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МатематикаРешебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 27
Задание 27
\[\boxed{\mathbf{27.}}\]
\[\textbf{а)}\ f(x) = \log_{2}x;\ \ x_{0} = 1\]
\[f^{'}(x) = \frac{1}{x\ln 2};\]
\[y_{0} = f(1) = \log_{2}1 = 0;\]
\[k = f^{'}(1) = \frac{1}{\ln 2};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 0 = \frac{1}{\ln 2}(x - 1)\]
\[y = \frac{1}{\ln 2}x - \frac{1}{\ln 2}.\]
\[Уравнение\ касательной:\]
\[y = \frac{1}{\ln 2}x - \frac{1}{\ln 2}.\]
\[\textbf{б)}\ f(x) = \log_{2}x;\ \ x_{0} = 2\]
\[f^{'}(x) = \frac{1}{x\ln 2};\]
\[y_{0} = f(2) = \log_{2}2 = 1;\]
\[k = f^{'}(2) = \frac{1}{2\ln 2};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 1 = \frac{1}{2\ln 2}(x - 2)\]
\[y - 1 = \frac{1}{2\ln 2}x - \frac{1}{\ln 2}\]
\[y = \frac{1}{2\ln 2}x - \frac{1}{\ln 2} + 1.\]
\[Уравнение\ касательной:\]
\[y = \frac{1}{2\ln 2}x - \frac{1}{\ln 2} + 1.\]
\[\textbf{в)}\ f(x) = \log_{2}x;\ \ x_{0} = 4\]
\[f^{'}(x) = \frac{1}{x\ln 2};\]
\[y_{0} = f(4) = \log_{2}4 =\]
\[= 2\log_{2}2 = 2;\]
\[k = f^{'}(4) = \frac{1}{4\ln 2};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 2 = \frac{1}{4\ln 2}(x - 4)\]
\[y - 2 = \frac{1}{4\ln 2}x - \frac{1}{\ln 2}\]
\[y = \frac{1}{4\ln 2}x - \frac{1}{\ln 2} + 2.\]
\[Уравнение\ касательной:\]
\[y = \frac{1}{4\ln 2}x - \frac{1}{\ln 2} + 2.\]
\[\textbf{г)}\ f(x) = \log_{2}x;\ \ x_{0} = 8\]
\[f^{'}(x) = \frac{1}{x\ln 2};\]
\[y_{0} = f(8) = \log_{2}8 =\]
\[= 3\log_{2}2 = 3;\]
\[k = f^{'}(8) = \frac{1}{8\ln 2};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 3 = \frac{1}{8\ln 2}(x - 8)\]
\[y - 3 = \frac{1}{8\ln 2}x - \frac{1}{\ln 2}\]
\[y = \frac{1}{8\ln 2}x - \frac{1}{\ln 2} + 3.\]
\[Уравнение\ касательной:\]
\[y = \frac{1}{8\ln 2}x - \frac{1}{\ln 2} + 3.\ \]
