Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1491

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

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

\[= \frac{1}{4} \bullet \left( - 2x^{- 3} \right) - \frac{1}{2}x^{- \frac{1}{2}} =\]

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

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

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

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

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

\[Ответ:\ \ a = - \frac{\pi}{4}.\]

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

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

\[tg\ a = f^{'}\left( \frac{1}{3} \right) = 3 \bullet \sqrt{\frac{1}{3}} = \sqrt{\frac{9}{3}} = \sqrt{3};\]

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

\[Ответ:\ \ a = \frac{\pi}{3}.\]