ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 149

\[1)\ y = x^{2} - 3x - 4\]

\[x_{0} = - \frac{- 3}{2 \bullet 1} = \frac{3}{2};\]

\[y_{0} = \frac{9}{4} - \frac{9}{2} - 4 = - \frac{25}{4}.\]

\[D(x) = R;\ \ \ \]

\[E(y) = \left\lbrack - \frac{25}{4};\ + \infty \right).\]

\[2)\ y = 3 - 2x - x^{2}\]

\[x_{0} = - \frac{- 2}{2 \bullet ( - 1)} = - 1;\]

\[y_{0} = 3 + 2 - 1 = 4.\]

\[D(x) = R;\ \ \ \]

\[E(y) = ( - \infty;\ 4\rbrack.\]

\[3)\ y = \frac{1}{x - 1}\]

\[\lim_{x \rightarrow \infty}\frac{1}{x - 1} = \lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{1 - \frac{1}{x}\ } =\]

\[= \frac{0}{1 - 0} = 0;\]

\[x - 1 \neq 0\]

\[x \neq 1.\]

\[D(x) = ( - \infty;\ 1) \cup (1;\ + \infty);\]

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

\[4)\ y = \frac{2 + x}{x + 1}\]

\[\lim_{x \rightarrow \infty}\frac{2 + x}{x + 1} = \lim_{x \rightarrow \infty}\frac{\frac{2}{x} + 1}{1 + \frac{1}{x}} =\]

\[= \frac{0 + 1}{1 + 0} = 1.\]

\[x + 1 \neq 0\]

\[x \neq - 1.\]

\[D(x) = ( - \infty;\ - 1) \cup ( - 1;\ + \infty);\]

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