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f (x) = 1/3x^3 + 3x2^ – 16x + 8
f ' (x) = x^2 + 6x – 16
so,
x^2 + 6x – 16 > 0
(x + 8)(x - 2) > 0
pembuat nol untuk x :
x = -8 atau x = 2
solusi = x < - 8 atau x > 2
f ' (x) = x² + 6x – 16
x² + 6x – 16 > 0
(x + 8)(x - 2) > 0
x < - 8 atau x > 2