ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 244

\[1)\ y = x^{2} - 2x,\ \ \ x_{0} = 3:\]

\[y(3) = 9 - 6 = 3;\]

\[y^{'}(x) = 2x - 2;\]

\[y^{'}(3) = 6 - 2 = 4;\]

\[y = 3 + 4(x - 3) = 4x - 9.\]

\[Ответ:\ \ y = 4x - 9.\]

\[2)\ y = x^{3} + 3x,\ \ \ x_{0} = 3:\]

\[y(3) = 27 + 9 = 36;\]

\[y^{'}(x) = 3x^{2} + 3;\]

\[y^{'}(3) = 27 + 3 = 30;\]

\[y = 36 + 30(x - 3) = 30x - 54.\]

\[Ответ:\ \ y = 30x - 54.\]

\[3)\ y = \sin x,\ \ \ x_{0} = \frac{\pi}{6}:\]

\[y\left( \frac{\pi}{6} \right) = \sin\frac{\pi}{6} = \frac{1}{2};\]

\[y^{'}(x) = \cos x;\]

\[y^{'}\left( \frac{\pi}{6} \right) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2};\]

\[y = \frac{1}{2} + \frac{\sqrt{3}}{2}\left( x - \frac{\pi}{6} \right) =\]

\[= \frac{\sqrt{3}}{2}x + \frac{1}{2} - \frac{\pi\sqrt{3}}{12}.\]

\[Ответ:\ \ y = \frac{\sqrt{3}}{2}x + \frac{1}{2} - \frac{\pi\sqrt{3}}{12}.\]

\[4)\ y = \cos x,\ \ \ x_{0} = \frac{\pi}{3}:\]

\[y\left( \frac{\pi}{3} \right) = \cos\frac{\pi}{3} = \frac{1}{2};\]

\[y^{'}(x) = - \sin x;\]

\[y^{'}\left( \frac{\pi}{3} \right) = - \cos\frac{\pi}{3} = - \frac{\sqrt{3}}{2};\]

\[y = \frac{1}{2} - \frac{\sqrt{3}}{2}\left( x - \frac{\pi}{3} \right) =\]

\[= - \frac{\sqrt{3}}{2}x + \frac{1}{2} + \frac{\pi\sqrt{3}}{6}.\]

\[Ответ:\ \ y = - \frac{\sqrt{3}}{2}x + \frac{1}{2} + \frac{\pi\sqrt{3}}{6}.\]