Найти наименьшее значение функции f (x) = x^2 (2x+3) + 6 (5-2x) на отрезке [-3; 3]

F' (x) = 2x (2x+3) + x²*2+6 * (-2) = 4x²+6x+2x²-12=6x²+6x-12=6 (x²+x-2) = 0

x1+x2=-1 U x1*x2=-2

x1=-2∈[-3; 3]

x2=1∈[-3; 3]

f (-3) = 9 * (-6+3) + 6 * (5+6) = - 27+66=39

f (-2) = 4 * (-4+3) + 6 * (5+4) = - 4+54=50

f (1) = 1 * (2+3) + 6 * (5-2) = 5+18=23 наим

f (3) = 9 * (6+3) + 6 * (5-6) = 81-6=75