Tutorial Exercise Find the absolute maximum and absolute minimum values of f on the given interval. f(t) 2 cos t+ sin 2t [0.5] Part 1 of 5 To find the absolute maximum and minimum for f(t) = 2 cos t+ sin 2t on the interval [0, 1/2], we must find the largest and smallest function values at the critical numbers and at the interval endpoints. First, we'll find f'(t) = (No Response) -2 sin t+ (No Response) 2 cos 2t. Part 2 of 5 Since cos 2t = 1- 2 sin² t, then f'(t)= -2 sin t + 2 cos 2t = -2(2 sin² t + sin t-1), which simplifies to f'(t) = -2( Part 3 of 5 x 2 sin (t) 1) (sin t + 2x1 1). We have 0 = f'(t) = -2(2 sin t - 1) (sin t+1) when 0-2 sin t - 1, which means t = is not in the interval. is a critical number in the interval, or when 0= sin t + 1. However, the only value of t between 0 and 2x where this is true is at t = which
Tutorial Exercise Find the absolute maximum and absolute minimum values of f on the given interval. f(t) 2 cos t+ sin 2t [0.5] Part 1 of 5 To find the absolute maximum and minimum for f(t) = 2 cos t+ sin 2t on the interval [0, 1/2], we must find the largest and smallest function values at the critical numbers and at the interval endpoints. First, we'll find f'(t) = (No Response) -2 sin t+ (No Response) 2 cos 2t. Part 2 of 5 Since cos 2t = 1- 2 sin² t, then f'(t)= -2 sin t + 2 cos 2t = -2(2 sin² t + sin t-1), which simplifies to f'(t) = -2( Part 3 of 5 x 2 sin (t) 1) (sin t + 2x1 1). We have 0 = f'(t) = -2(2 sin t - 1) (sin t+1) when 0-2 sin t - 1, which means t = is not in the interval. is a critical number in the interval, or when 0= sin t + 1. However, the only value of t between 0 and 2x where this is true is at t = which
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 35E
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![Tutorial Exercise
Find the absolute maximum and absolute minimum values of f on the given interval.
f(t) = 2 cost + sin 2t
[0,5]
Part 1 of 5
To find the absolute maximum and minimum for f(t) = 2 cos t + sin 2t on the interval [0, π/2], we must find the largest and smallest function values at the critical numbers and at the interval endpoints. First, we'll find
f '(t) = (No Response) -2 sin t+ (No Response) 2 cos 2t.
Part 2 of 5
Since cos 2t = 1 - 2 sin² t, then
f'(t) = -2 sin t + 2 cos 2t
which simplifies to
Part 3 of 5
= -2(2 sin² t + sin t - 1),
Submit
f'(t) = -2(
π
6
2 sin (t)
Skip (you cannot come back)
- 1)(sin t + 2 x
We have 0 = f'(t) = -2(2 sin t - 1)(sin t + 1) when 0 = 2 sin t - 1, which means t =
is not in the interval.
1).
is a critical number in the interval, or when 0 = sin t + 1. However, the only value of t between 0 and 2π where this is true is at t =
1
which](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcaec5776-4ce0-4870-90ba-519da2dda4dc%2F671ea8fd-9538-4748-827e-325fcdfbb395%2Fberhxh_processed.png&w=3840&q=75)
Transcribed Image Text:Tutorial Exercise
Find the absolute maximum and absolute minimum values of f on the given interval.
f(t) = 2 cost + sin 2t
[0,5]
Part 1 of 5
To find the absolute maximum and minimum for f(t) = 2 cos t + sin 2t on the interval [0, π/2], we must find the largest and smallest function values at the critical numbers and at the interval endpoints. First, we'll find
f '(t) = (No Response) -2 sin t+ (No Response) 2 cos 2t.
Part 2 of 5
Since cos 2t = 1 - 2 sin² t, then
f'(t) = -2 sin t + 2 cos 2t
which simplifies to
Part 3 of 5
= -2(2 sin² t + sin t - 1),
Submit
f'(t) = -2(
π
6
2 sin (t)
Skip (you cannot come back)
- 1)(sin t + 2 x
We have 0 = f'(t) = -2(2 sin t - 1)(sin t + 1) when 0 = 2 sin t - 1, which means t =
is not in the interval.
1).
is a critical number in the interval, or when 0 = sin t + 1. However, the only value of t between 0 and 2π where this is true is at t =
1
which
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