Using the First Derivative to Determine Intervals of Increase or Decrease

Introduction

Key Concepts

1. Understanding the First Derivative

The first derivative of a function, denoted as $f'(x)$ or $\frac{df}{dx}$, represents the rate at which the function's output changes with respect to its input. In geometric terms, it corresponds to the slope of the tangent line to the function at any given point. Formally, the first derivative is defined as: $$ f'(x) = \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h} $$ This limit, if it exists, provides the instantaneous rate of change of the function at point $x$.

2. Critical Points and Their Significance

Critical points are specific values of $x$ in the domain of the function where the first derivative is zero or undefined. These points are essential as they often correspond to local maxima, minima, or points of inflection. To find critical points, solve the equation: $$ f'(x) = 0 $$ or identify where $f'(x)$ does not exist. Once identified, these points help in partitioning the function's domain into intervals where the function is increasing or decreasing.

3. Determining Intervals of Increase and Decrease

To ascertain where a function is increasing or decreasing, follow these steps:

1. Find the First Derivative: Compute $f'(x)$ using differentiation rules.
2. Identify Critical Points: Solve $f'(x) = 0$ and determine where $f'(x)$ is undefined.
3. Choose Test Points: Select values within each interval defined by the critical points.
4. Evaluate the First Derivative: Substitute the test points into $f'(x)$ to determine the sign (positive or negative).
5. Conclude the Behavior: If $f'(x) > 0$ in an interval, the function is increasing there. If $f'(x) < 0$, it is decreasing.

Example: Consider the function $f(x) = x^3 - 3x^2 + 2x$.

1. First Derivative: $f'(x) = 3x^2 - 6x + 2$
2. Critical Points: Solve $3x^2 - 6x + 2 = 0$: $$ x = \frac{6 \pm \sqrt{(6)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = 1 \pm \frac{\sqrt{3}}{3} $$
3. Test Points: Choose $x = 0$, $x = 1$, and $x = 2$.
4. Evaluate $f'(x)$: <
   • At $x = 0$: $f'(0) = 2 > 0$ (Increasing)
   • At $x = 1$: $f'(1) = 3 - 6 + 2 = -1 < 0$ (Decreasing)
   • At $x = 2$: $f'(2) = 12 - 12 + 2 = 2 > 0$ (Increasing)
5. Conclusion:
   • Increasing on $(-\infty, 1 - \frac{\sqrt{3}}{3})$ and $(1 + \frac{\sqrt{3}}{3}, \infty)$
   • Decreasing on $(1 - \frac{\sqrt{3}}{3}, 1 + \frac{\sqrt{3}}{3})$

4. The First Derivative Test

The First Derivative Test is a method used to determine the nature of critical points. It involves analyzing the sign of $f'(x)$ around each critical point:

• If $f'(x)$ changes from positive to negative at a critical point $c$, then $f(c)$ is a local maximum.
• If $f'(x)$ changes from negative to positive at $c$, then $f(c)$ is a local minimum.
• If $f'(x)$ does not change signs, $c$ is neither a local maximum nor minimum (possible point of inflection).

Example: Using the previous function $f(x) = x^3 - 3x^2 + 2x$, analyze the critical points:

• At $x = 1 - \frac{\sqrt{3}}{3}$:
  • Test point to the left: $x = 0$ ($f'(0) = 2 > 0$)
  • Test point to the right: $x = 1$ ($f'(1) = -1 < 0$)
  • Sign changes from positive to negative: Local maximum at $x = 1 - \frac{\sqrt{3}}{3}$
• At $x = 1 + \frac{\sqrt{3}}{3}$:
  • Test point to the left: $x = 1$ ($f'(1) = -1 < 0$)
  • Test point to the right: $x = 2$ ($f'(2) = 2 > 0$)
  • Sign changes from negative to positive: Local minimum at $x = 1 + \frac{\sqrt{3}}{3}$

5. Applications in Real-World Problems

Determining intervals of increase and decrease is not merely an academic exercise; it has practical applications in various fields such as economics, physics, and engineering. For instance:

• Economics: Analyzing profit functions to determine ranges of production where profit increases or decreases.
• Physics: Studying velocity functions to identify periods when an object is speeding up or slowing down.
• Engineering: Optimizing designs by understanding the behavior of stress-strain curves.

**Example:** In economics, if a company's revenue function is $R(x) = 50x - x^2$, where $x$ represents the number of units sold, determining the intervals where revenue increases or decreases can guide production strategies.

1. First Derivative: $R'(x) = 50 - 2x$
2. Critical Point: $50 - 2x = 0$ leads to $x = 25$
3. Test Intervals:
   • For $x < 25$, say $x = 20$: $R'(20) = 10 > 0$ (Increasing)
   • For $x > 25$, say $x = 30$: $R'(30) = -10 < 0$ (Decreasing)
4. Conclusion:
   • Revenue increases when $0 < x < 25$ units.
   • Revenue decreases when $x > 25$ units.

6. Graphical Interpretation

Visualizing the function alongside its first derivative provides intuitive understanding:

• Positive Derivative: The graph of the function is rising; $f(x)$ is increasing.
• Negative Derivative: The graph of the function is falling; $f(x)$ is decreasing.
• Zero Derivative: Possible peaks, troughs, or inflection points.

Analyzing the graph helps in quickly identifying intervals of increase or decrease without extensive calculations.

7. Common Mistakes and How to Avoid Them

• Ignoring Undefined Derivatives: Critical points occur not only where $f'(x) = 0$ but also where $f'(x)$ is undefined.
• Incorrect Test Points: Selecting test points that do not lie within the intervals defined by critical points can lead to erroneous conclusions.
• Miscalculating the Derivative: Errors in differentiation can propagate through the entire analysis.
• Overlooking Endpoint Behavior: In some cases, especially with domain-restricted functions, endpoint behaviors are crucial.

To avoid these mistakes:

• Thoroughly solve $f'(x) = 0$ and identify all points where $f'(x)$ is undefined.
• Carefully choose test points that are within each interval.
• Double-check differentiation steps for accuracy.
• Consider the domain of the function when analyzing endpoints.

8. Advanced Considerations

While the first derivative provides insights into the increasing and decreasing behavior of functions, further analysis using the second derivative can uncover concavity and points of inflection. Additionally, understanding higher-order derivatives can lead to more nuanced interpretations of a function's behavior.

Moreover, in multivariable calculus, similar concepts extend to partial derivatives, enabling the analysis of functions with more than one variable.

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Summary and Key Takeaways