Question

Функция y=4x²+ 23 на отрезке [-2006; 2006] имеет наименьшее значение при х, равном..​

Answer 1

Ответ:

Объяснение:

Х=0

Answer 2

To find the value of x at which the function y = 4x² + 23 has the smallest value on the interval [-2006, 2006], we can use calculus. The minimum or maximum of a function occurs at critical points, where the derivative of the function is equal to zero or does not exist. In this case, we have a quadratic function, and its minimum occurs at the vertex of the parabola.

The equation y = 4x² + 23 represents a parabola that opens upwards because the coefficient of x² is positive (4 > 0). The x-coordinate of the vertex of the parabola can be found using the formula:

x_vertex = -b / 2a,

where a and b are the coefficients of x² and x, respectively. In this case, a = 4 and b = 0, as there is no linear term in the function.

x_vertex = -0 / 2(4) = 0.

So, the x-coordinate of the vertex is 0. However, we need to check if this value falls within the given interval [-2006, 2006].

Since 0 is within the interval [-2006, 2006], the minimum value of the function occurs at x = 0. Now, let's find the corresponding y-value:

y = 4(0)² + 23 = 0 + 23 = 23.

Therefore, the function y = 4x² + 23 has its minimum value of 23 at x = 0 on the interval [-2006, 2006].