Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 877

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

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

\[f^{'}(3) = 2 \bullet 3 - 2 = 6 - 2 = 4\]

\[f(3) = 3^{2} - 2 \bullet 3 = 9 - 6 = 3\]

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

\[= 3 + 4x - 12 = 4x - 9.\]

\[Ответ:\ \ y = 4x - 9.\]

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

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

\[f^{'}(3) = 3 \bullet 3^{2} + 3 = 3 \bullet 9 + 3 =\]

\[= 27 + 3 = 30\]

\[f(3) = 3^{3} + 3 \bullet 3 = 27 + 9 = 36\]

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

\[= 36 + 30x - 90 = 30x - 54.\]

\[Ответ:\ \ y = 30x - 54.\]

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

\[f^{'}(x) = \left( \sin x \right)^{'} = \cos x\]

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

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

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

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

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

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

\[f^{'}(x) = \left( \cos x \right)^{'} = - \sin x\]

\[f^{'}\left( \frac{\pi}{3} \right) = - \sin\frac{\pi}{3} = - \frac{\sqrt{3}}{2}\]

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

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

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

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