Question

Y=x^3-2x^2+3x-1 исследовать, построить график функции

Answer 1

ДАНО
Y(x) = x³-2x²+3x-1
ИССЛЕДОВАНИЕ.

1.Область определения D(x) - Х∈(-∞;+∞) - непрерывная.

2. Пересечение с осью Х. Y=0 при х = 3. 

3. Пересечение с осью У.  У(0) = ?. 

4. Поведение на бесконечности.limY(-∞) = - ∞  limY(+∞) = +∞ 

5. Исследование на чётность.Y(-x)  ≠ Y(x).

Функция ни чётная ни нечётная. 

6. Производная функции.Y'(x)= 3x²-4x+3. 

Корней нет.

7. Локальные экстремумы - нет.

8. Интервалы возрастания и убывания. Возрастает - Х∈(-∞;+∞)∪

9. Вторая производная - Y"(x) = 6x - 4 =0. 

Корень производной - точка перегиба Y"(x)= 2/3 ≈ 0.66. 

10. Выпуклая “горка» Х∈(-∞;2/3), Вогнутая – «ложка» Х∈(2/3;+∞).

11. График в приложении.

Answer 2

Function Analysis

To analyze the function Y = x^3 - 2x^2 + 3x - 1, we will examine its properties such as the domain, range, symmetry, intercepts, and critical points.

Domain and Range: The domain of a polynomial function is all real numbers. Therefore, the domain of this function is (-∞, ∞). The range of a cubic function can be any real number, so the range of this function is also (-∞, ∞).

Symmetry: To determine the symmetry of the function, we can check if it is an even or odd function. An even function satisfies the condition f(-x) = f(x), while an odd function satisfies the condition f(-x) = -f(x). Let's substitute -x into the function and see if it satisfies either condition:

f(-x) = (-x)^3 - 2(-x)^2 + 3(-x) - 1 = -x^3 - 2x^2 - 3x - 1

Since f(-x) ≠ f(x) and f(-x) ≠ -f(x), the function is neither even nor odd. Therefore, it does not exhibit any symmetry.

Intercepts: To find the x-intercepts, we set Y = 0 and solve for x:

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

Unfortunately, there is no simple algebraic solution to this equation. We can use numerical methods or graphing to approximate the x-intercepts.

To find the y-intercept, we substitute x = 0 into the function:

Y = (0)^3 - 2(0)^2 + 3(0) - 1 = -1

Therefore, the y-intercept is (0, -1).

Critical Points: To find the critical points, we need to find the values of x where the derivative of the function is equal to zero or undefined. Let's find the derivative of the function:

Y' = 3x^2 - 4x + 3

To find the critical points, we set Y' = 0 and solve for x:

3x^2 - 4x + 3 = 0

Again, there is no simple algebraic solution to this equation. We can use numerical methods or graphing to approximate the critical points.

Graph of the Function

To visualize the graph of the function Y = x^3 - 2x^2 + 3x - 1, we can plot some key points and sketch the curve.

Based on the analysis above, we know that the function has no symmetry, the y-intercept is at (0, -1), and there may be x-intercepts and critical points that need to be approximated.

Unfortunately, without the specific values of the x-intercepts and critical points, we cannot accurately plot the graph. However, we can use graphing software or online graphing tools to plot the function and obtain a more precise graph.

Please note that the graph of the function may have multiple turning points, depending on the values of the x-intercepts and critical points.