ГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 9

\[\boxed{\mathbf{9}\mathbf{.}}\]

\[\textbf{а)}\ y = e^{x^{2}} - ex^{2};\ \ \lbrack - e;e\rbrack\]

\[f^{'}(x) = 2x \cdot e^{x^{2}} - 2xe;\]

\[2x \cdot e^{x^{2}} - 2xe = 0\]

\[2x\left( e^{x^{2}} - e \right) = 0\]

\[x = 0;\]

\[e^{x^{2}} - e = 0\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[- 1;0;1 \in \lbrack - e;e\rbrack;\ \ так\ как\]

\[\ e \approx 2,72.\]

\[Ответ:\ - 1;0;\]

\[1 - критические\ точки.\]

\[\textbf{б)}\ y = e^{x^{2} - 2x};\ \ \lbrack - \pi;\ \pi\rbrack\]

\[f^{'}(x) = (2x - 2) \cdot e^{x^{2} - 2x};\]

\[(2x - 2) \cdot e^{x^{2} - 2x} = 0\]

\[2x - 2 = 0\]

\[2x = 2\]

\[x = 1.\]

\[e^{x^{2} - 2x} = 0\]

\[нет\ корней.\]

\[1 \in \lbrack - \pi;\ \pi\rbrack;так\ как\ \pi \approx 3,14.\]

\[Ответ:1 - критическая\ точка.\]

\[\textbf{в)}\ y = \frac{\ln x}{x};\ \ \ (0;\pi\rbrack\]

\[f^{'}(x) = \frac{\left( \ln x \right)^{'} \cdot x - \ln x \cdot x^{'}}{x^{2}} =\]

\[= \frac{\frac{1}{x} \cdot x - \ln x}{x^{2}} = \frac{1 - \ln x}{x^{2}};\ \]

\[\frac{1 - \ln x}{x^{2}} = 0\]

\[1 - \ln x = 0\]

\[\ln x = 1\]

\[\log_{e}x = 1\]

\[x = e.\]

\[e \in (0;\pi\rbrack.\]

\[Ответ:e - критическая\ точка.\]

\[\textbf{г)}\ y = \frac{e^{x}}{1 + x};\ \ ( - 1;\ \pi\rbrack\]

\[f^{'}(x) =\]

\[= \frac{\left( e^{x} \right)^{'} \cdot (1 + x) - e^{x} \cdot (1 + x)^{'}}{(1 + x)^{2}} =\]

\[= \frac{e^{x} \cdot (1 + x) - e^{x}}{(1 + x)^{2}} =\]

\[= \frac{e^{x}(1 + x - 1)}{(1 + x)^{2}} = \frac{xe^{x}}{(1 + x)^{2}};\]

\[\frac{xe^{x}}{(1 + x)^{2}} = 0\]

\[xe^{x} = 0\]

\[x = 0.\]

\[0 \in ( - 1;\ \pi\rbrack - критическая\]

\[\ точка.\]

\[Ответ:0 - критическая\ точка.\]