Question

Промежутки монотонности. y=0.5x^4-4x^2 Будет x=0,-2,2 Верно ли? Если нет то решите подробно

Answer 1

Верно, только промежутки надо записывать промежутками.
Производная функции равна f ' = 2x³ - 8x = 2x(x² - 4) = 2x(x-2)(x+2).
Критические точки находятся в точках, где производная равна 0 или не существует.
В данном случае таких точек 3:
2x(x-2)(x+2) = 0.
х₁ = 0,
х₂ = -2,
х₃ = 2.
Имеем  4 промежутка:
-∞ < x < -2,
-2 < x < 0,
0 < x < 2,
2 < x < +∞.

Для определения значений монотонности надо знать знак производной на данном промежутке: если производная положительна - функция возрастающая, если отрицательная - функция убывающая.

Надо определить знаки производной вблизи критических точек.
х      -3      -1       1       3
у '   -30      6      -6       30.

Отсюда ответ:
На промежутках:
-∞ < x < -2   функция убывающая,
-2 < x < 0     функция возрастающая,
0 < x < 2      функция убывающая,
2 < x < +∞   функция возрастающая.

Answer 2

Promoting Monotonicity of the Function

To determine the intervals of monotonicity for the function y = 0.5x^4 - 4x^2, we need to analyze the sign of its derivative. The derivative of a function gives us information about its slope and helps us identify where the function is increasing or decreasing.

Let's find the derivative of the function y = 0.5x^4 - 4x^2:

y' = d/dx (0.5x^4 - 4x^2)

To find the derivative, we can apply the power rule of differentiation. According to the power rule, if we have a function of the form f(x) = ax^n, where a is a constant and n is a real number, the derivative is given by f'(x) = anx^(n-1).

Applying the power rule to our function, we get:

y' = 4 * 0.5 * x^(4-1) - 2 * 4 * x^(2-1) = 2x^3 - 8x

Now that we have the derivative, we can analyze its sign to determine the intervals of monotonicity for the original function.

Analyzing the Sign of the Derivative

To analyze the sign of the derivative, we need to find the critical points of the function. Critical points occur where the derivative is equal to zero or undefined.

Setting the derivative equal to zero, we have:

2x^3 - 8x = 0

Factoring out 2x, we get:

2x(x^2 - 4) = 0

This equation is satisfied when either 2x = 0 or x^2 - 4 = 0.

Solving 2x = 0, we find x = 0.

Solving x^2 - 4 = 0, we get x = -2 and x = 2.

So, the critical points of the function are x = 0, x = -2, and x = 2.

Determining the Intervals of Monotonicity

To determine the intervals of monotonicity, we can create a sign chart using the critical points and test points within each interval.

Let's consider the intervals (-∞, -2), (-2, 0), (0, 2), and (2, +∞).

For the interval (-∞, -2), we can choose a test point x = -3. Plugging this value into the derivative, we get:

y'(-3) = 2(-3)^3 - 8(-3) = -54 + 24 = -30

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

For the interval (-2, 0), we can choose a test point x = -1. Plugging this value into the derivative, we get:

y'(-1) = 2(-1)^3 - 8(-1) = -2 + 8 = 6

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

For the interval (0, 2), we can choose a test point x = 1. Plugging this value into the derivative, we get:

y'(1) = 2(1)^3 - 8(1) = 2 - 8 = -6

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

For the interval (2, +∞), we can choose a test point x = 3. Plugging this value into the derivative, we get:

y'(3) = 2(3)^3 - 8(3) = 54 - 24 = 30

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

Conclusion

Based on the analysis of the derivative, we can conclude that the function y = 0.5x^4 - 4x^2 is decreasing in the intervals (-∞, -2) and (0, 2), and increasing in the intervals (-2, 0) and (2, +∞).

Therefore, the statement that x = 0, x = -2, and x = 2 are the intervals of monotonicity for the function y = 0.5x^4 - 4x^2 is incorrect. The correct intervals of monotonicity are (-∞, -2), (0, 2), (-2, 0), and (2, +∞).