Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1045

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

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

\[= \frac{3}{4} \bullet \left( - \frac{3}{2}x^{- \frac{5}{2}} \right) = - \frac{9}{8x^{2}\sqrt{x}};\]

\[f^{'}\left( \frac{1}{4} \right) = - \frac{9}{8 \bullet \left( \frac{1}{4} \right)^{2} \bullet \sqrt{\frac{1}{4}}} =\]

\[= - \frac{9}{8 \bullet \frac{1}{16} \bullet \frac{1}{2}} = - \frac{9}{\frac{1}{2} \bullet \frac{1}{2}} = - 36;\]

\[f\left( \frac{1}{4} \right) = \frac{3}{4 \bullet \frac{1}{4} \bullet \sqrt{\frac{1}{4}}} = \frac{3}{1 \bullet \frac{1}{2}} =\]

\[= 3 \bullet 2 = 6;\]

\[y = 6 - 36\left( x - \frac{1}{4} \right) =\]

\[= 6 - 36x + 9 = 15 - 36x.\]

\[Ответ:\ \ y = 15 - 36x.\]

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

\[f^{'}(x) = 2 \bullet 4x^{3} - 2x + 0 =\]

\[= 8x^{3} - 2x;\]

\[f^{'}( - 1) = 8 \bullet ( - 1)^{3} - 2 \bullet ( - 1) =\]

\[= - 8 + 2 = - 6;\]

\[f( - 1) = 2 \bullet ( - 1)^{4} - ( - 1)^{2} + 4 =\]

\[= 2 - 1 + 4 = 5;\]

\[y = 5 - 6(x + 1) = 5 - 6x - 6 =\]

\[= - 1 - 6x.\]

\[Ответ:\ \ y = - 1 - 6x.\]