Tentukan titik stasioner dari fungsi berikut. y = sin²(x+ π/4), 0 ≤ x ≤ 2π

Halo, Eni N. Kakak bantu jawab ya :) Jawabannya adalah (π/4, 1) dan (5π/4, 1). Perhatikan penjelasan berikut. INGAT! 1.) Jika y = sin²(ax + b) maka y' = 2a cos(ax + b) 2.) Jika mempunyai titik stationer maka y' = 0 3.) Jika cos x = cos a, maka x = ± a + k. 2π Maka: y = sin²(x+ π/4) y' = 2.1.cos(x+ π/4) y'= 2cos(x+ π/4) Sehingga: y' = 0 2cos(x+ π/4) = 0 kedua ruas dibagi 2 cos(x+ π/4) = 0 cos(x+ π/4) = cos π/2 ● x + π/4 = π/2 + k. 2π x = π/2 - π/4 + k. 2π x = 2π/4 - π/4 + k. 2π x = π/4 + k. 2π 0 ≤ x ≤ 2π k = 0 → x = π/4 + 0. 2π = π/4 + 0 = π/4 (memenuhi) k = 1 → x = π/4 + 1. 2π = π/4 + 2π = π/4 + 8π/4 = 9π/4 (tidak memenuhi) ● x + π/4 = -π/2 + k. 2π x = -π/2 - π/4 + k. 2π x = -2π/4 - π/4 + k. 2π x = -3π/4 + k. 2π 0 ≤ x ≤ 2π k = 0 → x = -3π/4 + 0. 2π = -3π/4 + 0 = -3π/4 (tidak memenuhi) k = 1 → x = -3π/4 + 1. 2π = -3π/4 + 2π = -3π/4 + 8π/4 = 5π/4 (memenuhi) k = 2 → x = -3π/4 + 2. 2π = -3π/4 + 4π = -3π/4 + 16π/4 = 13π/4 (tidak memenuhi) Substitusikan x = π/4 dan x = 5π/4 ke y = sin²(x+ π/4) x = π/4 → y = sin²(π/4 + π/4) =sin²(2π/4) = sin²(π/2) = (sin(π/2))² = 1² = 1 x = 5π/4 → y = sin²(5π/4 + π/4) =sin²(6π/4) = sin²(3π/2) = (sin(3π/2))² = (-1)² = 1 Jadi, titik stasioner dari fungsi tersebut adalah (π/4, 1) dan (5π/4, 1). Semoga membantu ya :)