Fungsi f (x)=2cos(2x-π/6),0<x<π mencapai maksimum pada saat x ...

Fungsi f (x)=2cos(2x-π/6),0<x<π mencapai maksimum pada saat x …

F'(x) = -2sin(2x-π/6) 2 = -4sin(2x-π/6)
f'(x) = 0
-4sin(2x-π/6) = 0
sin(2x-π/6) = sin 0°
sin(2x-30°) = sin 0°
2x-30° = 0° + k.360°
2x = 30° + k.360°
x = 15° + k.180°
k = 0 => x = 15°

f"(x) = -8cos(2x-π/6)
f"(15°) = -8cos(30°-30°) = -8cos0= -8 karena < 0
maka cekung ke bawah dan merupakan titik maksimum.
maka akan mencapai titik maksimum pada saat x = 15° atau π/12

maaf kalo salah

F(x) reach its maximum at
$cos(2x - frac{pi}{6} ) = 1$
$2x - frac{pi}{6} = 0 \ x = frac{pi}{12}$
or
$2x - frac{pi}{6} = 2pi \ x = frac{13pi}{12} : : (does : not : satisfy) \$
hence, f(x) reach its maximum at x = π/12