Question

f'(x)=0 найти корень уравнения

f(x)=x^3/3-x^2-3x+7​

Answer 1

Ответ:

f'(x)=x³/3-x²-3x+7 = 1/3*(x³)' - (x²)' -(3x)' +7' = x² - 2x - 3

x² - 2x - 3 = 0,

a = 1, b = -2, c = -3

D = (-2)² - 4 * 1 * (-3) = 16

x1 = -(-2) + √16/2*1 = 3,

x2 = -(-2) - √16/2*1 = -1

x1 = -1, x2 = 3.