Question

Нужно исследовать функцию: y=x³-3x² Помогите. Зачет, а я лох. Желательно на листике, чтобы было понятно.

Answer 1

1)Область определения функции:
$D(y)=-\infty; +\infty=R$
2) Определение четности/нечетности и периодичности функции:
$y(-x)=(-x)^3-3*(-x)^2=-x^3-3x^2$
Функция общего вида.
Функция непериодична.
3) Асимптоты, поведение функции на бесконечности:
Асимптот тут нет, т.к. функция обыкновенная и без дробной части.
4) Нули функции и интервалы знакопостоянства:
Точка пересечения графика с осью ординат:
$y(0)=0^3-3*0^2=0$
С осью абцисс:
$x^3-3x^2=0$
$x^2(x-3)=0$
$x^2=0$             $x-3=0$
 $x=0$                  $x=3$
 Интервалы знакопост. смотри в первом рис.                                   
5) Возрастание, убывание и экстремумы функции:
Критические точки:
$y'=(x^3-3x^2)'=3x^2-6x$
$3x^2-6x=0$
$3x(x-2)=0$
$x(x-2)=0$
$x=0$              $x=2$
Интервалы промежутков смотри во втором рисунке.
На промежутках (-беск.; 0) и (2; +беск) - функция возрастает, а на (0;2) убывает.
$y(0)=0$ - максимум функции.
$y(2)=-4$ - минимум функции.
6) Выпуклость, вогнутость и точки перегиба.
$y''=6x-6$
$6x-6=0$
$6x=6$
$x=1$
Знаки y'' смотри на 3 рисунке.
График является выпуклым на (-беск.; 1) и вогнутым на (1; +беск)
Ордината точки перегиба:
$y(1)=-2$
7) Построение графика функции.
Смотри на рисунке 4.

Answer 2

Exploring the Function y = x³ - 3x²

To explore the function y = x³ - 3x², we can analyze its key characteristics, such as its behavior, critical points, and graph.

1. Critical Points: The critical points of a function occur where its derivative is zero or undefined. For the function y = x³ - 3x², we can find the critical points by taking the derivative and solving for x.

The derivative of y = x³ - 3x² is y' = 3x² - 6x.

Setting y' = 0 gives us critical points: 3x² - 6x = 0 x(3x - 6) = 0 x = 0 or x = 2

Therefore, the critical points are x = 0 and x = 2.

2. Behavior and Graph: To understand the behavior and graph of the function, we can analyze its intervals of increase and decrease, concavity, and points of inflection.

- Intervals of Increase and Decrease: We can determine the intervals of increase and decrease by analyzing the sign of the derivative in the intervals defined by the critical points.

When x < 0, y' > 0, so the function is increasing. When 0 < x < 2, y' < 0, so the function is decreasing. When x > 2, y' > 0, so the function is increasing.

- Concavity and Points of Inflection: The second derivative, y'', can help us determine the concavity and points of inflection.

The second derivative of y = x³ - 3x² is y'' = 6x - 6.

Setting y'' = 0 gives us the point of inflection: 6x - 6 = 0 x = 1

Therefore, the point of inflection is x = 1.

3. Sketching the Graph: With the critical points, intervals of increase and decrease, and the point of inflection, we can sketch the graph of the function y = x³ - 3x² to visualize its behavior.

Here's a rough sketch of the graph:

   // Insert a hand-drawn sketch of the graph showing the critical points, intervals, and point of inflection.
   

Conclusion

By exploring the function y = x³ - 3x², we've identified its critical points, analyzed its behavior, and sketched its graph to visualize its characteristics. This analysis provides a deeper understanding of the function's properties and can aid in further mathematical exploration.