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.
(2/3)x⁻¹ᐟ³
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 4, Problem 4.3.42a
Identifying Extrema
In Exercises 41–52:
a. Identify the function’s local extreme values in the given domain, and say where they occur.
f(x) = (x + 1)², −∞ < x ≤ 0
Verified step by step guidance1
First, understand that local extrema refer to the local minimum or maximum values of a function within a given domain. For the function f(x) = (x + 1)^2, we need to find these values for -∞ < x ≤ 0.
To find the local extrema, we first need to find the derivative of the function f(x). The derivative, f'(x), will help us identify critical points where the slope of the tangent is zero or undefined.
Calculate the derivative of f(x) = (x + 1)^2. Using the power rule, the derivative is f'(x) = 2(x + 1).
Set the derivative equal to zero to find the critical points: 2(x + 1) = 0. Solve for x to find the critical point.
Evaluate the function f(x) at the critical point and at the endpoints of the domain (if applicable) to determine the local extrema. Since the domain is -∞ < x ≤ 0, consider the endpoint x = 0 as well.
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Video duration:
5mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Local Extrema
Local extrema refer to the points in a function where it reaches a local maximum or minimum within a specific interval. These are points where the function changes direction, and can be identified by analyzing the derivative of the function. In the context of the given domain, local extrema are found by evaluating the function's behavior at critical points and endpoints.
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Finding Extrema Graphically
Derivative and Critical Points
The derivative of a function provides information about its rate of change and helps identify critical points where the derivative is zero or undefined. These points are potential locations for local extrema. For the function f(x) = (x + 1)², the derivative f'(x) = 2(x + 1) can be used to find critical points by setting it equal to zero and solving for x.
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04:50Critical Points
Domain Considerations
The domain of a function specifies the set of input values for which the function is defined. In this problem, the domain is −∞ < x ≤ 0, which affects where extrema can occur. It's crucial to consider endpoints and any restrictions in the domain when identifying extrema, as they can influence the function's behavior and the location of extrema.
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Guided course
5:10Finding the Domain and Range of a Graph
Related Practice
Textbook Question
Roots (Zeros)
a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.
iii. y = x³ − 3x² + 4 = (x + 1)(x − 2)²
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Textbook Question
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
Textbook Question
Roots (Zeros)
a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.
i. y = x² − 4
Textbook Question
20.The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. a.What dimensions will give a box with a square end the largest possible volume?
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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 − 1)(x + 2)(x − 3)