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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 25
Задание 25
\[\boxed{\mathbf{25.}}\]
\[\textbf{а)}\ f(x) = ctg\ x;\ \ x_{0} = \frac{\pi}{2}\]
\[f^{'}(x) = - \frac{1}{\sin^{2}x};\]
\[y_{0} = f\left( \frac{\pi}{2} \right) = ctg\ \left( \frac{\pi}{2} \right) = 0;\]
\[k = f^{'}\left( \frac{\pi}{2} \right) = - \frac{1}{\sin^{2}\frac{\pi}{2}} =\]
\[= - \frac{1}{1} = - 1;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 0 = - 1\left( x - \frac{\pi}{2} \right)\]
\[y = - x + \frac{\pi}{2}.\]
\[Уравнение\ касательной:\ \ \]
\[y = - x + \frac{\pi}{2}.\]
\[\textbf{б)}\ f(x) = ctg\ x;\ \ x_{0} = - \frac{\pi}{2}\]
\[f^{'}(x) = - \frac{1}{\sin^{2}x};\]
\[y_{0} = f\left( - \frac{\pi}{2} \right) = ctg\ \left( - \frac{\pi}{2} \right) = 0;\]
\[k = f^{'}\left( - \frac{\pi}{2} \right) = - \frac{1}{\sin^{2}\left( - \frac{\pi}{2} \right)} =\]
\[= - \frac{1}{1} = - 1;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 0 = - 1\left( x + \frac{\pi}{2} \right)\]
\[y = - x - \frac{\pi}{2}.\]
\[Уравнение\ касательной:\ \ \]
\[y = - x - \frac{\pi}{2}.\]
\[\textbf{в)}\ f(x) = ctg\ x;\ \ x_{0} = \frac{\pi}{4}\]
\[f^{'}(x) = - \frac{1}{\sin^{2}x};\]
\[y_{0} = f\left( \frac{\pi}{4} \right) = ctg\ \left( \frac{\pi}{4} \right) = 1;\]
\[k = f^{'}\left( \frac{\pi}{4} \right) = - \frac{1}{\sin^{2}\frac{\pi}{4}} =\]
\[= - \frac{1}{\left( \frac{\sqrt{2}}{2} \right)^{2}} = - \frac{4}{2} = - 2;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 1 = - 2\left( x - \frac{\pi}{4} \right)\]
\[y - 1 = - 2x + \frac{\pi}{2}\]
\[y = - 2x + \frac{\pi}{2} + 1.\]
\[Уравнение\ касательной:\]
\[\ \ y = - 2x + \frac{\pi}{2} + 1.\]
\[\textbf{г)}\ f(x) = ctg\ x;\ \ x_{0} = - \frac{\pi}{6}\]
\[f^{'}(x) = - \frac{1}{\sin^{2}x};\]
\[y_{0} = f\left( - \frac{\pi}{6} \right) = ctg\ \left( - \frac{\pi}{6} \right) =\]
\[= - \sqrt{3};\]
\[k = f^{'}\left( - \frac{\pi}{6} \right) = - \frac{1}{\sin^{2}\left( - \frac{\pi}{6} \right)} =\]
\[= - \frac{1}{\left( \frac{1}{2} \right)^{2}} = - 4;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y + \sqrt{3} = - 4\left( x + \frac{\pi}{6} \right)\]
\[y + \sqrt{3} = - 4x - \frac{2\pi}{3}\]
\[y = - 4x - \frac{2\pi}{3} - \sqrt{3}.\]
\[Уравнение\ касательной:\ \ \]
\[y = - 4x - \frac{2\pi}{3} - \sqrt{3}.\]