найти точку максимума функции y=(11+4x-)-2

$y = log_3(11 + 4x - x^2) - 2\\\\ (f(g(x)))' = \frac{dg(x)}{dx}\frac{df(g(x))}{dg(x)} = g'(x)\frac{df(g(x))}{dg(x)}\\\\ y' = (4 - 2x) \frac{1}{(11 + 4x - x^2)ln(3)}\\\\ (4 - 2x) \frac{1}{(11 + 4x - x^2)ln(3)} = 0,\\\\ 11 + 4x - x^2 \ne 0,\\\\ 4 - 2x = 0, -2x = -4, \ \underline{x = 2}\\\\ 11 + 4x - x^2 = 0, \\\\\ D = 16 + 44 = 60\\\\ x_1 = -\frac{-4 - \sqrt{60}}{2} = 2 + \sqrt{15}\\\\ x_2 = -\frac{-4 + \sqrt{60}}{2} = 2 - \sqrt{15}$