ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 215

\[1)\ f(x) = x^{2} - 6x - 8\ln x;\]

\[f^{'}(x) = 2x - 6 - \frac{8}{x} = 0\]

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

\[D = 9 + 16 = 25\]

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

\[x_{2} = \frac{3 + 5}{2} = 4.\]

\[Ответ:\ \ 4.\]

\[2)\ f(x) = 2\sqrt{x} - 3\ln(x + 2);\]

\[f^{'}(x) = 2 \bullet \frac{1}{2\sqrt{x}} - 3 \bullet \frac{1}{x + 2} = 0\]

\[x + 2 - 3\sqrt{x} = 0\]

\[x - 3\sqrt{x} + 2 = 0\]

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

\[\sqrt{x_{1}} = \frac{3 - 1}{2} = 1;\text{\ \ }\]

\[\sqrt{x_{2}} = \frac{3 + 1}{2} = 2;\]

\[x_{1} = 1^{2} = 1;\ x_{2} = 2^{2} = 4.\]

\[Ответ:\ \ 1;\ 4.\]

\[3)\ f(x) = \sqrt{x + 1} - \ln(x - 2);\]

\[f^{'}(x) = \frac{1}{2\sqrt{x + 1}} - \frac{1}{x - 2} = 0\]

\[x - 2 - 2\sqrt{x + 1} = 0\]

\[(x + 1) - 2\sqrt{x + 1} - 3 = 0;\]

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

\[\sqrt{x_{1} + 1} = \frac{2 - 4}{2} = - 1;\text{\ \ }\]

\[\sqrt{x_{2} + 1} = \frac{2 + 4}{2} = 3;\]

\[x + 1 = 9\ \ \ \]

\[x = 8.\]

\[Ответ:\ \ 8.\]

\[4)\ f(x) = \ln(x - 1) + 2\ln(x + 2);\]

\[f^{'}(x) = \frac{1}{x - 1} + \frac{2}{x + 2} = 0\]

\[x + 2 + 2(x - 1) = 0\]

\[3x = 0\]

\[x = 0.\]

\[Ответ:\ \ корней\ нет.\]