Question

F'(x) =0, если f(x) =√x(x^2-3x-2)

Answer 1

Пошаговое объяснение:

1. Находим производную функции $f'(x) = (\sqrt{x})'(x^2-3x-2) + \sqrt{x}(x^2-3x-2)'=\frac{x^2-3x-2}{2\sqrt{x}} + (2x-3)\sqrt{x} =\\=\frac{x^2-3x-2 + 2x(2x-3)}{2\sqrt{x}} = \frac{5x^2-9x-2}{2\sqrt{x}}$
2. Приравниваем к нулю, ищем корни $x > 0:\\5x^2 - 9x - 2 = 0\\D = 81 + 40 = 121 = 11^2\\x_1 = \frac{9+11}{10} = 2\\x_2 = \frac{9 - 11}{10} = -0.2$
3. x = -0.2 не подходит под ОДЗ, значит производная равна 0 в точке x = 2

Answer 2

To find the critical points of the function f(x) = √(x(x^2-3x-2)), we need to find the values of x for which the derivative f'(x) equals zero.

First, let's find the derivative f'(x) of the function f(x):

f(x) = √(x(x^2-3x-2))

To find f'(x), we can use the chain rule:

f'(x) = d/dx[√(x(x^2-3x-2))] = (1/2) * (x(x^2-3x-2))^(-1/2) * (x^2-3x-2)'

Now, we find the derivative of (x^2-3x-2):

(x^2-3x-2)' = d/dx[x^2] - d/dx[3x] - d/dx[2] = 2x - 3

Substitute this back into f'(x):

f'(x) = (1/2) * (x(x^2-3x-2))^(-1/2) * (2x - 3)

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

(1/2) * (x(x^2-3x-2))^(-1/2) * (2x - 3) = 0

Now, we can ignore the (1/2) term since it's a constant and will not affect the critical points:

(x(x^2-3x-2))^(-1/2) * (2x - 3) = 0

Now, we can set the second part equal to zero:

2x - 3 = 0

Solve for x:

2x = 3

x = 3/2

Now, we need to check if x = 3/2 is a valid critical point. To do that, we need to make sure the original function f(x) is defined at x = 3/2. Let's verify:

f(x) = √(x(x^2-3x-2))

f(3/2) = √(3/2 * (9/4 - 9/2 - 2)) f(3/2) = √(3/2 * (9/4 - 18/4 - 8/4)) f(3/2) = √(3/2 * (-17/4)) f(3/2) = √(-51/8)

Since the result is a complex number, the function is not defined at x = 3/2. Hence, there are no critical points for the function f(x) = √(x(x^2-3x-2)), as the derivative does not equal zero for any value of x.