ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1044

\[1)\ f(x) = \frac{1}{4x^{2}} - \sqrt{x};\ \ x_{0} = 1:\]

\[f^{'}(x) = \frac{1}{4}\left( x^{- 2} \right)^{'} - \left( x^{\frac{1}{2}} \right)^{'} =\]

\[= \frac{- 2x^{- 3}}{4} - \frac{1}{2}x^{- \frac{1}{2}};\]

\[tg\ a = f^{'}(1) = \frac{- 2 \bullet 1^{- 3}}{4} - \frac{1}{2} \bullet 1^{- \frac{1}{2}} =\]

\[= - \frac{1}{2} - \frac{1}{2} = - 1;\]

\[a = arctg\ ( - 1) =\]

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

\[Ответ:\ \ a = - \frac{\pi}{4}.\]

\[2)\ f(x) = 2x\sqrt{x},\ \text{\ \ }x_{0} = \frac{1}{3};\]

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

\[= 2 \bullet \frac{3}{2} \bullet x^{\frac{1}{2}} = 3\sqrt{x};\]

\[tg\ a = f^{'}\left( \frac{1}{3} \right) = 3 \bullet \sqrt{\frac{1}{3}} =\]

\[= \sqrt{\frac{9}{3}} = \sqrt{3};\]

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

\[Ответ:\ \ a = \frac{\pi}{3}.\]