Problem 49 - University Calculus: Early Transcendentals

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 \( t = -2 \), absolute minimum \( -16 \), no absolute maximum.

Step 1: Find the Derivative

To identify local extreme values, first find the derivative of the function. For the function \( f(t) = 12t - t^3 \), the derivative is found using basic differentiation rules: \[ f'(t) = 12 - 3t^2. \]

Step 2: Solve for Critical Points

Determine the critical points by solving \( f'(t) = 0 \). Set the derivative to zero and solve for \( t \): \[ 12 - 3t^2 = 0 \] \[ 3t^2 = 12 \] \[ t^2 = 4 \] \[ t = \pm 2. \]Consider \( t = -2 \) since it falls within the domain constraints \(-3 \leq t < \infty\). Discard \( t = 2 \) as it's not within the suitable section of the domain but we'll keep note for contexts of extreme value assessments.

Step 3: Determine the Nature of Critical Points

Use the second derivative test to determine the nature of these critical points:Calculate the second derivative: \[ f''(t) = -6t. \]Evaluate at the critical point \( t = -2 \): \[ f''(-2) = -6(-2) = 12 > 0. \]Since \( f''(-2) > 0 \), this indicates a local minimum at \( t = -2 \).

Step 4: Evaluate Function at Critical Points and Domain Boundaries

Evaluate the function \( f(t) \) at \( t = -2 \) for local extrema, and consider the boundary \( t = -3 \): - \( f(-3) = 12(-3) - (-3)^3 = -36 + 27 = -9 \)- \( f(-2) = 12(-2) - (-2)^3 = -24 + 8 = -16 \).Since the domain is half-open and extends to \( \infty \), evaluate behavior as \( t \to \infty \): \[ \lim_{{t \to \infty}} (12t - t^3) = -\infty \].

Step 5: Determine Absolute Extrema Using Graphing Tool

Graph the function to visualize and verify extreme values. Notice as \( t \to \infty, f(t) \to -\infty \) and that \( f(t) \) decreases at the extremes. The highest value in this domain is reached at \( t = -2 \), confirming the absolute minimum of \( -16 \). Check bounds; there is no absolute maximum in this interval.

Key Concepts

Local Extrema

Local extrema refer to points in a function where it reaches a local minimum or maximum value within a given range. To identify these values, we closely examine intervals of the function to find tops and bottoms of curves that do not necessarily extend to the boundaries.

• Local Minimum: A point where the function value is lower than all nearby points.
• Local Maximum: A point where the function value is higher than all nearby points.

In the original exercise, the focus is on finding local extrema within the domain (-3, ∞). For the function \( f(t) = 12t - t^3 \), solving for the critical pointshelps identify these local extremes. The assessment of these points through further tests, like the second derivative, clarifies whether they are minima or maxima.
Understanding local extrema is crucial in calculus because they help in analyzing the behavior of the function in intervals and can also signal important transitional points across a graph.

Derivative

The derivative of a function is a tool that shows the rate at which the function's value changes as the input changes. It gives us insight into the slope of the function at any point. When a derivative equals zero or is undefined, it indicates critical points.For the function \( f(t) = 12t - t^3 \), the derivative is: \[ f'(t) = 12 - 3t^2 \]
Using basic differentiation rules, this derivative illustrates how the slope of the function changes. By analyzing the sign and value of \( f'(t) \), we can deduce how the function behaves across its domain:

• Positive Derivative: The function is increasing.
• Negative Derivative: The function is decreasing.

Derivatives are fundamental in calculus for understanding how functions vary, helping with tasks such as optimization and modeling real-world phenomena.

Critical Points

Critical points of a function are where the derivative is zero or undefined. These points hold potential for local minima, maxima, or points of inflection. Identifying these points is a fundamental step in analyzing any function.For \( f(t) = 12t - t^3 \), we find critical points by solving:\[ 12 - 3t^2 = 0 \]\[ t^2 = 4 \]\[ t = \pm 2 \]
Given the domain \(-3 \leq t < \infty\), we take \( t = -2 \) as a valid critical point. Analyzing these points helps in rebuilding how the function behaves in different segments of its range. Critical points often reveal spots where the function changes direction or rate of change — translating to local extrema or other such features.

Second Derivative Test

The second derivative test is a method for determining the nature of critical points identified by the first derivative being zero. By evaluating the second derivative, we ascertain if a critical point is a local maximum, minimum, or inconclusive.For instance, with the second derivative of \( f(t) = 12t - t^3 \) given by:\[ f''(t) = -6t \],we assess critical points.

• If \( f''(t) > 0 \), the point is a local minimum.
• If \( f''(t) < 0 \), the point is a local maximum.
• If \( f''(t) = 0 \), the test is inconclusive.

Applying this to \( t = -2 \):\[ f''(-2) = 12 \],indicating a local minimum. By effectively using this test, we gain deeper insight into the structure of the graph, understanding how and where significant changes happen.