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3.2 Composition of Functions & Transformations

Key Takeaways

  • Function composition (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)) evaluates the inner function 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))).

  • The domain of (f∘g)(f \circ g) is the set of all x∈Dom(g)x \in \text{Dom}(g) such that g(x)∈Dom(f)g(x) \in \text{Dom}(f); evaluating domain after algebraic simplification introduces extraneous domain points.

  • Rigid graph translations shift graphs horizontally (f(x−c)f(x - c) moves right by cc for c>0c > 0) and vertically (f(x)+cf(x) + c moves up by cc for c>0c > 0).

  • Horizontal scaling acts reciprocally on the xx-coordinates: f(cx)f(cx) horizontally compresses the graph by factor 1/c1/c when c>1c > 1 and stretches by 1/c1/c when 0<c<10 < c < 1.

  • When analyzing compound horizontal transformations f(bx−c)f(bx - c), always factor into f(b(x−c/b))f(b(x - c/b)) or apply input coordinate tracking xnew=(xold+c)/bx_{\text{new}} = (x_{\text{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→Bg: A \to B and f:B→Cf: B \to C, the composite function (f∘g):A→C(f \circ g): A \to C is defined by evaluating gg first, and then applying ff to that intermediate result:

(f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))

Here, g(x)g(x) is the inner function (input to ff) and ff is the outer function.

Non-Commutativity of Composition

Function composition is associative—f∘(g∘h)=(f∘g)∘hf \circ (g \circ h) = (f \circ g) \circ h—but it is not commutative. In general:

(f∘g)(x)≠(g∘f)(x)(f \circ g)(x) \neq (g \circ f)(x)

Algebraic Demonstration: Let f(x)=2x+5f(x) = 2x + 5 and g(x)=x2−1g(x) = x^2 - 1.

  • (f∘g)(x)=f(g(x))=2(x2−1)+5=2x2−2+5=2x2+3(f \circ g)(x) = f(g(x)) = 2(x^2 - 1) + 5 = 2x^2 - 2 + 5 = 2x^2 + 3
  • (g∘f)(x)=g(f(x))=(2x+5)2−1=4x2+20x+25−1=4x2+20x+24(g \circ f)(x) = g(f(x)) = (2x + 5)^2 - 1 = 4x^2 + 20x + 25 - 1 = 4x^2 + 20x + 24

Clearly, 2x2+3≠4x2+20x+242x^2 + 3 \neq 4x^2 + 20x + 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)), 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) consists strictly of all inputs xx that are valid in gg, and whose output g(x)g(x) is simultaneously valid in ff:

Dom(f∘g)={x∈R∣x∈Dom(g)andg(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)\}

Worked Example: False Simplification Fallacy

Let f(u)=u2f(u) = u^2 and g(x)=x−4g(x) = \sqrt{x - 4}.

  • Direct algebraic composition: f(g(x))=(x−4)2=x−4f(g(x)) = (\sqrt{x - 4})^2 = x - 4.
  • Looking solely at x−4x - 4, one might mistakenly assume the domain is all of R\mathbb{R}.
  • However, under the two-condition rule:
    1. x∈Dom(g)  ⟹  x−4≥0  ⟹  x≥4x \in \text{Dom}(g) \implies x - 4 \ge 0 \implies x \ge 4.
    2. g(x)∈Dom(f)  ⟹  x−4∈Rg(x) \in \text{Dom}(f) \implies \sqrt{x - 4} \in \mathbb{R}, which is true for all x≥4x \ge 4.
  • Thus, the true domain is [4,∞)[4, \infty), not R\mathbb{R}.

Advanced Domain Problem

Find the domain of (f∘g)(x)(f \circ g)(x) where f(u)=1u−2f(u) = \frac{1}{\sqrt{u - 2}} and g(x)=3x+1x−4g(x) = \frac{3x + 1}{x - 4}.

Condition 1 (Inner Domain): x∈Dom(g)  ⟺  x−4≠0  ⟺  x≠4x \in \text{Dom}(g) \iff x - 4 \neq 0 \iff x \neq 4.

Condition 2 (Outer Compatibility): g(x)∈Dom(f)g(x) \in \text{Dom}(f). The function f(u)f(u) requires u−2>0  ⟺  u>2u - 2 > 0 \iff u > 2. Therefore:

3x+1x−4>2  ⟺  3x+1x−4−2>0  ⟺  3x+1−2(x−4)x−4>0  ⟺  x+9x−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

Constructing a sign table for x+9x−4\frac{x + 9}{x - 4}:

  • Numerator root: x=−9x = -9
  • Denominator root: x=4x = 4
  • Test intervals: for x<−9x < -9, (−)/(−)=(+)(-)/(-) = (+); for −9<x<4-9 < x < 4, (+)/(−)=(−)(+)/(-) = (-); for x>4x > 4, (+)/(+)=(+)(+)/(+) = (+).

Thus, the inequality holds for x∈(−∞,−9)∪(4,∞)x \in (-\infty, -9) \cup (4, \infty). Both parts satisfy x≠4x \neq 4. Hence:

Dom(f∘g)=(−∞,−9)∪(4,∞)\text{Dom}(f \circ g) = (-\infty, -9) \cup (4, \infty)

3. Geometric Graph Transformations

Every functional modification of a parent function y=f(x)y = f(x) corresponds to a precise geometric transformation in the Cartesian plane. Transformations are categorized into vertical (acting on outputs yy) and horizontal (acting on inputs xx):

TransformationAlgebraic FormulaGeometric OperationCoordinate Mapping (x,y)↦(x, y) \mapsto
Vertical Shifty=f(x)+cy = f(x) + cShift up if c>0c > 0, down if c<0c < 0(x,y+c)(x, y + c)
Horizontal Shifty=f(x−c)y = f(x - c)Shift right if c>0c > 0, left if c<0c < 0(x+c,y)(x + c, y)
Vertical Stretch/Compressy=a⋅f(x)y = a \cdot f(x) (a>0a > 0)Stretch if a>1a > 1, compress if 0<a<10 < a < 1(x,a⋅y)(x, a \cdot y)
Horizontal Stretch/Compressy=f(b⋅x)y = f(b \cdot x) (b>0b > 0)Compress by 1/b1/b if b>1b > 1, stretch if 0<b<10 < b < 1(x/b,y)(x/b, y)
Vertical Reflectiony=−f(x)y = -f(x)Reflect across the xx-axis(x,−y)(x, -y)
Horizontal Reflectiony=f(−x)y = f(-x)Reflect across the yy-axis(−x,y)(-x, y)
Absolute Outputy=∣f(x)∣y = \vert f(x)\vertReflect portions below xx-axis upwards(x,∣y∣)(x, \vert y\vert)
Absolute Inputy=f(∣x∣)y = f(\vert x\vert)Delete x<0x < 0, reflect x≥0x \ge 0 over yy-axis(±x,y)(\pm x, y) for x≥0x \ge 0

The Inverse Behavior of Horizontal Operations

Notice the fundamental dichotomy in graphical transformations:

  • Vertical operations are intuitive: adding cc adds to yy; multiplying by aa multiplies yy.
  • Horizontal operations are counter-intuitive: replacing xx with x−cx - c shifts right (adds cc to the coordinates); replacing xx with 2x2x compresses coordinates horizontally by a factor of 12\frac{1}{2}.

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(2x+6)y = f(2x + 6):

  • Common Error: Students see +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)) 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 1/21/2.

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 (x0,y0)(x_0, y_0) on y=f(x)y = f(x), find the corresponding point (xnew,ynew)(x_{\text{new}}, y_{\text{new}}) on y=A⋅f(Bx+C)+Dy = A \cdot f(B x + C) + D:

  1. Horizontal coordinate: Set the inner expression equal to the parent input: Bxnew+C=x0  ⟹  xnew=x0−CBB x_{\text{new}} + C = x_0 \implies x_{\text{new}} = \frac{x_0 - C}{B}
  2. Vertical coordinate: Substitute y0=f(x0)y_0 = f(x_0) directly into the outer equation: ynew=A⋅y0+Dy_{\text{new}} = A \cdot y_0 + D

Worked Example: Coordinate Tracking

Suppose the point (−4,7)(-4, 7) lies on the graph of y=f(x)y = f(x). Determine the corresponding point that must lie on the graph of:

g(x)=−3f(2x−8)+5g(x) = -3 f(2x - 8) + 5

Applying the tracking formulas:

  • Horizontal coordinate: 2xnew−8=−4  ⟹  2xnew=4  ⟹  xnew=22x_{\text{new}} - 8 = -4 \implies 2x_{\text{new}} = 4 \implies x_{\text{new}} = 2.
  • Vertical coordinate: ynew=−3(7)+5=−21+5=−16y_{\text{new}} = -3(7) + 5 = -21 + 5 = -16.

Therefore, the point (2,−16)(2, -16) is guaranteed to lie on the graph of 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 yy-axis): f(−x)=f(x)∀x∈Dom(f)f(-x) = f(x) \quad \forall x \in \text{Dom}(f) Examples include x2,x4,∣x∣,cos⁡(x)x^2, x^4, |x|, \cos(x). Polynomials with only even powers of xx 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) Examples include x,x3,1x,sin⁡(x)x, x^3, \frac{1}{x}, \sin(x). Rotating the graph 180∘180^\circ around (0,0)(0, 0) leaves it unchanged.

6. Common Bocconi Traps & Exam Strategies

  1. Evaluating Domain on Simplified Forms: Never cancel factors before checking domain constraints. For instance, h(x)=(x−2)(x+1)x−2h(x) = \frac{(x-2)(x+1)}{x-2} has domain R∖{2}\mathbb{R} \setminus \{2\}, leaving a removable puncture (hole) at (2,3)(2, 3), not a continuous line across R\mathbb{R}.
  2. Failing to Factor Horizontal Arguments: In f(ax+b)f(ax + b), always factor out aa as f(a(x+b/a))f(a(x + b/a)) to read off the horizontal translation as −b/a-b/a.
  3. Order Inversion in Combined Vertical Transforms: For y=−2f(x)+4y = -2f(x) + 4, the graph is stretched vertically by 2 and reflected across the xx-axis first, then shifted up by 4. If you shift first, you get −2(f(x)+4)=−2f(x)−8-2(f(x) + 4) = -2f(x) - 8, which is entirely different.
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Function Transformation Decision Flowchart
Test Your Knowledge

What is the domain of the composite function (f ∘ g)(x) where f(u) = ln(u - 1) and g(x) = (2x + 8) / (x - 3)?

A

(-∞, 3) ∪ (3, ∞)

B

(-11, 3)

C

(-∞, -11) ∪ [3, ∞)

D

(-∞, -11) ∪ (3, ∞)

Test Your Knowledge

If the point (-6, 4) lies on the graph of y = f(x), which of the following points MUST lie on the graph of y = 2 f(-3x + 9) - 7?

A

(5, 1)

B

(-3, 1)

C

(5, 15)

D

(1, 1)

Test Your Knowledge

Let f(x) = x / (x + 1). What is the closed-form expression for the triple composition (f ∘ f ∘ f)(x)?

A

x / (x + 3)

B

x / (3x + 1)

C

3x / (x + 1)

D

x³ / (x + 1)³

Sections you finish are checked off in the contents.