Problem 27 - Calculus with Concepts in Calculus

Question

Use the Second Derivative Test to determine the relative extreme values (if any) of the function. $$ f(x)=x^{3}-3 x^{2}-24 x+1 $$

Step-by-Step Solution

Answer

The function has a local minimum at \( x = 4 \) and a local maximum at \( x = -2 \).

Step 1: Find the First Derivative

To find the critical points where the relative extrema might occur, we first need the derivative of the function. Differentiate the function \( f(x) = x^3 - 3x^2 - 24x + 1 \):\[ f'(x) = 3x^2 - 6x - 24 \].

Step 2: Solve for Critical Points

Set the first derivative equal to zero and solve for \( x \) to find the critical points:\[ 3x^2 - 6x - 24 = 0 \].Dividing the whole equation by 3 gives \( x^2 - 2x - 8 = 0 \). Factoring, we get \( (x - 4)(x + 2) = 0 \), which gives solutions \( x = 4 \) and \( x = -2 \). These are our critical points.

Step 3: Find the Second Derivative

Compute the second derivative of \( f(x) \):\[ f''(x) = 6x - 6 \].

Step 4: Apply the Second Derivative Test

Use the critical points found in Step 2 and plug them into the second derivative:- For \( x = 4 \): \[ f''(4) = 6(4) - 6 = 18 \], since \( 18 > 0 \), \( f(x) \) has a local minimum at \( x = 4 \).- For \( x = -2 \): \[ f''(-2) = 6(-2) - 6 = -18 \], since \( -18 < 0 \), \( f(x) \) has a local maximum at \( x = -2 \).

Key Concepts

Critical Points

Critical points are essential in calculus because they help us identify where functions may have relative extrema, meaning local maximum or minimum values. These points occur where the first derivative of a function is zero or undefined. To find critical points, follow these steps:

• Take the first derivative of the function.
• Set the first derivative equal to zero and solve for the variable.

For the function given in the exercise, the first derivative is differentiable everywhere, leading us to solve for occasions where it equals zero. By solving the equation obtained from the first derivative, we identify the critical points where potential relative extrema could occur.

First Derivative

The first derivative of a function provides us with crucial insights into the function's rate of change. It shows how the function's output changes with respect to its input. Here's why it's important:

• A positive first derivative suggests that the function is increasing.
• A negative first derivative suggests that the function is decreasing.
• Where the first derivative equals zero or is undefined is where critical points may be located.

In the given exercise, the first derivative is calculated as \( f'(x) = 3x^2 - 6x - 24 \). Solving \( f'(x) = 0 \), we determined the critical points at \( x = 4 \) and \( x = -2 \). At these points, the function could potentially have a relative extremum.

Second Derivative

The second derivative of a function tells us about the curvature or concavity of the function's graph. It plays a crucial role in using the Second Derivative Test to determine the nature of critical points. Here's what the second derivative reveals:

• If the second derivative is positive at a critical point, the function has a local minimum at that point.
• If the second derivative is negative at a critical point, the function has a local maximum at that point.
• If the second derivative is zero, the test is inconclusive, and the point might be an inflection or requires further analysis.

For the exercise function, the second derivative is \( f''(x) = 6x - 6 \). By evaluating the second derivative at \( x = 4 \) and \( x = -2 \), we confirmed the nature of these critical points.

Relative Extrema

Relative extrema refer to the local maximum or minimum values of a function. These points indicate where a function temporarily increases to its highest value or decreases to its lowest value within a particular region. Here's how we determine them:

• Find the critical points by setting the first derivative to zero.
• Use the Second Derivative Test at each critical point:
• A positive second derivative indicates a local minimum.
• A negative second derivative indicates a local maximum.

In the given problem, the application of this process revealed a local minimum at \( x = 4 \) and a local maximum at \( x = -2 \).