Tentukanlah nilai maksimum f(x) = x3 – 12X +3

F(x) = x^3 – 12x + 3
f'(x) = 3x^2 – 12 = 0
=> 3(x^2 – 4) = 0
=> 3(x + 2)(x – 2) = 0
=> x = -2 atau x = 2
+++ (-2) — (2) +++
Karena maksimum maka x = -2 (+++ —)
Nilai balik maksimumnya
f(-2) = (-2)^3 – 12(-2) + 3 = -8 + 24 + 3 = 19