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Learn identifying increasing, decreasing, and stationary functions using derivatives. Learn to analyze function behavior and determine monotonicity intervals.

---

## 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.

Visible text: - 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 $$y = f(x)$$ is continuous and differentiable over an interval.

Visible text: Suppose the function is continuous and differentiable over an interval.

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

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

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

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

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

- If <InlineMath math="f'(x) = 0" /> at a specific point, then 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 $$f(x) = x^3 - 3x$$ is increasing and decreasing.

Visible text: Determine the intervals for which the function is increasing and decreasing.

**Solution:**

**Step** $$1$$: Find the first derivative

Visible text: **Step** : Find the first derivative

First, we differentiate the function $$f(x)$$.

Visible text: First, we differentiate the function .

```math
f'(x) = 3x^2 - 3
```

**Step** $$2$$: Find the stationary points

Visible text: **Step** : Find the stationary points

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

Component: MathContainer
Children:

```math
3x^2 - 3 = 0
```

```math
x^2 - 1 = 0
```

```math
(x-1)(x+1) = 0
```

From this, we get the stationary points at $$x = 1$$ and $$x = -1$$.

Visible text: From this, we get the stationary points at and .

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

Visible text: **Step** : 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 $$x < -1$$:**

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

-   **Interval $$-1 < x < 1$$:**

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

-   **Interval $$x > 1$$:**

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

Visible text: - **Interval :**

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

- **Interval :**

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

- **Interval :**

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

**Step** $$4$$: Conclude the intervals

Visible text: **Step** : Conclude the intervals

Based on the tests, we can conclude:

-   The function is increasing on the intervals $$x < -1$$ or $$x > 1$$.

-   The function is decreasing on the interval $$-1 < x < 1$$.

Visible text: - The function is increasing on the intervals or .

- The function is decreasing on the interval .

Component: LineEquation
Props:
- title: Visualization of Function Monotonicity
- description: This graph illustrates the behavior of the function $$f(x) = x^3 - 3x$$.
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" />.
  Visible text: This graph illustrates the behavior of the function .
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" />.
- showZAxis: false
- cameraPosition: [0, 0, 10]
- data: [
{
points: Array.from({ length: 21 }, (_, i) => {
const x = -2 + i * 0.05;
const y = x ** 3 - 3 * x;
return { x, y, z: 0 };
}),
color: getColor("LIME"),
showPoints: false,
labels: [
{
text: "Increasing",
at: 5,
offset: [-1, 1, 0],
},
],
},
{
points: Array.from({ length: 41 }, (_, i) => {
const x = -1 + i * 0.05;
const y = x ** 3 - 3 * x;
return { x, y, z: 0 };
}),
color: getColor("ROSE"),
showPoints: false,
labels: [
{
text: "Decreasing",
at: 20,
offset: [0, -1, 0],
},
],
},
{
points: Array.from({ length: 21 }, (_, i) => {
const x = 1 + i * 0.05;
const y = x ** 3 - 3 * x;
return { x, y, z: 0 };
}),
color: getColor("LIME"),
showPoints: false,
labels: [
{
text: "Increasing",
at: 15,
offset: [1, 1, 0],
},
],
},
{
points: [{ x: -1, y: 2, z: 0 }],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{
text: "Stationary (-1, 2)",
offset: [-2.5, 0.5, 0],
},
],
},
{
points: [{ x: 1, y: -2, z: 0 }],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{
text: "Stationary (1, -2)",
offset: [2.5, -0.5, 0],
},
],
},
]

## Exercises

1.  Determine the intervals where the function is increasing and decreasing for the curve $$f(x) = x^3 - 3x^2 - 15$$.

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

3.  Determine the intervals where the function is increasing and decreasing if the curve is given by $$f(x) = \sin 2x \cos 2x$$!

Visible text: 1. Determine the intervals where the function is increasing and decreasing for the curve .

2. If the function is always increasing on the interval , determine the value of the coefficient !

3. Determine the intervals where the function is increasing and decreasing if the curve is given by !

### Answer Key

1.  **Solution:**

    The first derivative of $$f(x) = x^3 - 3x^2 - 15$$ is <InlineMath math="f'(x) = 3x^2 - 6x" />.

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

    
    
    ```math
    3x^2 - 6x = 0 \implies 3x(x - 2) = 0
    ```

    The stationary points are at $$x=0$$ and $$x=2$$.

    By testing the intervals on a number line:

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

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

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

    So, the function is increasing on $$x < 0$$ or $$x > 2$$, and decreasing on the interval $$0 < x < 2$$.

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 $$0 < x < 2$$, the factor $$x$$ is always positive. Therefore, for <InlineMath math="f'(x) \ge 0" />, the second factor, $$(3ax + 2)$$, must also be non-negative.

    
    
    ```math
    3ax + 2 \ge 0
    ```

    This inequality must hold for all values of $$x$$ in the interval $$0 < x < 2$$. Since $$h(x) = 3ax + 2$$ 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 $$a \ge 0$$, then $$3ax$$ is also non-negative, so $$3ax+2$$ will definitely be positive. This condition is met.

    -   If $$a < 0$$, then $$h(x)$$ is a decreasing function. Its smallest value will be at the right end of the interval ($$x=2$$). For $$h(x)$$ 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 $$x=2$$:

    <MathContainer>
        
        
        ```math
        3a(2) + 2 \ge 0
        ```

        
        
        ```math
        6a \ge -2
        ```

        
        
        ```math
        a \ge -1/3
        ```

    </MathContainer>

    Combining both cases, the condition for the function to be always increasing on the given interval is $$a \ge -1/3$$.

3.  **Solution:**

    Use the double angle trigonometric identity: $$\sin(2A) = 2\sin(A)\cos(A)$$.

    So, $$f(x) = \sin 2x \cos 2x = \frac{1}{2} \sin(4x)$$.

    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 $$2\cos(4x) > 0$$ or $$\cos(4x) > 0$$. This occurs in the first and fourth quadrants.

        <MathContainer>
            
        
        ```math
        -\frac{\pi}{2} + 2k\pi < 4x < \frac{\pi}{2} + 2k\pi
        ```

            
        
        ```math
        -\frac{\pi}{8} + \frac{k\pi}{2} < x < \frac{\pi}{8} + \frac{k\pi}{2}
        ```

        </MathContainer>

        This interval is valid for any integer $$k$$.

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

        <MathContainer>
            
        
        ```math
        \frac{\pi}{2} + 2k\pi < 4x < \frac{3\pi}{2} + 2k\pi
        ```

            
        
        ```math
        \frac{\pi}{8} + \frac{k\pi}{2} < x < \frac{3\pi}{8} + \frac{k\pi}{2}
        ```

        </MathContainer>

        This interval is valid for any integer $$k$$.

Visible text: 1. **Solution:**

 The first derivative of is <InlineMath math="f'(x) = 3x^2 - 6x" />.

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

 
 

 The stationary points are at and .

 By testing the intervals on a number line:

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

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

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

 So, the function is increasing on or , and decreasing on the interval .

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 , the factor is always positive. Therefore, for <InlineMath math="f'(x) \ge 0" />, the second factor, , must also be non-negative.

 
 

 This inequality must hold for all values of in the interval . Since 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 , then is also non-negative, so will definitely be positive. This condition is met.

 - If , then is a decreasing function. Its smallest value will be at the right end of the interval (). For 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 :

 <MathContainer>
 
 

 
 

 
 

 </MathContainer>

 Combining both cases, the condition for the function to be always increasing on the given interval is .

3. **Solution:**

 Use the double angle trigonometric identity: .

 So, .

 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 or . This occurs in the first and fourth quadrants.

 <MathContainer>
 
 

 
 

 </MathContainer>

 This interval is valid for any integer .

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

 <MathContainer>
 
 

 
 

 </MathContainer>

 This interval is valid for any integer .