Domanda del libro di testo
Finding Antiderivatives
In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
x⁻⁴ + 2x + 3
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- 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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 4, Problema 4.3.9c
Analyzing Functions from Derivatives
Answer the following questions about the functions whose derivatives are given in Exercises 1–14:
c. At what points, if any, does f assume local maximum or minimum values?
f′(x) = 1− 4/x², x ≠ 0
Guida verificata passo dopo passo1
First, identify the critical points of the function by setting the derivative f'(x) = 1 - 4/x² equal to zero and solving for x. This will help us find where the function might have local maxima or minima.
Solve the equation 1 - 4/x² = 0 for x. This involves rearranging the equation to find x² = 4, and then taking the square root of both sides to find the possible values of x.
The solutions to x² = 4 are x = 2 and x = -2. These are the critical points where the function might have local maxima or minima.
To determine whether these points are local maxima or minima, use the second derivative test. Compute the second derivative f''(x) and evaluate it at x = 2 and x = -2.
If f''(x) > 0 at a critical point, the function has a local minimum there. If f''(x) < 0, the function has a local maximum. If f''(x) = 0, the test is inconclusive, and further analysis is needed.
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Critical Points
Critical points of a function occur where its derivative is zero or undefined. These points are potential locations for local maxima or minima. In the given problem, setting f′(x) = 1 - 4/x² equal to zero helps identify critical points, which are essential for determining where the function might assume local extremum values.
Video consigliato:
04:50
Critical Points
Second Derivative Test
The second derivative test helps determine the nature of critical points. If the second derivative at a critical point is positive, the function has a local minimum there; if negative, a local maximum. Calculating the second derivative of f′(x) = 1 - 4/x² and evaluating it at critical points provides insight into whether these points are maxima or minima.
Video consigliato:
06:02The Second Derivative Test: Finding Local Extrema
Behavior at Infinity
Understanding the behavior of a function as x approaches infinity or negative infinity can reveal asymptotic tendencies and help identify global behavior. For f′(x) = 1 - 4/x², analyzing the limit as x approaches infinity or zero can provide additional context about the function's overall behavior, complementing the analysis of local extrema.
Video consigliato:
03:07Cases Where Limits Do Not Exist
Pratica correlata
Domanda del libro di testo
Theory and Examples
In Exercises 51 and 52, give reasons for your answers.
Let f(x) = |x³ − 9x|.
b. Does f'(-3) exist?
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Domanda del libro di testo
Analyzing Functions from Derivatives
Answer the following questions about the functions whose derivatives are given in Exercises 1–14:
c. At what points, if any, does f assume local maximum or minimum values?
f′(x) = (x − 1)²(x + 2)
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Domanda del libro di testo
Finding Antiderivatives
In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
−π csc (πx/2) cot (πx/2)
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Domanda del libro di testo
105. Motion Along a Line The graphs in Exercises 105 and 106 show the position s=f(t) of an object moving up and down on a coordinate line. At approximately what times is the (c) Acceleration equal to zero?
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Domanda del libro di testo
Applications
Suppose that f(x) = d/dx (1 − √x) and g(x) = d/dx (x + 2).
Find:
∫[−f(x)] dx
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