Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.5

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 guidance

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.

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.

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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Textbook Question

Finding Extreme Values

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.

y = 𝓍³ ― 2𝓍 + 4

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

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.

∫(√x + ³√x) dx

Textbook Question

37. What value of a makes f(x) = x^2 +(a/x) have

a. a local minimum at x = 2?

b. a point of inflection at x = 1?

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

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Key Concepts

Extreme Values

Critical Points

Natural Domain

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