Question

Приведите решение пожалуйста

Answer 1

$y=2sqrt2cosx+frac{24}{pi}{x+13; ,; ; [frac{3pi}{4};2pi ]$

$y'=-2sqrt2sinx+frac{24}{pi }=0$

$sinx=frac{24}{pi }cdot frac{1}{2sqrt2}=frac{6sqrt2}{pi } textgreater 1; ; to ; net; reshenij,; net; extremymov\\y(frac{3pi}{4})=2sqrt2cosfrac{3pi}{4}+frac{24}{pi}cdot frac{3pi}{4}+13=\\=2sqrt2cdot (-frac{sqrt2}{2})+18+13=2+18+13=33\\y(2pi )=2sqrt2cdot cos2pi +frac{24}{pi }cdot 2pi +13=2sqrt2+48+13=2sqrt2+61\\y_{naibol}=2sqrt2+61\\y_{naimen}=33$