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 4.2.24
Roots (Zeros)
Show that the functions in Exercises 19–26 have exactly one zero in the given interval.
r(θ) = 2θ − cos²θ + √2, (−∞, ∞)
Verified step by step guidance1
First, understand that finding the zero of a function means finding the value of θ for which r(θ) = 0. We need to show that there is exactly one such value in the interval (−∞, ∞).
Consider the function r(θ) = 2θ − cos²θ + √2. To find the zeros, set the equation to zero: 2θ − cos²θ + √2 = 0.
To determine if there is exactly one zero, analyze the behavior of the function. Start by finding the derivative r'(θ) to understand the function's monotonicity. The derivative is r'(θ) = 2 + 2cos(θ)sin(θ).
Evaluate the derivative r'(θ). Notice that r'(θ) = 2 + sin(2θ), which is always positive because sin(2θ) ranges from -1 to 1, making 2 + sin(2θ) always greater than 0. This indicates that r(θ) is a strictly increasing function.
Since r(θ) is strictly increasing, it can cross the x-axis at most once. Check the limits as θ approaches -∞ and ∞ to ensure the function crosses the x-axis. As θ approaches -∞, r(θ) approaches -∞, and as θ approaches ∞, r(θ) approaches ∞. Therefore, by the Intermediate Value Theorem, there is exactly one zero in the interval (−∞, ∞).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Intermediate Value Theorem
The Intermediate Value Theorem states that if a continuous function, f(x), takes on different signs at two points, a and b, then it must cross zero at some point between a and b. This theorem is crucial for proving the existence of a root within an interval, as it guarantees that a zero exists if the function changes sign.
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Fundamental Theorem of Calculus Part 1
Continuity of Functions
A function is continuous if there are no breaks, jumps, or holes in its graph. For the function r(θ) = 2θ − cos²θ + √2, continuity is essential to apply the Intermediate Value Theorem. Since polynomials, trigonometric functions, and their combinations are continuous over their domains, r(θ) is continuous over (−∞, ∞).
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05:34Intro to Continuity
Derivative and Monotonicity
The derivative of a function provides information about its monotonicity, indicating whether the function is increasing or decreasing. By analyzing the derivative of r(θ), we can determine if the function is strictly increasing or decreasing, which helps establish that there is exactly one zero in the interval by showing that the function does not change direction.
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05:44Derivatives
Related Practice
Textbook Question
Finding Critical Points
In Exercises 41–50, determine all critical points and all domain endpoints for each function.
f(x) = x(4 − x)³
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
Finding Extrema from Graphs
In Exercises 15–20, sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with Theorem 1.
g(x) = {−x, 0 ≤ x < 1
x − 1, 1 ≤ x ≤ 2
Textbook Question
Checking the Mean Value Theorem
Which of the functions in Exercises 7–12 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers.
f(x) = {2x − 3, 0 ≤ x ≤ 2
6x − x² − 7, 2 < x ≤ 3
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1