Question

Please help precalculus

Answer 1

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.  $f'(x)= \frac{(x^2+3)(0)-2(2x)}{(x^2+3)^2}$ which simplifies to $f'(x)= \frac{-4x}{(x^2+3)^2}$.  This derivative is equal to 0 when -4x is equal to 0.  If -4x = 0, then x = 0.  If we find f(0), then $f(0)= \frac{2}{(0)^2+3}$ and y here is 2/3.  So the max value occurs at (0, 2/3).  There you go!