Textbook Question
35. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.
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 Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 4, Problem 4.1.43
Finding Critical Points
In Exercises 41–50, determine all critical points and all domain endpoints for each function.
f(x) = x(4 − x)³
Verified step by step guidance1
First, understand that critical points occur where the derivative of the function is zero or undefined. Begin by finding the derivative of the function f(x) = x(4 - x)^3.
Apply the product rule to differentiate f(x) = x(4 - x)^3. The product rule states that if you have a function h(x) = u(x)v(x), then h'(x) = u'(x)v(x) + u(x)v'(x). Here, u(x) = x and v(x) = (4 - x)^3.
Differentiate u(x) = x to get u'(x) = 1. Next, differentiate v(x) = (4 - x)^3 using the chain rule. The chain rule states that if you have a composite function g(h(x)), then the derivative is g'(h(x)) * h'(x). Let g(y) = y^3 and h(x) = 4 - x, then g'(y) = 3y^2 and h'(x) = -1.
Combine the results from the product and chain rules to find f'(x). Substitute u'(x), v(x), u(x), and v'(x) into the product rule formula: f'(x) = 1 * (4 - x)^3 + x * 3(4 - x)^2 * (-1). Simplify this expression.
Set the derivative f'(x) equal to zero and solve for x to find the critical points. Also, consider the endpoints of the domain of f(x), which are determined by the context of the problem or any given interval.
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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 important because they can indicate local maxima, minima, or points of inflection. To find critical points, take the derivative of the function and solve for the values of x where the derivative equals zero or does not exist.
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04:50
Critical Points
Derivative
The derivative of a function represents the rate at which the function's value changes with respect to changes in its input. It is a fundamental tool in calculus for analyzing the behavior of functions. For the function f(x) = x(4 − x)³, use the product rule and chain rule to find its derivative, which is essential for identifying critical points.
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Domain Endpoints
Domain endpoints are the boundary values of the domain of a function, where the function is defined. These points are crucial when analyzing a function's behavior over its entire domain, especially when determining absolute extrema. For polynomial functions like f(x) = x(4 − x)³, the domain is typically all real numbers, but endpoints are considered in restricted domains.
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5:10Finding the Domain and Range of a Graph
Related Practice
Textbook Question
Estimate the open intervals on which the function y = ƒ(𝓍) is
a. increasing.
b. decreasing.
c. Use the given graph of ƒ' to indicate where any local extreme
values of the function occur, and whether each extreme
is a relative maximum or minimum.
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Textbook Question
Roots (Zeros)
Show that the functions in Exercises 19–26 have exactly one zero in the given interval.
r(θ) = 2θ − cos²θ + √2, (−∞, ∞)
Textbook Question
93. The accompanying figure shows a portion of the graph of a twice-differentiable function y=f(x). At each of the five labeled points, classify y' and \(\y\)'' as positive, negative, or zero.
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Textbook Question
32. Answer Exercise 31 if one piece is bent into a square and the other into a circle.
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1