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
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Chapter1: Functions And Models
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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.
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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