Jika f(x) 2×3+3×2-12x+4 tentukan nilai stationernya

Kelas 11 Matematika
Bab Turunan

f(x) = 2x³ + 3x² – 12x + 4
f'(x) = 0
6x² + 6x – 12 = 0
x² + x – 2 = 0 
(x + 2) (x – 1) = 0
x + 2 = 0
x = -2

f(-2) = 2 . (-2)³ + 3 . (-2) – 12 . (-2) + 4
f(-2) = 2 . (-8) – 6 + 24 + 4
f(-2) = -16 – 6 + 28
f(-2) = 6

x – 1 = 0
x = 1

f(1) = 2 . 1³ + 3 . 1² – 12 . 1 + 4
f(1) = 2 + 3 – 12 + 4
f(1) = -3

Nilai stationernya adalah -3 dan 6