Question

F(x) =2x^3-3x^2 свойства график ​

Answer 1

Дана функция F(x) =2x³-3x².

Полное исследование функций по схеме:

1. Область определения функции - ограничений нет, поэтому х ∈ Z.

2. Непрерывность функции, вертикальные асимптоты.

Разрывов функции нет - она непрерывна, асимптот нет.

3. Точки пересечения функции с осями координат.

- с осью Оу при х = 0 у = 0.

- с осью Оу при у = 0 надо решить уравнение 2x³-3x² = 0.

x²(2х - 3) = 0. Получаем 2 точки: х = 0 и х = 3/2.

4. Четность, нечетность - функция общего вида.

5. Периодичность - нет.

6. Промежутки возрастания, убывания, экстремумы функции.

Производная равна: y' = 6x² - 6x = 6x(x - 1) = 0.

Получаем 2 экстремума: х = 0 и х = 1.

Находим знаки производной на полученных промежутках.

x =    -1       0       0,5      1          2

y' =  12       0   -1,5      0   12 .

Максимум в точке х = 0 (переход с + на -), у = 0,

минимум в точке х = 1 (переход с - на +).

Функция возрастает на промежутках (-∞; 0) и (1; +∞),

убывает на промежутке (0; 1).

7. Промежутки выпуклости, вогнутости, точки перегиба.

Вторая производная равна: y'' = 12x - 6 = 6(2x - 1) = 0.

Перегиб в точке х = 1/2.

Функция выпуклая на промежутке (-∞; (1/2) и вогнута ((1/2); +∞).

8. Наклонные асимптоты - нет.

9. Построение графика по точкам:

x y

-2.0 -28

-1.5 -13.5

-1.0 -5

-0.5 -1

0 0

0.5 -0.5

1.0 -1

1.5 0

2.0 4

2.5 12.5

3.0 27

Answer 2

The function $F(x) = 2x^3 - 3x^2$ is a cubic function, and its graph exhibits several notable properties:

1. Degree: The function is a cubic function because the highest power of $x$ in the expression is 3. Cubic functions have graphs that appear like curves with both ends pointing upward or downward.

2. Roots: To find the roots (or zeros) of the function, set $F(x)$ equal to zero and solve for $x$: $2x^3 - 3x^2 = 0$

   Factoring out $x^2$ from the equation, we get: $x^2(2x - 3) = 0$

   So, the roots are: $x = 0$ (double root) and $x = \frac{3}{2}$

   These roots indicate the points where the graph intersects the x-axis.

3. Critical points: To find critical points (points where the slope is zero), take the derivative of $F(x)$ with respect to $x$ and set it equal to zero: $F'(x) = 6x^2 - 6x$ Setting $F'(x) = 0$ and solving for $x$, we get: $6x^2 - 6x = 0$ $6x(x - 1) = 0$ So, the critical points are $x = 0$ and $x = 1$.

4. Behavior around critical points: By analyzing the second derivative, $F''(x) = 12x - 6$, we can determine the concavity of the graph around the critical points. When $x < \frac{1}{2}$, $F''(x) < 0$, so the graph is concave down. When $x > \frac{1}{2}$, $F''(x) > 0$, so the graph is concave up. This means that $x = \frac{1}{2}$ is a local minimum point, and the concavity changes at this point.

5. Asymptotes: As $x$ approaches positive or negative infinity, the function grows without bound. Therefore, there are no horizontal asymptotes.

6. Symmetry: The function is neither even nor odd.

7. Y-intercept: To find the y-intercept, set $x = 0$ in the original function: $F(0) = 2(0)^3 - 3(0)^2 = 0$ So, the y-intercept is at the origin (0, 0).

8. Behavior at infinity: As $x$ approaches positive or negative infinity, $F(x)$ also approaches positive or negative infinity, respectively.

9. Graph shape: The graph of $F(x) = 2x^3 - 3x^2$ is a smooth curve that passes through the y-intercept at the origin and has turning points at $x = 0$ and $x = 1$. The graph is symmetric with respect to the vertical line $x = \frac{1}{2}$ and does not have any holes or breaks in its shape.

10. Increasing and decreasing intervals: By observing the sign of the first derivative $F'(x) = 6x^2 - 6x$, we can determine where the function is increasing or decreasing. It is increasing for $x < 0$ and $x > 1$ and decreasing for $0 < x < 1$.

Remember that these properties describe the general behavior of the graph; the exact appearance may vary based on the scale and size of the axes on the graph.