Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For each function: (a) Find all critical points on the specified interval. (b) Classify each critical point: Is it a local maximum, a local minimum, an absolute maximum, or an absolute minimum? (c) If the function attains an absolute maximum and/or minimum on the specified interval, what is the maximum and/or minimum value? \(f(x)=x^{3}-3 x+2\) on \([-5,5]\)

Step-by-Step Solution

Answer

There are two critical points: \(x = -1\) and \(x = 1\), where the point at \(x = -1\) is a local maximum and the point at \(x = 1\) is a local minimum. The function attains an absolute maximum value of \(108\) at \(x = 5\) and an absolute minimum value of \(-108\) at \(x = -5\).

Step 1: Calculate the derivative

Derivative of \(f(x)\) will be calculated as \(f'(x) = 3x^{2}-3\).

Step 2: Find the critical points

The critical points are found by setting the derivative equal to zero: \(f'(x) = 0\), which gives \(3x^{2}-3 = 0\) or \(x^{2} = 1\), thus the critical points are \(x = -1\) and \(x = 1\).

Step 3: Determine details about the critical points

Calculate the second derivative \(f''(x) = 6x\), and evaluate this at each critical point. \(f''(-1) = -6 < 0\), so \(x = -1\) is a local maximum. \(f''(1) = 6 > 0\), so \(x = 1\) is a local minimum.

Step 4: Find the absolute maximum and minimum values

Evaluate the function at the critical points and at the ends of the given interval, we get \(f(-5) = -108, f(-1) = 4, f(1) = 0, f(5) = 108\). So, the absolute maximum is \(108\) at \(x = 5\) and the absolute minimum is \(-108\) at \(x = -5\)

Key Concepts

Understanding Critical Points

In calculus, the concept of critical points is fundamental when studying the behavior of a function. A critical point occurs at a value of the variable where the derivative of the function is zero or undefined. For the function \(f(x) = x^3 - 3x + 2\), the derivative \(f'(x) = 3x^2 - 3\) is calculated to find these points.

By setting the derivative equal to zero, \(3x^2 - 3 = 0\), we solve for \(x\) and find the critical points at \(x = -1\) and \(x = 1\). These points are crucial because they indicate where the function changes direction, potentially revealing local maximum or minimum values.

Local Maximum and Minimum Explained

Local maxima and minima play a pivotal role in analyzing a function's behavior near critical points. Simply put, a local maximum is a point where the function value is higher than at any nearby points, whereas a local minimum is lower than at nearby points.

To classify the critical points of \(f(x)\), we use the second derivative test. The second derivative of \(f(x)\) is \(f''(x) = 6x\). Evaluating this at \(x = -1\) gives \(f''(-1) = -6\), indicating a local maximum since \(f''(x) < 0\) signifies a concave down curve. For \(x = 1\), \(f''(1) = 6\), and \(f''(x) > 0\) signals a concave up curve, making it a local minimum.

Absolute Maximum and Minimum Values

Absolute maximum and minimum values are the highest and lowest points a function reaches over its entire domain or a specified interval. For \(f(x) = x^3 - 3x + 2\) on the interval \([-5, 5]\), we examine the endpoints and critical points to determine these values.

Calculating \(f(x)\) at the critical points and boundary points, we find:

• At \(x = -5\), \(f(-5) = -108\)
• At \(x = 1\), \(f(1) = 0\)
• At \(x = -1\), \(f(-1) = 4\)
• At \(x = 5\), \(f(5) = 108\)

Thus, the absolute maximum is \(108\) at \(x = 5\), and the absolute minimum is \(-108\) at \(x = -5\).

Derivatives: The Building Blocks

Derivatives are fundamental tools in calculus used to analyze the rate of change of functions. They help identify important characteristics like critical points, inflection points, slopes, and more. For our function, \(f(x) = x^3 - 3x + 2\), we found the first derivative \(f'(x) = 3x^2 - 3\). This derivative tells us where the function's slope is zero.

The critical points, where \(f'(x) = 0\), give insight into potential locations for maximum and minimum values. Derivatives essentially allow us to "peek beneath the surface" of a function, turning complex curve analysis into manageable mathematics.

Second Derivative: A Deeper Insight

The second derivative of a function provides even more detailed information about the function's behavior. It examines how the slope of the original function's curve changes, offering insights into the concavity. By evaluating the second derivative, we determine whether a curve bends upwards (concave up) or downwards (concave down).

For \(f(x) = x^3 - 3x + 2\), the second derivative is \(f''(x) = 6x\). At \(x = -1\), the second derivative is negative, indicating a local maximum because the curve bends downwards. Conversely, at \(x = 1\), a positive second derivative reveals a local minimum as the curve bends upwards. This additional layer of derivative analysis is critical for accurately classifying critical points.