# Nakafa Framework: LLM

URL: https://nakafa.com/en/subject/high-school/12/mathematics/derivative-function/increasing-decreasing-and-stationary-function
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/12/mathematics/derivative-function/increasing-decreasing-and-stationary-function/en.mdx

Output docs content for large language models.

---

import { LineEquation } from "@repo/design-system/components/contents/line-equation";
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export const metadata = {
  title: "Increasing, Decreasing, and Stationary Functions",
  description: "Master identifying increasing, decreasing, and stationary functions using derivatives. Learn to analyze function behavior and determine monotonicity intervals.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/26/2025",
  subject: "Derivative Functions",
};

## Behavior of a Function and Its Derivative

Have you ever noticed how the graph of a function can move up, down, or even flatten out for a moment? This behavior, called the **monotonicity of a function**, is closely related to its first derivative.

Imagine you are walking along the curve of a graph from left to right.

-   When you are **climbing**, it means the function is **increasing**.

-   When you are **descending into a valley**, it means the function is **decreasing**.

-   When you are at the top of a hill or the bottom of a valley, you are at a **stationary** point.

Geometrically, the first derivative, <InlineMath math="f'(x)" />, is the gradient of the tangent line to the curve at that point. So, we can determine the function's behavior by looking at the sign of its gradient.

## Properties of Monotonicity

The relationship between the first derivative and the behavior of a function can be summarized by the following properties:

Suppose the function <InlineMath math="y = f(x)" /> is continuous and differentiable over an interval.

-   If <InlineMath math="f'(x) > 0" /> for all <InlineMath math="x" /> in that interval, then <InlineMath math="f(x)" /> is an **increasing function**.

-   If <InlineMath math="f'(x) < 0" /> for all <InlineMath math="x" /> in that interval, then <InlineMath math="f(x)" /> is a **decreasing function**.

-   If <InlineMath math="f'(x) = 0" /> at a specific point, then <InlineMath math="f(x)" /> has a **stationary point** there.

These stationary points are the key to finding where a function changes from increasing to decreasing, or vice versa.

## Analyzing Function Intervals

Let's break down a case to see how to determine the intervals where a function is increasing or decreasing.

Determine the intervals for which the function <InlineMath math="f(x) = x^3 - 3x" /> is increasing and decreasing.

**Solution:**

**Step 1: Find the first derivative**

First, we differentiate the function <InlineMath math="f(x)" />.

<BlockMath math="f'(x) = 3x^2 - 3" />

**Step 2: Find the stationary points**

Stationary points occur when <InlineMath math="f'(x) = 0" />.

<div className="flex flex-col gap-4">
    <BlockMath math="3x^2 - 3 = 0" />
    <BlockMath math="x^2 - 1 = 0" />
    <BlockMath math="(x-1)(x+1) = 0" />
</div>

From this, we get the stationary points at <InlineMath math="x = 1" /> and <InlineMath math="x = -1" />.

**Step 3: Create a number line and test intervals**

We place the stationary points on a number line. These points divide the line into three intervals. We take a test point from each interval to find the sign of <InlineMath math="f'(x)" /> (positive or negative).

-   **Interval <InlineMath math="x < -1" />:**
    
    Take <InlineMath math="x=-2" />. <InlineMath math="f'(-2) = 3(-2)^2 - 3 = 9" /> (Positive, function is increasing).

-   **Interval <InlineMath math="-1 < x < 1" />:**
    
    Take <InlineMath math="x=0" />. <InlineMath math="f'(0) = 3(0)^2 - 3 = -3" /> (Negative, function is decreasing).

-   **Interval <InlineMath math="x > 1" />:**
    
    Take <InlineMath math="x=2" />. <InlineMath math="f'(2) = 3(2)^2 - 3 = 9" /> (Positive, function is increasing).

**Step 4: Conclude the intervals**

Based on the tests, we can conclude:

-   The function is increasing on the intervals <InlineMath math="x < -1" /> or <InlineMath math="x > 1" />.

-   The function is decreasing on the interval <InlineMath math="-1 < x < 1" />.

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  title={<>Visualization of Function Monotonicity</>}
  description={
    <>
      This graph illustrates the behavior of the function <InlineMath math="f(x) = x^3 - 3x" />. 
      Notice how the curve increases when its derivative is positive, decreases when its derivative is negative, and flattens at the stationary points where <InlineMath math="f'(x)=0" />.
    </>
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      points: Array.from({ length: 21 }, (_, i) => {
        const x = -2 + i * 0.05;
        const y = x ** 3 - 3 * x;
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      color: getColor("LIME"),
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      labels: [
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          text: "Increasing",
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        const y = x ** 3 - 3 * x;
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          text: "Decreasing",
          at: 20,
          offset: [0, -1, 0],
        },
      ],
    },
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      points: Array.from({ length: 21 }, (_, i) => {
        const x = 1 + i * 0.05;
        const y = x ** 3 - 3 * x;
        return { x, y, z: 0 };
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      color: getColor("LIME"),
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          text: "Increasing",
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          offset: [1, 1, 0],
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      points: [{ x: -1, y: 2, z: 0 }],
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      labels: [
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          text: "Stationary (-1, 2)",
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## Exercises

1.  Determine the intervals where the function is increasing and decreasing for the curve <InlineMath math="f(x) = x^3 - 3x^2 - 15" />.

2.  If the function <InlineMath math="f(x) = ax^3 + x^2 + 5" /> is always increasing on the interval <InlineMath math="0 < x < 2" />, determine the value of the coefficient <InlineMath math="a" />!

3.  Determine the intervals where the function is increasing and decreasing if the curve is given by <InlineMath math="f(x) = \sin 2x \cos 2x" />!

### Answer Key

1.  **Solution:**

    The first derivative of <InlineMath math="f(x) = x^3 - 3x^2 - 15" /> is <InlineMath math="f'(x) = 3x^2 - 6x" />.
    
    Stationary points are found when <InlineMath math="f'(x) = 0" />.

    <BlockMath math="3x^2 - 6x = 0 \implies 3x(x - 2) = 0" />

    The stationary points are at <InlineMath math="x=0" /> and <InlineMath math="x=2" />.

    By testing the intervals on a number line:

    -   For <InlineMath math="x<0" />, <InlineMath math="f'(x)" /> is positive (increasing).

    -   For <InlineMath math="0<x<2" />, <InlineMath math="f'(x)" /> is negative (decreasing).

    -   For <InlineMath math="x>2" />, <InlineMath math="f'(x)" /> is positive (increasing).

    So, the function is increasing on <InlineMath math="x < 0" /> or <InlineMath math="x > 2" />, and decreasing on the interval <InlineMath math="0 < x < 2" />.

2.  **Solution:**

    For a function to be *always increasing* on an interval, its first derivative must be non-negative (<InlineMath math="f'(x) \ge 0" />) for every point within that interval.

    <BlockMath math="f'(x) = 3ax^2 + 2x = x(3ax + 2)" />
    
    In the interval <InlineMath math="0 < x < 2" />, the factor <InlineMath math="x" /> is always positive. Therefore, for <InlineMath math="f'(x) \ge 0" />, the second factor, <InlineMath math="(3ax + 2)" />, must also be non-negative.
    
    <BlockMath math="3ax + 2 \ge 0" />
    
    This inequality must hold for all values of <InlineMath math="x" /> in the interval <InlineMath math="0 < x < 2" />. Since <InlineMath math="h(x) = 3ax + 2" /> is a linear function, its behavior is monotonic. We only need to ensure it is non-negative at the most "critical" endpoint of the interval.

    -   If <InlineMath math="a \ge 0" />, then <InlineMath math="3ax" /> is also non-negative, so <InlineMath math="3ax+2" /> will definitely be positive. This condition is met.

    -   If <InlineMath math="a < 0" />, then <InlineMath math="h(x)" /> is a decreasing function. Its smallest value will be at the right end of the interval (<InlineMath math="x=2" />). For <InlineMath math="h(x)" /> to always be non-negative, we just need to ensure its minimum value is greater than or equal to zero.

    We test at the critical boundary <InlineMath math="x=2" />:

    <div className="flex flex-col gap-4">
        <BlockMath math="3a(2) + 2 \ge 0" />
        <BlockMath math="6a \ge -2" />
        <BlockMath math="a \ge -1/3" />
    </div>

    Combining both cases, the condition for the function to be always increasing on the given interval is <InlineMath math="a \ge -1/3" />.

3.  **Solution:**

    Use the double angle trigonometric identity: <InlineMath math="\sin(2A) = 2\sin(A)\cos(A)" />.

    So, <InlineMath math="f(x) = \sin 2x \cos 2x = \frac{1}{2} \sin(4x)" />.
    
    Its first derivative is:

    <BlockMath math="f'(x) = \frac{1}{2} \cdot \cos(4x) \cdot 4 = 2\cos(4x)" />
    
    -   **The function is increasing** when <InlineMath math="f'(x) > 0" />, which is <InlineMath math="2\cos(4x) > 0" /> or <InlineMath math="\cos(4x) > 0" />. This occurs in the first and fourth quadrants.

        <div className="flex flex-col gap-4">
            <BlockMath math="-\frac{\pi}{2} + 2k\pi < 4x < \frac{\pi}{2} + 2k\pi" />
            <BlockMath math="-\frac{\pi}{8} + \frac{k\pi}{2} < x < \frac{\pi}{8} + \frac{k\pi}{2}" />
        </div>

        This interval is valid for any integer <InlineMath math="k" />.

    -   **The function is decreasing** when <InlineMath math="f'(x) < 0" />, which is <InlineMath math="\cos(4x) < 0" />. This occurs in the second and third quadrants.

        <div className="flex flex-col gap-4">
            <BlockMath math="\frac{\pi}{2} + 2k\pi < 4x < \frac{3\pi}{2} + 2k\pi" />
            <BlockMath math="\frac{\pi}{8} + \frac{k\pi}{2} < x < \frac{3\pi}{8} + \frac{k\pi}{2}" />
        </div>

        This interval is valid for any integer <InlineMath math="k" />.