Question

Найти наибольшеезначение функции f(x)=2x^3+3x^2-36x на отрезке [-4;3]

Answer 1

f'(x) = 6x²+6x-36
6x²+6x-36 = 0    ÷6
x²+x-6 = 0
D = 1+4*6 = 1+24 = 25
x1 = (-1-5)/2 = -6/2 = -3
x2 = (-1+5)/2 = 4/2 = 2
f(-3) = 2*(-3)³+3*(-3)²-36*(-3) = -54+27+108 = 81
f(2) = 2*2³+3*2²-36*2 = 16+12-72 = -44
f(-4) = 2*(-4)³+3*(-4)²-36*(-4) = -128+48+144 = 64
f(3) = 2*3³+3*3²-36*3 = 54+27-108 = -27

Наибольшее: f(x) = 81 при x = -3
Наименьшее: f(x) = -44 при x = 2