Textbook Question
Identifying Extrema
In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.
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 4.1.47
Finding Critical Points
In Exercises 41–50, determine all critical points and all domain endpoints for each function.
y = x² − 32√x
Verified step by step guidance1
First, identify the domain of the function y = x² − 32√x. Since the square root function √x is only defined for x ≥ 0, the domain of the function is x ≥ 0.
Next, find the derivative of the function y with respect to x. The derivative of y = x² is 2x, and the derivative of −32√x is −16/√x. Therefore, the derivative y' is y' = 2x − 16/√x.
Set the derivative y' equal to zero to find the critical points: 2x − 16/√x = 0. Solve this equation for x to find the critical points.
To solve 2x − 16/√x = 0, multiply through by √x to eliminate the fraction: 2x√x = 16. Then, solve for x by isolating x on one side of the equation.
Finally, evaluate the function at the endpoints of the domain and at any critical points found to determine the behavior of the function at these points. The endpoint in this case is x = 0, and any critical points found from the previous step.
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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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04:50
Critical Points
Derivative
The derivative of a function represents the rate at which the function's value changes with respect to changes in its input. It is a fundamental tool in calculus for analyzing the behavior of functions. For the function y = x² − 32√x, the derivative is found using the power rule and the chain rule, resulting in y' = 2x - 16x^(-1/2).
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05:44Derivatives
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 y = x² − 32√x, the domain is restricted to x ≥ 0 because the square root function is only defined for non-negative numbers. Identifying the domain is crucial for determining endpoints and ensuring the function is evaluated correctly.
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5:10Finding the Domain and Range of a Graph
Related Practice
Textbook Question
Absolute Extrema on Finite Closed Intervals
In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.
f(x) = (2/3)x − 5, −2 ≤ x ≤ 3
Textbook Question
117. Suppose that the second derivative of the function y = f(x) isy" =(x+1)(x-2).
For what x-values does the graph of f have an inflection point?
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
dv/dt = (1/2)sec t tan t, v(0) = 1
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) = x³ / (3x² + 1)
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Textbook Question
Absolute Extrema on Finite Closed Intervals
In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.
f(t) = 2 − |t|, −1 ≤ t ≤ 3