nilai minimum fungsi f(x)=2x³-6x²-48x+5 dalama interval -3<_x<_4 adalah

f'(x) = 0
6x^2 - 12x - 48 = 0
x^2 - 2x - 8 = 0
(x - 4)(x + 2) = 0

x = 4 ; x = -2

uji turunan kedua
f"(x) = 12x - 12
f"(4) = 12(4) - 12 = 48-12 = 36 ( >0) min

fmin
f(4) = 2(4)^3 - 6(4)^2 - 48(4) + 5
f(4) = 128 - 96- 192 + 5
f(4) = -155