Question

Дослідіть функцію у = х³ - 3х² +2 за алгоритмом (1. Область визначення; 2. Парність, непарність; 3. Нулі функції; 4. Проміжки зростання, спадання; 5. Точки екстремуму та екстремуми функції. 6. Інші особливості) та побудуйте її графік

Answer 1

Відповідь: на фото

Answer 2

Ответ:

Во вложениях

Объяснение:

Во вложениях

Answer 3

Function Analysis: у = х³ - 3х² + 2

To analyze the function у = х³ - 3х² + 2, we will follow the given algorithm and examine the following aspects:

1. Domain of the Function 2. Parity 3. Zeros of the Function 4. Intervals of Increase and Decrease 5. Extremum Points and Extrema 6. Other Characteristics

Let's go through each of these aspects step by step.

1. Domain of the Function

The domain of a function refers to the set of all possible input values for which the function is defined. In this case, since there are no restrictions mentioned, the domain of the function у = х³ - 3х² + 2 is all real numbers.

2. Parity

The parity of a function determines whether it is even, odd, or neither.

To determine the parity of the function у = х³ - 3х² + 2, we need to check if it satisfies the following conditions:

- If у(-х) = у(х), the function is even. - If у(-х) = -у(х), the function is odd.

Let's substitute -х into the function and check if it satisfies either of the conditions:

у(-х) = (-х)³ - 3(-х)² + 2 = -х³ - 3х² + 2

Since у(-х) is not equal to у(х) or -у(х), the function у = х³ - 3х² + 2 is neither even nor odd.

3. Zeros of the Function

The zeros of a function are the values of x for which the function equals zero. To find the zeros of the function у = х³ - 3х² + 2, we need to solve the equation у = 0.

х³ - 3х² + 2 = 0

Unfortunately, there is no simple algebraic solution to this equation. We can use numerical methods or graphing to approximate the zeros of the function.

4. Intervals of Increase and Decrease

To determine the intervals of increase and decrease, we need to find the critical points of the function. Critical points occur where the derivative of the function is either zero or undefined.

The derivative of у = х³ - 3х² + 2 is given by:

у' = 3х² - 6х

Setting the derivative equal to zero and solving for x, we get:

3х² - 6х = 0 3х(х - 2) = 0

From this equation, we find two critical points: x = 0 and x = 2.

To determine the intervals of increase and decrease, we can use a sign chart or test points in each interval.

For x < 0, we can choose x = -1 as a test point: у'(-1) = 3(-1)² - 6(-1) = 3 + 6 = 9 > 0

For 0 < x < 2, we can choose x = 1 as a test point: у'(1) = 3(1)² - 6(1) = 3 - 6 = -3 < 0

For x > 2, we can choose x = 3 as a test point: у'(3) = 3(3)² - 6(3) = 27 - 18 = 9 > 0

From this analysis, we can conclude that the function у = х³ - 3х² + 2 is increasing on the interval (-∞, 0) and (2, +∞), and decreasing on the interval (0, 2).

5. Extremum Points and Extrema

To find the extremum points and extrema of the function, we need to examine the behavior of the function at the critical points and the endpoints of the intervals.

At x = 0, we have: у(0) = 0³ - 3(0)² + 2 = 2

At x = 2, we have: у(2) = 2³ - 3(2)² + 2 = 0

Therefore, the function у = х³ - 3х² + 2 has a minimum point at (0, 2) and a maximum point at (2, 0).

6. Other Characteristics

- The function у = х³ - 3х² + 2 is continuous and smooth, as it is a polynomial function. - The graph of the function will be symmetric with respect to the y-axis due to the absence of odd powers of x. - The graph will approach positive infinity as x approaches positive or negative infinity. - The graph will intersect the x-axis at the zeros of the function, which we were unable to determine algebraically.

Graph of the Function

Unfortunately, I am unable to generate a graph here. However, you can use graphing software or online graphing tools to plot the function у = х³ - 3х² + 2 and visualize its shape based on the information provided.

I hope this analysis helps you understand the function у = х³ - 3х² + 2 better. Let me know if you have any further questions!