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МатематикаРешебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 23
Задание 23
\[\boxed{\mathbf{23}\mathbf{.}}\]
\[\textbf{а)}\ y = \cos x;\ \ x_{0} = 0\]
\[f^{'}(x) = - \sin x;\]
\[y_{0} = f(0) = \cos 0 = 1;\]
\[k = f^{'}(0) = - \sin 0 = 0;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 1 = 0 \cdot (x - 0)\]
\[y = 1.\]
\[Уравнение\ касательной:y = 1.\]
\[\textbf{б)}\ y = \cos x;\ \ x_{0} = \frac{\pi}{2}\]
\[f^{'}(x) = - \sin x;\]
\[y_{0} = f\left( \frac{\pi}{2} \right) = \cos\frac{\pi}{2} = 0;\]
\[k = f^{'}\left( \frac{\pi}{2} \right) = - \sin\frac{\pi}{2} = - 1;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 0 = - 1 \cdot \left( x - \frac{\pi}{2} \right)\]
\[y = - x + \frac{\pi}{2}.\]
\[Уравнение\ касательной:\]
\[y = - x + \frac{\pi}{2}.\]
\[\textbf{в)}\ y = \cos x;\ \ x_{0} = - \ \frac{\pi}{2}\]
\[f^{'}(x) = - \sin x;\]
\[y_{0} = f\left( - \frac{\pi}{2} \right) = \cos\frac{\pi}{2} = 0;\]
\[k = f^{'}\left( - \frac{\pi}{2} \right) = \sin\frac{\pi}{2} = 1;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 0 = 1 \cdot \left( x + \frac{\pi}{2} \right)\]
\[y = x + \frac{\pi}{2}.\]
\[Уравнение\ касательной:\]
\[y = x + \frac{\pi}{2}.\]
\[\textbf{г)}\ y = \cos x;\ \ x_{0} = - \pi\]
\[f^{'}(x) = - \sin x;\]
\[y_{0} = f( - \pi) = \cos\pi = - 1;\]
\[k = f^{'}( - \pi) = \sin\pi = 0;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y + 1 = 0 \cdot \left( x + \frac{\pi}{2} \right)\]
\[y = - 1.\]
\[Уравнение\ касательной:\]
\[y = - 1.\]
