ГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 22

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\[\textbf{а)}\ f(x) = \sin x;\ \ x_{0} = 0\]

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

\[y_{0} = f(0) = \sin 0 = 0;\]

\[k = f^{'}(0) = \cos 0 = 1;\]

\[y - y_{0} = k\left( x - x_{0} \right)\]

\[y - 0 = 1 \cdot (x - 0)\]

\[y = x.\]

\[Уравнение\ касательной:y = x.\]

\[\textbf{б)}\ f(x) = \sin x;\ \ x_{0} = \frac{\pi}{2}\]

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

\[y_{0} = f\left( \frac{\pi}{2} \right) = \sin\frac{\pi}{2} = 1;\]

\[k = f^{'}\left( \frac{\pi}{2} \right) = \cos\frac{\pi}{2} = 0;\]

\[y - y_{0} = k\left( x - x_{0} \right)\]

\[y - 1 = 0 \cdot \left( x - \frac{\pi}{2} \right)\]

\[y = 1.\]

\[Уравнение\ касательной:y = 1.\]

\[\textbf{в)}\ f(x) = \sin x;\ \ x_{0} = - \frac{\pi}{2}\]

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

\[y_{0} = f\left( - \frac{\pi}{2} \right) = \sin\left( - \frac{\pi}{2} \right) =\]

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

\[k = f^{'}\left( - \frac{\pi}{2} \right) = \cos\left( - \frac{\pi}{2} \right) =\]

\[= \cos\frac{\pi}{2} = 0;\]

\[y - y_{0} = k\left( x - x_{0} \right)\]

\[y + 1 = 0 \cdot \left( x + \frac{\pi}{2} \right)\]

\[y = - 1.\]

\[Уравнение\ касательной:\]

\[y = - 1.\]

\[\textbf{г)}\ f(x) = \sin x;\ \ x_{0} = \pi\]

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

\[y_{0} = f(\pi) = \sin(\pi) = 0;\]

\[k = f^{'}(\pi) = \cos(\pi) = - 1;\]

\[y - y_{0} = k\left( x - x_{0} \right)\]

\[y - 0 = - 1 \cdot (x - \pi)\]

\[y = - x + \pi.\]

\[Уравнение\ касательной:\]

\[y = - x + \pi.\]