Question

Провести полное исследование функции(на выпуклость и тд,найти ассимптомы графика функции) и построить график y=x(^3)+4/x^2

Answer 1

ДАНО
Y = x³ + 4*x⁻²
Исследование.
1. Область определения.
Х∈(R: x≠0)
2. Область значений. 
Y∈(-∞;+∞).
3. Пересечение с осью Ох.
Х = -2²/⁵ ≈ -1,3195
4. Функция ни чётная ни нечётная.
5. Первая производная.
Y'(x) = 3*x² - 8*x⁻³.
6. Экстремумы -  в корнях производной.
Y'(x) = 0  при х1= 2³/⁵ * 3⁻⁵ ≈ 1,5157/1,2457 ≈1,217 - минимум
Возрастает - Х∈(-∞;0)∪(х1;+∞)
Убывает - Х∈(0;х1).
7. Вторая производная.
Y"(x) = 6*x + 24*x⁻⁴.
8. Вогнутая  на области определения. точек перегиба - нет.
9. Асимптота - Y = x³.
10. Графики в приложении.

Answer 2

Function Analysis

The given function is: f(x) = x^3 + 4/x^2 To analyze this function, we first need to find its domain. Since we are dividing by x^2, x cannot equal 0. Therefore, the domain of the function is: D = {x | x ≠ 0} Next, we analyze the characteristics of the function: - To find the asymptotes, we see that as x approaches ±∞, the x^3 term dominates and the function approaches ±∞. Therefore, the function has horizontal asymptotes at y = ±∞. - To find the intercepts (where the graph crosses the x- or y-axes), we set f(x) equal to 0 and solve: f(x) = x^3 + 4/x^2 = 0. This has no real solutions, so there are no intercepts. - To find the stationary points (maximums/minimums), we take the derivative: f'(x) = 3x^2 - 8/x^3. Setting this equal to 0 and solving, we get a stationary point at x = ±2. The second derivative test shows that these are a local minimum and maximum respectively.

Graphing the Function

With the above information, we can now graph the function: It has asymptotes at y = ±∞. There are stationary points at (-2, 0) and (2, 8) which are a local min and max. The graph is always above the x-axis since the x^3 term dominates as x gets large. A graph of the function is shown below: ``` / \ / \ / \ (-2, 0)--------(2, 8) \ / \ / \ / ```