Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 867

\[\boxed{\mathbf{867}\mathbf{.}}\]

\[y = \frac{x + 2}{x - 2}\ \ и\ \ a = - \frac{\pi}{4}:\]

\[k = tg\ a = - tg\frac{\pi}{4} = - 1\]

\[y^{'}(x) =\]

\[= \frac{(x + 2)^{'} \bullet (x - 2) - (x + 2) \bullet (x - 2)'}{(x - 2)^{2}} =\]

\[= \frac{1 \bullet (x - 2) - (x + 2) \bullet 1}{(x - 2)^{2}} =\]

\[= \frac{x - 2 - x - 2}{(x - 2)^{2}} =\]

\[= - \frac{4}{(x - 2)^{2}}.\]

\[k = - \frac{4}{(x - 2)^{2}} = - 1\]

\[\frac{4}{(x - 2)^{2}} = 1\]

\[4 = (x - 2)^{2}\]

\[x^{2} - 4x + 4 - 4 = 0\]

\[x^{2} - 4x = 0\]

\[x(x - 4) = 0\]

\[x = 0\ или\ x = 4.\]

\[f(0) = \frac{0 + 2}{0 - 2} = \frac{2}{- 2} = - 1\]

\[f(4) = \frac{4 + 2}{4 - 2} = \frac{6}{2} = 3.\]

\[Ответ:\ \ (0\ - 1)\text{\ \ }(4\ 3).\]