pembahasan :

f(x) = cos 2x, untuk 0° ≤x≤ 360°.

syarat maksimum minimum f'(x) = 0

f(x) = cos 2x

f'(x) = ( -sin 2x).2

= -2sin 2x

-2sin 2x = 0

sin 2x = 0 , 2x = 360 , 2x = 180

2x = 0 x = 180⁰ x = 90⁰

x = 0⁰

2x = 540

x = 270⁰

2x = 720

x = 360⁰

jadi x = {0⁰, 90⁰, 180⁰, 270⁰, 360⁰}

cek maksimum minimum:

f(x) = cos2x

x = 0 f(0) = cos2(0) = cos0 = 1 (max)

x = 90 f(90) = cos2(90) = cos180=-1(min)

x=180 f(180) = cos360 = 1 (max)

x = 270 f(270) = cos540= -1(min)

x =360 f(360) = cos720 =1 (max)

koordinat titik balik maksimum:

(0⁰, 1)

(180⁰, 1)

(360⁰, 1)

koordinat titik balik minimum:

(90⁰, -1)

(270⁰, -1)

NILAI MINIMUM = -1

NILAI MAKSIMUM = 1