Найдите наименьшее значение функции y=x^3-x^2-40x на отрезке [3; 8]

Y'=3x²-2x-40=0

D=4+480=484

x1 = (2-22) / 6=-10/3∉[[3; 8]

x2 = (2+22) / 6=4∈[3; 8]

y (3) = 27-9-120=-102

y (4) = 64-16-160=-112 наим

y (8) = 512-64-320=128

Y ' = 3x^2 - 2x - 40

y ' = 0

3x^2 - 2x - 40 = 0

D = 4 + 4*40*3 = 484 = 22^2

x1 = (2 + 22) / 6 = 24/6 = 4 ∈ [3; 8]

x2 = (2 - 22) / 6 = - 20/6 = - 10/3 = - 3 ц 1/3 ∉ [3; 8]

y (3) = 27 - 9 - 120 = - 102

y (4) = 64 - 16 - 160 = - 112

y (8) = 512 - 64 - 320 = 128

Ответ

y (4) = - 112