2.2 Monotonicity, Parity, and Periodicity of Functions

2.2 Monotonicity, Parity, and Periodicity of Functions

1. Function Monotonicity (函数的单调性)

Monotonicity describes how a function's output changes as its input increases across a specified interval $I \subseteq D_f$.

Formal Definition

Let $f(x)$ be defined on an interval $I \subseteq D_f$.

• Strictly Increasing (单调递增): For any $x_1, x_2 \in I$ with $x_1 < x_2$, we have $f(x_1) < f(x_2)$, or equivalently: $\frac{f(x_1) - f(x_2)}{x_1 - x_2} > 0$
• Strictly Decreasing (单调递减): For any $x_1, x_2 \in I$ with $x_1 < x_2$, we have $f(x_1) > f(x_2)$, or equivalently: $\frac{f(x_1) - f(x_2)}{x_1 - x_2} < 0$

Derivative Criteria for Monotonicity

For a differentiable function $f(x)$ on an open interval $(a, b)$:

1. If $f'(x) \ge 0$ for all $x \in (a, b)$ and $f'(x)$ is not identically zero on any subinterval, then $f(x)$ is strictly increasing on $(a, b)$.
2. If $f'(x) \le 0$ for all $x \in (a, b)$ and $f'(x)$ is not identically zero on any subinterval, then $f(x)$ is strictly decreasing on $(a, b)$.

Composite Function Monotonicity Rule ("同增异减")

For a composite function $y = f(g(x))$, where $u = g(x)$ and $y = f(u)$:

• If $f(u)$ and $g(x)$ have the same monotonicity (both increasing or both decreasing), then $f(g(x))$ is strictly increasing.
• If $f(u)$ and $g(x)$ have opposite monotonicity (one increasing, one decreasing), then $f(g(x))$ is strictly decreasing.

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2. Function Parity (函数的奇偶性)

Parity characterizes the algebraic symmetry of a function with respect to the origin or the $y$-axis.

Prerequisites and Definitions

A necessary prerequisite for parity analysis is that the domain $D_f$ must be symmetric about the origin ($x \in D_f \iff -x \in D_f$).

• Even Function (偶函数): $f(-x) = f(x)$ for all $x \in D_f$. Graph is symmetric with respect to the $y$-axis.
• Odd Function (奇函数): $f(-x) = -f(x)$ for all $x \in D_f$. Graph is symmetric with respect to the origin $(0,0)$.

Crucial Theorem for Odd Functions: If $f(x)$ is an odd function and $0 \in D_f$, then $f(0) = 0$. This property provides a fast method to determine unknown coefficients in Gaokao problems.

Operations and Parity Rules

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3. Function Periodicity and Structural Identities

A function $f(x)$ is periodic if there exists a non-zero constant $T$ such that $f(x + T) = f(x)$ for all $x \in D_f$. The smallest positive such $T$ is the fundamental period (最小正周期).

Essential Functional Equation Identities

In Gaokao problems, periodicity is frequently expressed through functional equations:

Dual Symmetry and Periodicity Theorems

Geometric symmetry in $y=f(x)$ implies fundamental periodicity:

1. Two Axes of Symmetry: If $y=f(x)$ is symmetric about $x=a$ and $x=b$ ($a \neq b$), then $f(x)$ is periodic with period $T = 2|b - a|$.
2. Two Centers of Symmetry: If $y=f(x)$ is symmetric about points $(a, 0)$ and $(b, 0)$ ($a \neq b$), then $f(x)$ is periodic with period $T = 2|b - a|$.
3. One Axis and One Center: If $y=f(x)$ is symmetric about axis $x=a$ and center $(b, 0)$ ($a \neq b$), then $f(x)$ is periodic with period $T = 4|b - a|$.

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4. Worked Gaokao Exam Examples

Example 1: Solving Inequalities Using Parity and Monotonicity

Problem: Let $f(x)$ be an odd function defined on $\mathbb{R}$ that is strictly decreasing on $(0, +\infty)$. If $f(1) = 0$, solve the inequality $f(\ln x) > 0$.

Solution:

1. Determine function properties:
   • Since $f(x)$ is odd and $0 \in \mathbb{R}$, we have $f(0) = 0$.
   • Since $f(x)$ is odd and strictly decreasing on $(0, +\infty)$, it must also be strictly decreasing on $(-\infty, 0)$ and overall strictly decreasing on $\mathbb{R}$.
2. Determine zero values:
   • Given $f(1) = 0$, by odd parity $f(-1) = -f(1) = 0$.
   • Combining with $f(0) = 0$, the function zeros are at $x = -1, 0, 1$.
3. Analyze inequality $f(t) > 0$ where $t = \ln x$:
   • Since $f(x)$ is strictly decreasing on $\mathbb{R}$ and $f(0) = 0$, $f(t) > 0 = f(0) \iff t < 0$.
4. Solve for $x$:
   • $\ln x < 0 \implies 0 < x < 1$. Conclusion: The solution set of the inequality is $(0, 1)$.

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Example 2: Determining Period and Value Evaluation

Problem: Let $f(x)$ be an even function on $\mathbb{R}$ satisfying $f(x+2) = -\frac{1}{f(x)}$ for all $x$. If $f(x) = 2x - 1$ for $x \in [0, 1]$, calculate $f(11.5)$.

Solution:

1. Periodicity deduction:
   • From $f(x+2) = -\frac{1}{f(x)}$, we find $f(x+4) = f((x+2)+2) = -\frac{1}{f(x+2)} = -\frac{1}{-1/f(x)} = f(x)$.
   • Thus $f(x)$ is periodic with period $T = 4$.
2. Reduce argument $11.5$ into standard base interval:
   • $f(11.5) = f(11.5 - 3 \times 4) = f(11.5 - 12) = f(-0.5)$.
3. Apply parity and domain definition:
   • Since $f(x)$ is even, $f(-0.5) = f(0.5)$.
   • Since $0.5 \in [0, 1]$, we use $f(x) = 2x - 1$ to calculate $f(0.5) = 2(0.5) - 1 = 0$. Conclusion: $f(11.5) = 0$.