Question

Найдите наименьшее и наибольшее значения функции:y=2x^2-3x^2+15

Answer 1

Находим производную минимальное значение 15, а максимальное безконечность

Answer 2

To find the minimum and maximum values of the given function y = 2x^2 - 3x + 15, we first need to determine whether it opens upwards (has a minimum) or downwards (has a maximum).

The function y = 2x^2 - 3x + 15 is a quadratic function in the form of ax^2 + bx + c, where a = 2, b = -3, and c = 15. Since the coefficient of x^2 (a) is positive, the parabola opens upwards, indicating that it has a minimum value.

To find the minimum value, we can use the formula for the x-coordinate of the vertex of the parabola, which is given by -b / (2a). Once we find the x-coordinate of the vertex, we can plug it back into the original function to get the corresponding y-coordinate.

1. Finding the x-coordinate of the vertex: x_vertex = -(-3) / (2 * 2) = 3/4

2. Finding the y-coordinate of the vertex: y_vertex = 2 * (3/4)^2 - 3 * (3/4) + 15

Now, let's calculate the y-coordinate:

y_vertex = 2 * (9/16) - 9/4 + 15 y_vertex = 18/16 - 9/4 + 15 y_vertex = (18 - 36 + 60) / 16 y_vertex = 42 / 16 y_vertex = 21/8

So, the minimum value of the function y = 2x^2 - 3x + 15 is 21/8.

Since the parabola opens upwards, there is no maximum value; the function goes on indefinitely in the positive y-direction.