Problem 37 - Calculus: Early Transcendentals

Question

Describe the concavity of the functions below. $$ y=x^{3}-9 x^{2}+24 x $$

Step-by-Step Solution

Answer

The function is concave down on \((-\infty, 3)\) and concave up on \((3, \infty)\).

Step 1: Find the Second Derivative

To determine the concavity of a function, we first find the second derivative.Start by finding the first derivative of the function:\[ y = x^3 - 9x^2 + 24x \]The first derivative is:\[ y' = 3x^2 - 18x + 24 \]Now differentiate the first derivative to find the second derivative:\[ y'' = 6x - 18 \].

Step 2: Determine the Second Derivative's Sign

The concavity of the function depends on the sign of the second derivative:Set the second derivative equal to zero to find critical points:\[ 6x - 18 = 0 \].Solve for \( x \):\[ x = 3 \].The sign of \( y'' \) changes at these points, which can help us identify intervals of concavity.

Step 3: Analyze Intervals of Concavity

Test intervals around the critical point \( x = 3 \) to determine the concavity in each interval:1. Choose a test point from the interval \( x < 3 \), e.g., \( x = 2 \): \[ y''(2) = 6(2) - 18 = -6 \] Since \( y''(2) < 0 \), the function is concave down on \( (-\infty, 3) \).2. Choose a test point from the interval \( x > 3 \), e.g., \( x = 4 \): \[ y''(4) = 6(4) - 18 = 6 \] Since \( y''(4) > 0 \), the function is concave up on \( (3, \infty) \).

Key Concepts

Second Derivative

To understand the concavity of a function, finding the second derivative is a crucial step. The second derivative, denoted as \( y'' \), is the derivative of the first derivative of a function. It provides insight into how the slope of the function is changing. For the given function \( y = x^3 - 9x^2 + 24x \), we begin by finding the first derivative \( y' = 3x^2 - 18x + 24 \). Differentiating once more gives us the second derivative:

• \( y'' = 6x - 18 \)

The second derivative tells us whether the slope of \( y' \) is increasing or decreasing, which directly affects the concavity of the function. If the second derivative is positive, the original function slopes upwards; if negative, it slopes downwards. Thus, the sign of the second derivative indicates the concavity.

Critical Points

Critical points play a significant role in analyzing the behavior of mathematical functions. For the second derivative, critical points are found where it equals zero or does not exist. In this case, we set the second derivative \( y'' = 6x - 18 \) to zero to find the critical points:

• Setting \( 6x - 18 = 0 \)
• Solve for \( x \) to get \( x = 3 \)

This critical point indicates where there could be a change in the concavity of the function. The sign of \( y'' \) on either side of this point will tell us more about whether the function transitions from concave up to concave down, or vice versa. Identifying and analyzing these points help in mapping out the curvature of the function.

Intervals of Concavity

The sign of the second derivative across different intervals determines the concavity of a function. To find these intervals for our function, we consider the critical point \( x = 3 \) identified earlier. We test the sign of the second derivative \( y'' = 6x - 18 \) in intervals around this point:

• For \( x < 3 \), choose a point like \( x = 2 \). \( y''(2) = -6 \), indicating the function is concave down.
• For \( x > 3 \), choose a point like \( x = 4 \). \( y''(4) = 6 \), indicating the function is concave up.

By evaluating test points in these intervals, we determine that the function is concave down on \( (-\infty, 3) \) and concave up on \( (3, \infty) \). Thus, the function's concavity changes at \( x = 3 \), confirming our critical point analysis.

Concave Up and Down

Concavity influences how a function curves, and understanding this concept is essential for interpreting graphs. A function is described as concave up when it curves upwards, resembling a "U" shape, and concave down when it curves downwards, like an "n" shape. From our analysis:

• The function is concave down where \( y'' < 0 \). For our example, this occurs on the interval \( (-\infty, 3) \).
• The function is concave up where \( y'' > 0 \). In our case, this occurs on the interval \( (3, \infty) \).

Recognizing these patterns helps in predicting the general shape of a graph. On a graph, points like \( x = 3 \) where concavity switches are inflection points. Here, the function shifts from being concave down to concave up, signalling a change in curvature direction.