Question

Дана функция y=2x^3-3x^2-36x+40. При каких значениях х: y'(x)=0

Answer 1

y'(x) = 6x^2 - 6x - 36 = 0

6(x^2 - x - 6) = 0

6(x - 3)(x + 2) = 0

x1 = -2; y(-2) = 2(-8) - 3*4 - 36(-2) + 40 = -16 - 12 + 72 + 40 = 84

x2 = 3; y(3) = 2*27 - 3*9 - 36*3 + 40 = 54 - 27 - 108 + 40 = -41