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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 4.3.48a
Chapter 4, Problem 4.3.48a

Identifying Extrema


In Exercises 41–52:


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


k(x) = x³ + 3x² + 3x + 1, −∞ < x ≤ 0

Verified step by step guidance
1
To find the local extrema of the function k(x) = x³ + 3x² + 3x + 1 on the domain (-∞, 0], we first need to find the critical points. This involves taking the derivative of k(x) and setting it equal to zero.
Calculate the derivative of k(x): k'(x) = 3x² + 6x + 3. This derivative will help us find the critical points where the slope of the tangent is zero or undefined.
Set the derivative equal to zero to find the critical points: 3x² + 6x + 3 = 0. Solve this quadratic equation for x to find the critical points within the given domain.
Use the quadratic formula x = (-b ± √(b² - 4ac)) / 2a, where a = 3, b = 6, and c = 3, to solve for x. This will give you the potential critical points.
Evaluate the function k(x) at the critical points and at the endpoint x = 0 to determine the local extrema. Compare these values to identify the local maximum and minimum values within the domain (-∞, 0].

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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 (maximum or minimum values). For the function k(x) = x³ + 3x² + 3x + 1, finding the derivative and solving k'(x) = 0 will help identify critical points within the given domain.
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Critical Points

First Derivative Test

The First Derivative Test helps determine whether a critical point is a local maximum, minimum, or neither. By analyzing the sign changes of the derivative around the critical points, we 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's a local minimum.
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The First Derivative Test: Finding Local Extrema

Domain Consideration

Understanding the domain is crucial when identifying extrema, as it limits where extrema can occur. For k(x) = x³ + 3x² + 3x + 1, the domain is −∞ < x ≤ 0, meaning we only consider critical points and endpoints within this range. The endpoint x = 0 must also be evaluated to determine if it is an extremum.
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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 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

Finding Antiderivatives

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

−πsin πx

Textbook Question

Analyzing Functions from Derivatives


Answer the following questions about the functions whose derivatives are given in Exercises 1–14:


a. What are the critical points of f?


f′(x) = x(x − 1)

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.

sec²x

2
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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).