To find the maximum and minimum values of the function Y=(4x-1)/(x^2+3) on the interval [-1, 3], we first need to find the critical points of the function.

To find the critical points, we first need to find the derivative of the function:

Y' = [(4)(x^2 + 3) - (4x - 1)(2x)] / (x^2 + 3)^2
Y' = (4x^2 + 12 - 8x^2 + 2x) / (x^2 + 3)^2
Y' = (-4x^2 + 2x + 12) / (x^2 + 3)^2

Next, we need to set the derivative equal to 0 and solve for x to find the critical points:

-4x^2 + 2x + 12 = 0
4x^2 - 2x - 12 = 0
2x^2 - x - 6 = 0
(2x + 3)(x - 2) = 0

So, x = -3/2 or x = 2.

Now we need to evaluate the function at the critical points and at the endpoints of the interval [-1, 3]:

Y(-1) = (4(-1) - 1) / ((-1)^2 + 3) = (-4 - 1) / (1 + 3) = -5/4
Y(-3/2) = (4(-3/2) - 1) / ((-3/2)^2 + 3) = (-6 - 1)/((9/4) + 3) = -7/(36/4) = -7/9
Y(2) = (4(2) - 1) / (2^2 + 3) = (8 - 1) / 7 = 7/7 = 1
Y(3) = (4(3) - 1) / (3^2 + 3) = (12 - 1) / 12 = 11/12

Now we compare the values we found to determine the maximum and minimum values on the interval:

The minimum value is -7/9 and the maximum value is 1.