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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.8
Chapter 4, Problem 4.8

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.
__________
y = √ 3 + 2𝓍 ―𝓍²

Verified step by step guidance
1
Identify the function: \( y = \sqrt{3 + 2x - x^2} \). The domain of this function is determined by the expression inside the square root being non-negative, i.e., \( 3 + 2x - x^2 \geq 0 \).
Find the critical points by first determining the derivative of the function. Use the chain rule to differentiate \( y = \sqrt{3 + 2x - x^2} \).
Set the derivative equal to zero to find critical points. Solve the equation \( \frac{dy}{dx} = 0 \) to find the values of \( x \) where the slope of the tangent is zero.
Analyze the critical points and endpoints of the domain to determine the extreme values. Evaluate the function \( y \) at these points to find the local and absolute extrema.
Verify the nature of each critical point using the second derivative test or by analyzing the sign changes of the first derivative around these points to confirm whether they are maxima or minima.

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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 of a function refer to the maximum and minimum values that the function can attain. These can be absolute (global) extremes, which are the highest or lowest values over the entire domain, or local (relative) extremes, which are the highest or lowest values within a specific interval. Identifying these values involves analyzing the function's behavior and its critical points.
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Average Value of a Function

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are potential candidates for local extreme values. To find them, take the derivative of the function, set it equal to zero, and solve for the variable. Additionally, check where the derivative does not exist, as these points may also indicate extremes.
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Critical Points

Natural Domain

The natural domain of a function is the set of all possible input values (𝓍) for which the function is defined. For the function y = √(3 + 2𝓍 - 𝓍²), the expression under the square root must be non-negative, as square roots of negative numbers are not real. Thus, determining the natural domain involves solving inequalities to ensure the function remains real-valued.
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Related Practice
Textbook Question

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


y = (𝓍 + 1) / (𝓍² + 2𝓍 + 2)

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Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1

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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

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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?

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Textbook Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

80. y' = 1 - cot²θ, for 0 < θ < π

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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.


∫(1 + cos 4t)/2 dt

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