Найдите значение функции y=8+6x(3+x)-2x^3 d точке максимума

y = - 2x^3 + 6x^2 + 18x + 8
y ' = - 6x^2 + 12x + 18

y ' = 0
- 6x^2 + 12x + 18 = 0 /:(-6)
x^2 - 2x - 3 = 0
D = 4 + 4*3 = 16
x1 = (2 + 4)/2 = 6 /2 = 3
x2 = (2 - 4)/2 = - 2/2 = - 1

---------( - 1) --------------------( 3) ------------> x

x = 3 max

y(3) = 8 + 6*18 - 2*27 = 62