值域zhíyù

Core Concept

The range of a function $f$ is the set of all possible output values (y-values) that the function can produce.

Mathematical Definition

$\text{Range}(f) = \{y \mid y = f(x) \text{ for some } x \in \text{Domain}(f)\}$

The range is also called the codomain image or simply the image of the function.

Domain vs. Range

Concept	Symbol	Description
Domain	$D_f$	Set of all valid input values (x)
Range	$R_f$	Set of all output values (y)

Key relationship: The range depends on both the function rule AND the domain.

Methods to Find Range

Method 1: Direct Analysis (观察法)

For simple functions, analyze the behavior directly.

Example: $f(x) = x^2$, $x \in \mathbb{R}$

Since $x^2 \geq 0$ for all real $x$, and $x^2$ can be arbitrarily large:

Range: $[0, +\infty)$

Method 2: Inverse Function Method (反函数法)

1. Write $y = f(x)$
2. Solve for $x$ in terms of $y$
3. Find values of $y$ for which $x$ is defined

Example: $f(x) = \dfrac{2x + 1}{x - 1}$, $x \neq 1$

Let $y = \dfrac{2x + 1}{x - 1}$

Solve for $x$: $y(x - 1) = 2x + 1$ $xy - y = 2x + 1$ $xy - 2x = y + 1$ $x(y - 2) = y + 1$ $x = \dfrac{y + 1}{y - 2}$

For $x$ to exist, $y \neq 2$.

Range: $\{y \mid y \neq 2\} = (-\infty, 2) \cup (2, +\infty)$

Method 3: Monotonicity Method (单调性法)

Use the function's monotonicity to find range from domain.

Example: $f(x) = 2^x$, $x \in [-1, 2]$

Since $f(x) = 2^x$ is strictly increasing:

• Minimum: $f(-1) = 2^{-1} = \dfrac{1}{2}$
• Maximum: $f(2) = 2^2 = 4$

Range: $[\dfrac{1}{2}, 4]$

Method 4: Completing the Square (配方法)

For quadratic functions $f(x) = ax^2 + bx + c$.

Example: $f(x) = x^2 - 4x + 5$, $x \in \mathbb{R}$

Complete the square: $f(x) = (x - 2)^2 + 1$

Since $(x - 2)^2 \geq 0$, the minimum is 1 at $x = 2$.

Range: $[1, +\infty)$

Method 5: Substitution (换元法)

Example: $f(x) = x + \sqrt{x - 1}$, $x \geq 1$

Let $t = \sqrt{x - 1}$, where $t \geq 0$

Then $x = t^2 + 1$, so: $f = t^2 + 1 + t = t^2 + t + 1 = (t + \dfrac{1}{2})^2 + \dfrac{3}{4}$

Since $t \geq 0$, the minimum occurs at $t = 0$: $f_{\min} = 0 + 0 + 1 = 1$

Range: $[1, +\infty)$

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus.

Example 1: Basic (Difficulty ★★☆☆☆)

Find the range of $f(x) = 3x - 2$, $x \in [0, 4]$.

Solution: Linear function is strictly increasing.

• At $x = 0$: $f(0) = -2$
• At $x = 4$: $f(4) = 10$

Answer: $[-2, 10]$

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Example 2: Intermediate (Difficulty ★★★☆☆)

Find the range of $f(x) = x^2 - 2x + 3$, $x \in [-1, 2]$.

Solution:

Complete the square: $f(x) = (x - 1)^2 + 2$

Vertex at $x = 1$ (within domain), minimum = 2

Check endpoints:

• $f(-1) = 1 + 2 + 3 = 6$
• $f(2) = 4 - 4 + 3 = 3$

Answer: $[2, 6]$

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Example 3: Advanced (Difficulty ★★★★☆)

Find the range of $f(x) = \dfrac{x^2 + 1}{x^2 + 2}$, $x \in \mathbb{R}$.

Solution:

Let $y = \dfrac{x^2 + 1}{x^2 + 2}$

Cross multiply: $y(x^2 + 2) = x^2 + 1$

$yx^2 + 2y = x^2 + 1$

$x^2(y - 1) = 1 - 2y$

$x^2 = \dfrac{1 - 2y}{y - 1}$

For real $x$, need $x^2 \geq 0$: $\dfrac{1 - 2y}{y - 1} \geq 0$

This requires $(1 - 2y)$ and $(y - 1)$ to have the same sign.

• Case 1: Both positive: $y < \dfrac{1}{2}$ and $y > 1$ → impossible
• Case 2: Both negative: $y > \dfrac{1}{2}$ and $y < 1$ → $\dfrac{1}{2} < y < 1$

Also, as $x^2 \to +\infty$, $y \to 1$ (never equals 1). At $x = 0$: $y = \dfrac{1}{2}$ (achievable).

Answer: $[\dfrac{1}{2}, 1)$

Common Mistakes

❌ Mistake 1: Ignoring Domain Restrictions

Wrong: Range of $f(x) = \sqrt{x}$ is $\mathbb{R}$ ✗

Correct: Range of $f(x) = \sqrt{x}$ is $[0, +\infty)$ ✓

❌ Mistake 2: Using Wrong Method for Bounded Domain

Wrong: For $f(x) = x^2$, $x \in [1, 3]$, range is $[0, 9]$ ✗

Correct: Range is $[1, 9]$ (minimum at $x = 1$, not $x = 0$) ✓

❌ Mistake 3: Forgetting to Check Endpoints

Always verify function values at domain boundaries.

Study Tips

1. ✅ Identify function type first: Linear, quadratic, rational, etc.
2. ✅ Check if domain is bounded: Use monotonicity if bounded
3. ✅ For quadratics, locate vertex: Is it inside the domain?
4. ✅ For fractions, use inverse method: Solve for $x$ in terms of $y$

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💡 Exam Tip: For bounded domains, always check both the vertex (for quadratics) AND the endpoints!