Question

решить уравнение f '(x)=0 если f(x)=2x^3 - 3x^2 - 12x -3

Answer 1

f '(x) = 2 × 3x² - 3 × 2x - 12 * 1 - 0 = 6x² - 6x - 12

6x² - 6x - 12 = 0

D = 36 - 4 × 6 × (-12) = 36 + 288 = 324

√D = 18

$x_{1} = \frac{6+18}{12} =\frac{24}{12} =2\\ x_{2} =\frac{6-18}{12} =\frac{-12}{12} =-1$

Ответ: 2; -1.