Use the second derivative test to find the relative extrema of ƒ (x) = √√2x + 2 cos x on the interval [0, 2π]. ㅠ Relative Maximum at x = ☎ because ƒ¹ (7) = 0 and ƒ/1 (7) > 0. Relative Minimum at x = Relative Minimum at x = 4 Relative Maximum at x = O Relative Maximum at x = Relative Minimum at x = Relative Maximum at x = 3π 4 Relative Minimum at x = because ƒ¹ (7) = 0 and ƒ11 (7) < 0. because f(³) = 0 and ƒ11 (³) > 0. 3π 4 5п 4 3π 4 because f1 (³) = 0 and f11 (³1) < 0. 4 3π 4 because ƒ1 (57) because ƒ1 (3³) = = 0 and f11 (5) < 0. = 0 and f11 (³″) > 0. because f¹ (1) = 0 and ƒ11 (7) < 0. because f(3) = 0 and f/l (3) > 0.
Use the second derivative test to find the relative extrema of ƒ (x) = √√2x + 2 cos x on the interval [0, 2π]. ㅠ Relative Maximum at x = ☎ because ƒ¹ (7) = 0 and ƒ/1 (7) > 0. Relative Minimum at x = Relative Minimum at x = 4 Relative Maximum at x = O Relative Maximum at x = Relative Minimum at x = Relative Maximum at x = 3π 4 Relative Minimum at x = because ƒ¹ (7) = 0 and ƒ11 (7) < 0. because f(³) = 0 and ƒ11 (³) > 0. 3π 4 5п 4 3π 4 because f1 (³) = 0 and f11 (³1) < 0. 4 3π 4 because ƒ1 (57) because ƒ1 (3³) = = 0 and f11 (5) < 0. = 0 and f11 (³″) > 0. because f¹ (1) = 0 and ƒ11 (7) < 0. because f(3) = 0 and f/l (3) > 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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![Use the second derivative test to find the relative extrema of
f(x) = √√2x + 2 cos x on the interval [0, 2π].
Relative Maximum at x =
Relative Minimum at x
=
Relative Minimum at x =
Relative Maximum at x =
Relative Maximum at x =
Relative Minimum at x =
Relative Maximum at x =
Relative Minimum at x =
3π
4
4
because f() = 0 and fll (7) < 0.
3π
because f1 (³) = 0 and f11 (³)
3π
4
5π
4
3π
4
because f1 (1) = 0 and ƒ/1 (7) > 0.
because f(³) = 0 and f11 (³) < 0.
3π
4
because ƒ/ (57) = 0 and f11 (5¹) < 0.
4
4
3π
because f1 (³) = 0 and f11 (³1) > 0.
because f1 (1)
because f1 (
3π
> 0.
=
0 and fl() <0.
=
3π
0 and fll (³) > 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7353de8a-3834-465a-af3b-558129de7e04%2F074e325f-6763-4387-aeb8-c42bf7f258c8%2Fosskp6m_processed.png&w=3840&q=75)
Transcribed Image Text:Use the second derivative test to find the relative extrema of
f(x) = √√2x + 2 cos x on the interval [0, 2π].
Relative Maximum at x =
Relative Minimum at x
=
Relative Minimum at x =
Relative Maximum at x =
Relative Maximum at x =
Relative Minimum at x =
Relative Maximum at x =
Relative Minimum at x =
3π
4
4
because f() = 0 and fll (7) < 0.
3π
because f1 (³) = 0 and f11 (³)
3π
4
5π
4
3π
4
because f1 (1) = 0 and ƒ/1 (7) > 0.
because f(³) = 0 and f11 (³) < 0.
3π
4
because ƒ/ (57) = 0 and f11 (5¹) < 0.
4
4
3π
because f1 (³) = 0 and f11 (³1) > 0.
because f1 (1)
because f1 (
3π
> 0.
=
0 and fl() <0.
=
3π
0 and fll (³) > 0.
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