Textbook Question
Sketch the graphs of the rational functions in Exercises 53–60.
𝓍⁴ ― 1
y = ------------------
𝓍²
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 4.1.48
Finding Critical Points
In Exercises 41–50, determine all critical points and all domain endpoints for each function.
g(x) = √(2x − x²)
Verified step by step guidance1
First, identify the domain of the function g(x) = √(2x − x²). The expression inside the square root, 2x − x², must be greater than or equal to zero. Solve the inequality 2x − x² ≥ 0 to find the domain.
Rewrite the inequality 2x − x² ≥ 0 as x(2 − x) ≥ 0. This is a quadratic inequality, and we can solve it by finding the roots of the equation x(2 − x) = 0, which are x = 0 and x = 2.
Determine the intervals where the inequality holds by testing values in the intervals (−∞, 0), (0, 2), and (2, ∞). The inequality x(2 − x) ≥ 0 is satisfied in the interval [0, 2]. Thus, the domain of g(x) is [0, 2].
Find the critical points by taking the derivative of g(x). First, express g(x) in a form suitable for differentiation: g(x) = (2x − x²)^(1/2). Use the chain rule to differentiate: g'(x) = (1/2)(2x − x²)^(-1/2) * (2 - 2x).
Set the derivative g'(x) equal to zero to find critical points: (1/2)(2x − x²)^(-1/2) * (2 - 2x) = 0. Solve 2 - 2x = 0 to find x = 1. Check if x = 1 is within the domain [0, 2]. Since it is, x = 1 is a critical point. Also, consider the endpoints of the domain, x = 0 and x = 2, as potential critical points.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Critical Points
Critical points of a function occur where its derivative is zero or undefined. These points are important because they can indicate local maxima, minima, or points of inflection. To find them, take the derivative of the function and solve for the values of x where the derivative equals zero or does not exist.
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Critical Points
Domain of a Function
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function g(x) = √(2x − x²), the expression inside the square root must be non-negative, as square roots of negative numbers are not real. Solving 2x − x² ≥ 0 will give the domain of the function.
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5:10Finding the Domain and Range of a Graph
Derivative of a Function
The derivative of a function represents the rate of change of the function with respect to its variable. For g(x) = √(2x − x²), use the chain rule to differentiate. The derivative helps identify critical points and analyze the behavior of the function, such as increasing or decreasing intervals.
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06:30Derivatives of Other Trig Functions
Related Practice
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
ds/dt = 1 + cos t, s(0) = 4
Textbook Question
Theory and Examples
In Exercises 53 and 54, show that the function has neither an absolute minimum nor an absolute maximum on its natural domain.
y = 3x + tan x
Textbook Question
41. Among all triangles in the first quadrant formed by the x-axis, the y-axis, and tangent lines to the graph of y=3x-x^2, what is the smallest possible area?
Textbook Question
22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.
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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.
∫(8y − 2 / y¹ᐟ⁴) dy