Question

F'(x)=0, F(x)=x3-3x2

Answer 1

Решение задания смотри на фотографии

Answer 2

To find the critical points of the function F(x) = x^3 - 3x^2, we need to find the values of x where the derivative F'(x) is equal to zero.

Let's start by finding the derivative of F(x). Using the power rule, we can differentiate each term:

F(x) = x^3 - 3x^2

F'(x) = 3x^2 - 6x

To find the critical points, we set F'(x) equal to zero and solve for x:

3x^2 - 6x = 0

Factor out the common factor of 3x:

3x(x - 2) = 0

Now, we have two possibilities:

1. 3x = 0 x = 0

2. x - 2 = 0 x = 2

So the critical points of the function F(x) = x^3 - 3x^2 are x = 0 and x = 2.