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

[latex]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=4^2-2\cdot(-19)=16+38=54, \\ x=\frac{-4\pm3\sqrt{6}}{2}, \\ x\in(-2-\frac{3\sqrt{6}}{2};-2+\frac{3\sqrt{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, \ y\nearrow, \\ x\ \textgreater \ -2, \ y'\ \textless \ 0, \ y\searrow, \\ x_{max}=-2, \ y_{max}=\log_3(19-8\cdot(-2)-2\cdot(-2)^2)=\\=\log_3 27= 3[/latex]