The second derivative f" is defined on the closed interval [0, 8n]. Now test f" in the intervals (0, 4m) and (4m, 8T). Intervals: 0 < x < 4m 4n < x < 8T Test Value: X = 2n X = 6n Sign of f"(x): f"(27) < 0 f"(6n) > 0 Conclusion: ---Select--- ---Select---
The second derivative f" is defined on the closed interval [0, 8n]. Now test f" in the intervals (0, 4m) and (4m, 8T). Intervals: 0 < x < 4m 4n < x < 8T Test Value: X = 2n X = 6n Sign of f"(x): f"(27) < 0 f"(6n) > 0 Conclusion: ---Select--- ---Select---
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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![X= 47.
.
47
f(4m) = sin
Therefore, the possible point of inflection is as follows.
(x, y) = ( 4x.0
47, 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, 8T]. Now test f" in the intervals (0, 4T) and (4m, 8m).
Intervals:
0<X<4元
4T <x<8元
Test Value:
X= 2元
Sign of f"(x):
f"(27) < 0
f"(67) > 0
Conclusion:
--Select---
---Select---
Submit Skip (you cannot come back)
omit Answer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13febeea-9b27-425f-8228-94390452e7cf%2Fab183994-5697-4eaf-9bf9-43ff941fa6c7%2F9jxybvw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:X= 47.
.
47
f(4m) = sin
Therefore, the possible point of inflection is as follows.
(x, y) = ( 4x.0
47, 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, 8T]. Now test f" in the intervals (0, 4T) and (4m, 8m).
Intervals:
0<X<4元
4T <x<8元
Test Value:
X= 2元
Sign of f"(x):
f"(27) < 0
f"(67) > 0
Conclusion:
--Select---
---Select---
Submit Skip (you cannot come back)
omit Answer
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