Textbook Question
101. In Exercises 101 and 102, the graph of f' is given. Determine x-values corresponding to local minima, local maxima, and inflection points for the graph of f.
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 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
Chapter 4, Problem 4.4.4
Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
4. y=9/14x^(1/3)(x^2-7)
Verified step by step guidance1
To find the inflection points, local maxima, and minima, start by taking the first derivative of the function y = \(\frac{9}{14}\)x^{\(\frac{1}{3}\)}(x^2 - 7). Use the product rule: if y = u*v, then y' = u'v + uv'.
Calculate the first derivative: Let u = \(\frac{9}{14}\)x^{\(\frac{1}{3}\)} and v = (x^2 - 7). Find u' and v'. u' = \(\frac{3}{14}\)x^{-\(\frac{2}{3}\)} and v' = 2x.
Substitute u, u', v, and v' into the product rule formula to find y': y' = \(\frac{3}{14}\)x^{-\(\frac{2}{3}\)}(x^2 - 7) + \(\frac{9}{14}\)x^{\(\frac{1}{3}\)}(2x).
To find critical points, set y' = 0 and solve for x. These points will help identify local maxima and minima. Analyze the sign changes of y' around these points to determine the nature of each critical point.
To find inflection points, take the second derivative of y and set it equal to zero. Solve for x to find where the concavity changes. Analyze the intervals where the second derivative is positive (concave up) and negative (concave down).
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Video duration:
10mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inflection Points
Inflection points occur where the concavity of a function changes from concave up to concave down or vice versa. To find these points, we need to determine where the second derivative of the function equals zero or is undefined, and verify a change in concavity. These points are crucial for understanding the overall shape and behavior of the graph.
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04:50
Critical Points
Local Maxima and Minima
Local maxima and minima are points where a function reaches a highest or lowest value, respectively, within a certain interval. These can be found by setting the first derivative to zero and using the second derivative test to confirm the nature of the critical points. Identifying these points helps in understanding the peaks and troughs of the graph.
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07:09The First Derivative Test: Finding Local Extrema
Concavity and Differentiability
Concavity describes the direction a graph curves, either upwards (concave up) or downwards (concave down). A function is differentiable on intervals where it is smooth and continuous, without sharp corners or cusps. Analyzing concavity and differentiability provides insights into the function's behavior and helps identify intervals of interest for further analysis.
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05:59Determining Concavity Given a Function
Related Practice
Textbook Question
99. In Exercises 99 and 100, the graph of f' is given. Determine x-values corresponding to inflection points for the graph of f.
Textbook Question
Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
7. y=sin|x|, -2π≤x≤2π
Textbook Question
Find values of a and b such that the function
ƒ(𝓍) = (a𝓍 + b) / 𝓍² ―1)
has a local extreme value of 1 at 𝓍 = 3. Is this extreme value a local maximum or a local minimum? Give reasons for your answer.
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
dy/dx = 3x⁻²ᐟ³, y(−1) = −5
Textbook Question
Theory and Examples
[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.
h(x) = |x + 2| − |x − 3|, −∞ < x < ∞