Question

помогите пожалуйста
(на луче [0;+∞))​

Answer 1

$y = \frac{x}{16+x^2}\\ y' = \frac{16+x^2 - x*2x}{(x^2+16)^2} = \frac{16-x^2}{(x^2+16)^2}\\y' = 0 => \frac{16-x^2}{(x^2+16)^2} = 0 => 16-x^2 = 0 => x = \pm4 => x = 4\\y(4) = y_{max} = \frac{4}{32} = \frac{1}{8} \\ Answer: \frac{1}{8}$