Question

Найдите наибольшее и наименьшее значение функции y=3+4x³+1 на промежутке [-2;1]Нужно подробное решение

Answer 1

1)Находим производную
y' = 12x^3 + 12x^2
3)Находим стационарные точки
 12x^3 + 12x^2 = 0
x^3 + x^2 = 0
x^2(x + 1) = 0
x = 0            или                  x = -1
Отрезку [-2;1] принадлежат обе точки.
Вычислим значения самой функции в них, а также в концах отрезка.
y(0) = 0 + 0 + 1 = 1
y(-1) = 3 - 4 + 1 = 0
y(-2) = 48 - 32 + 1 = 16 + 1 = 17
y(1) = 3 + 4 + 1 = 8
И теперь выберем среди полученных значений функции наибольшее и наименьшее.
Таким образом, наибольшее значение функции равно 17, а наименьшее значение функции на этом отрезке равно 0.

Answer 2

Finding the Maximum and Minimum Values of the Function y=3+4x³+1 on the Interval [-2;1]

To find the maximum and minimum values of the function y=3+4x³+1 on the interval [-2;1], we can use the first and second derivative tests to locate the critical points and determine whether they correspond to maximum or minimum values.

1. Finding the Critical Points: - To find the critical points, we need to take the derivative of the function and set it equal to zero to solve for x.

- The derivative of the function y=3+4x³+1 is y' = 12x². - Setting y' equal to zero gives us 12x² = 0, which implies x = 0.

- Therefore, the critical point is x = 0.

2. Determining the Nature of the Critical Point: - To determine whether the critical point corresponds to a maximum or minimum value, we can use the second derivative test.

- The second derivative of the function y=3+4x³+1 is y'' = 24x.

- Substituting x = 0 into the second derivative gives us y''(0) = 0.

- Since the second derivative test is inconclusive at x = 0, we need to consider the behavior of the function around this point.

3. Evaluating the Function at the Endpoints of the Interval: - We also need to evaluate the function at the endpoints of the interval, which are x = -2 and x = 1.

- When x = -2, y = 3+4(-2)³+1 = 3-32+1 = -28.

- When x = 1, y = 3+4(1)³+1 = 3+4+1 = 8.

- Therefore, the function takes the value -28 at x = -2 and the value 8 at x = 1.

4. Conclusion: - The function y=3+4x³+1 has a maximum value of 8 at x = 1 and a minimum value of -28 at x = -2 on the interval [-2;1].

- Therefore, the maximum value of the function is 8, and the minimum value is -28.

Summary:

The maximum value of the function y=3+4x³+1 on the interval [-2;1] is 8, and the minimum value is -28.