ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 147

\[1)\ y = 2^{\frac{x^{2} - 4}{x + 2}};\ \ A(2;\ 1):\]

\[y(2) = 2^{\frac{4 - 4}{2 + 2}} = 2^{0} = 1.\]

\[Ответ:\ \ да.\]

\[2)\ y = tg\left( \frac{x}{3} + \frac{\pi}{4} \right);\ A\left( \frac{3\pi}{4};\ 0 \right):\]

\[y\left( \frac{3\pi}{4} \right) = tg\left( \frac{\pi}{4} + \frac{\pi}{4} \right) = tg\frac{\pi}{2}.\]

\[Ответ:\ \ нет.\]

\[3)\ y = 3^{\frac{x - \sqrt{2}}{x^{2} - 2}};\ \ \ A\left( - \sqrt{2};\ 1 \right):\]

\[y\left( - \sqrt{2} \right) = 3^{\frac{- \sqrt{2} - \sqrt{2}}{2 - 2}} = 3^{- \frac{2\sqrt{2}}{0}}.\]

\[Ответ:\ \ нет.\]

\[4)\ y = ctg\left( \frac{2x}{3} - \frac{\pi}{6} \right);\]

\[A\left( \frac{3\pi}{4};\ \frac{\sqrt{3}}{3} \right):\]

\[y\left( \frac{3\pi}{4} \right) = ctg\left( \frac{\pi}{2} - \frac{\pi}{6} \right) =\]

\[= ctg\frac{\pi}{3} = \frac{\sqrt{3}}{3}.\]

\[Ответ:\ \ да.\]