Applying the Candidates Test to Determine Absolute Extrema on a Closed Interval

Introduction

Key Concepts

Understanding Absolute Extrema

Absolute extrema refer to the highest and lowest values that a function attains over a specific interval. On a closed interval $[a, b]$, a function $f(x)$ may have absolute maximum and minimum values. Identifying these points is crucial for analyzing the function's overall behavior.

The Candidates Test Explained

The Candidates Test is a systematic method used to determine the absolute extrema of a continuous function on a closed interval. The test involves evaluating the function at critical points within the interval and at the endpoints. The highest and lowest values obtained from these evaluations are the absolute maximum and minimum, respectively.

Steps to Apply the Candidates Test

1. Ensure Continuity: Verify that the function $f(x)$ is continuous on the closed interval $[a, b]$.
2. Find Critical Points: Determine the critical points by finding where the first derivative $f'(x)$ is zero or undefined within $(a, b)$.
3. Evaluate the Function: Calculate the function values at each critical point and at the endpoints $a$ and $b$.
4. Compare Values: Identify the largest and smallest values from these evaluations to determine the absolute maximum and minimum.

Mathematical Formulation

Given a continuous function $f$ on a closed interval $[a, b]$, the absolute extrema are found by:

• Computing $f(a)$ and $f(b)$.
• Finding all $x$ in $(a, b)$ where $f'(x) = 0$ or $f'(x)$ does not exist.
• Evaluating $f(x)$ at each critical point.
• Comparing all evaluated values to determine the absolute maximum and minimum.

Example Application

Consider the function $f(x) = x^3 - 3x^2 + 4$ on the interval $[0, 3]$.

1. Find $f'(x)$: $f'(x) = 3x^2 - 6x$.
2. Set $f'(x) = 0$: $3x^2 - 6x = 0 \Rightarrow x(3x - 6) = 0 \Rightarrow x = 0$ or $x = 2$.
3. Determine Critical Points: $x = 0$ and $x = 2$. Note that $x = 0$ is also an endpoint.
4. Evaluate $f(x)$ at Critical Points and Endpoints:
   • $f(0) = 0^3 - 3 \cdot 0^2 + 4 = 4$
   • $f(2) = 2^3 - 3 \cdot 2^2 + 4 = 8 - 12 + 4 = 0$
   • $f(3) = 3^3 - 3 \cdot 3^2 + 4 = 27 - 27 + 4 = 4$
5. Determine Absolute Extrema: The absolute maximum is 4 at $x = 0$ and $x = 3$, and the absolute minimum is 0 at $x = 2$.

Graphical Interpretation

Visualizing the function's graph can provide additional insights. The Candidates Test identifies points where the function reaches its peaks and troughs within the interval. By plotting $f(x)$, students can see how the absolute extrema correspond to the highest and lowest points on the graph over the specified interval.

Importance in Optimization Problems

Absolute extrema are pivotal in optimization problems, where the goal is to find the maximum or minimum values of a function under given constraints. The Candidates Test simplifies this process by narrowing down the possible points where these extrema can occur, thereby facilitating efficient problem-solving.

Common Mistakes to Avoid

• Ignoring Endpoints: Always evaluate the function at the endpoints of the interval, as absolute extrema can occur there.
• Overlooking Critical Points: Ensure all critical points where $f'(x) = 0$ or $f'(x)$ is undefined are considered.
• Assuming Extrema Without Verification: Always compare function values at critical points and endpoints to confirm absolute extrema.

Advanced Applications

Beyond basic optimization, the Candidates Test is foundational in fields such as physics, economics, and engineering. For instance, determining the maximum profit or the minimum cost involves identifying absolute extrema of relevant functions, leveraging the principles outlined in the Candidates Test.

Comparison Table

Summary and Key Takeaways