ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1016

\[1)\ y = \sqrt{x^{2} - 9x - 10}\]

\[y^{'}(x) = \frac{\left( x^{2} \right)^{'} - (9x)^{'} - 10^{'}}{2\sqrt{x^{2} - 9x - 10}} =\]

\[= \frac{2x - 9}{2\sqrt{x^{2} - 9x - 10}} \geq 0;\]

\[2x - 9 \geq 0\]

\[2x \geq 9\]

\[x \geq 4,5.\]

\[Область\ определения:\]

\[x^{2} - 9x - 10 \geq 0\]

\[D = 81 + 40 = 121\]

\[x_{1} = \frac{9 - 11}{2} = - 1;\]

\[x_{2} = \frac{9 + 11}{2} = 10;\]

\[(x + 1)(x - 10) \geq 0\]

\[x \leq - 1;\ \ \ x \geq 10.\]

\[Ответ:\ \ x \in \lbrack 10;\ + \infty).\]

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

\[y^{'}(x) = \frac{7^{'} - (6x)^{'} - \left( x^{2} \right)^{'}}{2\sqrt{7 - 6x - x^{2}}} =\]

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

\[- 6 - 2x \geq 0\]

\[2x \leq - 6\]

\[x \leq - 3.\]

\[Область\ определения:\]

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

\[x^{2} + 6x - 7 \leq 0\]

\[D = 36 + 28 = 64\]

\[x_{1} = \frac{- 6 - 8}{2} = - 7;\]

\[x_{2} = \frac{- 6 + 8}{2} = 1;\]

\[(x + 7)(x - 1) \leq 0\]

\[- 7 \leq x \leq 1.\]

\[Ответ:\ \ x \in \lbrack - 7;\ - 3\rbrack.\]

\[3)\ y = \frac{x - 3}{x - 4}\]

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

\[= \frac{(x - 3)^{'}(x - 4) - (x - 3)(x - 4)^{'}}{(x - 4)^{2}} =\]

\[= \frac{(x - 4) - (x - 3)}{(x - 4)^{2}} =\]

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

\[Ответ:\ \ x \in \varnothing.\]

\[4)\ y = \frac{x - 5}{x - 6}\]

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

\[= \frac{(x - 5)^{'}(x - 6) - (x - 5)(x - 6)^{'}}{(x - 6)^{2}} =\]

\[= \frac{(x - 6) - (x - 5)}{(x - 6)^{2}} =\]

\[= \frac{- 1}{(x - 6)^{2}} < 0.\]

\[Ответ:\ \ x \in \varnothing.\]

\[5)\ y = \frac{1}{x^{2} - 4} = \left( x^{2} - 4 \right)^{- 1}\]

\[y^{'}(x) = - \left( x^{2} - 4 \right)^{- 2} \bullet 2x =\]

\[= - \frac{2x}{\left( x^{2} - 4 \right)^{2}} \geq 0;\]

\[- 2x \geq 0\]

\[x \leq 0.\]

\[Область\ определения:\]

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

\[x^{2} \neq 4\]

\[x \neq \pm 2.\]

\[Ответ:\ \ x \in ( - \infty;\ - 2) \cup ( - 2;\ 0\rbrack.\]

\[6)\ y = \frac{3}{x^{2} + 3x - 4} =\]

\[= 3\left( x^{2} + 3x - 4 \right)^{- 1}\]

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

\[= - 3\left( x^{2} + 3x - 4 \right)^{- 2} \bullet (2x + 3) =\]

\[= \frac{- 3(2x + 3)}{\left( x^{2} + 3x - 4 \right)^{2}} \geq 0;\]

\[2x + 3 \leq 0\]

\[2x \leq - 3\]

\[x \leq - 1,5.\]

\[Область\ определения:\]

\[x^{2} + 3x - 4 \neq 0\]

\[D = 9 + 16 = 25\]

\[x_{1} \neq \frac{- 3 - 5}{2} = - 4;\]

\[x_{2} \neq \frac{- 3 + 5}{2} = 1.\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 4) \cup ( - 4;\ - 1,5\rbrack.\]