Problem 33 - Thomas Calculus

Question

In Exercises \(29-36 :\) a. Identify the function's local extreme values in the given domain, and say where they are assumed. 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

The function has a local minimum at \( t = -2 \) and a local (and absolute) maximum at \( t = 2 \). There is no absolute minimum.

Step 1: Find the Derivative

To find the local extreme values of the function, we first need to find its critical points. Start by computing the derivative of the function: Given \( f(t) = 12t - t^3 \).The derivative is: \( f'(t) = 12 - 3t^2 \).

Step 2: Find Critical Points

To find critical points, set the derivative to zero and solve:\[ 0 = 12 - 3t^2 \] \[ 3t^2 = 12 \] \[ t^2 = 4 \] \[ t = \pm 2 \].Thus, the critical points are \( t = 2 \) and \( t = -2 \).

Step 3: Determine Local Extrema

Evaluate the second derivative to apply the second derivative test:Compute \( f''(t) = -6t \). - At \( t = 2 \): \( f''(2) = -12 \) which is negative, indicating a local maximum.- At \( t = -2 \): \( f''(-2) = 12 \) which is positive, indicating a local minimum.

Step 4: Evaluate Function at Critical Points

Evaluate the function at the critical points:- \( f(-2) = 12(-2) - (-2)^3 = -24 + 8 = -16 \) - \( f(2) = 12(2) - (2)^3 = 24 - 8 = 16 \)

Step 5: Consider the Endpoints and Infinity

In the domain \(-3 \leq t < \infty\): - Check at the endpoint: \( f(-3) = 12(-3) - (-3)^3 = -36 + 27 = -9 \) - As \( t \to \infty \), \( f(t) = 12t - t^3 \), and as \( t^3 \) dominates \( 12t \), \( f(t) \to -\infty \).

Step 6: Determine Absolute Extreme Values

Compare the function values: - Local minimum at \( t = -2 \) gives \( f(-2) = -16 \) - Local maximum at \( t = 2 \) gives \( f(2) = 16 \)The absolute maximum in the considered domain is at \( t = 2 \), and there is no absolute minimum because the function approaches \(-\infty\) as \( t \to \infty \).

Step 7: Graphical Confirmation

Use a graphing calculator or software to plot \( f(t) = 12t - t^3 \) over the domain \(-3 \leq t < \infty\). The graph should show a peak at \( t = 2 \) corresponding to a local (and absolute) maximum, and a valley at \( t = -2 \) corresponding to a local minimum. As \( t \to \infty \), the function decreases without bounds.

Key Concepts

Critical Points

In calculus, critical points are the points on the graph of a function where its derivative is zero or undefined. These points are significant because they indicate where the function's slope is zero, meaning the function could potentially have a local maximum or minimum.
To find the critical points of a function, we first find the derivative. For the function given, \( f(t) = 12t - t^3 \), the derivative is \( f'(t) = 12 - 3t^2 \). We then set this derivative equal to zero:
\[ 0 = 12 - 3t^2 \]
Solving this equation, we find the critical points to be at \( t = \pm 2 \).

• At \( t = 2 \), \( f'(t) = 0 \).
• At \( t = -2 \), \( f'(t) = 0 \).

These critical points help us identify possible local extrema, which we explore further in other concepts.

Local Extreme Values

Local extreme values are the highest or lowest values of a function within a certain interval and occur at critical points. These are known as local maxima and minima. To determine these, we use the first and second derivative tests.
From our given function, \( f(t) = 12t - t^3 \), once we have the critical points at \( t = \pm 2 \), we need to analyze these points:

• For \( t = 2 \), we suspect a local maximum as the slope changes from positive to negative.
• For \( t = -2 \), the slope likely changes from negative to positive, indicating a local minimum.

Remember, **local maxima** means the function values are higher than those around them, and **local minima** means the function values are lower than nearby points. The actual values can be calculated by substituting back into the original function:

• Local maximum: \( f(2) = 16 \).
• Local minimum: \( f(-2) = -16 \).

Second Derivative Test

The second derivative test is a method used to determine whether a critical point is a local maximum, minimum, or neither based on the concavity of the function. We compute the second derivative and evaluate it at the critical points.
For our function \( f(t) = 12t - t^3 \), the second derivative is found as \( f''(t) = -6t \). We then determine the concavity at the critical points:

• At \( t = 2 \): \( f''(2) = -12 \), which is negative, indicating the function is concave down and hence has a **local maximum** at this point.
• At \( t = -2 \): \( f''(-2) = 12 \), which is positive, indicating the function is concave up and has a **local minimum** at this point.

This test is particularly useful as it confirms the nature of the critical points, providing certainty about the location of local maxima and minima.

Absolute Extreme Values

Absolute extreme values (or global extrema) are the highest and lowest values a function can attain over its entire domain. To identify these, we need to check the critical points and endpoints, as well as consider the behavior as \( t \to \infty \).
For the function \( f(t) = 12t - t^3 \) within the domain \( -3 \leq t < \infty \), we examined:

• The local maximum at \( t = 2 \) gives \( f(2) = 16 \). This is the highest value in the domain, thus the **absolute maximum**.
• The function decreases without bounds as \( t \to \infty \), indicating there is no **absolute minimum** since \( f(t) \to -\infty \).

We also evaluate endpoints:

• At \( t = -3 \), \( f(-3) = -9 \).

Comparing these values confirms the absolute extremes within the given domain.