ГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 27

\[\boxed{\mathbf{27.}}\]

\[\textbf{а)}\ f(x) = \log_{2}x;\ \ x_{0} = 1\]

\[f^{'}(x) = \frac{1}{x\ln 2};\]

\[y_{0} = f(1) = \log_{2}1 = 0;\]

\[k = f^{'}(1) = \frac{1}{\ln 2};\]

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

\[y - 0 = \frac{1}{\ln 2}(x - 1)\]

\[y = \frac{1}{\ln 2}x - \frac{1}{\ln 2}.\]

\[Уравнение\ касательной:\]

\[y = \frac{1}{\ln 2}x - \frac{1}{\ln 2}.\]

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

\[f^{'}(x) = \frac{1}{x\ln 2};\]

\[y_{0} = f(2) = \log_{2}2 = 1;\]

\[k = f^{'}(2) = \frac{1}{2\ln 2};\]

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

\[y - 1 = \frac{1}{2\ln 2}(x - 2)\]

\[y - 1 = \frac{1}{2\ln 2}x - \frac{1}{\ln 2}\]

\[y = \frac{1}{2\ln 2}x - \frac{1}{\ln 2} + 1.\]

\[Уравнение\ касательной:\]

\[y = \frac{1}{2\ln 2}x - \frac{1}{\ln 2} + 1.\]

\[\textbf{в)}\ f(x) = \log_{2}x;\ \ x_{0} = 4\]

\[f^{'}(x) = \frac{1}{x\ln 2};\]

\[y_{0} = f(4) = \log_{2}4 =\]

\[= 2\log_{2}2 = 2;\]

\[k = f^{'}(4) = \frac{1}{4\ln 2};\]

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

\[y - 2 = \frac{1}{4\ln 2}(x - 4)\]

\[y - 2 = \frac{1}{4\ln 2}x - \frac{1}{\ln 2}\]

\[y = \frac{1}{4\ln 2}x - \frac{1}{\ln 2} + 2.\]

\[Уравнение\ касательной:\]

\[y = \frac{1}{4\ln 2}x - \frac{1}{\ln 2} + 2.\]

\[\textbf{г)}\ f(x) = \log_{2}x;\ \ x_{0} = 8\]

\[f^{'}(x) = \frac{1}{x\ln 2};\]

\[y_{0} = f(8) = \log_{2}8 =\]

\[= 3\log_{2}2 = 3;\]

\[k = f^{'}(8) = \frac{1}{8\ln 2};\]

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

\[y - 3 = \frac{1}{8\ln 2}(x - 8)\]

\[y - 3 = \frac{1}{8\ln 2}x - \frac{1}{\ln 2}\]

\[y = \frac{1}{8\ln 2}x - \frac{1}{\ln 2} + 3.\]

\[Уравнение\ касательной:\]

\[y = \frac{1}{8\ln 2}x - \frac{1}{\ln 2} + 3.\ \]