Step 5 Now evaluate f(x) for x = 4m. (4) = sin 4 Therefore, the possible point of inflection is as follows. (X, y) = (47.0 4a, 0) We must check that concavity changes here to confirm that it is an inflection point. Step 6 The second derivative f" is defined on the closed interval [0, 8m). Now test f" in the intervals (0, 4m) and (4n, 8n). Intervals: 0 0 Concave downward Concave upward Conclusion: Concave downward Concave upward Step 7 Therefore, we conclude that the possible point of inflection (4, 0) is an actual point of inflection, and intervals on which the function is concave upward and concave downward are as follows. (Enter your answers using interval notation.) concave upward 0,4n concave downward 4x.8n
Step 5 Now evaluate f(x) for x = 4m. (4) = sin 4 Therefore, the possible point of inflection is as follows. (X, y) = (47.0 4a, 0) We must check that concavity changes here to confirm that it is an inflection point. Step 6 The second derivative f" is defined on the closed interval [0, 8m). Now test f" in the intervals (0, 4m) and (4n, 8n). Intervals: 0 0 Concave downward Concave upward Conclusion: Concave downward Concave upward Step 7 Therefore, we conclude that the possible point of inflection (4, 0) is an actual point of inflection, and intervals on which the function is concave upward and concave downward are as follows. (Enter your answers using interval notation.) concave upward 0,4n concave downward 4x.8n
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
Related questions
Question
What is the correct answer for step 7
![Step 5
Now evaluate f(x) for x = 4x.
4T
r(41) = sin
Therefore, the possible point of inflection is as follows.
(X, y) = (47.0
4x, 0
We must check that concavity changes here to confirm that it is an inflection point.
Step 6
The second derivative f" is defined on the closed interval [0, 8n). Now test f" in the intervals (0, 4n) and
(4n, Bn).
Intervals:
0 <x < 4m
4TT <X < 8T
Test Value:
Sign of f"(x):
F"(2n) < 0
(6m)> 0
Concave downward
Concave upward v
Conclusion:
Concave downward
Concave upward
Step 7
Therefore, we conclude that the possible point of infiection (4, 0) is an actual point of inflection, and
intervals on which the function is concave upward and concave downward are as follows. (Enter your answers
using interval notation.)
concave upward
0,4n
concave downward
4x.8n
Submit
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Transcribed Image Text:Step 5
Now evaluate f(x) for x = 4x.
4T
r(41) = sin
Therefore, the possible point of inflection is as follows.
(X, y) = (47.0
4x, 0
We must check that concavity changes here to confirm that it is an inflection point.
Step 6
The second derivative f" is defined on the closed interval [0, 8n). Now test f" in the intervals (0, 4n) and
(4n, Bn).
Intervals:
0 <x < 4m
4TT <X < 8T
Test Value:
Sign of f"(x):
F"(2n) < 0
(6m)> 0
Concave downward
Concave upward v
Conclusion:
Concave downward
Concave upward
Step 7
Therefore, we conclude that the possible point of infiection (4, 0) is an actual point of inflection, and
intervals on which the function is concave upward and concave downward are as follows. (Enter your answers
using interval notation.)
concave upward
0,4n
concave downward
4x.8n
Submit
Skip (yosi cannot come back)
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