Найти найбольшее и найменьшее значение функции f (x) = 1/3x3-4x, [0; 3]

F ' (x) = (x^3 - 2*x^2 + x + 3) ' = 3x^2 - 4x + 1;

f ' (x) = 0;

3x^2 - 4x + 1 = 0;

D = 16 - 12 = 4

x1 = (4 + 2) / 6 = 1 ∈ [ 0; 1,5]

x2 = (4 - 2) / 6 = 1/3 ∈ [0; 1.5]

y (0) = 3; min

y (1/3) = (1/3) ^3 - 2 * (1/3) ^2 + 1/3 + 3 = 1/27 - 2/9 + 1/3 + 3 =

= (1 - 2*3 + 9 + 3*27) / 27 = 85/27 = 3, (148)

y (1) = 1 - 2 + 4 = 3; min

y (3/2) = (3/2) ^3 - 2 * (3/2) ^2 + 3/2 + 3 = 27/8 - 18/4 + 3/2 + 3 =

= 3,375 - 4,5 + 4,5 = 3,375 max