Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 264

\[y = \frac{k}{x};\ \ x_{0};\text{\ \ \ k} > 0.\]

\[y\left( x_{0} \right) = \frac{k}{x_{0}};\]

\[y^{'}(x) = k \bullet \left( - \frac{1}{x^{2}} \right);\]

\[y^{'}\left( x_{0} \right) = - \frac{k}{x_{0}^{2}}.\]

\[y = \frac{k}{x_{0}} - \frac{k}{x_{0}^{2}}\left( x - x_{0} \right) =\]

\[= \frac{k}{x_{0}} - \frac{\text{kx}}{x_{0}^{2}} + \frac{k}{x_{0}} = \frac{2k}{x_{0}} - \frac{\text{kx}}{x_{0}^{2}}.\]

\[1)\ \frac{2k}{x_{0}} - \frac{\text{kx}}{x_{0}^{2}} = 0\text{\ \ }\]

\[x = 2x_{0};\]

\[y = \frac{2k}{x_{0}} - \frac{k \bullet 0}{x_{0}^{2}} = \frac{2k}{x_{0}}.\]

\[S = \frac{1}{2}xy = \frac{1}{2} \bullet 2x_{0} \bullet \frac{2k}{x_{0}} = 2k.\]

\[2)\ y\left( x_{0} \right) = \frac{2k}{x_{0}} - \frac{k \bullet x_{0}}{x_{0}^{2}} =\]

\[= \frac{2k}{x_{0}} - \frac{k}{x_{0}} = \frac{k}{x_{0}};\]

\[y\left( 2x_{0} \right) = \frac{2k}{x_{0}} - \frac{k \bullet 2x_{0}}{x_{0}^{2}} =\]

\[= \frac{2k}{x_{0}} - \frac{2k}{x_{0}} = 0.\]

\(Ответ:\ \ 2k.\)