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Magazine Chapter 4: Problem 33
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. $$ f(t)=12 t-t^{3}, \quad-3 \leq t<\infty $$
Short Answer
Expert verified
The function has a local minimum at \( t = -2 \) and a local (and absolute) maximum at \( t = 2 \). There is no absolute minimum.
Step by step solution
01
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: \( f'(t) = 12 - 3t^2 \).
02
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 \( t = -2 \).
03
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.
04
Evaluate Function at Critical Points
Evaluate the function at the critical points:- \( f(-2) = 12(-2) - (-2)^3 = -24 + 8 = -16 \) - \( f(2) = 12(2) - (2)^3 = 24 - 8 = 16 \)
05
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 \( 12t \), \( f(t) \to -\infty \).
06
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 \( t \to \infty \).
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( t = \pm 2 \).
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 \( t = -2 \), \( f'(t) = 0 \).
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:
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.
- 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:
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.
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:
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 \( f(t) \to -\infty \).
- At \( t = -3 \), \( f(-3) = -9 \).
