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)= ((x^2+3)(0)-2(2x))/((x^2+3)^2)` which simplifies to
`f'(x)= (-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)= (2)/((0)^2+3)` and y here is 2/3. So the max value occurs at (0, 2/3). There you go!