Question

Найдите промежутки возрастания и убывания и точки экстремума функции f(x) =x+2x^2-x^3 с появлением пожалуйста)​

Answer 1

Ответ:

Точки экстремума: min 0, max 4/3

Объяснение:

Точки экстремума находят с помощью производной функции.

Это когда производная равна нулю,то есть ее корни.

1) Найдем производную

f'(x) = (x+2x^2-x^3)' = 4x-3x^2 = x(4-3x)

2) Найдем корни производной

x(4-3x) = 0

x1=0; x2=4/3

3) Определим в каких промежутках функция убывает и возрастает

f'(-1) = -1(4-3*(-1))=-7

f'(1) = 1(4-3*1)=1

f'(2) = 2(4-3*2)=-4

В промежутке до нуля функция убывает, так как производная меньше нуля. В промежутке от 0 до 4/3 функция возрастает, так как производная больше нуля. В промежутке от 4/3 функция убывает, так как производная меньше нуля.

Из этого следует что точка минимума 0, а точка максимума 4/3

Answer 2

Finding the Intervals of Increase and Decrease and the Extrema of the Function

To find the intervals of increase and decrease and the extrema of the function f(x) = x + 2x^2 - x^3, we need to analyze the first derivative and the second derivative of the function.

Let's start by finding the first derivative of the function:

f'(x) = 1 + 4x - 3x^2.

To find the intervals of increase and decrease, we need to determine where the first derivative is positive or negative. We can do this by finding the critical points and testing the intervals between them.

To find the critical points, we set the first derivative equal to zero and solve for x:

1 + 4x - 3x^2 = 0.

Simplifying the equation, we get:

3x^2 - 4x - 1 = 0.

Using the quadratic formula, we can solve for x:

x = (-b ± √(b^2 - 4ac)) / (2a).

Plugging in the values, we get:

x = (4 ± √(16 + 12)) / (6).

Simplifying further, we have:

x = (4 ± √(28)) / (6).

Therefore, the critical points are:

x = (2 ± √(7)) / (3).

Now, let's test the intervals between the critical points and determine where the first derivative is positive or negative.

For x < (2 - √(7)) / 3, we can choose a test point, such as x = 0. Plugging this value into the first derivative, we get:

f'(0) = 1 + 4(0) - 3(0)^2 = 1.

Since the first derivative is positive, the function is increasing in this interval.

For (2 - √(7)) / 3 < x < (2 + √(7)) / 3, we can choose a test point, such as x = 1. Plugging this value into the first derivative, we get:

f'(1) = 1 + 4(1) - 3(1)^2 = 2.

Since the first derivative is positive, the function is increasing in this interval as well.

For x > (2 + √(7)) / 3, we can choose a test point, such as x = 2. Plugging this value into the first derivative, we get:

f'(2) = 1 + 4(2) - 3(2)^2 = -5.

Since the first derivative is negative, the function is decreasing in this interval.

Therefore, we can conclude that the function f(x) = x + 2x^2 - x^3 is increasing on the intervals (-∞, (2 - √(7)) / 3] and [(2 + √(7)) / 3, +∞), and decreasing on the interval [(2 - √(7)) / 3, (2 + √(7)) / 3].

To find the extrema of the function, we need to analyze the second derivative.

The second derivative of the function is:

f''(x) = 4 - 6x.

To find the points of inflection, we set the second derivative equal to zero and solve for x:

4 - 6x = 0.

Simplifying the equation, we get:

6x = 4.

Dividing both sides by 6, we have:

x = 2/3.

Therefore, the point of inflection is x = 2/3.

Now, let's analyze the concavity of the function.

For x < 2/3, we can choose a test point, such as x = 0. Plugging this value into the second derivative, we get:

f''(0) = 4 - 6(0) = 4.

Since the second derivative is positive, the function is concave up in this interval.

For x > 2/3, we can choose a test point, such as x = 1. Plugging this value into the second derivative, we get:

f''(1) = 4 - 6(1) = -2.

Since the second derivative is negative, the function is concave down in this interval.

Therefore, we can conclude that the function f(x) = x + 2x^2 - x^3 is concave up on the interval (-∞, 2/3) and concave down on the interval (2/3, +∞).

To summarize:

- The function is increasing on the intervals (-∞, (2 - √(7)) / 3] and [(2 + √(7)) / 3, +∞). - The function is decreasing on the interval [(2 - √(7)) / 3, (2 + √(7)) / 3]. - The function is concave up on the interval (-∞, 2/3). - The function is concave down on the interval (2/3, +∞).

Please let me know if there's anything else I can help you with!