Problem 45 - Thomas Calculus in SI Units

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. $$f(t)=12 t-t^{3}, \quad-3 \leq t<\infty$$

Step-by-Step Solution

Answer

Local minimum at \( t = -2 \), local maximum at \( t = 2 \). Absolute maximum is \( f(2) = 16 \), no absolute minimum.

Step 1: Find the Critical Points

To identify the local extreme values, we first need to find the critical points of the function. We start by differentiating the function. The derivative of the function is \( f'(t) = 12 - 3t^2 \). Set the derivative equal to zero to find critical points: \( 12 - 3t^2 = 0 \). Solve for \( t \) to find \( t^2 = 4 \), thus \( t = 2 \) and \( t = -2 \). Since the domain is \( -3 \leq t < \infty \), both critical points \( t = 2 \) and \( t = -2 \) are included.

Step 2: Determine Local Extremes Using Second Derivative Test

We use the second derivative test to determine if these critical points are maxima, minima, or neither. The second derivative is \( f''(t) = -6t \). Evaluate the second derivative at the critical points: \( f''(-2) = 12 \), which is positive, indicating a local minimum at \( t = -2 \); \( f''(2) = -12 \), which is negative, indicating a local maximum at \( t = 2 \).

Step 3: Evaluate Boundary and Critical Values

Evaluate the function \( f(t) \) at the critical points and boundaries of the domain. At \( t = -3 \), \( f(-3) = 12(-3) - (-3)^3 = -9 \). At \( t = -2 \), \( f(-2) = 12(-2) - (-2)^3 = -8 \). At \( t = 2 \), \( f(2) = 12(2) - 2^3 = 16 \). As \( t \to \infty \), \( f(t) \to -\infty \).

Step 4: Identify Absolute Extremes

Compare the function values from the previous step: \( f(-3) = -9 \), \( f(-2) = -8 \), \( f(2) = 16 \). The absolute maximum in the domain \( -3 \leq t < \infty \) is \( f(2) = 16 \). There is no absolute minimum since \( f(t) \to -\infty \) as \( t \to \infty \).

Step 5: Graph the Function for Confirmation

Using a graphing tool, plot the function \( f(t) = 12t - t^3 \). Observe the local minimum at \( t = -2 \), local maximum at \( t = 2 \), and the decrease towards \( -\infty \) as \( t \) increases. The graph confirms our analytical findings.

Key Concepts

Critical Points

Critical points are vital when determining where a function's slope is zero, leading to potential maxima or minima. To find them, differentiate the given function and set the derivative to zero. This pinpoints where the slope flattens out. In our example, the function is

• Differentiate: The derivative of the function \( f(t) = 12t - t^3 \) is \( f'(t) = 12 - 3t^2 \).
• Set to zero: Solve \( 12 - 3t^2 = 0 \) to find \( t^2 = 4 \).
• Find solutions: The solutions, \( t = 2 \) and \( t = -2 \), are critical points within the domain \( -3 \leq t < \infty \).

These critical points help us determine where the function might reach local extrema.

Local Extrema

Local extrema refer to the points where a function reaches a local maximum or minimum. These occur at critical points, which is why finding critical points is an essential first step. Once critical points are identified, the next task is to determine what type of extremum they represent.In our function, we have identified the critical points at \( t = 2 \) and \( t = -2 \). Evaluating the function at these points:

• \( f(-2) = 12(-2) - (-2)^3 = -8 \): Local minimum.
• \( f(2) = 12(2) - 2^3 = 16 \): Local maximum.

Understanding local extrema is essential in analyzing and graphing the behavior of functions within their domain.

Second Derivative Test

The second derivative test is a powerful way to determine the nature of critical points. By looking at the sign of the second derivative at these points, we can easily identify if the critical point is a maximum, a minimum, or neither. Here's how it works:

• Compute the second derivative: For \( f(t) = 12t - t^3 \), the second derivative is \( f''(t) = -6t \).
• Evaluate at critical points:
  • At \( t = -2 \), \( f''(-2) = 12 \). Since \( 12 > 0 \), it's a local minimum.
  • At \( t = 2 \), \( f''(2) = -12 \). Since \( -12 < 0 \), it's a local maximum.

This test quickly tells us what the curve is doing at these critical points without having to graph the entire function.

Graphing Calculator

A graphing calculator can visually confirm calculations done analytically, offering a helpful cross-reference. In the case of our function \( f(t) = 12t - t^3 \), using a graphing calculator involves several handy steps:

• Input the function: Enter \( f(t) = 12t - t^3 \) into the calculator.
• Set the domain: Ensure the graph shows \( -3 \leq t < \infty \).
• Analyze features: Look for peaks and troughs corresponding to local extrema.
• Confirm analytical solutions: Notice the local minimum at \( t = -2 \) and maximum at \( t = 2 \).

This visual aid reinforces understanding and highlights any differences between predicted and actual behavior, making it a valuable tool for calculus students.