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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.3.45a
Chapter 4, Problem 4.3.45a

Identifying Extrema


In Exercises 41–52:


a. Identify the function’s local extreme values in the given domain, and say where they occur.


f(t) = 12t − t³, −3 ≤ t < ∞

Verified step by step guidance
1
To find the local extrema of the function \( f(t) = 12t - t^3 \), we first need to find its critical points. This involves taking the derivative of the function and setting it equal to zero.
Calculate the derivative: \( f'(t) = \frac{d}{dt}(12t - t^3) = 12 - 3t^2 \).
Set the derivative equal to zero to find critical points: \( 12 - 3t^2 = 0 \). Solve for \( t \) to find the critical points.
Solve the equation \( 12 - 3t^2 = 0 \) to get \( t^2 = 4 \), which gives \( t = 2 \) and \( t = -2 \). These are the critical points within the domain \( -3 \leq t < \infty \).
Evaluate the function \( f(t) \) at the critical points and endpoints of the domain to determine the local extrema. Compare the values of \( f(t) \) at \( t = -3, -2, \) and \( 2 \) to identify the local maximum and minimum values.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are potential locations for local extrema, as they indicate where the function's slope changes direction. To find critical points, compute the derivative of the function and solve for values of the variable where the derivative equals zero or does not exist.
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Critical Points

First Derivative Test

The First Derivative Test helps determine whether a critical point is a local maximum or minimum. By analyzing the sign of the derivative before and after the critical point, one can infer the behavior of the function. If the derivative changes from positive to negative, the point is a local maximum; if it changes from negative to positive, it is a local minimum.
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The First Derivative Test: Finding Local Extrema

Domain Considerations

Understanding the domain of a function is crucial when identifying extrema, as it defines the range of input values to consider. In this problem, the domain is given as −3 ≤ t < ∞, meaning the function is evaluated from t = -3 onwards. This affects where extrema can occur, especially at the boundary points, which must be checked separately.
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Finding the Domain and Range of a Graph
Related Practice
Textbook Question

53. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day.

a. Start counting time with t=0 at noon and express the distance s between the ships as a function of t.

Textbook Question

Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

−3x⁻⁴

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Textbook Question

Identifying Extrema


In Exercises 53–60:


a. Find the local extrema of each function on the given interval, and say where they occur.


f(x) = sin 2x, 0 ≤ x ≤ π

Textbook Question

Finding displacement from an antiderivative of velocity


a. Suppose that the velocity of a body moving along the s-axis is


ds/dt = v = 9.8t − 3.


i. Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0.

Textbook Question

Identifying Extrema


In Exercises 53–60:


a. Find the local extrema of each function on the given interval, and say where they occur.


f(x) = csc²x − 2cot x, 0 < x < π

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Textbook Question

25. Paper folding A rectangular sheet of 8.5-in.-by-11-in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper.

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a. Show that L^2=2x^3/(2x-8.5).