Question

Найти наибольшее и наименьшее значения функции y=x^3+3/x на отрезке [1/2;2]

Answer 1

Ответ:

Объяснение:

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Answer 2

Finding the Maximum and Minimum Values of the Function

To find the maximum and minimum values of the function y = x^3 + 3/x on the interval [1/2, 2], we can use calculus to find the critical points and then test the values to determine the maximum and minimum.

First, let's find the derivative of the function to locate the critical points.

The derivative of the function y = x^3 + 3/x is given by: y' = 3x^2 - 3/x^2

Now, let's find the critical points by setting the derivative equal to zero and solving for x:

3x^2 - 3/x^2 = 0

Solving for x, we get: x^4 = 1 x = ±1

Since the interval is [1/2, 2], we only consider the critical point x = 1.

Next, we need to test the critical point and the endpoints of the interval to determine the maximum and minimum values.

When x = 1, y = 1^3 + 3/1 = 4.

When x = 1/2, y = (1/2)^3 + 3/(1/2) = 1/8 + 6 = 49/8.

When x = 2, y = 2^3 + 3/2 = 11.

Therefore, the maximum value of the function on the interval [1/2, 2] is 11 and the minimum value is 49/8.

This information is based on the mathematical calculations and analysis of the given function and interval.