ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 230

\[1)\ y = 8 - x;\ \ \ y = 4\sqrt{x + 4}:\]

\[8 - x = 4\sqrt{x + 4}\]

\[64 - 16x + x^{2} = 16(x + 4)\]

\[64 - 16x + x^{2} = 16x + 64\]

\[x^{2} - 32x = 0\]

\[x(x - 32) = 0\]

\[x_{1} = 0;\text{\ \ \ }\]

\[x_{2} = 32.\]

\[8 - x > 0;\ \ \ x < 8;\]

\[4 + x > 0;\text{\ \ \ x} > - 4.\]

\[Угол\ между\ кривыми:\]

\[y^{'}(0) = - 1;\]

\[y^{'}(0) = \frac{4}{2\sqrt{4 + x}} = \frac{2}{\sqrt{4}} = 1;\]

\[a = - \frac{\pi}{4};\text{\ \ \ }\]

\[b = \frac{\pi}{4};\]

\[c = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]

\[2)\ y = \frac{1}{2}(1 + x)^{2};\ y = \frac{1}{2}(1 - x)^{2}:\]

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

\[(1 + x)^{2} = (1 - x)^{2}\]

\[1 + 2x + x^{2} = 1 - 2x + x^{2}\]

\[4x = 0\]

\[x = 0.\]

\[Угол\ между\ кривыми:\]

\[y^{'}(0) = \frac{1}{2} \bullet 2(1 + x) = 1;\]

\[y^{'}(0) = - \frac{1}{2} \bullet 2(1 - x) = - 1;\]

\[a = \frac{\pi}{4};\text{\ \ \ }\]

\[b = - \frac{\pi}{4};\]

\[c = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]

\[3)\ y = \ln(1 + x);\ y = \ln(1 - x):\]

\[\ln(1 + x) = \ln(1 - x)\]

\[1 + x = 1 - x\]

\[2x = 0\]

\[x = 0.\]

\[Угол\ между\ кривыми:\]

\[y^{'}(0) = \frac{1}{1 + x} = \frac{1}{1} = 1;\]

\[y^{'}(0) = - \frac{1}{1 - x} = - \frac{1}{1} = - 1;\]

\[a = \frac{\pi}{4};\text{\ \ \ }\]

\[b = - \frac{\pi}{4};\]

\[c = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]

\[4)\ y = e^{x};\ \ \ y = e^{- x}:\]

\[e^{x} = e^{- x}\]

\[x = - x\]

\[2x = 0\]

\[x = 0.\]

\[Угол\ между\ кривыми:\]

\[y^{'}(0) = e^{x} = e^{0} = 1;\]

\[y^{'}(0) = - e^{- x} = - e^{0} = - 1;\]

\[a = \frac{\pi}{4};\text{\ \ \ }\]

\[b = - \frac{\pi}{4};\]

\[c = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]