The max value can be found by setting the derivative equal to 0 and solving for x, then subbing that value for x back into the original function and solving for y. [tex]f'(x)= \frac{(x^2+3)(0)-2(2x)}{(x^2+3)^2} [/tex] which simplifies to [tex]f'(x)= \frac{-4x}{(x^2+3)^2} [/tex]. This derivative is equal to 0 when -4x is equal to 0. If -4x = 0, then x = 0. If we find f(0), then [tex]f(0)= \frac{2}{(0)^2+3} [/tex] and y here is 2/3. So the max value occurs at (0, 2/3). There you go!
