ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 90

\[y = tg\ x.\]

\[1)\ x \in \left\lbrack - \frac{\pi}{4};\ \frac{\pi}{3} \right\rbrack:\]

\[y\left( - \frac{\pi}{4} \right) = tg\left( - \frac{\pi}{4} \right) =\]

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

\[y\left( \frac{\pi}{3} \right) = tg\left( \frac{\pi}{3} \right) = \sqrt{3};\]

\[E(y) = \left\lbrack - 1;\ \sqrt{3} \right\rbrack.\]

\[2)\ x \in \left( \frac{3\pi}{4};\ \frac{3\pi}{2} \right):\]

\[y\left( \frac{3\pi}{4} \right) = tg\frac{3\pi}{4} = - tg\frac{\pi}{4} = - 1;\]

\[E(y) = ( - 1;\ + \infty).\]

\[3)\ x \in (0;\ \pi):\]

\[y(0) = tg\ 0 = 0;\]

\[y(\pi) = tg\ \pi = 0;\]

\[E(y) = ( - \infty;\ 0) \cup (0;\ + \infty).\]

\[4)\ x \in \left\lbrack \frac{\pi}{4};\ \frac{3\pi}{4} \right\rbrack:\]

\[y\left( \frac{\pi}{4} \right) = tg\frac{\pi}{4} = 1;\]

\[y\left( \frac{3\pi}{4} \right) = tg\frac{3\pi}{4} = - tg\frac{\pi}{4} = - 1;\]

\[E(y) = ( - \infty;\ - 1\rbrack \cup \lbrack 1;\ + \infty).\]