Determining Intervals of Increase and Decrease

Introduction

Key Concepts

1. Understanding Function Behavior

Functions describe the relationship between variables, where each input has a corresponding output. Analyzing the behavior of functions involves determining how the output changes as the input varies. Specifically, identifying intervals where the function is increasing or decreasing provides insights into the function's overall shape and potential maxima or minima.

2. Definitions

• Interval of Increase: An interval on the domain of the function where, as \(x\) increases, \(f(x)\) also increases.
• Interval of Decrease: An interval on the domain of the function where, as \(x\) increases, \(f(x)\) decreases.
• Critical Points: Points on the graph of a function where the first derivative is zero or undefined. These points are potential candidates for local maxima or minima.

3. First Derivative and Its Role

The first derivative of a function, denoted as \(f'(x)\), represents the rate of change or the slope of the function at any given point. By analyzing \(f'(x)\), we can determine where the function is increasing or decreasing:

• If \(f'(x) > 0\) on an interval, the function is increasing on that interval.
• If \(f'(x) < 0\) on an interval, the function is decreasing on that interval.

Critical points occur where \(f'(x) = 0\) or \(f'(x)\) is undefined. These points are crucial in identifying changes in the behavior of the function.

4. Finding the First Derivative

To determine intervals of increase and decrease, follow these steps:

1. Find the first derivative \(f'(x)\): Differentiate the given function with respect to \(x\).
2. Solve \(f'(x) = 0\): Find the critical points by setting the first derivative equal to zero and solving for \(x\).
3. Determine where \(f'(x)\) is positive or negative: Test intervals around the critical points to see where the derivative is positive (increasing) or negative (decreasing).

5. Example: Quadratic Function

Consider the quadratic function \(f(x) = x^2 - 4x + 3\).

1. Find the first derivative: $$ f'(x) = 2x - 4 $$
2. Solve \(f'(x) = 0\): $$ 2x - 4 = 0 \implies x = 2 $$
3. Determine the sign of \(f'(x)\) around \(x = 2\):
   • For \(x < 2\), choose \(x = 1\): \(f'(1) = 2(1) - 4 = -2 < 0\).
   • For \(x > 2\), choose \(x = 3\): \(f'(3) = 2(3) - 4 = 2 > 0\).

Conclusion:

• The function is decreasing on \( (-\infty, 2) \).
• The function is increasing on \( (2, \infty) \).

6. Example: Cubic Function

Consider the cubic function \(f(x) = x^3 - 3x^2 + 2x\).

1. Find the first derivative: $$ f'(x) = 3x^2 - 6x + 2 $$
2. Solve \(f'(x) = 0\): $$ 3x^2 - 6x + 2 = 0 \implies x = \frac{6 \pm \sqrt{(6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{12}}{6} = 1 \pm \frac{\sqrt{3}}{3} $$
3. Determine the sign of \(f'(x)\) around the critical points:
   • For \(x < 1 - \frac{\sqrt{3}}{3}\), choose \(x = 0\): \(f'(0) = 2 > 0\).
   • For \(1 - \frac{\sqrt{3}}{3} < x < 1 + \frac{\sqrt{3}}{3}\), choose \(x = 1\): \(f'(1) = 3 - 6 + 2 = -1 < 0\).
   • For \(x > 1 + \frac{\sqrt{3}}{3}\), choose \(x = 2\): \(f'(2) = 12 - 12 + 2 = 2 > 0\).

Conclusion:

• The function is increasing on \( (-\infty, 1 - \frac{\sqrt{3}}{3}) \) and \( (1 + \frac{\sqrt{3}}{3}, \infty) \).
• The function is decreasing on \( (1 - \frac{\sqrt{3}}{3}, 1 + \frac{\sqrt{3}}{3}) \).

7. Higher-Order Functions and Rational Functions

For higher-degree polynomials and rational functions, the process remains similar but may involve more critical points and complex intervals. For example, consider the rational function \(f(x) = \frac{x^2 - 1}{x - 2}\).

1. Find the first derivative using the quotient rule: $$ f'(x) = \frac{(2x)(x - 2) - (x^2 - 1)(1)}{(x - 2)^2} = \frac{2x(x - 2) - (x^2 - 1)}{(x - 2)^2} = \frac{2x^2 - 4x - x^2 + 1}{(x - 2)^2} = \frac{x^2 - 4x + 1}{(x - 2)^2} $$
2. Solve \(f'(x) = 0\): $$ x^2 - 4x + 1 = 0 \implies x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3} $$
3. Determine the sign of \(f'(x)\) around the critical points and vertical asymptote at \(x = 2\):
   • For \(x < 2 - \sqrt{3}\), choose \(x = 0\): \(f'(0) = \frac{0 - 0 + 1}{4} = \frac{1}{4} > 0\).
   • For \(2 - \sqrt{3} < x < 2\), choose \(x = 1\): \(f'(1) = \frac{1 - 4 + 1}{1} = -2 < 0\).
   • For \(2 < x < 2 + \sqrt{3}\), choose \(x = 3\): \(f'(3) = \frac{9 - 12 + 1}{1} = -2 < 0\).
   • For \(x > 2 + \sqrt{3}\), choose \(x = 4\): \(f'(4) = \frac{16 - 16 + 1}{4} = \frac{1}{4} > 0\).

Conclusion:

• The function is increasing on \( (-\infty, 2 - \sqrt{3}) \) and \( (2 + \sqrt{3}, \infty) \).
• The function is decreasing on \( (2 - \sqrt{3}, 2) \) and \( (2, 2 + \sqrt{3}) \).

8. Application: Optimization Problems

Determining intervals of increase and decrease is not just an academic exercise; it has practical applications in optimization problems. For instance, finding the dimensions that maximize area or minimize cost often involves identifying critical points and determining where functions increase or decrease.

Example:

• Problem: Find the dimensions of a rectangle with a fixed perimeter of 20 units that maximize the area.
• Solution:
  1. Let \(x\) be the length and \(y\) be the width. Given \(2x + 2y = 20 \implies y = 10 - x\).
  2. Area Function: \(A(x) = x \cdot y = x(10 - x) = 10x - x^2\).
  3. First Derivative: $$ A'(x) = 10 - 2x $$
  4. Set \(A'(x) = 0\) to find critical points: $$ 10 - 2x = 0 \implies x = 5 $$
  5. Determine intervals:
     • For \(x < 5\), \(A'(x) > 0\) (increasing).
     • For \(x > 5\), \(A'(x) < 0\) (decreasing).
  6. Conclusion: The area is maximized when \(x = 5\) and \(y = 5\), forming a square.

9. Testing Intervals Using the First Derivative Test

The First Derivative Test is a method to determine whether a critical point is a local maximum, local minimum, or neither. After identifying critical points, the sign of the first derivative changes around these points to classify them:

• Local Maximum: The function changes from increasing to decreasing.
• Local Minimum: The function changes from decreasing to increasing.
• Neither: The function does not change its increasing/decreasing behavior.

Applying this test to critical points helps in understanding the function's graph and behavior comprehensively.

10. Higher-Order Derivatives and Concavity

While determining intervals of increase and decrease primarily involves the first derivative, understanding concavity through the second derivative can provide additional insights:

• Second Derivative \(f''(x)\): Represents the rate of change of the first derivative. It indicates the concavity of the function.
• Relationship with First Derivative: If the second derivative is positive, the first derivative is increasing; if negative, the first derivative is decreasing.

This relationship is useful in more advanced analysis but extends the foundational concepts of increasing and decreasing intervals.

11. Graphical Interpretation

Visualizing functions helps in intuitively understanding intervals of increase and decrease. The slope of the tangent line to the graph at any point corresponds to the first derivative:

• Positive Slope: Indicates increasing function.
• Negative Slope: Indicates decreasing function.

Graphing both the function and its derivative can facilitate a deeper comprehension of the function's behavior across different intervals.

12. Common Mistakes to Avoid

• Incorrect Derivative Calculation: Ensure accurate differentiation to avoid flawed conclusions.
• Neglecting Undefined Derivatives: Critical points can also occur where the derivative does not exist, not just where it is zero.
• Mistaking Intervals: Carefully test each interval around critical points to determine the correct behavior.
• Ignoring Domain Restrictions: Functions may have domain restrictions that impact the intervals of increase and decrease.

13. Practice Problems

Engaging with practice problems is vital for mastering the determination of intervals of increase and decrease. Here are a few examples:

• Problem 1: Determine the intervals where the function \( f(x) = -2x^3 + 9x^2 - 12x + 4 \) is increasing or decreasing.
• Problem 2: For the function \( f(x) = \frac{3x - 1}{x + 2} \), find the intervals of increase and decrease.
• Problem 3: Given \( f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1 \), identify the intervals where the function is increasing and decreasing.

Attempting these problems reinforces the concepts and techniques essential for determining intervals of increase and decrease.

14. Real-World Applications

Understanding intervals of increase and decrease is crucial in various real-world contexts, such as:

• Economics: Determining profit maximization and cost minimization points.
• Engineering: Analyzing the performance of systems and optimizing designs.
• Physics: Describing motion, where velocity and acceleration relate to increasing and decreasing positions.

These applications highlight the practical importance of mathematical concepts in solving tangible problems.

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Summary and Key Takeaways