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Core Concept

Monotonicity describes whether a function is consistently increasing or decreasing over an interval. A function is said to be monotonic on an interval if it is either entirely increasing or entirely decreasing throughout that interval.

Definitions

Increasing Function (递增函数)

A function $f$ is increasing on interval $I$ if: $\forall x_1, x_2 \in I: \quad x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$

Equivalently: larger input → larger output.

Decreasing Function (递减函数)

A function $f$ is decreasing on interval $I$ if: $\forall x_1, x_2 \in I: \quad x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$

Equivalently: larger input → smaller output.

Non-strict Monotonicity

• Non-decreasing: $x_1 < x_2 \Rightarrow f(x_1) \leq f(x_2)$
• Non-increasing: $x_1 < x_2 \Rightarrow f(x_1) \geq f(x_2)$

Methods to Determine Monotonicity

Method 1: Definition Method (定义法)

1. Take any $x_1, x_2$ in the interval with $x_1 < x_2$
2. Calculate $f(x_1) - f(x_2)$
3. Determine the sign of the difference:
   • If always negative → increasing
   • If always positive → decreasing

Example: Prove $f(x) = x^2$ is increasing on $(0, +\infty)$.

For $0 < x_1 < x_2$: $f(x_1) - f(x_2) = x_1^2 - x_2^2 = (x_1 - x_2)(x_1 + x_2)$

Since $x_1 < x_2$: $(x_1 - x_2) < 0$ Since $x_1, x_2 > 0$: $(x_1 + x_2) > 0$

Therefore: $f(x_1) - f(x_2) < 0$, meaning $f(x_1) < f(x_2)$.

Conclusion: $f(x) = x^2$ is increasing on $(0, +\infty)$.

Method 2: Derivative Method (导数法)

For differentiable functions:

• $f'(x) > 0$ on interval $I$ → $f$ is increasing on $I$
• $f'(x) < 0$ on interval $I$ → $f$ is decreasing on $I$

Example: Find monotonic intervals of $f(x) = x^3 - 3x$.

$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$

• $f'(x) > 0$ when $x < -1$ or $x > 1$
• $f'(x) < 0$ when $-1 < x < 1$

Monotonic intervals:

• Increasing: $(-\infty, -1)$ and $(1, +\infty)$
• Decreasing: $(-1, 1)$

Common Functions and Their Monotonicity

Function	Increasing Interval	Decreasing Interval
$y = kx + b$ ($k > 0$)	$(-\infty, +\infty)$	—
$y = kx + b$ ($k < 0$)	—	$(-\infty, +\infty)$
$y = x^2$	$[0, +\infty)$	$(-\infty, 0]$
$y = \dfrac{1}{x}$	—	$(-\infty, 0)$, $(0, +\infty)$
$y = \sqrt{x}$	$[0, +\infty)$	—
$y = a^x$ ($a > 1$)	$(-\infty, +\infty)$	—
$y = a^x$ ($0 < a < 1$)	—	$(-\infty, +\infty)$

CSCA Practice Problems

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

Example 1: Basic (Difficulty ★★☆☆☆)

Determine the monotonic intervals of $f(x) = -x^2 + 4x$.

Solution:

$f(x) = -(x^2 - 4x) = -(x - 2)^2 + 4$

This is a downward parabola with vertex at $x = 2$.

• Increasing: $(-\infty, 2]$
• Decreasing: $[2, +\infty)$

Answer: Increasing on $(-\infty, 2]$, decreasing on $[2, +\infty)$

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

Prove that $f(x) = \dfrac{x}{x+1}$ is increasing on $(-1, +\infty)$.

Solution:

Let $-1 < x_1 < x_2$.

$f(x_1) - f(x_2) = \dfrac{x_1}{x_1+1} - \dfrac{x_2}{x_2+1}$

$= \dfrac{x_1(x_2+1) - x_2(x_1+1)}{(x_1+1)(x_2+1)}$

$= \dfrac{x_1x_2 + x_1 - x_1x_2 - x_2}{(x_1+1)(x_2+1)}$

$= \dfrac{x_1 - x_2}{(x_1+1)(x_2+1)}$

Since $x_1 < x_2$: numerator $< 0$ Since $x_1, x_2 > -1$: $(x_1+1)(x_2+1) > 0$

Therefore: $f(x_1) - f(x_2) < 0$, so $f(x_1) < f(x_2)$.

Conclusion: $f(x)$ is increasing on $(-1, +\infty)$.

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

If $f(x) = x^2 - 2ax + 1$ is increasing on $[1, +\infty)$, find the range of $a$.

Solution:

$f(x) = (x - a)^2 + 1 - a^2$

The vertex is at $x = a$.

For $f(x)$ to be increasing on $[1, +\infty)$, the vertex must be at or to the left of $x = 1$.

Therefore: $a \leq 1$

Answer: $a \leq 1$, or $(-\infty, 1]$

Properties of Monotonic Functions

1. Composition Rules

$f$	$g$	$f \circ g$
Increasing	Increasing	Increasing
Increasing	Decreasing	Decreasing
Decreasing	Increasing	Decreasing
Decreasing	Decreasing	Increasing

Memory aid: "Same → Increasing, Different → Decreasing"

2. Inverse Function

If $f$ is strictly monotonic, then $f^{-1}$ exists and has the same monotonicity as $f$.

Common Mistakes

❌ Mistake 1: Confusing Monotonic Interval with Domain

Wrong: $y = \dfrac{1}{x}$ is decreasing on $(-\infty, +\infty)$ ✗

Correct: $y = \dfrac{1}{x}$ is decreasing on $(-\infty, 0)$ AND on $(0, +\infty)$ separately ✓

❌ Mistake 2: Combining Disconnected Intervals

Wrong: $y = \dfrac{1}{x}$ is decreasing on $(-\infty, 0) \cup (0, +\infty)$ ✗

Correct: State intervals separately: decreasing on $(-\infty, 0)$ and on $(0, +\infty)$ ✓

❌ Mistake 3: Ignoring Boundary in Proof

When proving monotonicity, ensure $x_1 < x_2$ are both within the specified interval.

Study Tips

1. ✅ Master the definition: $x_1 < x_2$ implies what about $f(x_1)$ vs $f(x_2)$?
2. ✅ Know common functions: Memorize monotonicity of standard functions
3. ✅ Use derivatives when applicable: Faster for complex functions
4. ✅ Never combine disconnected intervals: List each interval separately

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💡 Exam Tip: For quadratic functions, always find the vertex first. The monotonicity changes at the vertex!