Tentukan nilai minimum fungsi f(x) = 1/3 x3 + 3x2 - 16x + 8

Jawab:

Diketahui : f(x) = 1/3x³ + 3x² - 16x + 8 Ditanya : nilai minimum = ... ? Maka: f(x) = 1/3x³ + 3x² - 16x + 8 f(x) = 1/3 . x³ + 3 . x² - 16 . x + 8 f'(x) = 1/3 . 3 . x³¯¹ + 3 . 2 . x²¯¹ - 16 + 0 f'(x) = 1 . x²+ 6 . x¹ - 16 f'(x) = x² + 6x - 16 Sehingga: f'(x) = 0 x² + 6x - 16 = 0 (x + 8)(x - 2) = 0 pembuat nol: x + 8 = 0 → x = -8 x - 2 = 0 → x = 2 saat x = -8 f(x) = 1/3x³ + 3x² - 16x + 8 f(-8) = 1/3(-8)³ + 3(-8)² - 16(-8) + 8 f(-8) = 1/3(-512) + 3(64) + 128 + 8 f(-8) = -512/3 + 192 + 128 + 8 f(-8) = -512/3 + 328 f(-8) = -512/3 + 984/3 f(-8) = 472/3 saat x = 2 f(x) = 1/3x³ + 3x² - 16x + 8 f(2) = 1/3(2)³ + 3(2)² - 16(2) + 8 f(2) = 1/3(8) + 2(4) - 32 + 8 f(2) = 8/3 + 8 - 32 + 8 f(2) = 8/3 - 16 f(2) = 8/3 - 48/3 f(2) = -40/3