+400 If $y = \frac{x}{x^2 + 1},$ and $x$ is any real number, then what is the sum of the maximum and minimum possible values of $y \, ?$
+129850 y = x (x^2 +1)^(-1) take the derivative and set to 0
y' = (x^2 + 1)^(-1) - 2x*x (x^2+ 1)^(-2) =0
( x^2 + 1)(-2) ( x^2 + 1 - 2x^2) = 0
-x^2 + 1 = 0
x^2 -1 = 0
(x - 1) ( x + 1) = 0
x -1 = 0
=1 y = 1 / (1^2 + 1) = 1/2 = max y
x + 1 = 0
x = -1 y = -1 / ((-1)^2 + 1) = -1/2 = min y
The sum of max + min y's = 0
