(I can see no reason at all for taking the second derivative.)
f(x) = x/(x^2+4)
0 = plus or minus 2
2 is in the period.
Therefore, you test on either side of 2.
I chose 1 and 3 for test values.
At 1, the slope of the tangent line to the function is positive, meaning that the function is increasing there. (You can draw this out and see it. You might already know that. :P)
At 3, f'(x) (the slope of the tangent line to the function) is negative, meaning that the function is decreasing there.
Therefore, there is a max at 2. There is no min.
If you plug 2 back into the original function...
y = x/(x^2+4)
y = x/(x^2+4) = 2/8 = .25
Therefore, there is a min in the period at (2, .25).