Problem 2 - Calculus

Question

Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points give a local maximum and which give a local minimum. $$ f(x)=x^{3}-12 x+\pi $$

Step-by-Step Solution

Answer

Local max at \(x = -2\), local min at \(x = 2\).

Step 1: Find the First Derivative

To find the critical points, start by differentiating the function. Given \( f(x) = x^3 - 12x + \pi \), the first derivative is \( f'(x) = 3x^2 - 12 \).

Step 2: Solve for Critical Points

Set \( f'(x) = 0 \) to find critical points. So, \( 3x^2 - 12 = 0 \). Simplify to get \( x^2 = 4 \), leading to \( x = 2 \) and \( x = -2 \) as critical points.

Step 3: Apply the First Derivative Test

Evaluate the sign of \( f'(x) = 3x^2 - 12 \) around the critical points. For \( x = -3 \), \( f'(x) = 27 - 12 = 15 > 0 \). For \( x = 0 \), \( f'(x) = -12 < 0 \). For \( x = 3 \), \( f'(x) = 27 - 12 = 15 > 0 \). The function changes from increasing to decreasing at \( x = 2 \) and from decreasing to increasing at \( x = -2 \). Therefore, \( x = 2 \) is a local maximum and \( x = -2 \) is a local minimum.

Step 4: Find the Second Derivative

Differentiate \( f'(x) = 3x^2 - 12 \) to find the second derivative: \( f''(x) = 6x \).

Step 5: Apply the Second Derivative Test

Check the second derivative at each critical point: At \( x = -2 \), \( f''(-2) = 6(-2) = -12 < 0 \), confirming a local maximum. At \( x = 2 \), \( f''(2) = 6(2) = 12 > 0 \), confirming a local minimum.

Key Concepts

First Derivative Test

The First Derivative Test is a fundamental method in calculus for analyzing functions and locating their local maxima and minima. This test involves assessing how the function changes direction around its critical points.

To begin, we determine the critical points by finding where the first derivative of the function is zero or undefined. In our exercise, the first derivative of the function \( f(x) = x^3 - 12x + \pi \) is \( f'(x) = 3x^2 - 12 \). Setting this equal to zero gives us the critical points \( x = 2 \) and \( x = -2 \).

Next, we examine the sign of the first derivative before and after each critical point to understand how the function's slope changes. For \( x = -3 \) (a point before \( x = -2 \)), \( f'(x) = 15 \), indicating the function is increasing. For \( x = 0 \), \( f'(x) = -12 \), indicating the function is decreasing. For \( x = 3 \) (a point after \( x = 2 \)), \( f'(x) = 15 \), showing that the function is increasing once again.

These sign changes tell us:

• The function transitions from increasing to decreasing at \( x = 2 \), suggesting a local maximum.
• The function transitions from decreasing to increasing at \( x = -2 \), suggesting a local minimum.

The First Derivative Test efficiently helps us identify local extrema by analyzing these directional changes.

Second Derivative Test

The Second Derivative Test is a more refined method to confirm if the critical points are local maxima or minima. This test utilizes the concavity of the function, as determined by the second derivative at critical points.

To apply this test, differentiate the first derivative of the function again to obtain the second derivative. In our scenario, the second derivative is \( f''(x) = 6x \). Once we have this, we evaluate it at each critical point we found earlier.

• At \( x = -2 \), \( f''(-2) = 6(-2) = -12 \). A negative second derivative indicates that the function is concave down, confirming \( x = -2 \) as a local maximum.


• At \( x = 2 \), \( f''(2) = 6(2) = 12 \). A positive second derivative indicates that the function is concave up, confirming \( x = 2 \) as a local minimum.

This method provides an additional layer of verification of the results obtained using the First Derivative Test, by looking at the curvature at those points. It is particularly useful when the first derivative test indicates equivocal results.

Local Maximum and Minimum

Identifying points of local maxima and minima is a key part of understanding the behavior of a function. These points give insights into where the function attains its highest and lowest values in the neighborhood around specific points.

A **local maximum** occurs at a critical point if the function transitions from increasing to decreasing at that point. This is observed for \( x = 2 \) using the First Derivative Test in our exercise, where the function switches its direction from going upwards to going downwards.

A **local minimum**, conversely, is at a critical point if the function transitions from decreasing to increasing. This occurs for \( x = -2 \), where the function changes from a downward slope to an upward slope.

Understanding these concepts not only helps in solving mathematical functions but also in applying these principles to real-world situations where optimization is required—such as calculating costs, areas, or even maximizing efficiency in different systems.