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)

Answer 2

To solve the inequality f'(x) < 0, we need to find the derivative of f(x) and determine where it is negative.

Taking the derivative of f(x) using the power rule, we get:

f'(x) = 12x^3 - 12x^2 - 24x

Now we need to find where f'(x) is negative:

f'(x) < 0

12x^3 - 12x^2 - 24x < 0

Dividing both sides by 12x:

x^2 - x - 2 < 0

Factoring the quadratic expression:

(x - 2)(x + 1) < 0

This quadratic expression is negative for values of x between -1 and 2, exclusive. Therefore, the solution to the inequality f'(x) < 0 is:

-1 < x < 2

In other words, the derivative of f(x) is negative for values of x between -1 and 2, exclusive.