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The graph of the second derivative f" of a function fis shown. State the x-coordinates of the inflection points of f. Part 1 of 5 JA We know that possible inflection points of f(x) occur where F"(x) = 0. These will be the points where the graph of f "(x) crosses the x-axis. The x-values of these points are: 3 (smallest value) 5 9 (largest value) Part 2 of 5 We know that f"(x) = 0 at x = 3, x = 5, and x= 9. However, this alone is not enough to indicate an inflection point. There must be a concavity change in f(x), which is indicated by a sign change in F"(x). For example, at x=3 we see that f"(x) is below✔ below the x-axis before x = 3, which means that f(x) is negative ✔ negative Part 3 of 5 After x = 3, f"(x) is above ✔✔ an inflection point. Part 4 of 5 x= 10 Using this logic, we see that "(x) does not ✓ Part 5 of 5 Finally, at x = 9, we see f"(x) changes sign from positive ✔✔✔an inflection point. is So, the x-coordinates of the inflections points are: above the x-axis, indicating that f"(x) is positive (smaller value) (larger value) Submit Skip.(you cannot come back) positive. This means that f(x) changes concavity from concave down✔ does not change sign at x = 5, and so x = 5 is not✔ is not an inflection point. before x = 9 to negative ✔✔✔afterward. Therefore, f(x) changes concavity from concave up down before x = 3 to concave up✔ before x = 9 to concave down up afterward, and so x= 3 is ✔✔✔afterward, and so x=9

The graph of the second derivative f" of a function fis shown. State the x-coordinates of the inflection points of f. Part 1 of 5 JA We know that possible inflection points of f(x) occur where F"(x) = 0. These will be the points where the graph of f "(x) crosses the x-axis. The x-values of these points are: 3 (smallest value) 5 9 (largest value) Part 2 of 5 We know that f"(x) = 0 at x = 3, x = 5, and x= 9. However, this alone is not enough to indicate an inflection point. There must be a concavity change in f(x), which is indicated by a sign change in F"(x). For example, at x=3 we see that f"(x) is below✔ below the x-axis before x = 3, which means that f(x) is negative ✔ negative Part 3 of 5 After x = 3, f"(x) is above ✔✔ an inflection point. Part 4 of 5 x= 10 Using this logic, we see that "(x) does not ✓ Part 5 of 5 Finally, at x = 9, we see f"(x) changes sign from positive ✔✔✔an inflection point. is So, the x-coordinates of the inflections points are: above the x-axis, indicating that f"(x) is positive (smaller value) (larger value) Submit Skip.(you cannot come back) positive. This means that f(x) changes concavity from concave down✔ does not change sign at x = 5, and so x = 5 is not✔ is not an inflection point. before x = 9 to negative ✔✔✔afterward. Therefore, f(x) changes concavity from concave up down before x = 3 to concave up✔ before x = 9 to concave down up afterward, and so x= 3 is ✔✔✔afterward, and so x=9

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Tutorial Exercise The graph of the second derivative f" of a function f is shown. State the x-coordinates of the inflection points of f. Part 1 of 5 We know that possible inflection points of f(x) occur where f"(x) = 0. These will be the points where the graph of f "(x) crosses the x-axis. The x-values of these points are: (smallest value) Part 2 of 5 Part 3 of 5 X = Part 4 of 5 X = Sony 5 6 7 8 9 10 X We know that f "(x) = 0 at x = 3, x = 5, and x = 9. However, this alone is not enough to indicate an inflection point. There must be a concavity change in f(x), which is indicated by a sign change in f "(x). For example, at x = 3 we see that f "(x) is below✔ below the x-axis before x = 3, which means that f "(x) is negative ✔ negative. Submit ✓ After x = 3, f"(x) is above ✔ an inflection point. 9 (largest value) X = Using this logic, we see that f"(x) does not ✔ above the x-axis, indicating that f"(x) is positive Part 5 of 5 Finally, at x = 9, we see f "(x) changes sign from [positive is v ✓ an inflection point. So, the x-coordinates of the inflections points are: (smaller value) (larger value) Skip (you cannot come back) does not change sign at x = 5, and so x = 5 positive. This means that f(x) changes concavity from concave down ✓ before x = 9 to negative ✓ is not is not an inflection point. afterward. Therefore, f(x) changes concavity from concave up down before x = 3 to concave up before x = 9 to concave down up afterward, and so x = 3 is afterward, and so x = 9
Transcribed Image Text:Tutorial Exercise The graph of the second derivative f" of a function f is shown. State the x-coordinates of the inflection points of f. Part 1 of 5 We know that possible inflection points of f(x) occur where f"(x) = 0. These will be the points where the graph of f "(x) crosses the x-axis. The x-values of these points are: (smallest value) Part 2 of 5 Part 3 of 5 X = Part 4 of 5 X = Sony 5 6 7 8 9 10 X We know that f "(x) = 0 at x = 3, x = 5, and x = 9. However, this alone is not enough to indicate an inflection point. There must be a concavity change in f(x), which is indicated by a sign change in f "(x). For example, at x = 3 we see that f "(x) is below✔ below the x-axis before x = 3, which means that f "(x) is negative ✔ negative. Submit ✓ After x = 3, f"(x) is above ✔ an inflection point. 9 (largest value) X = Using this logic, we see that f"(x) does not ✔ above the x-axis, indicating that f"(x) is positive Part 5 of 5 Finally, at x = 9, we see f "(x) changes sign from [positive is v ✓ an inflection point. So, the x-coordinates of the inflections points are: (smaller value) (larger value) Skip (you cannot come back) does not change sign at x = 5, and so x = 5 positive. This means that f(x) changes concavity from concave down ✓ before x = 9 to negative ✓ is not is not an inflection point. afterward. Therefore, f(x) changes concavity from concave up down before x = 3 to concave up before x = 9 to concave down up afterward, and so x = 3 is afterward, and so x = 9
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