Study guide
8.1 Function Notation, Transformations, Compositions & Inverses
ACCUPLACER Suite Study Guide Exam
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1 Chapter 1: ACCUPLACER Suite Overview & Testing Mechanics
1.1 The ACCUPLACER Testing System & Computer-Adaptive Mechanics 1.2 Placement Policies, Score Scales & Institutional Cut Scores 1.3 Test-Taking Strategies, Tool Mechanics & Pacing
2 Chapter 2: Next-Generation Reading
2.1 The Reading Test Blueprint: Set-Based and Discrete Items 2.2 Information and Ideas: Close Reading & Central Themes 2.3 Summarizing and Understanding Relationships 2.4 Rhetoric, Text Structure & Author's Purpose 2.5 Analyzing Point of View and Authorial Purpose 2.6 Analyzing Arguments: Claims, Counterclaims, and Evidence 2.7 Synthesis & Paired Text Comparative Analysis 2.8 Vocabulary in Context & Figurative Language 2.9 Passage Genre, Content Area, Mode, and Text Complexity 2.10 Answer-Choice Traps and Evidence Discipline
3 Chapter 3: Next-Generation Writing: Expression of Ideas
3.1 The Writing Test Blueprint: Passage Sets and Content Distribution 3.2 Development: Proposition and Supporting Evidence 3.3 Focus: Adding, Retaining, and Deleting for Relevance 3.4 Organization: Logical Sequence and Paragraph Order 3.5 Introductions, Conclusions, and Transitional Cohesion 3.6 Effective Language Use: Precision and Word Choice 3.7 Concision: Eliminating Wordiness and Redundancy 3.8 Style, Tone, and Voice Consistency 3.9 Syntax: Sentence Combining for Rhetorical Purpose
4 Chapter 4: Next-Generation Writing: Standard English Conventions
4.1 Sentence Boundaries: Fragments, Run-ons, and Comma Splices 4.2 Subordination and Coordination 4.3 Parallel Structure 4.4 Modifier Placement: Misplaced and Dangling Modifiers 4.5 Inappropriate Shifts in Verb Tense 4.6 Inappropriate Shifts in Verb Voice and Mood 4.7 Inappropriate Shifts in Pronoun Person and Number 4.8 Subject-Verb Agreement, Pronoun Case, and Verb Forms 4.9 Possessive Determiners, Contractions, and Confusable Adverbs 4.10 Pronoun Clarity and Pronoun-Antecedent Agreement 4.11 Frequently Confused Words, Noun Agreement, and Logical Comparison 4.12 Within-Sentence Punctuation: Semicolons, Colons, Dashes, and Apostrophes 4.13 Items in a Series and Nonrestrictive Elements 4.14 End-of-Sentence Punctuation, Hyphenation, and Unnecessary Punctuation
5 Chapter 5: Next-Generation Arithmetic
5.1 Whole Number Operations, Estimation & Multi-Step Applications 5.2 Fractions, Mixed Numbers & Operations 5.3 Decimal Operations & Place-Value Conversions 5.4 Percent Calculations, Increase, Decrease, and Applications 5.5 Number Comparisons and Equivalents Across Formats
6 Chapter 6: Next-Generation Quantitative Reasoning, Algebra, and Statistics (QAS)
6.1 Rational Numbers, Absolute Value, and Order of Operations 6.2 Exponents, Radicals, and Scientific Notation 6.3 Ratio, Proportional Relationships, and Unit Conversions 6.4 Creating and Evaluating Algebraic Expressions 6.5 Linear Equations, Inequalities, and Systems of Two Equations 6.6 Linear Applications, Functions, and Graphs 6.7 Probability and Sets 6.8 Descriptive Statistics: Center, Spread, and Graphical Displays 6.9 Geometry for Pre-Algebra and Algebra 1
7 Chapter 7: Next-Generation Advanced Algebra, Quadratics & Polynomials
7.1 Advanced Linear Equations, Systems, and Coordinate Geometry 7.2 Factoring Methods for Quadratics, Cubics, and Polynomials 7.3 Quadratic Equations, Vertex Form & Parabolic Functions 7.4 Polynomial Equations, Roots, and Graphs 7.5 Radical and Rational Equations and Functions 7.6 Geometry for Algebra 1 and Algebra 2
8 Chapter 8: Functions, Exponentials, Logarithms & Trigonometry
8.1 Function Notation, Transformations, Compositions & Inverses 8.2 Domain, Range, and Interpreting Functions in Context 8.3 Exponential & Logarithmic Functions, Laws & Modeling 8.4 Right-Triangle Trigonometry, the Unit Circle, and Radian Measure 8.5 Law of Sines, Law of Cosines, and Trigonometric Equations 8.6 Choosing Your Math Test: Arithmetic, QAS, or AAF
9 Chapter 9: WritePlacer & WritePlacer ESL Essays
9.1 WritePlacer Prompt Format, Planning, and the Writing Process 9.2 The WritePlacer Scoring Rubric & 6 Core Dimensions 9.3 Reading the WritePlacer Score Scale: What Each Band Demonstrates 9.4 Essay Architecture: Thesis, Evidence Integration & Structure 9.5 WritePlacer ESL Scoring Standards & Synthesis
10 Chapter 10: ACCUPLACER ESL Placement Suite
10.1 The ESL Placement Suite: Structure, Scoring, and Score Interpretation 10.2 ESL Reading Skills & Passage Comprehension 10.3 ESL Sentence Meaning & Context Clues 10.4 ESL Language Use & Grammatical Mechanics 10.5 ESL Listening Comprehension & Spoken Contexts
8.1 Function Notation, Transformations, Compositions & Inverses
Key Takeaways Explain This Topic AI
A relation constitutes a function if and only if each input element in the domain maps to exactly one output value in the range; geometrically, this is confirmed when every vertical line intersects the graph at most once (Vertical Line Test).
Domain restrictions for real-valued functions occur when real outputs cannot be produced: denominators cannot equal zero (division by zero is undefined), and radicands of even-indexed roots must be non-negative (to avoid non-real complex numbers).
Function transformations systematically modify parent graphs: f(x) ± k shifts vertically, f(x ± h) shifts horizontally (left for +h, right for -h), a·f(x) stretches or compresses vertically, f(bx) stretches or compresses horizontally by factor 1/|b|, -f(x) reflects across the x-axis, and f(-x) reflects across the y-axis.
The composition (f ∘ g)(x) = f(g(x)) evaluates the inner function g(x) first and maps the result through the outer function f; its domain requires x to be in the domain of g and the output g(x) to reside within the domain of f.
An inverse function f^(-1)(x) reverses the input-output mapping of a one-to-one function (verified by the Horizontal Line Test), satisfying f(f^(-1)(x)) = x and f^(-1)(f(x)) = x; finding f^(-1)(x) algebraically requires swapping x and y and isolating y, representing a reflection across the diagonal line y = x.
Last updated: August 2026
Function Notation, Transformations, Compositions & Inverses
Core Advanced Algebra: Functions form the foundational language of advanced mathematics on the ACCUPLACER Advanced Algebra and Functions (AAF) assessment. Mastery requires analyzing input-output mappings, establishing analytical domain constraints, executing algebraic combinations and compositions, applying multi-step parent function transformations, and deriving inverse functions algebraically and graphically.
Function Fundamentals, Relations & The Vertical Line Test
A relation is any set of ordered pairs ( x , y ) (x, y) ( x , y ) establishing a relationship between an input set and an output set. A function is a specialized relation with a strict rule: each input element x x x in the domain maps to exactly one output value y y y in the range . FUNCTION (Valid Mapping) NOT A FUNCTION (Invalid Mapping)
- Domain (Inputs) Range (Outputs) Domain (Inputs) Range (Outputs)
- [ 1 ] ─────────> [ 4 ] [ 1 ] ─────────> [ 4 ]
- [ 2 ] ─────────> [ 5 ] [ 2 ] ──┬──────> [ 5 ]
- [ 3 ] ─────────> [ 6 ] └──────> [ 6 ] (One input maps
- (Each input has exactly one output) to two outputs!)
- The Vertical Line Test (VLT)
Geometrically, a curve plotted in the Cartesian coordinate plane represents y y y as a function of x x x if and only if no vertical line intersects the curve at more than one point .
If a vertical line x = c x = c x = c intersects a curve at two or more distinct points ( c , y 1 ) (c, y_1) ( c , y 1 ) and ( c , y 2 ) (c, y_2) ( c , y 2 ) where y 1 ≠ y 2 y_1 \neq y_2 y 1 = y 2 , the single input c c c produces multiple outputs, violating the definition of a function.
Example: The circle x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25 fails the vertical line test (at x = 0 x = 0 x = 0 , y = 5 y = 5 y = 5 and y = − 5 y = -5 y = − 5 ). However, the upper semicircle y = 25 − x 2 y = \sqrt{25 - x^2} y = 25 − x 2 passes the vertical line test and is a valid function.
Determining Domain & Range Restrictions
The domain of a real-valued function f ( x ) f(x) f ( x ) is the complete set of all real input values x x x for which the function produces a real output. The range is the resulting set of all real output values y = f ( x ) y = f(x) y = f ( x ) .
The Two Universal Algebraic Domain Restrictions
When working in the real number system ( R \mathbb{R} R ), domain restrictions arise primarily from two algebraic conditions:
Rational Denominator Constraint (Division by Zero):
Division by zero is mathematically undefined.
For any rational algebraic expression f ( x ) = p ( x ) q ( x ) f(x) = \frac{p(x)}{q(x)} f ( x ) = q ( x ) p ( x ) , set the denominator equal to zero ( q ( x ) = 0 q(x) = 0 q ( x ) = 0 ) and exclude those roots from the domain.
Even Radical Constraint (Negative Radicands):
Even-indexed roots ( x , x 4 , x 6 \sqrt{\phantom{x}}, \sqrt[4]{\phantom{x}}, \sqrt[6]{\phantom{x}} x , 4 x , 6 x ) of negative numbers produce imaginary numbers and are excluded from real-valued domains.
For f ( x ) = g ( x ) 2 n f(x) = \sqrt[2n]{g(x)} f ( x ) = 2 n g ( x ) , enforce the inequality g ( x ) ≥ 0 g(x) \ge 0 g ( x ) ≥ 0 .
Note: Odd-indexed roots ( x 3 , x 5 \sqrt[3]{\phantom{x}}, \sqrt[5]{\phantom{x}} 3 x , 5 x ) are defined for all real numbers (e.g., − 8 3 = − 2 \sqrt[3]{-8} = -2 3 − 8 = − 2 ), so odd radicals introduce no domain restrictions .
Combined Denominator and Radical Constraint:
If an even radical appears in a denominator, f ( x ) = 1 g ( x ) f(x) = \frac{1}{\sqrt{g(x)}} f ( x ) = g ( x ) 1 , the radicand cannot be negative AND cannot equal zero, establishing the strict inequality g ( x ) > 0 g(x) > 0 g ( x ) > 0 .
Worked Example: Determining Analytical Domain
Find the domain of the function f ( x ) = 2 x + 10 x 2 − 2 x − 15 f(x) = \frac{\sqrt{2x + 10}}{x^2 - 2x - 15} f ( x ) = x 2 − 2 x − 15 2 x + 10 in interval notation.
Step 1 (Radicand constraint): The numerator contains a square root, requiring the radicand to be non-negative: 2 x + 10 ≥ 0 ⟹ 2 x ≥ − 10 ⟹ x ≥ − 5 2x + 10 \ge 0 \implies 2x \ge -10 \implies x \ge -5 2 x + 10 ≥ 0 ⟹ 2 x ≥ − 10 ⟹ x ≥ − 5
Step 2 (Denominator constraint): The denominator cannot equal zero: x 2 − 2 x − 15 ≠ 0 ⟹ ( x − 5 ) ( x + 3 ) ≠ 0 ⟹ x ≠ 5 and x ≠ − 3 x^2 - 2x - 15 \neq 0 \implies (x - 5)(x + 3) \neq 0 \implies x \neq 5 \quad \text{and} \quad x \neq -3 x 2 − 2 x − 15 = 0 ⟹ ( x − 5 ) ( x + 3 ) = 0 ⟹ x = 5 and x = − 3
Step 3 (Intersection of conditions): Combine x ≥ − 5 x \ge -5 x ≥ − 5 with the exclusions x ≠ − 3 x \neq -3 x = − 3 and x ≠ 5 x \neq 5 x = 5 : Domain = [ − 5 , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ ) \text{Domain} = [-5, -3) \cup (-3, 5) \cup (5, \infty) Domain = [ − 5 , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ )
Function Evaluation & The Algebra of Functions
Function evaluation involves substituting a specified numerical value or algebraic expression into every occurrence of the independent variable x x x .
Evaluating Algebraic Inputs & The Difference Quotient
Given f ( x ) = 3 x 2 − 4 x + 1 f(x) = 3x^2 - 4x + 1 f ( x ) = 3 x 2 − 4 x + 1 :
f ( − 2 ) = 3 ( − 2 ) 2 − 4 ( − 2 ) + 1 = 3 ( 4 ) + 8 + 1 = 21 f(-2) = 3(-2)^2 - 4(-2) + 1 = 3(4) + 8 + 1 = 21 f ( − 2 ) = 3 ( − 2 ) 2 − 4 ( − 2 ) + 1 = 3 ( 4 ) + 8 + 1 = 21
f ( a + 2 ) = 3 ( a + 2 ) 2 − 4 ( a + 2 ) + 1 = 3 ( a 2 + 4 a + 4 ) − 4 a − 8 + 1 = 3 a 2 + 12 a + 12 − 4 a − 7 = 3 a 2 + 8 a + 5 f(a + 2) = 3(a + 2)^2 - 4(a + 2) + 1 = 3(a^2 + 4a + 4) - 4a - 8 + 1 = 3a^2 + 12a + 12 - 4a - 7 = 3a^2 + 8a + 5 f ( a + 2 ) = 3 ( a + 2 ) 2 − 4 ( a + 2 ) + 1 = 3 ( a 2 + 4 a + 4 ) − 4 a − 8 + 1 = 3 a 2 + 12 a + 12 − 4 a − 7 = 3 a 2 + 8 a + 5
The Difference Quotient measures the average rate of change across an interval of width h h h : f ( x + h ) − f ( x ) h ( h ≠ 0 ) \frac{f(x + h) - f(x)}{h} \quad (h \neq 0) h f ( x + h ) − f ( x ) ( h = 0 ) [ 3 ( x + h ) 2 − 4 ( x + h ) + 1 ] − [ 3 x 2 − 4 x + 1 ] h = 3 x 2 + 6 x h + 3 h 2 − 4 x − 4 h + 1 − 3 x 2 + 4 x − 1 h = 6 x h + 3 h 2 − 4 h h = 6 x + 3 h − 4 \frac{[3(x+h)^2 - 4(x+h) + 1] - [3x^2 - 4x + 1]}{h} = \frac{3x^2 + 6xh + 3h^2 - 4x - 4h + 1 - 3x^2 + 4x - 1}{h} = \frac{6xh + 3h^2 - 4h}{h} = 6x + 3h - 4 h [ 3 ( x + h ) 2 − 4 ( x + h ) + 1 ] − [ 3 x 2 − 4 x + 1 ] = h 3 x 2 + 6 x h + 3 h 2 − 4 x − 4 h + 1 − 3 x 2 + 4 x − 1 = h 6 x h + 3 h 2 − 4 h = 6 x + 3 h − 4
Operations on Functions
For functions f ( x ) f(x) f ( x ) and g ( x ) g(x) g ( x ) , standard arithmetic operations produce new functions:
Operation Algebraic Definition Domain Rule
Sum ( f + g ) ( x ) = f ( x ) + g ( x ) (f + g)(x) = f(x) + g(x) ( f + g ) ( x ) = f ( x ) + g ( x ) Domain ( f ) ∩ Domain ( g ) \text{Domain}(f) \cap \text{Domain}(g) Domain ( f ) ∩ Domain ( g )
Difference ( f − g ) ( x ) = f ( x ) − g ( x ) (f - g)(x) = f(x) - g(x) ( f − g ) ( x ) = f ( x ) − g ( x ) Domain ( f ) ∩ Domain ( g ) \text{Domain}(f) \cap \text{Domain}(g) Domain ( f ) ∩ Domain ( g )
Product ( f g ) ( x ) = f ( x ) ⋅ g ( x ) (fg)(x) = f(x) \cdot g(x) ( f g ) ( x ) = f ( x ) ⋅ g ( x ) Domain ( f ) ∩ Domain ( g ) \text{Domain}(f) \cap \text{Domain}(g) Domain ( f ) ∩ Domain ( g )
Quotient ( f g ) ( x ) = f ( x ) g ( x ) \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} ( g f ) ( x ) = g ( x ) f ( x ) { x ∈ Domain ( f ) ∩ Domain ( g ) ∣ g ( x ) ≠ 0 } \{x \in \text{Domain}(f) \cap \text{Domain}(g) \mid g(x) \neq 0\} { x ∈ Domain ( f ) ∩ Domain ( g ) ∣ g ( x ) = 0 }
Composite Functions: Evaluation & Domain Restrictions
A composite function ( f ∘ g ) ( x ) (f \circ g)(x) ( f ∘ g ) ( x ) applies the output of an inner function g ( x ) g(x) g ( x ) as the direct input to an outer function f ( x ) f(x) f ( x ) : ( f ∘ g ) ( x ) = f ( g ( x ) ) ("f of g of x") (f \circ g)(x) = f(g(x)) \quad \text{("f of g of x")} ( f ∘ g ) ( x ) = f ( g ( x )) ("f of g of x") Composite Function Pipeline
- Input x ───> [ Inner Function g(x) ] ───> Output g(x) ───> [ Outer Function f(u) ] ───> f(g(x))
- Non-Commutative Property of Composition
In general, function composition is not commutative : ( f ∘ g ) ( x ) ≠ ( g ∘ f ) ( x ) (f \circ g)(x) \neq (g \circ f)(x) ( f ∘ g ) ( x ) = ( g ∘ f ) ( x ) .
Determining the Domain of a Composite Function ( f ∘ g ) ( x ) (f \circ g)(x) ( f ∘ g ) ( x )
The domain of ( f ∘ g ) ( x ) (f \circ g)(x) ( f ∘ g ) ( x ) consists of all real numbers x x x satisfying two strict criteria:
x x x must be in the domain of the inner function g ( x ) g(x) g ( x ) .
The output g ( x ) g(x) g ( x ) must be in the domain of the outer function f ( x ) f(x) f ( x ) .
Worked Example: Composite Function and Composite Domain
Let f ( x ) = 4 x − 3 f(x) = \frac{4}{x - 3} f ( x ) = x − 3 4 and g ( x ) = x + 2 g(x) = \sqrt{x + 2} g ( x ) = x + 2 . Find ( f ∘ g ) ( x ) (f \circ g)(x) ( f ∘ g ) ( x ) and state its domain.
Step 1 (Form composite equation): ( f ∘ g ) ( x ) = f ( g ( x ) ) = f ( x + 2 ) = 4 x + 2 − 3 (f \circ g)(x) = f(g(x)) = f(\sqrt{x + 2}) = \frac{4}{\sqrt{x + 2} - 3} ( f ∘ g ) ( x ) = f ( g ( x )) = f ( x + 2 ) = x + 2 − 3 4
Step 2 (Inner function domain): For g ( x ) = x + 2 g(x) = \sqrt{x + 2} g ( x ) = x + 2 , the radicand must be non-negative: x + 2 ≥ 0 ⟹ x ≥ − 2 x + 2 \ge 0 \implies x \ge -2 x + 2 ≥ 0 ⟹ x ≥ − 2 .
Step 3 (Outer function restriction): For f ( u ) = 4 u − 3 f(u) = \frac{4}{u - 3} f ( u ) = u − 3 4 , the denominator cannot equal zero: u ≠ 3 ⟹ g ( x ) ≠ 3 u \neq 3 \implies g(x) \neq 3 u = 3 ⟹ g ( x ) = 3 . x + 2 ≠ 3 ⟹ x + 2 ≠ 9 ⟹ x ≠ 7 \sqrt{x + 2} \neq 3 \implies x + 2 \neq 9 \implies x \neq 7 x + 2 = 3 ⟹ x + 2 = 9 ⟹ x = 7
Step 4 (Combine restrictions): Combine x ≥ − 2 x \ge -2 x ≥ − 2 with x ≠ 7 x \neq 7 x = 7 : Domain ( f ∘ g ) = [ − 2 , 7 ) ∪ ( 7 , ∞ ) \text{Domain}(f \circ g) = [-2, 7) \cup (7, \infty) Domain ( f ∘ g ) = [ − 2 , 7 ) ∪ ( 7 , ∞ )
Master Reference: Function Transformations
Transformations modify parent functions geometrically through shifts, scalings, and reflections. The generalized transformation format for any parent function f ( x ) f(x) f ( x ) is: g ( x ) = a ⋅ f ( b ( x − h ) ) + k g(x) = a \cdot f(b(x - h)) + k g ( x ) = a ⋅ f ( b ( x − h )) + k
Transformation Rules Reference Table
Transformation Type Algebraic Form Geometric Effect on Graph Coordinate Point Mapping
Vertical Shift Up y = f ( x ) + k y = f(x) + k y = f ( x ) + k ( k > 0 k > 0 k > 0 ) Shifts entire graph upward by k k k units ( x , y ) → ( x , y + k ) (x, y) \to (x, y + k) ( x , y ) → ( x , y + k )
Vertical Shift Down y = f ( x ) − k y = f(x) - k y = f ( x ) − k ( k > 0 k > 0 k > 0 ) Shifts entire graph downward by k k k units ( x , y ) → ( x , y − k ) (x, y) \to (x, y - k) ( x , y ) → ( x , y − k )
Horizontal Shift Right y = f ( x − h ) y = f(x - h) y = f ( x − h ) ( h > 0 h > 0 h > 0 ) Shifts entire graph rightward by h h h units ( x , y ) → ( x + h , y ) (x, y) \to (x + h, y) ( x , y ) → ( x + h , y )
Horizontal Shift Left y = f ( x + h ) y = f(x + h) y = f ( x + h ) ( h > 0 h > 0 h > 0 ) Shifts entire graph leftward by h h h units ( x , y ) → ( x − h , y ) (x, y) \to (x - h, y) ( x , y ) → ( x − h , y )
Vertical Stretch y = a ⋅ f ( x ) y = a \cdot f(x) y = a ⋅ f ( x ) ( ∣ a ∣ > 1 \vert a\vert> 1 ∣ a ∣ > 1 ) Stretches vertically by factor of ∣ a ∣ \vert a\vert ∣ a ∣ (away from x x x -axis) ( x , y ) → ( x , a ⋅ y ) (x, y) \to (x, a \cdot y) ( x , y ) → ( x , a ⋅ y )
Vertical Compression y = a ⋅ f ( x ) y = a \cdot f(x) y = a ⋅ f ( x ) ( 0 < ∣ a ∣ < 1 0 < \vert a\vert< 1 0 < ∣ a ∣ < 1 ) Compresses vertically by factor of ∣ a ∣ \vert a\vert ∣ a ∣ (toward x x x -axis) ( x , y ) → ( x , a ⋅ y ) (x, y) \to (x, a \cdot y) ( x , y ) → ( x , a ⋅ y )
Horizontal Compression y = f ( b ⋅ x ) y = f(b \cdot x) y = f ( b ⋅ x ) ( ∣ b ∣ > 1 \vert b\vert> 1 ∣ b ∣ > 1 ) Compresses horizontally by factor of 1 ∣ b ∣ \frac{1}{\vert b\vert} ∣ b ∣ 1 (toward y y y -axis) ( x , y ) → ( x b , y ) (x, y) \to \left(\frac{x}{b}, y\right) ( x , y ) → ( b x , y )
Horizontal Stretch y = f ( b ⋅ x ) y = f(b \cdot x) y = f ( b ⋅ x ) ( 0 < ∣ b ∣ < 1 0 < \vert b\vert< 1 0 < ∣ b ∣ < 1 ) Stretches horizontally by factor of 1 ∣ b ∣ \frac{1}{\vert b\vert} ∣ b ∣ 1 (away from y y y -axis) ( x , y ) → ( x b , y ) (x, y) \to \left(\frac{x}{b}, y\right) ( x , y ) → ( b x , y )
Reflection over x x x -axis y = − f ( x ) y = -f(x) y = − f ( x ) Reflects graph vertically across the x x x -axis ( x , y ) → ( x , − y ) (x, y) \to (x, -y) ( x , y ) → ( x , − y )
Reflection over y y y -axis y = f ( − x ) y = f(-x) y = f ( − x ) Reflects graph horizontally across the y y y -axis ( x , y ) → ( − x , y ) (x, y) \to (-x, y) ( x , y ) → ( − x , y )
The Standard Order of Transformations
When applying multiple transformations to a parent function, execute operations in the following order:
- Horizontal Shifts ( x − h x - h x − h )
- Horizontal Stretch / Compression / Reflection ( b ⋅ x b \cdot x b ⋅ x and negative sign inside)
- Vertical Stretch / Compression / Reflection ( a ⋅ f ( u ) a \cdot f(u) a ⋅ f ( u ) and negative sign outside)
- Vertical Shifts ( + k + k + k )
- One-to-One Functions, Horizontal Line Test & Inverse Functions
A function f f f is one-to-one (injective) if no two distinct inputs produce the same output: f ( x 1 ) = f ( x 2 ) ⟹ x 1 = x 2 f(x_1) = f(x_2) \implies x_1 = x_2 f ( x 1 ) = f ( x 2 ) ⟹ x 1 = x 2 .
The Horizontal Line Test (HLT)
A function f ( x ) f(x) f ( x ) possesses an inverse function f − 1 ( x ) f^{-1}(x) f − 1 ( x ) if and only if no horizontal line intersects its graph at more than one point .
Quadratic parent f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 fails the HLT (horizontal line y = 4 y = 4 y = 4 intersects at x = − 2 x = -2 x = − 2 and x = 2 x = 2 x = 2 ). It has no inverse over ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ ) unless its domain is restricted (e.g., x ≥ 0 x \ge 0 x ≥ 0 ).
Cubic parent f ( x ) = x 3 f(x) = x^3 f ( x ) = x 3 passes the HLT and is one-to-one over its entire real domain.
Fundamental Properties of Inverse Functions
Cancellation Identity: f ( f − 1 ( x ) ) = x f(f^{-1}(x)) = x f ( f − 1 ( x )) = x for all x x x in the domain of f − 1 f^{-1} f − 1 , and f − 1 ( f ( x ) ) = x f^{-1}(f(x)) = x f − 1 ( f ( x )) = x for all x x x in the domain of f f f .
Domain-Range Reversal: Domain ( f − 1 ) = Range ( f ) and Range ( f − 1 ) = Domain ( f ) \text{Domain}(f^{-1}) = \text{Range}(f) \quad \text{and} \quad \text{Range}(f^{-1}) = \text{Domain}(f) Domain ( f − 1 ) = Range ( f ) and Range ( f − 1 ) = Domain ( f )
Geometric Symmetry: The graph of y = f − 1 ( x ) y = f^{-1}(x) y = f − 1 ( x ) is the reflection of the graph of y = f ( x ) y = f(x) y = f ( x ) across the identity line y = x y = x y = x . Every coordinate pair ( a , b ) (a, b) ( a , b ) on f f f becomes ( b , a ) (b, a) ( b , a ) on f − 1 f^{-1} f − 1 .
The 4-Step Algebraic Procedure for Finding f − 1 ( x ) f^{-1}(x) f − 1 ( x )
Replace f ( x ) f(x) f ( x ) with y y y .
Swap variables x x x and y y y (interchange every x x x with y y y and every y y y with x x x ).
Solve the resulting equation algebraically for y y y .
Replace y y y with the formal inverse notation f − 1 ( x ) f^{-1}(x) f − 1 ( x ) .
Worked Example: Finding the Inverse of a Rational Function
Find the inverse function f − 1 ( x ) f^{-1}(x) f − 1 ( x ) for f ( x ) = 3 x + 2 5 x − 4 f(x) = \frac{3x + 2}{5x - 4} f ( x ) = 5 x − 4 3 x + 2 where x ≠ 4 5 x \neq \frac{4}{5} x = 5 4 .
Step 1 (Set y = f ( x ) y = f(x) y = f ( x ) ): y = 3 x + 2 5 x − 4 y = \frac{3x + 2}{5x - 4} y = 5 x − 4 3 x + 2
Step 2 (Swap x x x and y y y ): x = 3 y + 2 5 y − 4 x = \frac{3y + 2}{5y - 4} x = 5 y − 4 3 y + 2
Step 3 (Clear denominator): Multiply both sides by ( 5 y − 4 ) (5y - 4) ( 5 y − 4 ) : x ( 5 y − 4 ) = 3 y + 2 ⟹ 5 x y − 4 x = 3 y + 2 x(5y - 4) = 3y + 2 \implies 5xy - 4x = 3y + 2 x ( 5 y − 4 ) = 3 y + 2 ⟹ 5 x y − 4 x = 3 y + 2
Step 4 (Collect y y y -terms): Move all terms containing y y y to the left and non- y y y terms to the right: 5 x y − 3 y = 4 x + 2 5xy - 3y = 4x + 2 5 x y − 3 y = 4 x + 2
Step 5 (Factor and isolate y y y ): y ( 5 x − 3 ) = 4 x + 2 ⟹ y = 4 x + 2 5 x − 3 y(5x - 3) = 4x + 2 \implies y = \frac{4x + 2}{5x - 3} y ( 5 x − 3 ) = 4 x + 2 ⟹ y = 5 x − 3 4 x + 2
Step 6 (Formal notation): f − 1 ( x ) = 4 x + 2 5 x − 3 f^{-1}(x) = \frac{4x + 2}{5x - 3} f − 1 ( x ) = 5 x − 3 4 x + 2 with domain x ≠ 3 5 x \neq \frac{3}{5} x = 5 3 .