Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 26

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

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

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

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

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

\[y - 0 = 1(x - 1)\]

\[y = x - 1.\]

\[Уравнение\ касательной:\ \ \]

\[y = x - 1.\]

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

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

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

\[k = f^{'}(2) = \frac{1}{2} = 0,5;\]

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

\[y - \ln 2 = 0,5(x - 2)\]

\[y = 0,5x - 1 + \ln 2.\]

\[Уравнение\ касательной:\ \ \]

\[y = 0,5x - 1 + \ln 2.\]

\[\textbf{в)}\ f(x) = \ln x;\ \ x_{0} = 3\]

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

\[y_{0} = f(3) = \ln 3;\]

\[k = f^{'}(3) = \frac{1}{3};\]

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

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

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

\[Уравнение\ касательной:\ \ \]

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

\[\textbf{г)}\ f(x) = \ln x;\ \ x_{0} = e\]

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

\[y_{0} = f(e) = \ln e = 1;\]

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

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

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

\[y - 1 = \frac{1}{x}e - 1\]

\[y = \frac{1}{x}\text{e.}\]

\[Уравнение\ касательной:\ \ \]

\[y = \frac{1}{x}\text{e.}\]