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