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Monotonicity of Functions

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Monotonicity of Functions

1. Concept

**Monotonicity** describes the trend of a function's value as the independent variable increases. It is the mathematical expression for whether a graph is "rising" or "falling".

Let the domain of function $f(x)$ be $D$, and let $I \subseteq D$ be an interval.

* **Monotonically Increasing**: If for **any** $x_1, x_2 \in I$ with $x_1 < x_2$, we have $f(x_1) < f(x_2)$. The graph **rises** from left to right.

* **Monotonically Decreasing**: If for **any** $x_1, x_2 \in I$ with $x_1 < x_2$, we have $f(x_1) > f(x_2)$. The graph **falls** from left to right.

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**Note**: Monotonicity is a local property defined on an **interval**. A function might not be monotonic over its entire domain but can be monotonic on specific sub-intervals.

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2. Methods of Determination

#### (1) Derivative Method (Most Common)

For a differentiable function $f(x)$, analyzing the sign of the derivative $f'(x)$ is the most efficient method.

* If $f'(x) > 0$ on interval $I$, then $f(x)$ is **increasing** on $I$.

* If $f'(x) < 0$ on interval $I$, then $f(x)$ is **decreasing** on $I$.

* If $f'(x) = 0$ on interval $I$, then $f(x)$ is **constant** on $I$.

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#### (2) Definition Method

Pick arbitrary $x_1, x_2 \in I$ with $x_1 < x_2$, and check the sign of the difference $f(x_1) - f(x_2)$.

* Difference $< 0$ implies increasing.

* Difference $> 0$ implies decreasing.

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3. Typical Examples

**Example 1: Finding Monotonic Intervals using Derivatives**

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

**Solution**:

1. **Domain**: $x \in \mathbb{R}$.

2. **Derivative**: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$.

3. **Solve Inequalities**:

* Set $f'(x) > 0$: $(x-1)(x+1) > 0 \Rightarrow x > 1$ or $x < -1$.

* Set $f'(x) < 0$: $(x-1)(x+1) < 0 \Rightarrow -1 < x < 1$.

**Conclusion**:

* Increasing intervals: $(-\infty, -1]$ and $[1, +\infty)$.

* Decreasing interval: $[-1, 1]$.

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**Example 2: Application of Monotonicity**

Given that $f(x)$ is increasing on $[0, +\infty)$ and $f(2)=5$, solve the inequality $f(x) < 5$.

**Solution**:

Since $f(2)=5$, the inequality is equivalent to $f(x) < f(2)$.

Due to the increasing property, a smaller function value implies a smaller input value, so $x < 2$.

Also considering the domain constraint $x \in [0, +\infty)$.

The solution set is $\{ x | 0 \le x < 2 \}$, or $[0, 2)$.

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4. Common Pitfalls

* **Ignoring Domain**: Forgetting constraints (like non-negative radicands) when solving inequalities.

* **Incorrect Union Notation**: If a function decreases on separate intervals A and B, you list them separately. Writing "decreasing on $A \cup B$" is mathematically incorrect if the property doesn't hold across the gap.