Question

F (x) = 2x (в Кубе) (1-2х), f (x) < 0

Answer 1

Решение задания смотри на фотографии

Answer 2

To find the values of x for which f(x) < 0, we need to evaluate the given function f(x) = 2x^3(1 - 2x) and find the values of x that satisfy f(x) < 0.

Step 1: Find the critical points of f(x). Critical points occur when the derivative of the function is equal to zero or does not exist.

f(x) = 2x^3(1 - 2x)

Take the derivative with respect to x:

f'(x) = d/dx [2x^3(1 - 2x)] f'(x) = 2[3x^2(1 - 2x) + x^3(-2)] f'(x) = 2[3x^2 - 6x^3 - 2x^3] f'(x) = 2(3x^2 - 8x^3)

Set f'(x) equal to zero to find the critical points:

2(3x^2 - 8x^3) = 0 3x^2 - 8x^3 = 0

Step 2: Solve for x.

Factor out x:

x(3x - 8) = 0

Set each factor to zero:

1. x = 0
2. 3x - 8 = 0

Solve for x:

2. 3x = 8 x = 8/3

So, the critical points are x = 0 and x = 8/3.

Step 3: Determine the intervals where f(x) may be negative.

To determine the intervals where f(x) may be negative, we can create a sign chart and test the intervals around the critical points. Choose test points within each interval and evaluate f(x) at those points to find the sign.

Consider three intervals: (-∞, 0), (0, 8/3), and (8/3, ∞).

Test point for interval (-∞, 0): x = -1 f(-1) = 2(-1)^3(1 - 2(-1)) = 2(-1)(1 + 2) = -6 < 0

Test point for interval (0, 8/3): x = 1 f(1) = 2(1)^3(1 - 2(1)) = 2(1)(1 - 2) = -2 < 0

Test point for interval (8/3, ∞): x = 3 f(3) = 2(3)^3(1 - 2(3)) = 2(27)(1 - 6) = -108 < 0

Step 4: State the solution.

Based on the sign chart and the test points, the function f(x) = 2x^3(1 - 2x) is negative (less than zero) in the intervals (-∞, 0) and (8/3, ∞). In other words, the values of x that satisfy f(x) < 0 are x < 0 and x > 8/3.