Question

locate the absolute extrema of y=(2x)/(x2+1) on the closed interval [-2,2]. Please show the work so i can undertand. I will give 80 points

Answer 1

The absolute extrema of the function y=(2x)/(x^2+1) on the closed interval [-2, 2] are 0.8 (absolute maximum) at x=2 and -0.8 (absolute minimum) at x=-2.

Step-by-step explanation:

To locate the absolute extrema of the function y = (2x)/(x^2+1) on the closed interval [-2, 2], we need to find the highest and lowest points on the graph within this interval.

We can start by finding the critical points of the function, which are the points where the derivative equals zero or does not exist.

Taking the derivative of the function and setting it equal to zero, we find that the critical point is x=0.

We also need to check the endpoints of the interval.

Evaluating the function at x=-2, 0, 2, we find y(-2) = -0.8, y(0) = 0, and y(2) = 0.8.

Therefore, the absolute maximum value is 0.8 and it occurs at x=2, while the absolute minimum value is -0.8 and it occurs at x=-2.