Crosswalk Notes

Cross-version relationships are drafted in data/curricula/crosswalks/csee-basic-mathematics-2005-to-mathematics-2023.json. Partial and 2005-only mappings remain reviewable.