Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 888

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\[1)\ y = 2\sqrt{x}\text{\ \ }и\ \ y = 2\sqrt{6 - x}\]

\[2\sqrt{x} = 2\sqrt{6 - x}\]

\[x = 6 - x\]

\[2x = 6\]

\[x = 3.\]

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

\[= 2 \bullet \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}}\]

\[k = y^{'}(3) = \frac{1}{\sqrt{3}}\]

\[a = arctg\frac{1}{\sqrt{3}} = \frac{\pi}{6}.\]

\[2)\ y^{'}(x) = 2 \bullet {(6 - x)^{\frac{1}{2}}}^{'} =\]

\[= 2 \bullet \frac{1}{2} \bullet ( - 1) \bullet (6 - x)^{- \frac{1}{2}} =\]

\[= - \frac{1}{\sqrt{6 - x}}\]

\[k = y^{'}(3) = - \frac{1}{\sqrt{6 - 3}} = - \frac{1}{\sqrt{3}}\]

\[a = - arctg\frac{1}{\sqrt{3}} = - \frac{\pi}{6}.\]

\[3)\ \beta = \frac{\pi}{6} - \left( - \frac{\pi}{6} \right) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}.\]

\[Ответ:\ \ \frac{\pi}{3}.\]

\[2)\ y = \sqrt{2x + 1}\text{\ \ }и\ \ y = 1\]

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

\[2x + 1 = 1\]

\[2x = 0\]

\[x = 0.\]

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

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

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

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

\[2)\ y^{'}(x) = (1)^{'} = 0\]

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

\[a = arctg\ 0 = 0.\]

\[3)\ \beta = \frac{\pi}{4} - 0 = \frac{\pi}{4}.\]

\[Ответ:\ \ \frac{\pi}{4}.\]