Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1036

\[1)\ y = \sqrt{6x - 7} - 2x;\]

\[y^{'}(x) = \frac{6}{2\sqrt{6x - 7}} - 2 \geq 0;\]

\[3 - 2\sqrt{6x - 7} \geq 0\]

\[2\sqrt{6x - 7} \leq 3\]

\[4(6x - 7) \leq 9\]

\[24x - 28 \leq 9\]

\[24x \leq 37\]

\[x \leq \frac{37}{24}.\]

\[y\left( \frac{37}{24} \right) = \sqrt{\frac{37}{4} - 7} - \frac{37}{12} =\]

\[= \sqrt{\frac{9}{4}} - \frac{37}{12} = \frac{3}{2} - \frac{37}{12} =\]

\[= \frac{18 - 37}{12} = - \frac{19}{12}.\]

\[Ответ:\ \ E(y) = \left( - \infty;\ - \frac{19}{12} \right\rbrack.\]

\[2)\ y = \sqrt{x^{2} - 4x - 5};\]

\[y^{'}(x) = \frac{2x - 4}{2\sqrt{x^{2} - 4x - 5}} \geq 0;\]

\[2x - 4 \geq 0\]

\[2x \geq 4\]

\[x \geq 2\]

\[y(2) = \sqrt{4 - 8 - 5} = \sqrt{- 9}.\]

\[Ответ:\ \ E(y) = \lbrack 0;\ + \infty).\]