Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 793

\[\boxed{\mathbf{793}\mathbf{.}}\]

\[1)\ f(x) = x^{6}\text{\ \ }x_{0} = \frac{1}{2}:\]

\[f^{'}(x) = 6 \bullet x^{6 - 1} = 6x^{5}\]

\[f^{'}\left( \frac{1}{2} \right) = 6 \bullet \left( \frac{1}{2} \right)^{5} = \frac{6}{32} = \frac{3}{16}.\]

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

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

\[= - 2 \bullet x^{- 3} = - \frac{2}{x^{3}}\]

\[f^{'}(3) = - \frac{2}{3^{3}} = - \frac{2}{27}.\]

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

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

\[f^{'}(4) = \frac{1}{2\sqrt{4}} = \frac{1}{2 \bullet 2} = \frac{1}{4}.\]

\[4)\ f(x) = \sqrt[3]{x} = x^{\frac{1}{3}}\text{\ \ }x_{0} = 8:\]

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

\[f^{'}(8) = \frac{1}{3\sqrt[3]{8^{2}}} = \frac{1}{3\sqrt[3]{64}} = \frac{1}{3 \bullet 4} = \frac{1}{12}.\]

\[5)\ f(x) = \sqrt{5 - 4x} = (5 - 4x)^{\frac{1}{2}}\]

\[x_{0} = 1:\]

\[f^{'}(x) = \frac{1}{2} \bullet ( - 4) \bullet (5 - 4x)^{\frac{1}{2} - 1} =\]

\[= - 2 \bullet (5 - 4x)^{- \frac{1}{2}} = - \frac{2}{\sqrt{5 - 4x}}\]

\[f^{'}(1) = - \frac{2}{\sqrt{5 - 4 \bullet 1}} =\]

\[= - \frac{2}{\sqrt{1}} = - 2.\]

\[6)\ f(x) = \frac{1}{\sqrt{3x + 1}} = \frac{1}{(3x + 1)^{\frac{1}{2}}} =\]

\[= (3x + 1)^{- \frac{1}{2}}\text{\ \ }x_{0} = 1:\]

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

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

\[= - \frac{3}{2\sqrt{(3x + 1)^{3}}}\]

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

\[= - \frac{3}{2\sqrt{4^{3}}} = - \frac{3}{2\sqrt{64}} =\]

\[= - \frac{3}{2 \bullet 8} = - \frac{3}{16}.\]