2.1 Concept, Domain, Range, and Mapping of Functions

2.1 Concept, Domain, Range, and Mapping of Functions

1. Mathematical Definition and Essential Elements of a Function

In high school mathematics and the Gaokao examination framework, a function is formally defined using set theory. Let $A$ and $B$ be non-empty sets of real numbers ($A, B \subseteq \mathbb{R}$). A function $f$ from $A$ to $B$, denoted as $f: A \to B$, is a rule that assigns to each element $x \in A$ exactly one unique element $y \in B$.

The element $x$ is called the independent variable, and the set $A$ is called the domain (定义域) of the function $f$, denoted $D_f = A$. The corresponding value $y = f(x)$ is called the image (象) of $x$, and the set of all images $R_f = \{f(x) \mid x \in A\} \subseteq B$ is called the range (值域) of the function.

The Three Essential Elements (函数的三要素)

A function is completely determined by three core components:

1. Domain ($D_f$): The set of all admissible input values for $x$.
2. Corresponding Rule ($f$): The specific algebraic operation or mapping rule that transforms $x$ into $f(x)$.
3. Range ($R_f$): The set of all actual output values resulting from applying $f$ to elements of $A$.

Identical Functions Criterion (相同函数的判定): Two functions $f(x)$ and $g(x)$ are identical if and only if they have the same domain ($D_f = D_g$) and the same corresponding rule ($f(x) = g(x)$ for all $x \in D_f$). The choice of variable letter (e.g., $f(x)$ versus $f(t)$) does not affect function identity.

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2. Determination of Function Domain

The domain of a function is the foundation of all function analysis. In Gaokao problems, domain determination falls into two primary categories: natural algebraic domains and abstract function domains.

A. Natural Algebraic Domains (自然定义域)

When a function is defined by an explicit algebraic expression without specified constraints, its domain is the set of all real numbers for which the expression is mathematically meaningful. The fundamental algebraic constraints are summarized below:

When a function consists of multiple algebraic terms, the overall domain is the intersection ($\cap$) of the domain constraints of each individual term.

B. Abstract Function Domains (抽象函数的定义域)

An abstract function is expressed without a specific algebraic formula (e.g., $f(x)$). Solving abstract domain problems relies on one core principle: The input argument inside the function wrapper $f(\cdot)$ must fall within the identical set of values.

1. Given domain of $f(x)$ is $[a, b]$, find domain of $f(g(x))$: Set $a \le g(x) \le b$ and solve the inequality for the independent variable $x$.
2. Given domain of $f(g(x))$ is $[a, b]$, find domain of $f(x)$: Determine the set of values attained by $g(x)$ as $x$ ranges over $[a, b]$. That set of values is the domain of $f(x)$.

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3. Systematic Methods for Finding Function Range

Determining the range of a function requires selecting the appropriate algebraic tool based on the structural form of $f(x)$:

Method 1: Substitution / Variable Change (换元法)

Useful for algebraic expressions involving radicals or quadratic forms. For instance, given $f(x) = x + 2\sqrt{1-x}$, set $t = \sqrt{1-x} \ge 0$. Then $x = 1 - t^2$. The function transforms into a bounded quadratic $g(t) = (1-t^2) + 2t = -t^2 + 2t + 1 = -(t-1)^2 + 2$ for $t \ge 0$. The maximum occurs at $t=1$, yielding Range $=(-\infty, 2]$.

Method 2: Monotonicity Analysis (单调性法)

If $f(x)$ is continuous and strictly monotonic on a closed interval $[a, b]$, the range is simply bounded by the endpoint values: $[f(a), f(b)]$ if increasing, or $[f(b), f(a)]$ if decreasing.

Method 3: Fundamental Inequalities / AM-GM (基本不等式法)

For expressions of the form $f(x) = x + \frac{k}{x}$ ($k > 0$), apply the Arithmetic Mean-Geometric Mean inequality $a + b \ge 2\sqrt{ab}$ (for $a, b > 0$). Equality holds if and only if $a = b$. Remember to verify that the equality condition $x = \frac{k}{x}$ is attainable within the domain!

Method 4: Quadratic Quotient / Discriminant Method (判别式法)

For rational functions of the form $y = \frac{ax^2 + bx + c}{dx^2 + ex + f}$, clear the denominator to obtain a quadratic equation in $x$: $A(y)x^2 + B(y)x + C(y) = 0$. Since $x \in \mathbb{R}$ is real, the discriminant $\Delta = [B(y)]^2 - 4A(y)C(y) \ge 0$ must hold, yielding an inequality for $y$ that defines the range.

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4. Concept of Mapping and Classification

A mapping $f: A \to B$ is a generalization of a function where $A$ and $B$ can be any non-empty sets (not restricted to sets of real numbers).

Key Properties of Mappings:

• Existence & Uniqueness: Every element $x \in A$ must have a unique corresponding element $y \in B$.
• Directionality: A mapping from $A$ to $B$ is distinct from a mapping from $B$ to $A$.
• Surjection, Injection, Bijection:
  • Injection (单射): Distinct elements in $A$ map to distinct elements in $B$.
  • Surjection (满射): Every element in $B$ is the image of at least one element in $A$ (Range $= B$).
  • Bijection (双射): Both injective and surjective (one-to-one correspondence). An inverse mapping exists if and only if $f$ is a bijection.

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5. Worked Gaokao Exam Examples

Example 1: Domain of Abstract Composite Functions

Problem: Suppose the domain of the function $f(x)$ is $[-2, 4]$. Find the domain of the function $g(x) = \frac{f(2x - 1)}{\sqrt{2 - x}}$.

Solution: To find the domain of $g(x)$, we must satisfy two simultaneous conditions:

1. The argument of the abstract function $f$ must lie within its given domain $[-2, 4]$: $$-2 \le 2x - 1 \le 4$$ Adding $1$ throughout: $-1 \le 2x \le 5 \implies -\frac{1}{2} \le x \le \frac{5}{2}$.
2. The square root in the denominator must be strictly positive (since denominator $\neq 0$): $$2 - x > 0 \implies x < 2$$

Taking the intersection of these two solution sets:

$$D_g = \left[-\frac{1}{2}, \frac{5}{2}\right] \cap (-\infty, 2) = \left[-\frac{1}{2}, 2\right)$$

Conclusion: The domain of $g(x)$ is $\left[-\frac{1}{2}, 2\right)$.

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Example 2: Range Determination by Substitution and Bounded Quadratic

Problem: Find the range of the function $f(x) = 2x - \sqrt{4x - 1}$.

Solution: First, determine the domain from the radicand constraint $4x - 1 \ge 0 \implies x \ge \frac{1}{4}$. Let $t = \sqrt{4x - 1} \ge 0$. Squaring both sides gives $t^2 = 4x - 1$, so $x = \frac{t^2 + 1}{4}$. Substitute $x$ into $f(x)$:

$$g(t) = 2\left(\frac{t^2 + 1}{4}\right) - t = \frac{t^2 + 1}{2} - t = \frac{1}{2}t^2 - t + \frac{1}{2}$$

Complete the square for the quadratic function in $t$ (where $t \ge 0$):

$$g(t) = \frac{1}{2}(t^2 - 2t + 1) = \frac{1}{2}(t - 1)^2$$

Since $(t - 1)^2 \ge 0$ for all real $t$, and $t = 1$ is within the domain $t \ge 0$ (corresponding to $x = \frac{1^2 + 1}{4} = \frac{1}{2}$): The minimum value of $g(t)$ is $0$ at $t = 1$. As $t \to +\infty$, $g(t) \to +\infty$. Conclusion: The range of $f(x)$ is $[0, +\infty)$.