Question

Y=(4x-1)/(x^2+3) [-1;3]

Answer 1

Пересечение с осью ординат (0;- 1/3)
корень (1/4;0)
область определения x принадлежит R

Answer 2

To find the maximum and minimum values of the function Y=(4x-1)/(x^2+3) on the interval [-1, 3], we need to follow these steps:

1. Find the critical points of the function on the interval [-1, 3]. These are the points where the derivative of the function is zero or undefined.

2. Evaluate the function at the critical points and at the endpoints of the interval.

3. Compare the values obtained in step 2 to determine the maximum and minimum values.

Let's start by finding the derivative of the function:

Y' = [(4(x^2 + 3) - (4x - 1)(2x))/ (x^2 + 3)^2]

Simplifying the numerator:

Y' = [(8x^2 + 12 - 8x^2 + 2x)/ (x^2 + 3)^2]

Y' = [2x + 12/ (x^2 + 3)^2]

The critical points are the solutions to the equation Y' = 0 or Y' undefined.

Y' = [2x + 12/ (x^2 + 3)^2] = 0

2x + 12 = 0

x = -6/2 = -3 (rejected because it's outside the interval [-1, 3])

The only critical point within the interval is when the denominator is 0:

x^2 + 3 = 0

x^2 = -3 (rejected because it has no real solutions)

Therefore, there are no critical points within the interval [-1, 3].

Next, let's evaluate the function at the endpoints of the interval:

Y(-1) = (4(-1) - 1)/((-1)^2 + 3) = -5/4 Y(3) = (4(3) - 1)/(3^2 + 3) = 11/12

Finally, we compare the values obtained to determine the maximum and minimum values:

The maximum value is 11/12 at x=3, and the minimum value is -5/4 at x=-1.

Therefore, the maximum value of Y on the interval [-1, 3] is 11/12, and the minimum value is -5/4.