Question

Решите неравенство f'(x)<0, если f(x)=3x^4-4x^3-12x^2+3

Answer 1

f(x) = 3x⁴ - 4x³ - 12x² + 3
f '(x) = 3(x⁴)' - 4(x³)' - 12(x²)' + 3' = 12x³ - 12x² - 24x

f '(x) < 0
12x³ - 12x² - 24x < 0
x³ - x² - 2x < 0
x(x² - x - 2) < 0
x(x - 2)(x + 1) < 0
          -                        +                 -                      +

____________₀__________₀__________₀____________
                       - 1                      0                    2


x ∈ (- ∞ ; - 1) ∪ (0 ; 2)