F(x)=x^3+3x^2+3x+2, f '(x)=0

Предыдущий
вопрос Следующий
вопрос

To find when f'(x) = 0, we first need to find the derivative of the function f(x).

Given that F(x) = x^3 + 3x^2 + 3x + 2, we can find f'(x) by taking the derivative of each term:
f'(x) = 3x^2 + 6x + 3

Now we need to find when f'(x) = 0:
3x^2 + 6x + 3 = 0

To solve this quadratic equation, we can first divide through by 3 to simplify:
x^2 + 2x + 1 = 0

Now, we can factor this quadratic equation:
(x + 1)^2 = 0

Setting each factor to zero:
x + 1 = 0
x = -1

Therefore, the value of x when f'(x) = 0 is x = -1.