Study guide
3.2 Composition of Functions & Transformations
Bocconi Admission Test Online Bocconi Test for Bachelor Programs
Track your progress as you read
Finished sections get a check mark.
1 Chapter 1: Exam Structure, Scoring Mechanics & Test Strategy
1.1 Test Architecture, Timing & the Sequential Test Interface 1.2 Scoring System, Penalty Mathematics & Strategic Answering
2 Chapter 2: Mathematics — Algebra, Equations & Inequalities
2.1 Algebraic Expressions, Polynomials, Powers & Radicals 2.2 Linear & Quadratic Equations and Systems 2.3 Linear & Quadratic Inequalities and Sign Charts 2.4 Equations and Inequalities with Absolute Values
3 Chapter 3: Mathematics — Functions, Exponents & Logarithms
3.1 Function Concepts, Domain, Range & Inverses 3.2 Composition of Functions & Transformations 3.3 Exponential Functions, Properties & Equations 3.4 Logarithmic Functions, Laws & Equations
4 Chapter 4: Mathematics — Geometry, Analytic Geometry & Trigonometry
4.1 Plane Geometry: Polygons, Circles & Area Theorems 4.2 Analytic Geometry: Lines, Slopes & Distances in the Cartesian Plane 4.3 Conic Sections: Circles & Parabolas in the Coordinate Plane 4.4 Trigonometric Functions, Identities & Right-Triangle Applications
5 Chapter 5: Mathematics — Combinatorics, Probability, Statistics & Set Theory
5.1 Set Theory, Operations & Venn Diagrams 5.2 Combinatorics: Permutations, Combinations & Factorials 5.3 Probability Theory: Independent Events, Conditional Probability & Bayes' Basics 5.4 Descriptive Statistics: Central Tendency, Dispersion & Frequency Distributions
6 Chapter 6: Reading Comprehension — Business, Economic & Academic Analysis
6.1 Main Idea, Text Structure & Passage Mapping 6.2 Explicit Information Retrieval & Detail Verification 6.3 Inference, Implicit Meaning & Authorial Purpose 6.4 Tone, Rhetorical Devices & Vocabulary in Academic Context
7 Chapter 7: Numerical Reasoning — Charts, Tables & Data Interpretation
7.1 Data Tables & Multi-Series Matrices 7.2 Graphical Interpretation: Bar, Line, Pie & Composite Charts 7.3 Percentage Calculations, Changes & Financial Metrics 7.4 Ratios, Proportions, Rates & Weighted Averages
8 Chapter 8: Logic & Critical Thinking — Formal & Deductive Reasoning
8.1 Propositional Logic, Truth Tables & Logical Connectives 8.2 Conditional Reasoning, Contrapositives & Modus Ponens/Tollens 8.3 Syllogisms, Quantifiers & Venn Logic
9 Chapter 9: Critical Thinking — Argument Analysis, Assumptions & the Official Item Types
9.1 Argument Components, Premises & Unstated Assumptions 9.2 Strengthening, Weakening & Flaw Identification 9.3 Constraint Deduction Drills: Ordering, Arrangement & Assignment 9.4 The Two Official Critical-Thinking Item Types: Data Propositions & Deducibility
3.2 Composition of Functions & Transformations
Key Takeaways Explain This Topic AI
Function composition ( f ∘ g ) ( x ) = f ( g ( x ) ) (f \circ g)(x) = f(g(x)) ( f ∘ g ) ( x ) = f ( g ( x )) evaluates the inner function g ( x ) g(x) g ( x ) first; it is associative but generally non-commutative ( f ( g ( x ) ) ≠ g ( f ( x ) ) f(g(x)) \neq g(f(x)) f ( g ( x )) = g ( f ( x )) ).
The domain of ( f ∘ g ) (f \circ g) ( f ∘ g ) is the set of all x ∈ Dom ( g ) x \in \text{Dom}(g) x ∈ Dom ( g ) such that g ( x ) ∈ Dom ( f ) g(x) \in \text{Dom}(f) g ( x ) ∈ Dom ( f ) ; evaluating domain after algebraic simplification introduces extraneous domain points.
Rigid graph translations shift graphs horizontally ( f ( x − c ) f(x - c) f ( x − c ) moves right by c c c for c > 0 c > 0 c > 0 ) and vertically ( f ( x ) + c f(x) + c f ( x ) + c moves up by c c c for c > 0 c > 0 c > 0 ).
Horizontal scaling acts reciprocally on the x x x -coordinates: f ( c x ) f(cx) f ( c x ) horizontally compresses the graph by factor 1 / c 1/c 1/ c when c > 1 c > 1 c > 1 and stretches by 1 / c 1/c 1/ c when 0 < c < 1 0 < c < 1 0 < c < 1 .
When analyzing compound horizontal transformations f ( b x − c ) f(bx - c) f ( b x − c ) , always factor into f ( b ( x − c / b ) ) f(b(x - c/b)) f ( b ( x − c / b )) or apply input coordinate tracking x new = ( x old + c ) / b x_{\text{new}} = (x_{\text{old}} + c)/b x new = ( x old + c ) / b .
Last updated: September 2026
3.2 Composition of Functions & Transformations
Composite functions and graphical transformations are heavily featured on the Bocconi test. They test whether candidates can manipulate abstract algebraic operations and predict the resulting geometric shifts in the Cartesian plane. Speed and precision depend on recognizing operational hierarchy and avoiding common domain-simplification traps.
1. Composition of Functions: Definition & Non-Commutativity
Given two real functions g : A → B g: A \to B g : A → B and f : B → C f: B \to C f : B → C , the composite function ( f ∘ g ) : A → C (f \circ g): A \to C ( f ∘ g ) : A → C is defined by evaluating g g g first, and then applying f f f to that intermediate result: ( f ∘ g ) ( x ) = f ( g ( x ) ) (f \circ g)(x) = f(g(x)) ( f ∘ g ) ( x ) = f ( g ( x ))
Here, g ( x ) g(x) g ( x ) is the inner function (input to f f f ) and f f f is the outer function .
Non-Commutativity of Composition
Function composition is associative — f ∘ ( g ∘ h ) = ( f ∘ g ) ∘ h f \circ (g \circ h) = (f \circ g) \circ h f ∘ ( g ∘ h ) = ( f ∘ g ) ∘ h —but it is not commutative . In general: ( f ∘ g ) ( x ) ≠ ( g ∘ f ) ( x ) (f \circ g)(x) \neq (g \circ f)(x) ( f ∘ g ) ( x ) = ( g ∘ f ) ( x )
Algebraic Demonstration: Let f ( x ) = 2 x + 5 f(x) = 2x + 5 f ( x ) = 2 x + 5 and g ( x ) = x 2 − 1 g(x) = x^2 - 1 g ( x ) = x 2 − 1 .
( f ∘ g ) ( x ) = f ( g ( x ) ) = 2 ( x 2 − 1 ) + 5 = 2 x 2 − 2 + 5 = 2 x 2 + 3 (f \circ g)(x) = f(g(x)) = 2(x^2 - 1) + 5 = 2x^2 - 2 + 5 = 2x^2 + 3 ( f ∘ g ) ( x ) = f ( g ( x )) = 2 ( x 2 − 1 ) + 5 = 2 x 2 − 2 + 5 = 2 x 2 + 3
( g ∘ f ) ( x ) = g ( f ( x ) ) = ( 2 x + 5 ) 2 − 1 = 4 x 2 + 20 x + 25 − 1 = 4 x 2 + 20 x + 24 (g \circ f)(x) = g(f(x)) = (2x + 5)^2 - 1 = 4x^2 + 20x + 25 - 1 = 4x^2 + 20x + 24 ( g ∘ f ) ( x ) = g ( f ( x )) = ( 2 x + 5 ) 2 − 1 = 4 x 2 + 20 x + 25 − 1 = 4 x 2 + 20 x + 24
Clearly, 2 x 2 + 3 ≠ 4 x 2 + 20 x + 24 2x^2 + 3 \neq 4x^2 + 20x + 24 2 x 2 + 3 = 4 x 2 + 20 x + 24 . The two expressions yield entirely different degrees, coefficients, and geometric graphs.
2. The Strict Domain of a Composite Function
A critical Bocconi exam trap is computing the algebraic formula of f ( g ( x ) ) f(g(x)) f ( g ( x )) , simplifying it, and then deducing the domain from the simplified formula. This procedure is mathematically flawed and regularly produces incorrect answers.
The Two-Condition Domain Rule
The domain of ( f ∘ g ) (f \circ g) ( f ∘ g ) consists strictly of all inputs x x x that are valid in g g g , and whose output g ( x ) g(x) g ( x ) is simultaneously valid in f f f : Dom ( f ∘ g ) = { x ∈ R ∣ x ∈ Dom ( g ) and g ( x ) ∈ Dom ( f ) } \text{Dom}(f \circ g) = \{x \in \mathbb{R} \mid x \in \text{Dom}(g) \quad \text{and} \quad g(x) \in \text{Dom}(f)\} Dom ( f ∘ g ) = { x ∈ R ∣ x ∈ Dom ( g ) and g ( x ) ∈ Dom ( f )}
Worked Example: False Simplification Fallacy
Let f ( u ) = u 2 f(u) = u^2 f ( u ) = u 2 and g ( x ) = x − 4 g(x) = \sqrt{x - 4} g ( x ) = x − 4 .
Direct algebraic composition: f ( g ( x ) ) = ( x − 4 ) 2 = x − 4 f(g(x)) = (\sqrt{x - 4})^2 = x - 4 f ( g ( x )) = ( x − 4 ) 2 = x − 4 .
Looking solely at x − 4 x - 4 x − 4 , one might mistakenly assume the domain is all of R \mathbb{R} R .
However, under the two-condition rule:
x ∈ Dom ( g ) ⟹ x − 4 ≥ 0 ⟹ x ≥ 4 x \in \text{Dom}(g) \implies x - 4 \ge 0 \implies x \ge 4 x ∈ Dom ( g ) ⟹ x − 4 ≥ 0 ⟹ x ≥ 4 .
g ( x ) ∈ Dom ( f ) ⟹ x − 4 ∈ R g(x) \in \text{Dom}(f) \implies \sqrt{x - 4} \in \mathbb{R} g ( x ) ∈ Dom ( f ) ⟹ x − 4 ∈ R , which is true for all x ≥ 4 x \ge 4 x ≥ 4 .
Thus, the true domain is [ 4 , ∞ ) [4, \infty) [ 4 , ∞ ) , not R \mathbb{R} R .
Advanced Domain Problem
Find the domain of ( f ∘ g ) ( x ) (f \circ g)(x) ( f ∘ g ) ( x ) where f ( u ) = 1 u − 2 f(u) = \frac{1}{\sqrt{u - 2}} f ( u ) = u − 2 1 and g ( x ) = 3 x + 1 x − 4 g(x) = \frac{3x + 1}{x - 4} g ( x ) = x − 4 3 x + 1 .
Condition 1 (Inner Domain): x ∈ Dom ( g ) ⟺ x − 4 ≠ 0 ⟺ x ≠ 4 x \in \text{Dom}(g) \iff x - 4 \neq 0 \iff x \neq 4 x ∈ Dom ( g ) ⟺ x − 4 = 0 ⟺ x = 4 .
Condition 2 (Outer Compatibility): g ( x ) ∈ Dom ( f ) g(x) \in \text{Dom}(f) g ( x ) ∈ Dom ( f ) . The function f ( u ) f(u) f ( u ) requires u − 2 > 0 ⟺ u > 2 u - 2 > 0 \iff u > 2 u − 2 > 0 ⟺ u > 2 . Therefore: 3 x + 1 x − 4 > 2 ⟺ 3 x + 1 x − 4 − 2 > 0 ⟺ 3 x + 1 − 2 ( x − 4 ) x − 4 > 0 ⟺ x + 9 x − 4 > 0 \frac{3x + 1}{x - 4} > 2 \iff \frac{3x + 1}{x - 4} - 2 > 0 \iff \frac{3x + 1 - 2(x - 4)}{x - 4} > 0 \iff \frac{x + 9}{x - 4} > 0 x − 4 3 x + 1 > 2 ⟺ x − 4 3 x + 1 − 2 > 0 ⟺ x − 4 3 x + 1 − 2 ( x − 4 ) > 0 ⟺ x − 4 x + 9 > 0
Constructing a sign table for x + 9 x − 4 \frac{x + 9}{x - 4} x − 4 x + 9 :
- Numerator root: x = − 9 x = -9 x = − 9
- Denominator root: x = 4 x = 4 x = 4
Test intervals: for x < − 9 x < -9 x < − 9 , ( − ) / ( − ) = ( + ) (-)/(-) = (+) ( − ) / ( − ) = ( + ) ; for − 9 < x < 4 -9 < x < 4 − 9 < x < 4 , ( + ) / ( − ) = ( − ) (+)/(-) = (-) ( + ) / ( − ) = ( − ) ; for x > 4 x > 4 x > 4 , ( + ) / ( + ) = ( + ) (+)/(+) = (+) ( + ) / ( + ) = ( + ) .
Thus, the inequality holds for x ∈ ( − ∞ , − 9 ) ∪ ( 4 , ∞ ) x \in (-\infty, -9) \cup (4, \infty) x ∈ ( − ∞ , − 9 ) ∪ ( 4 , ∞ ) . Both parts satisfy x ≠ 4 x \neq 4 x = 4 . Hence: Dom ( f ∘ g ) = ( − ∞ , − 9 ) ∪ ( 4 , ∞ ) \text{Dom}(f \circ g) = (-\infty, -9) \cup (4, \infty) Dom ( f ∘ g ) = ( − ∞ , − 9 ) ∪ ( 4 , ∞ )
3. Geometric Graph Transformations
Every functional modification of a parent function y = f ( x ) y = f(x) y = f ( x ) corresponds to a precise geometric transformation in the Cartesian plane. Transformations are categorized into vertical (acting on outputs y y y ) and horizontal (acting on inputs x x x ):
Transformation Algebraic Formula Geometric Operation Coordinate Mapping ( x , y ) ↦ (x, y) \mapsto ( x , y ) ↦
Vertical Shift y = f ( x ) + c y = f(x) + c y = f ( x ) + c Shift up if c > 0 c > 0 c > 0 , down if c < 0 c < 0 c < 0 ( x , y + c ) (x, y + c) ( x , y + c )
Horizontal Shift y = f ( x − c ) y = f(x - c) y = f ( x − c ) Shift right if c > 0 c > 0 c > 0 , left if c < 0 c < 0 c < 0 ( x + c , y ) (x + c, y) ( x + c , y )
Vertical Stretch/Compress y = a ⋅ f ( x ) y = a \cdot f(x) y = a ⋅ f ( x ) ( a > 0 a > 0 a > 0 ) Stretch if a > 1 a > 1 a > 1 , compress if 0 < a < 1 0 < a < 1 0 < a < 1 ( x , a ⋅ y ) (x, a \cdot y) ( x , a ⋅ y )
Horizontal Stretch/Compress y = f ( b ⋅ x ) y = f(b \cdot x) y = f ( b ⋅ x ) ( b > 0 b > 0 b > 0 ) Compress by 1 / b 1/b 1/ b if b > 1 b > 1 b > 1 , stretch if 0 < b < 1 0 < b < 1 0 < b < 1 ( x / b , y ) (x/b, y) ( x / b , y )
Vertical Reflection y = − f ( x ) y = -f(x) y = − f ( x ) Reflect across the x x x -axis ( x , − y ) (x, -y) ( x , − y )
Horizontal Reflection y = f ( − x ) y = f(-x) y = f ( − x ) Reflect across the y y y -axis ( − x , y ) (-x, y) ( − x , y )
Absolute Output y = ∣ f ( x ) ∣ y = \vert f(x)\vert y = ∣ f ( x ) ∣ Reflect portions below x x x -axis upwards ( x , ∣ y ∣ ) (x, \vert y\vert) ( x , ∣ y ∣ )
Absolute Input y = f ( ∣ x ∣ ) y = f(\vert x\vert) y = f ( ∣ x ∣ ) Delete x < 0 x < 0 x < 0 , reflect x ≥ 0 x \ge 0 x ≥ 0 over y y y -axis ( ± x , y ) (\pm x, y) ( ± x , y ) for x ≥ 0 x \ge 0 x ≥ 0
The Inverse Behavior of Horizontal Operations
Notice the fundamental dichotomy in graphical transformations:
Vertical operations are intuitive: adding c c c adds to y y y ; multiplying by a a a multiplies y y y .
Horizontal operations are counter-intuitive: replacing x x x with x − c x - c x − c shifts right (adds c c c to the coordinates); replacing x x x with 2 x 2x 2 x compresses coordinates horizontally by a factor of 1 2 \frac{1}{2} 2 1 .
4. Multi-Step Transformations & Order of Operations
When multiple transformations occur simultaneously, their execution order is critical. While horizontal and vertical operations are completely independent of each other, the order within horizontal or vertical sequences matters.
Factored Form vs. Unfactored Form for Horizontal Shifts
Consider the function y = f ( 2 x + 6 ) y = f(2x + 6) y = f ( 2 x + 6 ) :
Common Error: Students see + 6 +6 + 6 and assume the graph shifts left by 6 units.
Correct Algebraic Factoring: Factor out the horizontal compression factor: y = f ( 2 ( x + 3 ) ) y = f(2(x + 3)) y = f ( 2 ( x + 3 )) This reveals that the horizontal shift is actually 3 units to the left , followed by (or preceded by) a horizontal compression by a factor of 1 / 2 1/2 1/2 .
The Coordinate Tracking Method (Bocconi Speed Technique)
The most reliable, error-free method for Bocconi test questions is direct coordinate tracking .
Given a known point ( x 0 , y 0 ) (x_0, y_0) ( x 0 , y 0 ) on y = f ( x ) y = f(x) y = f ( x ) , find the corresponding point ( x new , y new ) (x_{\text{new}}, y_{\text{new}}) ( x new , y new ) on y = A ⋅ f ( B x + C ) + D y = A \cdot f(B x + C) + D y = A ⋅ f ( B x + C ) + D :
Horizontal coordinate: Set the inner expression equal to the parent input: B x new + C = x 0 ⟹ x new = x 0 − C B B x_{\text{new}} + C = x_0 \implies x_{\text{new}} = \frac{x_0 - C}{B} B x new + C = x 0 ⟹ x new = B x 0 − C
Vertical coordinate: Substitute y 0 = f ( x 0 ) y_0 = f(x_0) y 0 = f ( x 0 ) directly into the outer equation: y new = A ⋅ y 0 + D y_{\text{new}} = A \cdot y_0 + D y new = A ⋅ y 0 + D
Worked Example: Coordinate Tracking
Suppose the point ( − 4 , 7 ) (-4, 7) ( − 4 , 7 ) lies on the graph of y = f ( x ) y = f(x) y = f ( x ) . Determine the corresponding point that must lie on the graph of: g ( x ) = − 3 f ( 2 x − 8 ) + 5 g(x) = -3 f(2x - 8) + 5 g ( x ) = − 3 f ( 2 x − 8 ) + 5
Applying the tracking formulas:
Horizontal coordinate: 2 x new − 8 = − 4 ⟹ 2 x new = 4 ⟹ x new = 2 2x_{\text{new}} - 8 = -4 \implies 2x_{\text{new}} = 4 \implies x_{\text{new}} = 2 2 x new − 8 = − 4 ⟹ 2 x new = 4 ⟹ x new = 2 .
Vertical coordinate: y new = − 3 ( 7 ) + 5 = − 21 + 5 = − 16 y_{\text{new}} = -3(7) + 5 = -21 + 5 = -16 y new = − 3 ( 7 ) + 5 = − 21 + 5 = − 16 .
Therefore, the point ( 2 , − 16 ) (2, -16) ( 2 , − 16 ) is guaranteed to lie on the graph of g ( x ) g(x) g ( x ) .
5. Symmetry Properties: Even and Odd Functions
Understanding algebraic symmetry allows candidates to eliminate options immediately on coordinate questions:
Even Function (Symmetric about the y y y -axis): f ( − x ) = f ( x ) ∀ x ∈ Dom ( f ) f(-x) = f(x) \quad \forall x \in \text{Dom}(f) f ( − x ) = f ( x ) ∀ x ∈ Dom ( f ) Examples include x 2 , x 4 , ∣ x ∣ , cos ( x ) x^2, x^4, |x|, \cos(x) x 2 , x 4 , ∣ x ∣ , cos ( x ) . Polynomials with only even powers of x x x are even.
Odd Function (Symmetric about the Origin): f ( − x ) = − f ( x ) ∀ x ∈ Dom ( f ) f(-x) = -f(x) \quad \forall x \in \text{Dom}(f) f ( − x ) = − f ( x ) ∀ x ∈ Dom ( f ) Examples include x , x 3 , 1 x , sin ( x ) x, x^3, \frac{1}{x}, \sin(x) x , x 3 , x 1 , sin ( x ) . Rotating the graph 180 ∘ 180^\circ 18 0 ∘ around ( 0 , 0 ) (0, 0) ( 0 , 0 ) leaves it unchanged.
6. Common Bocconi Traps & Exam Strategies
Evaluating Domain on Simplified Forms: Never cancel factors before checking domain constraints. For instance, h ( x ) = ( x − 2 ) ( x + 1 ) x − 2 h(x) = \frac{(x-2)(x+1)}{x-2} h ( x ) = x − 2 ( x − 2 ) ( x + 1 ) has domain R ∖ { 2 } \mathbb{R} \setminus \{2\} R ∖ { 2 } , leaving a removable puncture (hole) at ( 2 , 3 ) (2, 3) ( 2 , 3 ) , not a continuous line across R \mathbb{R} R .
Failing to Factor Horizontal Arguments: In f ( a x + b ) f(ax + b) f ( a x + b ) , always factor out a a a as f ( a ( x + b / a ) ) f(a(x + b/a)) f ( a ( x + b / a )) to read off the horizontal translation as − b / a -b/a − b / a .
Order Inversion in Combined Vertical Transforms: For y = − 2 f ( x ) + 4 y = -2f(x) + 4 y = − 2 f ( x ) + 4 , the graph is stretched vertically by 2 and reflected across the x x x -axis first , then shifted up by 4. If you shift first, you get − 2 ( f ( x ) + 4 ) = − 2 f ( x ) − 8 -2(f(x) + 4) = -2f(x) - 8 − 2 ( f ( x ) + 4 ) = − 2 f ( x ) − 8 , which is entirely different.