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1.3 Functions, Domain/Range, Graphs & Transformations

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Ateneo College Entrance Test ACET

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1 Chapter 1: Mathematics

1.1 Real Numbers, Ratios, Rates & Percentages 1.2 Algebraic Expressions, Polynomials & Linear Equations 1.3 Functions, Domain/Range, Graphs & Transformations 1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations 1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area 1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections 1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities 1.8 Arithmetic & Geometric Sequences, Series & Word Patterns 1.9 Permutations, Combinations, Probability & Statistics

2 Chapter 2: English Proficiency

2.1 Subject-Verb Agreement, Verb Tenses & Aspect 2.2 Pronoun-Antecedent Agreement, Case, Misplaced & Dangling Modifiers 2.3 Sentence Structure, Run-Ons, Fragments & Parallelism 2.4 Contextual Vocabulary, Word Roots, Prefixes & Suffixes 2.5 Advanced Synonyms, Antonyms & Nuanced Vocabulary 2.6 Verbal Analogies: Relationship Types & Solving Strategies 2.7 Punctuation, Capitalization, Mechanics & Common Errors 2.8 Diction, Idiomatic Expressions, Prepositions & Frequently Confused Words

3 Chapter 3: Reading Comprehension

3.1 Identifying Main Ideas, Central Themes & Paragraph Summaries 3.2 Locating Explicit Details, Supporting Evidence & Data 3.3 Drawing Logical Inferences & Unstated Conclusions 3.4 Analyzing Author's Purpose, Point of View, Tone & Attitude 3.5 Interpreting Figurative Language, Tone Words & Context Clues 3.6 Text Structure, Rhetorical Devices & Evaluating Arguments

4 Chapter 4: Logical and Abstract Reasoning

4.1 Number & Letter Series Patterns & Formulas 4.2 Abstract Reasoning: Matrix Patterns & Figure Rotations 4.3 Formal Syllogisms, Categorical Logic & Deductive Proofs 4.4 Coding-Decoding, Cipher Patterns & Symbol Operations 4.5 Spatial Visualization, Paper Folding & Venn Diagram Logic

5 Chapter 5: General Knowledge

5.1 Philippine History, Government, Constitution & Civics 5.2 Integrated Science: Biology, Chemistry, Physics & Earth Science 5.3 Philippine & World Geography, Basic Economics & Cultural Facts 5.4 Current Events, World Landmarks, Literature & General Trivia

6 Chapter 6: ACET Essay Writing & Exam Strategies

6.1 ACET Essay: Thesis Development, Structuring & Argumentation 6.2 Time Allocation, Pacing Strategies & Subtest Tactics 6.3 Guessing Strategies (No Penalty), Test Anxiety & Exam Day Readiness

1.3 Functions, Domain/Range, Graphs & Transformations

Key Takeaways Explain This Topic AI

Determine domain restrictions by excluding values that cause zero denominators, negative radical radicands, or non-positive logarithmic arguments.

Apply the vertical line test to establish function status and the horizontal line test to verify one-to-one invertibility.

Calculate function compositions f(g(x)) by evaluating internal functions first and maintaining strict domain compatibility.

Master rigid shifts, reflections, and non-rigid stretch transformations to graph functions rapidly without point plotting.

Last updated: July 2026

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.

1. Formal Definition of a Function & Mapping Rules

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

Graphical & Relation Tests

Vertical Line Test (VLT): A curve in the Cartesian plane represents a function of x x x if and only if no vertical line intersects the graph more than once.

Horizontal Line Test (HLT): A function f f f is one-to-one (injective) and possesses a valid inverse function f − 1 f^{-1} 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.

  • 2. Determining Algebraic Domain & Range
  • The Three Core Domain Restriction Rules

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

Denominator Non-Zero Rule: For rational functions f ( x ) = P ( x ) Q ( x ) f(x) = \frac{P(x)}{Q(x)} f ( x ) = Q ( x ) P ( x ) ​ , set the denominator Q ( x ) ≠ 0 Q(x) \neq 0 Q ( x )  = 0 .

Even Radical Radicand Rule: For radical functions f ( x ) = g ( x ) n f(x) = \sqrt[n]{g(x)} f ( x ) = n g ( x ) ​ where n n n is an even integer, set the radicand g ( x ) ≥ 0 g(x) \ge 0 g ( x ) ≥ 0 .

Logarithmic Argument Rule: For logarithmic functions f ( x ) = log ⁡ b ( h ( x ) ) f(x) = \log_b(h(x)) f ( x ) = lo g b ​ ( h ( x )) , set the argument h ( x ) > 0 h(x) > 0 h ( x ) > 0 .

Detailed Worked Domain Example

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

Radical requirement: x + 5 ≥ 0 ⟹ x ≥ − 5 x + 5 \ge 0 \implies x \ge -5 x + 5 ≥ 0 ⟹ x ≥ − 5 .

Denominator requirement: x 2 − 9 ≠ 0 ⟹ ( x − 3 ) ( x + 3 ) ≠ 0 ⟹ x ≠ 3 x^2 - 9 \neq 0 \implies (x-3)(x+3) \neq 0 \implies x \neq 3 x 2 − 9  = 0 ⟹ ( x − 3 ) ( x + 3 )  = 0 ⟹ x  = 3 and x ≠ − 3 x \neq -3 x  = − 3 .

Combined Domain: [ − 5 , − 3 ) ∪ ( − 3 , 3 ) ∪ ( 3 , ∞ ) [-5, -3) \cup (-3, 3) \cup (3, \infty) [ − 5 , − 3 ) ∪ ( − 3 , 3 ) ∪ ( 3 , ∞ ) .

3. Function Operations, Composition & Domain Evaluation

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

Sum / Difference: ( f ± g ) ( x ) = f ( x ) ± g ( x ) (f \pm g)(x) = f(x) \pm g(x) ( f ± g ) ( x ) = f ( x ) ± g ( x )

Product: ( f ⋅ g ) ( x ) = f ( x ) ⋅ g ( x ) (f \cdot g)(x) = f(x) \cdot g(x) ( f ⋅ g ) ( x ) = f ( x ) ⋅ g ( x )

Quotient: ( f / g ) ( x ) = f ( x ) g ( x ) (f / g)(x) = \frac{f(x)}{g(x)} ( f / g ) ( x ) = g ( x ) f ( x ) ​ where g ( x ) ≠ 0 g(x) \neq 0 g ( x )  = 0

  • Function Composition: ( f ∘ g ) ( x ) = f ( g ( x ) ) (f \circ g)(x) = f(g(x)) ( f ∘ g ) ( x ) = f ( g ( x ))
  • Determining the Domain of Composite Functions

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

x x x is in the domain of the inner function g ( x ) g(x) g ( x ) .

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

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

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

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

Replace f ( x ) f(x) f ( x ) with y y y .

Swap all variables: interchange x x x and y y y .

Solve the resulting algebraic equation explicitly for y y y .

Replace y y y with f − 1 ( x ) f^{-1}(x) f − 1 ( x ) .

  • Essential Properties of Inverse Functions
  • Domain ( f − 1 ) = Range ( f ) \text{Domain}(f^{-1}) = \text{Range}(f) Domain ( f − 1 ) = Range ( f )
  • Range ( f − 1 ) = Domain ( f ) \text{Range}(f^{-1}) = \text{Domain}(f) Range ( f − 1 ) = Domain ( f )

f ( f − 1 ( x ) ) = x f(f^{-1}(x)) = x f ( f − 1 ( x )) = x for all x x x in Domain ( f − 1 ) \text{Domain}(f^{-1}) Domain ( f − 1 )

f − 1 ( f ( x ) ) = x f^{-1}(f(x)) = x f − 1 ( f ( x )) = x for all x x x in Domain ( f ) \text{Domain}(f) Domain ( f )

The graph of f − 1 ( x ) f^{-1}(x) f − 1 ( x ) is a symmetric reflection of f ( x ) f(x) f ( x ) across the line y = x y = x y = x .

5. Symmetry of Functions: Even, Odd & Neither

Function Type Algebraic Test Geometric Symmetry Structural Example

Even Function f ( − x ) = f ( x ) f(-x) = f(x) f ( − x ) = f ( x ) Symmetric about the y-axis f ( x ) = x 4 − 3 x 2 + 5 f(x) = x^4 - 3x^2 + 5 f ( x ) = x 4 − 3 x 2 + 5

Odd Function f ( − x ) = − f ( x ) f(-x) = -f(x) f ( − x ) = − f ( x ) Symmetric about the origin f ( x ) = 2 x 3 − 5 x f(x) = 2x^3 - 5x f ( x ) = 2 x 3 − 5 x

Neither f ( − x ) ≠ ± f ( x ) f(-x) \neq \pm f(x) f ( − x )  = ± f ( x ) No standard axis/origin symmetry f ( x ) = x 2 + 2 x + 1 f(x) = x^2 + 2x + 1 f ( x ) = x 2 + 2 x + 1

6. Complete Graph Transformation Matrix

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

Transformation Operation Formula Geometric Effect on Graph

Vertical Shift Up y = f ( x ) + k ( k > 0 ) y = f(x) + k \quad (k > 0) y = f ( x ) + k ( k > 0 ) Shift graph UP by k k k units

Vertical Shift Down y = f ( x ) − k ( k > 0 ) y = f(x) - k \quad (k > 0) y = f ( x ) − k ( k > 0 ) Shift graph DOWN by k k k units

Horizontal Shift Right y = f ( x − c ) ( c > 0 ) y = f(x - c) \quad (c > 0) y = f ( x − c ) ( c > 0 ) Shift graph RIGHT by c c c units

Horizontal Shift Left y = f ( x + c ) ( c > 0 ) y = f(x + c) \quad (c > 0) y = f ( x + c ) ( c > 0 ) Shift graph LEFT by c c c units

  • Reflection over x-axis y = − f ( x ) y = -f(x) y = − f ( x ) Reflect graph vertically across the x-axis
  • Reflection over y-axis y = f ( − x ) y = f(-x) y = f ( − x ) Reflect graph horizontally across the y-axis

Vertical Stretch y = a f ( x ) ( a > 1 ) y = a f(x) \quad (a > 1) y = a f ( x ) ( a > 1 ) Stretch graph vertically by factor a a a

Vertical Compression y = a f ( x ) ( 0 < a < 1 ) y = a f(x) \quad (0 < a < 1) y = a f ( x ) ( 0 < a < 1 ) Compress graph vertically by factor a a a

Horizontal Compression y = f ( b x ) ( b > 1 ) y = f(b x) \quad (b > 1) y = f ( b x ) ( b > 1 ) Compress graph horizontally by factor 1 / b 1/b 1/ b

Horizontal Stretch y = f ( b x ) ( 0 < b < 1 ) y = f(b x) \quad (0 < b < 1) y = f ( b x ) ( 0 < b < 1 ) Stretch graph horizontally by factor 1 / b 1/b 1/ b

  • Worked Step-by-Step ACET Exam Problems
  • Problem 1: Finding an Inverse Function with Domain & Range Analysis

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

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

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

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

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

Denominator of f − 1 f^{-1} f − 1 cannot be zero: x ≠ 2 ⟹ Domain ( f − 1 ) = R ∖ { 2 } x \neq 2 \implies \text{Domain}(f^{-1}) = \mathbb{R} \setminus \{2\} x  = 2 ⟹ Domain ( f − 1 ) = R ∖ { 2 } .

Therefore, Range ( f ) = Domain ( f − 1 ) = R ∖ { 2 } \text{Range}(f) = \text{Domain}(f^{-1}) = \mathbb{R} \setminus \{2\} Range ( f ) = Domain ( f − 1 ) = R ∖ { 2 } , and Range ( f − 1 ) = Domain ( f ) = R ∖ { 4 } \text{Range}(f^{-1}) = \text{Domain}(f) = \mathbb{R} \setminus \{4\} Range ( f − 1 ) = Domain ( f ) = R ∖ { 4 } .

Problem 2: Multi-Step Graph Transformation Tracking

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

Step 1: Trace horizontal coordinate changes. The horizontal transformation f ( x − 1 ) f(x - 1) f ( x − 1 ) shifts the graph RIGHT by 1 unit. x new = x old + 1 = 3 + 1 = 4 x_{\text{new}} = x_{\text{old}} + 1 = 3 + 1 = 4 x new ​ = x 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 new = − 2 ⋅ y old + 5 = − 2 ( − 4 ) + 5 = 8 + 5 = 13 y_{\text{new}} = -2 \cdot y_{\text{old}} + 5 = -2(-4) + 5 = 8 + 5 = 13 y new ​ = − 2 ⋅ y old ​ + 5 = − 2 ( − 4 ) + 5 = 8 + 5 = 13 .

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

ACET Speed Tactics & Traps Summary

Problem Type Common Student Trap ACET Speed Tactic

Horizontal Shift Moving f ( x − c ) f(x-c) f ( x − c ) left instead of right Remember: Horizontal transformations operate inversely

Function Inverse Forgetting to swap x x x and y y y before solving Swap variables first, then isolate y y y

Symmetry Test Testing only positive values for x x x Evaluate f ( − x ) f(-x) f ( − x ) algebraically and compare with − f ( x ) -f(x) − f ( x )

Composition Domain Finding domain of f ( g ( x ) ) f(g(x)) f ( g ( x )) using only f ( x ) f(x) f ( x ) Check domain of inner function g ( x ) g(x) g ( x ) first

Piecewise Boundaries Evaluating boundary points in both branches Check open vs. closed interval inequalities (< vs <=)