Question

y=log3(19-8x-2x^) найти наибольшее значение функции

Answer 1

$y=log_3(19-8x-2x^2), \ 19-8x-2x^2 textgreater 0, \ 2x^2+8x-19 textless 0, \ 2x^2+8x-19=0, \ D_1=4^2-2cdot(-19)=16+38=54, \ x=frac{-4pm3sqrt{6}}{2}, \ xin(-2-frac{3sqrt{6}}{2};-2+frac{3sqrt{6}}{2}), \ y'=(log_3(19-8x-2x^2))'=frac{1}{(19-8x-2x^2)ln3}cdot(19-8x-2x^2)'=\=frac{-8-4x}{(19-8x-2x^2)ln3}, \ y'=0, -8-4x=0, x=-2, \ 19-8x-2x^2 textgreater 0, ln3 textgreater 0, \ x textless -2, y' textgreater 0, ynearrow, \ x textgreater -2, y' textless 0, ysearrow, \ x_{max}=-2, y_{max}=log_3(19-8cdot(-2)-2cdot(-2)^2)=\=log_3 27= 3$