Ch. 4 - Applications of Derivatives

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

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.8

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

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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.

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.

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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Textbook Question

The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. How far from the stronger light is the total illumination least?

Textbook Question

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = x¹ᐟ³(x + 8)

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.

69. y' = x(x - 3)²

Textbook Question

Initial Value Problems

Find the curve y = f(x) in the xy-plane that passes through the point (9,4) and whose slope at each point is 3√x.

Textbook Question

Checking Antiderivative Formulas

Right, or wrong? Give a brief reason why.

∫−15(x + 3)² / (x − 2)⁴ dx = ((x + 3)/(x − 2))³ + C

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Textbook Question

Initial Value Problems

Solve the initial value problems in Exercises 71–90.