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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 24
Задание 24
\[\boxed{\mathbf{24}\mathbf{.}}\]
\[\textbf{а)}\ y = tg\ x;\ \ x_{0} = 0\]
\[f'(x) = \frac{1}{\cos^{2}x};\]
\[y_{0} = f(0) = tg\ 0 = 0;\]
\[k = f^{'}(0) = \frac{1}{\cos^{2}0} = 1;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 0 = 1 \cdot (x - 0)\]
\[y = x.\]
\[Уравнение\ касательной:y = x.\]
\[\textbf{б)}\ y = tg\ x;\ \ x_{0} = \frac{\pi}{4}\]
\[f'(x) = \frac{1}{\cos^{2}x};\]
\[y_{0} = f\left( \frac{\pi}{4} \right) = tg\ \frac{\pi}{4} = 1;\]
\[k = f^{'}\left( \frac{\pi}{4} \right) = \frac{1}{\cos^{2}\frac{\pi}{4}} = \frac{1}{\left( \frac{\sqrt{2}}{2} \right)^{2}} =\]
\[= 1\ :\frac{2}{4} = \frac{4}{2} = 2;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 1 = 2 \cdot \left( x - \frac{\pi}{4} \right)\]
\[y - 1 = 2x - \frac{\pi}{2}\]
\[y = 2x - \frac{\pi}{2} + 1.\]
\[Уравнение\ касательной:\]
\[y = 2x - \frac{\pi}{2} + 1.\]
\[\textbf{в)}\ y = tg\ x;\ \ x_{0} = - \frac{\pi}{4}\]
\[f'(x) = \frac{1}{\cos^{2}x};\]
\[y_{0} = f\left( - \frac{\pi}{4} \right) = tg\ \left( - \frac{\pi}{4} \right) = - 1;\]
\[k = f^{'}\left( - \frac{\pi}{4} \right) = \frac{1}{\cos^{2}\left( - \frac{\pi}{4} \right)} =\]
\[= \frac{1}{\left( \frac{\sqrt{2}}{2} \right)^{2}} = 1\ :\frac{2}{4} = \frac{4}{2} = 2;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y + 1 = 2 \cdot \left( x + \frac{\pi}{4} \right)\]
\[y + 1 = 2x + \frac{\pi}{2}\]
\[y = 2x + \frac{\pi}{2} - 1.\]
\[Уравнение\ касательной:\]
\[y = 2x + \frac{\pi}{2} - 1.\]
\[\textbf{г)}\ y = tg\ x;\ \ x_{0} = \frac{\pi}{6}\]
\[f'(x) = \frac{1}{\cos^{2}x};\]
\[y_{0} = f\left( \frac{\pi}{6} \right) = tg\ \frac{\pi}{6} = \frac{\sqrt{3}}{3};\]
\[k = f^{'}\left( \frac{\pi}{6} \right) = \frac{1}{\cos^{2}\frac{\pi}{6}} = \frac{1}{\left( \frac{\sqrt{3}}{2} \right)^{2}} =\]
\[= 1\ :\frac{3}{4} = \frac{4}{3};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - \frac{\sqrt{3}}{3} = \frac{4}{3} \cdot \left( x - \frac{\pi}{6} \right)\]
\[y - \frac{\sqrt{3}}{3} = \frac{4}{3}x - \frac{2\pi}{9}\]
\[y = \frac{4}{3}x - \frac{2\pi}{9} + \frac{\sqrt{3}}{3}.\]
\[Уравнение\ касательной:\]
\[y = \frac{4}{3}x - \frac{2\pi}{9} + \frac{\sqrt{3}}{3}.\]