A function f and its derivative f' are shown below. Assume a is a nonzero constant. Note: a is a lowercase letter. f(x) = ax² - 7x².In(x) + 7x² 2 2 4 (a) Determine the exact value of the critical point of f. Your answer may contain a. X = (b) Determine f"(x). f'(x) = ax - 7xln(x) f"(x) = (c) Evaluate f" at the critical point you found in part (a). f" (critical pt) =
A function f and its derivative f' are shown below. Assume a is a nonzero constant. Note: a is a lowercase letter. f(x) = ax² - 7x².In(x) + 7x² 2 2 4 (a) Determine the exact value of the critical point of f. Your answer may contain a. X = (b) Determine f"(x). f'(x) = ax - 7xln(x) f"(x) = (c) Evaluate f" at the critical point you found in part (a). f" (critical pt) =
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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![A function f and its derivative f' are shown below. Assume a is a nonzero constant.
Note: a is a lowercase letter.
f(x) = ax² - 7x².In(x)+7x²
2
X =
(b) Determine f"(x).
(a) Determine the exact value of the critical point of f. Your answer may contain a.
f"(x)
=
2
f" (critical pt)
4
f'(x) = ax - 7xln(x)
(c) Evaluate f" at the critical point you found in part (a).
=
(d) Apply the 2nd derivative test to the critical point found in part (a).
At the critical point, the graph of f is ---Select---
because f" is ---Select--- ✓ there.
Therefore, the critical point is a ---Select---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a3448c4-c3c6-49e9-9c9d-a456f7983c24%2Fee898b37-75ce-4869-9854-9273cbb6de14%2F5cntm4b_processed.png&w=3840&q=75)
Transcribed Image Text:A function f and its derivative f' are shown below. Assume a is a nonzero constant.
Note: a is a lowercase letter.
f(x) = ax² - 7x².In(x)+7x²
2
X =
(b) Determine f"(x).
(a) Determine the exact value of the critical point of f. Your answer may contain a.
f"(x)
=
2
f" (critical pt)
4
f'(x) = ax - 7xln(x)
(c) Evaluate f" at the critical point you found in part (a).
=
(d) Apply the 2nd derivative test to the critical point found in part (a).
At the critical point, the graph of f is ---Select---
because f" is ---Select--- ✓ there.
Therefore, the critical point is a ---Select---
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