Textbook Question
Roots (Zeros)
Show that the functions in Exercises 19–26 have exactly one zero in the given interval.
f(x) = x³ + 4x² + 7, (−∞, 0)
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 4.5
In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.
______
y = √𝓍² ― 1
Verified step by step guidance1
Identify the natural domain of the function y = √(x² - 1). The expression under the square root, x² - 1, must be greater than or equal to zero for y to be real. Solve the inequality x² - 1 ≥ 0 to find the domain.
Solve the inequality x² - 1 ≥ 0. This can be rewritten as x² ≥ 1, which implies x ≤ -1 or x ≥ 1. Therefore, the natural domain of the function is x ∈ (-∞, -1] ∪ [1, ∞).
Find the critical points by taking the derivative of the function. The derivative of y = √(x² - 1) is y' = (1/2)(x² - 1)^(-1/2) * 2x = x / √(x² - 1). Set y' = 0 to find critical points.
Solve the equation x / √(x² - 1) = 0. This implies x = 0. However, x = 0 is not in the domain of the function, so there are no critical points within the domain.
Evaluate the function at the endpoints of the domain, x = -1 and x = 1, to find the extreme values. Calculate y(-1) and y(1) to determine the absolute minimum and maximum values of the function over its natural domain.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Extreme Values
Extreme values refer to the maximum and minimum values of a function within a given domain. Absolute extreme values are the highest and lowest points over the entire domain, while local extreme values are the highest or lowest points within a specific interval. Identifying these values often involves analyzing the function's critical points and endpoints.
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Average Value of a Function
Critical Points
Critical points are values in the domain of a function where the derivative is either zero or undefined. These points are essential for finding local extrema, as they indicate where the function's slope changes, potentially leading to local maxima or minima. To find critical points, one typically takes the derivative of the function and solves for when it equals zero.
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04:50Critical Points
Natural Domain
The natural domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function y = √(x² - 1), the natural domain is determined by ensuring the expression under the square root is non-negative, leading to the condition x² - 1 ≥ 0. This results in the domain being x ≤ -1 or x ≥ 1.
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Related Practice
Textbook Question
In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
53. y = x * √(8 - x²)
Textbook Question
Finding Critical Points
In Exercises 41–50, determine all critical points and all domain endpoints for each function.
y = x² + 2/x
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d²r/dt² = 2/t³; dr/dt|ₜ ₌ ₁ =1, r(1) = 1
Textbook Question
113. If b, c, and d are constants, for what value of b will the curve y = x^3 + bx^2 + cx + d have a
point of inflection at x = 1? Give reasons for your answer.
Textbook Question
Finding Indefinite Integrals
In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
∫(−2cost) dt