1.3 Functions, Domain/Range, Graphs & Transformations

Functions, Domain/Range, Graphs & Transformations

Understanding functions is fundamental to achieving high scores on the Ateneo College Entrance Test (ACET) Mathematics section. Questions on the ACET range from formal functional definition tests to multi-step functional composition, algebraic domain restriction identification, piecewise functions, inverse derivation, and rapid visual graph transformations.

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1. Formal Definition of a Function & Mapping Rules

A function $f: X \rightarrow Y$ is a mathematical rule that assigns to each element $x$ in the domain $X$ exactly one element $y$ in the codomain $Y$. In coordinate geometry, a relation represents a function if no single input value $x$ generates multiple distinct output values $y$.

Graphical & Relation Tests

• Vertical Line Test (VLT): A curve in the Cartesian plane represents a function of $x$ if and only if no vertical line intersects the graph more than once.
• Horizontal Line Test (HLT): A function $f$ is one-to-one (injective) and possesses a valid inverse function $f^{-1}$ if and only if no horizontal line intersects its graph more than once.
• Mapping Diagrams: If an element in the domain branches into two or more distinct target elements in the codomain, the relation is not a function.

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2. Determining Algebraic Domain & Range

The Three Core Domain Restriction Rules

When finding the real-valued domain of a function $f(x)$, exclude all real numbers that violate fundamental algebraic operations:

1. Denominator Non-Zero Rule: For rational functions $f(x) = \frac{P(x)}{Q(x)}$, set the denominator $Q(x) \neq 0$.
2. Even Radical Radicand Rule: For radical functions $f(x) = \sqrt[n]{g(x)}$ where $n$ is an even integer, set the radicand $g(x) \ge 0$.
3. Logarithmic Argument Rule: For logarithmic functions $f(x) = \log_b(h(x))$, set the argument $h(x) > 0$.

Detailed Worked Domain Example

Find the domain of $f(x) = \frac{\sqrt{x + 5}}{x^2 - 9}$.

• Radical requirement: $x + 5 \ge 0 \implies x \ge -5$.
• Denominator requirement: $x^2 - 9 \neq 0 \implies (x-3)(x+3) \neq 0 \implies x \neq 3$ and $x \neq -3$.
• Combined Domain: $[-5, -3) \cup (-3, 3) \cup (3, \infty)$.

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3. Function Operations, Composition & Domain Evaluation

Given two functions $f(x)$ and $g(x)$:

• Sum / Difference: $(f \pm g)(x) = f(x) \pm g(x)$
• Product: $(f \cdot g)(x) = f(x) \cdot g(x)$
• Quotient: $(f / g)(x) = \frac{f(x)}{g(x)}$ where $g(x) \neq 0$
• Function Composition: $(f \circ g)(x) = f(g(x))$

Determining the Domain of Composite Functions

The domain of $(f \circ g)(x) = f(g(x))$ consists of all real numbers $x$ such that:

1. $x$ is in the domain of the inner function $g(x)$.
2. The evaluated output $g(x)$ is in the domain of the outer function $f(x)$.

Example: If $f(x) = \frac{1}{x - 4}$ and $g(x) = \sqrt{x}$, the domain of $g(x)$ is $x \ge 0$. The outer function requires $g(x) \neq 4$, meaning $\sqrt{x} \neq 4 \implies x \neq 16$. Thus, the domain of $f(g(x))$ is $[0, 16) \cup (16, \infty)$.

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4. Inverse Functions ($f^{-1}$) & Step-by-Step Derivation

To find the inverse function equation $f^{-1}(x)$ for a one-to-one function $y = f(x)$:

1. Replace $f(x)$ with $y$.
2. Swap all variables: interchange $x$ and $y$.
3. Solve the resulting algebraic equation explicitly for $y$.
4. Replace $y$ with $f^{-1}(x)$.

Essential Properties of Inverse Functions

• $\text{Domain}(f^{-1}) = \text{Range}(f)$
• $\text{Range}(f^{-1}) = \text{Domain}(f)$
• $f(f^{-1}(x)) = x$ for all $x$ in $\text{Domain}(f^{-1})$
• $f^{-1}(f(x)) = x$ for all $x$ in $\text{Domain}(f)$
• The graph of $f^{-1}(x)$ is a symmetric reflection of $f(x)$ across the line $y = x$.

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5. Symmetry of Functions: Even, Odd & Neither

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6. Complete Graph Transformation Matrix

Starting from a base function $y = f(x)$, transformations alter the visual graph according to clear algebraic rules:

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Worked Step-by-Step ACET Exam Problems

Problem 1: Finding an Inverse Function with Domain & Range Analysis

Find the inverse function $f^{-1}(x)$ for $f(x) = \frac{2x + 3}{x - 4}$, and state its domain and range.

Step 1: State domain of $f(x)$. Denominator cannot be zero: $x - 4 \neq 0 \implies \text{Domain}(f) = \mathbb{R} \setminus {4}$.

Step 2: Swap $x$ and $y$ to set up inverse. Set $x = \frac{2y + 3}{y - 4}$.

Step 3: Solve algebraically for $y$. Multiply both sides by $(y - 4)$: $x(y - 4) = 2y + 3 \implies xy - 4x = 2y + 3$ Rearrange terms containing $y$ to one side: $xy - 2y = 4x + 3 \implies y(x - 2) = 4x + 3$ Divide by $(x - 2)$: $y = f^{-1}(x) = \frac{4x + 3}{x - 2}$

Step 4: Determine domain and range of $f^{-1}(x)$.

• Denominator of $f^{-1}$ cannot be zero: $x \neq 2 \implies \text{Domain}(f^{-1}) = \mathbb{R} \setminus {2}$.
• Therefore, $\text{Range}(f) = \text{Domain}(f^{-1}) = \mathbb{R} \setminus {2}$, and $\text{Range}(f^{-1}) = \text{Domain}(f) = \mathbb{R} \setminus {4}$.

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Problem 2: Multi-Step Graph Transformation Tracking

The point $(3, -4)$ lies on the graph of $y = f(x)$. What are the coordinates of the corresponding point on the transformed graph $g(x) = -2 f(x - 1) + 5$?

Step 1: Trace horizontal coordinate changes. The horizontal transformation $f(x - 1)$ shifts the graph RIGHT by 1 unit. $x_{\text{new}} = x_{\text{old}} + 1 = 3 + 1 = 4$.

Step 2: Trace vertical coordinate changes. Vertical operations execute in standard operational order: vertical stretch by 2, vertical reflection across x-axis (negation), and vertical shift up by 5. $y_{\text{new}} = -2 \cdot y_{\text{old}} + 5 = -2(-4) + 5 = 8 + 5 = 13$.

Step 3: State final coordinate pair. The transformed point is $(4, 13)$.

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