Question

Решите уравнение F'(x) =0, если F(x) = x^3 - 3x^2

Answer 1

Ответ:

x1=0;

x2=2

Объяснение:

1) Найдём производную функции f(x):

$f(x) = {x}^{3} - 3 {x}^{2}$

f'(x)=

$= 3 {x}^{2} - 6x$

2)Решим уравнение f'(x)=0:

$3 {x}^{2} - 6x = 0 \\ x(3x - 6) = 0 \\ x_1 = 0 \\ x_2 = 6 \div 3 = 2$

Answer 2

To find the critical points of the function F(x), we need to find where the derivative F'(x) is equal to zero.

Given that F(x) = x^3 - 3x^2, we can find the derivative as follows:

F'(x) = d/dx (x^3 - 3x^2) = 3x^2 - 6x

Now, let's set F'(x) = 0 and solve for x:

3x^2 - 6x = 0

Factoring out 3x, we have:

3x(x - 2) = 0

Setting each factor equal to zero, we get two solutions:

1. 3x = 0 x = 0

2. x - 2 = 0 x = 2

So, the critical points of F(x) occur at x = 0 and x = 2, where F'(x) equals zero.