3.1 Function Concepts, Domain, Range & Inverses

3.1 Function Concepts, Domain, Range & Inverses

On the Bocconi Admission Test, functions form the algebraic architecture underlying higher-level calculus, optimization, and economic modeling. A solid grasp of function definitions, operational restrictions, and invertibility conditions enables candidates to solve multi-step problems in under 90 seconds without a calculator.

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1. Mathematical Definition of a Function

A binary relation $R$ between two sets $X$ and $Y$ is a subset of their Cartesian product: $R \subseteq X \times Y$. A relation is classified as a function $f: X \to Y$ if and only if it satisfies the strict condition of deterministic single-valued mapping:

$\forall x \in X, \quad \exists! \, y \in Y \quad \text{such that} \quad (x, y) \in f \quad [\text{commonly written } y = f(x)]$

Here, $X$ is the domain (input set), $Y$ is the codomain (target set), and the unique element $y \in Y$ associated with $x$ is the image of $x$ under $f$.

The Vertical Line Test (VLT)

In Cartesian coordinates, a geometric curve represents $y$ as a function of $x$ if and only if every vertical line $x = c$ (for $c \in X$) intersects the curve at most once.

• A circle $x^2 + y^2 = r^2$ fails the VLT because vertical lines with $-r < x < r$ intersect the curve twice ($y = \pm \sqrt{r^2 - x^2}$); hence, it is a relation, not a function.
• Isolating the positive square root $y = +\sqrt{r^2 - x^2}$ yields a legitimate function representing the upper semicircle.

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2. Determining Natural Domains

When a function formula $y = f(x)$ is specified without an explicit domain, candidates must determine its natural domain (or maximal domain): the largest subset of $\mathbb{R}$ for which the algebraic expression evaluates to a real number. In the Bocconi syllabus, domain restrictions arise from three non-negotiable algebraic operations:

Compound Domain Example

Determine the natural domain of the compound function:

$f(x) = \frac{\sqrt{16 - x^2}}{\ln(x + 1)} + \frac{1}{x - 3}$

To find the admissible domain $\mathcal{D}$, we establish the system of simultaneous inequalities:

1. Even radical condition: $16 - x^2 \ge 0 \iff x^2 \le 16 \iff -4 \le x \le 4$.
2. Logarithm argument condition: $x + 1 > 0 \iff x > -1$.
3. Denominator non-zero conditions:
   • $\ln(x + 1) \neq 0 \iff x + 1 \neq e^0 = 1 \iff x \neq 0$.
   • $x - 3 \neq 0 \iff x \neq 3$.

Intersecting these sets on the real number line:

$\mathcal{D} = [-4, 4] \cap (-1, \infty) \setminus \{0, 3\} = (-1, 0) \cup (0, 3) \cup (3, 4]$

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3. Codomain vs. Range (Image)

The codomain $Y$ is the formal set into which all outputs must fall (typically $\mathbb{R}$ unless restricted). The range (or image set) $\text{Ran}(f)$ is the set of values actually attained by $f$:

$\text{Ran}(f) = \{y \in Y \mid \exists x \in \text{Dom}(f), \, f(x) = y\}$

Finding the Range Algebraically

To find the range of a function $y = f(x)$, solve the equation for $x$ in terms of $y$, and determine the values of $y$ that produce valid $x \in \text{Dom}(f)$.

Example: Find the range of the homographic function $f(x) = \frac{2x - 5}{x + 3}$ defined on $\mathbb{R} \setminus {-3}$. Set $y = \frac{2x - 5}{x + 3}$ and isolate $x$:

$y(x + 3) = 2x - 5 \implies yx + 3y = 2x - 5 \implies x(y - 2) = -3y - 5 \implies x = \frac{-3y - 5}{y - 2}$

The expression for $x$ is well-defined for all $y \in \mathbb{R}$ except where the denominator vanishes: $y - 2 = 0 \iff y = 2$. Furthermore, substituting $x = -3$ yields $-3(y - 2) = -3y - 5 \implies -3y + 6 = -3y - 5 \implies 6 = -5$ (impossible). Hence, the range is:

$\text{Ran}(f) = \mathbb{R} \setminus \{2\}$

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4. Mapping Classifications: Injective, Surjective & Bijective

Understanding structural mappings is essential for determining whether an inverse function exists:

1. Injective (One-to-One): A function $f: X \to Y$ is injective if distinct domain elements map to distinct images: $\forall x_1, x_2 \in X, \quad f(x_1) = f(x_2) \implies x_1 = x_2$ Geometric Test (Horizontal Line Test - HLT): Every horizontal line $y = k$ intersects the graph of an injective function at most once. Any function that is strictly increasing or strictly decreasing across an interval is guaranteed to be injective on that interval.

2. Surjective (Onto): A function $f: X \to Y$ is surjective if every element of the codomain is mapped to by at least one element of the domain: $\forall y \in Y, \quad \exists x \in X \quad \text{such that} \quad f(x) = y \iff \text{Ran}(f) = Y$ Geometric Test: Every horizontal line $y = k$ (for $k \in Y$) intersects the graph at least once.

3. Bijective (One-to-One Correspondence): A function is bijective if and only if it is both injective and surjective. Every horizontal line $y = k$ intersects the graph exactly once. Bijectivity is the necessary and sufficient condition for invertibility.

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5. The Inverse Function $f^{-1}(x)$

If $f: X \to Y$ is bijective, there exists a unique inverse function $f^{-1}: Y \to X$ defined such that:

$(f^{-1} \circ f)(x) = x, \quad \forall x \in X \qquad \text{and} \qquad (f \circ f^{-1})(y) = y, \quad \forall y \in Y$

The Fundamental Swap of Domain and Range

Because $f^{-1}$ reverses the direction of mapping:

$\text{Dom}(f^{-1}) = \text{Ran}(f) \qquad \text{and} \qquad \text{Ran}(f^{-1}) = \text{Dom}(f)$

Four-Step Algebraic Algorithm for Inversion

1. Establish Domains: Write $y = f(x)$ and explicitly record $\text{Dom}(f)$ and $\text{Ran}(f)$.
2. Swap Variables: Exchange $x$ and $y$ to reflect the mapping across $y = x$: $x = f(y)$.
3. Solve for $y$: Manipulate the equation algebraically to isolate $y$ in terms of $x$.
4. State the Inverse: Replace $y$ with $f^{-1}(x)$ and restrict its domain to $\text{Dom}(f^{-1}) = \text{Ran}(f)$.

Worked Example 1: Inverting a Homographic Function

Consider $f(x) = \frac{ax + b}{cx + d}$ with $ad - bc \neq 0$ and $c \neq 0$.

1. Set $x = \frac{ay + b}{cy + d}$.
2. Multiply out: $x(cy + d) = ay + b \implies cxy + dx = ay + b$.
3. Group $y$ terms: $y(cx - a) = -dx + b \implies y = \frac{-dx + b}{cx - a}$.

Thus, the general inverse of a homographic function is:

$f^{-1}(x) = \frac{-dx + b}{cx - a}$

Bocconi Exam Shortcut: When $a = -d$, the function is self-inverse (an involution), satisfying $f^{-1}(x) = f(x)$. For instance, $f(x) = \frac{3x + 1}{2x - 3}$ satisfies $f(f(x)) = x$ directly without calculation!

Worked Example 2: Inverting a Restricted Quadratic

Find the inverse of $f(x) = 2x^2 - 12x + 23$ defined on the restricted domain $[3, \infty)$.

Step 1: Determine the Range. Complete the square: $f(x) = 2(x^2 - 6x) + 23 = 2(x - 3)^2 - 18 + 23 = 2(x - 3)^2 + 5$ For $x \ge 3$, $(x - 3)^2 \ge 0$, meaning the minimum occurs at the vertex $(3, 5)$. Hence, $\text{Ran}(f) = [5, \infty)$.

Step 2: Swap Variables and Solve for $y$. $x = 2(y - 3)^2 + 5 \implies x - 5 = 2(y - 3)^2 \implies (y - 3)^2 = \frac{x - 5}{2}$ Taking the square root yields $y - 3 = \pm \sqrt{\frac{x - 5}{2}}$. Since the original domain was $x \ge 3$, the new range must satisfy $y \ge 3$. We must select the positive branch:

$y = 3 + \sqrt{\frac{x - 5}{2}}$

Thus, $f^{-1}(x) = 3 + \sqrt{\frac{x - 5}{2}}$ with domain $[5, \infty)$.

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6. Geometric Symmetry Across $y = x$

The graph of $y = f^{-1}(x)$ is the mirror image of $y = f(x)$ across the line of identity $y = x$.

• If point $(a, b)$ lies on the graph of $f$, then $(b, a)$ lies on the graph of $f^{-1}$.
• Intersection Rule: If $f(x)$ is a strictly increasing function, any points of intersection between $f$ and $f^{-1}$ must lie on the line $y = x$. Therefore, rather than solving the complex equation $f(x) = f^{-1}(x)$, candidates can simply solve the much faster linear equation:

$f(x) = x$

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7. Common Bocconi Traps & Exam Strategies

1. Confusing Notation: $f^{-1}(x)$ denotes the functional inverse under composition, never the reciprocal: $f^{-1}(x) \neq \frac{1}{f(x)} = [f(x)]^{-1}$.
2. Omitting Branch Restrictions: Inverting even-degree polynomials ($x^2, x^4$) over their full domain is impossible because they fail the Horizontal Line Test. When an interval restriction is specified, always verify that the $\pm$ sign in the inverse matches the restricted domain.
3. Assuming Semicircle Domain is Full $\mathbb{R}$: When dealing with functions such as $f(x) = \sqrt{9 - x^2}$, the domain is strictly $[-3, 3]$ and range is $[0, 3]$.