Problem 46 - Thomas Calculus

Question

a. Identify the function's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? c. Support your findings with a graphing calculator or computer grapher. $$ f(t)=t^{3}-3 t^{2}, \quad-\infty< t \leq 3 $$

Step-by-Step Solution

Answer

Local maximum at \( t = 0, f(0) = 0 \); local minimum at \( t = 2, f(2) = -4 \); absolute minimum is \( -4 \); absolute maximum is \( 0 \).

Step 1: Find the First Derivative

First, we differentiate the function \( f(t) = t^3 - 3t^2 \) with respect to \( t \). Using the power rule, the derivative is \( f'(t) = 3t^2 - 6t \).

Step 2: Set the Derivative Equal to Zero

To find the critical points, set the derivative \( f'(t) = 3t^2 - 6t \) equal to zero and solve for \( t \). This gives us the equation \( 3t(t-2) = 0 \). Thus, the critical points are \( t = 0 \) and \( t = 2 \).

Step 3: Determine the Nature of the Critical Points

Use the second derivative test to determine whether these critical points are local minima or maxima. First, find the second derivative \( f''(t) = 6t - 6 \). At \( t=0 \), \( f''(0) = -6 \), indicating a local maximum. At \( t=2 \), \( f''(2) = 6 \), indicating a local minimum.

Step 4: Evaluate the Function at Critical Points and Endpoint

Evaluate \( f(t) \) at the critical points and the boundary of the domain. We have \( f(0) = 0^3 - 3 imes 0^2 = 0 \), \( f(2) = 2^3 - 3 imes 2^2 = -4 \), and \( f(3) = 3^3 - 3 imes 3^2 = 0 \).

Step 5: Identify Absolute Extreme Values

Compare the function values at \( t = 0 \), \( t = 2 \), and \( t = 3 \). The smallest value is \( -4 \) at \( t = 2 \), making it the absolute minimum in the given domain. The absolute maximum value is \( 0 \) and occurs at both \( t = 0 \) and \( t = 3 \).

Step 6: Graphical Verification

Using a graphing calculator or computer grapher, plot the function \( f(t) = t^3 - 3t^2 \) over \( t = [-fty, 3] \). Verify visually that \( t = 2 \) is a local and absolute minimum, \( t = 0 \) is a local maximum, and \( t = 3 \) corresponds to a maximum value on the endpoint.

Key Concepts

First Derivative

The first derivative of a function gives us important information about the function's rate of change. When we want to understand where a function has peaks or valleys, we look to its derivative. In this exercise, we have the function \( f(t) = t^3 - 3t^2 \). We first find the derivative, noted as \( f'(t) \), to locate the critical points. Calculating the derivative involves using basic differentiation rules, particularly the power rule, which is a method used to find the derivative of power functions.
For \( f(t) = t^3 - 3t^2 \), the first derivative is \( f'(t) = 3t^2 - 6t \). By setting \( f'(t) \) to zero, we solve the equation \( 3t^2 - 6t = 0 \) and get the critical points, which are \( t = 0 \) and \( t = 2 \).
These critical points are essential as they highlight where changes in the function's behavior occur, such as turning points which might be local maxima or minima.

Second Derivative Test

Once we have the critical points from the first derivative, the second derivative test helps us determine the nature of these points. The second derivative, \( f''(t) \), represents the rate of change of the rate of change of the function, essentially describing the curvature or concavity of the graph.
For \( f(t) = t^3 - 3t^2 \), the second derivative is \( f''(t) = 6t - 6 \). This derivative tells us how the slope behaves around any point under consideration. Here’s how the test works:

• If \( f''(t) > 0 \), the function is concave up at that point, indicating a local minimum.
• If \( f''(t) < 0 \), the function is concave down, showing a local maximum.

Applying the test:

• At \( t = 0 \), \( f''(0) = -6 \), indicating that the point is a local maximum since the graph is concave down.
• At \( t = 2 \), \( f''(2) = 6 \), indicating that the point is a local minimum as it is concave up.

Local Maxima and Minima

Local maxima and minima are points in a function where the function reaches a high (or low) value on a small interval around the point. They are identified using the first and second derivative tests. In this exercise, we found:

• The local maximum occurs at \( t = 0 \) with a value of \( f(0) = 0 \).
• The local minimum occurs at \( t = 2 \) with a value of \( f(2) = -4 \).

Local extrema reflect significant points where the function changes direction. Graphically, these are points where the tangent to the curve is horizontal, and they give critical insights into the behavior and structure of the graph.

Absolute Extrema

Absolute extrema are the highest or lowest point across the entire domain of the function. Unlike local extrema, absolute extrema compare values across all points in the domain, including endpoints. For this specific function and domain \( -\infty < t \leq 3 \), we checked the critical points and endpoint of our domain.
By examining \( f(t) \) at \( t = 0 \), \( t = 2 \), and \( t = 3 \), we find:

• At \( t = 2 \), the function has an absolute minimum value of \( f(2) = -4 \).
• The absolute maximum value of \( 0 \) occurs at both \( t = 0 \) and \( t = 3 \).

Absolute extrema provide a comprehensive view of a function's behavior over its entire domain, which is crucial in understanding the overall landscape of the function graph.