ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 859

\[\boxed{\mathbf{859}\mathbf{.}}\]

\[k = tg\ a\]

\[a = arctg\ k.\]

\[1)\ f(x) = \frac{1}{3}x^{3}\text{\ \ }и\ \ x_{0} = 1:\]

\[f^{'}(x) = \frac{1}{3} \bullet \left( x^{3} \right)^{'} = \frac{1}{3} \bullet 3x^{2} = x^{2}\]

\[k = f^{'}(1) = 1^{2} = 1\]

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

\[Ответ:\ \ a = \frac{\pi}{4}.\]

\[2)\ f(x) = \frac{1}{x}\text{\ \ }и\ \ x_{0} = 1:\]

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

\[k = f^{'}(1) = - \frac{1}{1^{2}} = - 1\]

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

\[Ответ:\ \ a = - \frac{\pi}{4}.\]

\[3)\ f(x) = 2\sqrt{x}\text{\ \ }и\ \ x_{0} = 3:\]

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

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

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

\[Ответ:\ \ a = \frac{\pi}{6}.\]

\[4)\ f(x) = \frac{18}{\sqrt{x}}\text{\ \ }и\ \ x_{0} = 3:\]

\[f^{'}(x) = 18 \bullet \left( x^{- \frac{1}{2}} \right)^{'} =\]

\[= 18 \bullet \left( - \frac{1}{2} \right) \bullet x^{- \frac{3}{2}} = - \frac{9}{x\sqrt{x}}\]

\[k = f^{'}(3) = - \frac{9}{3\sqrt{3}} =\]

\[= - \frac{3}{\sqrt{3}} = - \sqrt{3}\]

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

\[Ответ:\ \ a = - \frac{\pi}{3}.\]

\[5)\ f(x) = e^{\frac{3x + 1}{2}}\text{\ \ }и\ \ x_{0} = 0:\]

\[f^{'}(x) = \left( e^{\frac{3}{2}x + \frac{1}{2}} \right)^{'} = \frac{3}{2}e^{\frac{3x + 1}{2}}\]

\[k = f^{'}(0) = \frac{3}{2}e^{\frac{3 \bullet 0 + 1}{2}} =\]

\[= \frac{3}{2}e^{\frac{1}{2}} = \frac{3\sqrt{e}}{2}\]

\[a = arctg\frac{3\sqrt{e}}{2}.\]

\[Ответ:\ \ a = arctg\ \frac{3\sqrt{e}}{2}.\]

\[6)\ f(x) = \ln(2x + 1)\ и\ \ x_{0} = 2:\]

\[f^{'}(x) = \left( \ln(2x + 1) \right)^{'} = \frac{2}{2x + 1}\]

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

\[= \frac{2}{4 + 1} = \frac{2}{5}\]

\[a = arctg\frac{2}{5}.\]

\[Ответ:\ \ a = arctg\frac{2}{5}.\]