Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 864

\[\boxed{\mathbf{864}\mathbf{.}}\]

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

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

\[64 - 16x + x^{2} - 16(x + 4) = 0\]

\[64 - 16x + x^{2} - 16x - 64 = 0\]

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

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

\[x_{1} = 0\ \ и\ \ x_{2} = 32.\]

\[Выражение\ имеет\ смысл:\]

\[x + 4 \geq 0\]

\[x \geq - 4\]

\[8 - x \geq 0\]

\[x \leq 8.\]

\[1)\ y^{'}(x) = (8 - x)^{'} = - 1\]

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

\[a = - arctg\ 1 = - \frac{\pi}{4}.\]

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

\[= 4 \bullet \frac{1}{2} \bullet (x + 4)^{- \frac{1}{2}} = \frac{2}{\sqrt{x + 4}}\]

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

\[= \frac{2}{\sqrt{4}} = \frac{2}{2} = 1\]

\[a = atctg\ 1 = \frac{\pi}{4}.\]

\[3)\ \beta = \frac{\pi}{4} - \left( - \frac{\pi}{4} \right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]

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

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

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

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

\[4x = 0\]

\[x = 0.\]

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

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

\[k = y^{'}(0) = 0 + 1 = 1\]

\[a = arctg\ 1 = \frac{\pi}{4}.\]

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

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

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

\[a = - arctg\ 1 = - \frac{\pi}{4}.\]

\[3)\ \beta = \frac{\pi}{4} - \left( - \frac{\pi}{4} \right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]

\[3)\ y = \ln(1 + x)\text{\ \ }y = \ln(1 - x)\]

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

\[1 + x = 1 - x\]

\[2x = 0\]

\[x = 0.\]

\[1)\ y^{'}(x) = \left( \ln(1 + x) \right)^{'} = \frac{1}{1 + x}\]

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

\[a = arctg\ 1 = \frac{\pi}{4}.\]

\[2)\ y^{'}(x) = \left( \ln(1 - x) \right)^{'} =\]

\[= - \frac{1}{1 - x}\]

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

\[a = - arctg\ 1 = - \frac{\pi}{4}.\]

\[3)\ \beta = \frac{\pi}{4} - \left( - \frac{\pi}{4} \right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]

\[4)\ y = e^{x}\text{\ \ }и\ \ y = e^{- x}\]

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

\[x = - x\]

\[2x = 0\]

\[x = 0.\]

\[1)\ y^{'}(x) = \left( e^{x} \right)^{'} = e^{x}\]

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

\[a = arctg\ 1 = \frac{\pi}{4}.\]

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

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

\[a = - arctg\ 1 = - \frac{\pi}{4}.\]

\[3)\ \beta = \frac{\pi}{4} - \left( - \frac{\pi}{4} \right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]