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 = √ 3 + 2𝓍 ―𝓍²
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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 4.5.37
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?
Verified step by step guidance1
To find the value of 'a' that makes f(x) = x^2 + (a/x) have a local minimum at x = 2, first find the first derivative f'(x) and set it equal to zero to find critical points. The derivative is f'(x) = 2x - a/x^2.
Substitute x = 2 into the derivative equation: 2(2) - a/(2^2) = 0. Solve this equation for 'a' to find the value that makes x = 2 a critical point.
To ensure that x = 2 is a local minimum, check the second derivative f''(x) = 2 + 2a/x^3. Substitute x = 2 and the value of 'a' found in the previous step into f''(x) to verify that f''(2) > 0.
For the point of inflection at x = 1, find the second derivative f''(x) = 2 + 2a/x^3 and set it equal to zero. Substitute x = 1 into the equation: 2 + 2a/(1^3) = 0. Solve this equation for 'a' to find the value that makes x = 1 a point of inflection.
Verify that the sign of f''(x) changes around x = 1 by checking the values of f''(x) just before and after x = 1 with the found value of 'a'. This confirms the point of inflection.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Local Minimum
A local minimum of a function occurs at a point where the function value is lower than at nearby points. To find a local minimum, we use the first derivative test: set the derivative equal to zero to find critical points, and then use the second derivative to determine if the point is a minimum (second derivative > 0) or maximum (second derivative < 0).
Recommended video:
06:02
The Second Derivative Test: Finding Local Extrema
Point of Inflection
A point of inflection is where the function changes concavity, from concave up to concave down or vice versa. It is identified by setting the second derivative equal to zero and confirming a change in sign around the point. This indicates a transition in the curvature of the graph, but not necessarily a local extremum.
Recommended video:
04:50Critical Points
Derivative Calculation
Calculating derivatives is essential for analyzing the behavior of functions. The first derivative, f'(x), provides information on the slope and critical points, while the second derivative, f''(x), helps determine concavity and points of inflection. For f(x) = x^2 + (a/x), apply differentiation rules to find these derivatives and solve for the conditions given in the problem.
Recommended video:
05:44Derivatives
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)
212
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Textbook Question
6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.
213
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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
23
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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
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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