Object Height. Suppose an object is thrown straight up from the ground. The height after t seconds is given by the formula h(t) = -5t3 + 89t² + 182 (a) The time in seconds, rounded to 4 decimal places, when the object reached the highest point was O 182 s O 8.9 s None of the other answers O os O 26.7 s O 11.8667 s (b) The height is maximized at the critical point x = a because the second derivative test found O f"(a) > 0 O f'(a) was negative to the left of x = a and positive to the right O f'(a) = 0 O f"(a) < 0 O f'(a) was positive to the left of x = a and negative to the right O f"(a) = 0
Object Height. Suppose an object is thrown straight up from the ground. The height after t seconds is given by the formula h(t) = -5t3 + 89t² + 182 (a) The time in seconds, rounded to 4 decimal places, when the object reached the highest point was O 182 s O 8.9 s None of the other answers O os O 26.7 s O 11.8667 s (b) The height is maximized at the critical point x = a because the second derivative test found O f"(a) > 0 O f'(a) was negative to the left of x = a and positive to the right O f'(a) = 0 O f"(a) < 0 O f'(a) was positive to the left of x = a and negative to the right O f"(a) = 0
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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![Object Height. Suppose an object is thrown straight up from the ground. The height after t seconds is given by the formula
h(t) = -5t3 + 89t2 + 182
(a) The time in seconds, rounded to 4 decimal places, when the object reached the highest point was
182 s
O 8.9 s
None of the other answers
O Os
O 26.7 s
O 11.8667 s
(b) The height is maximized at the critical point x = a because the second derivative test found
O f"(a) > 0
O f'(a) was negative to the left of x = a and positive to the right
O f'(a) = 0
O f"(a) < 0
O f'(a) was positive to the left of x = a and negative to the right
O f"(a) = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87f9aa08-5432-4b39-9583-9e3af51a652b%2F3cd9c652-8330-4a2e-8592-465a30198ff0%2Fakoj67c_processed.png&w=3840&q=75)
Transcribed Image Text:Object Height. Suppose an object is thrown straight up from the ground. The height after t seconds is given by the formula
h(t) = -5t3 + 89t2 + 182
(a) The time in seconds, rounded to 4 decimal places, when the object reached the highest point was
182 s
O 8.9 s
None of the other answers
O Os
O 26.7 s
O 11.8667 s
(b) The height is maximized at the critical point x = a because the second derivative test found
O f"(a) > 0
O f'(a) was negative to the left of x = a and positive to the right
O f'(a) = 0
O f"(a) < 0
O f'(a) was positive to the left of x = a and negative to the right
O f"(a) = 0
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