ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 913

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\[1)\ y = \frac{2 + x^{2}\ }{x} = \frac{2}{x} + x\]

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

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

\[Стационарные\ точки:\]

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

\[\frac{2}{x^{2}} = 1\]

\[x^{2} = 2\]

\[x = \pm \sqrt{2}.\]

\[Ответ:\ \ x_{1} = - \sqrt{2};\ \ x_{2} = \sqrt{2}.\]

\[2)\ y = \frac{x^{2} + 3}{2x} = \frac{x}{2} + \frac{3}{2x}\]

\[y^{'}(x) = \frac{1}{2} \bullet (x)^{'} + \frac{3}{2} \bullet \left( \frac{1}{x} \right)^{'} =\]

\[= \frac{1}{2} - \frac{3}{2x^{2}}.\]

\[Стационарные\ точки:\]

\[\frac{1}{2} - \frac{3}{2x^{2}} = 0\]

\[x^{2} - 3 = 0\]

\[x^{2} = 3\]

\[x = \pm \sqrt{3}.\]

\[Ответ:\ \ x_{1} = - \sqrt{3};\ \ x_{2} = \sqrt{3}.\]

\[3)\ y = e^{x^{2} - 1}\]

\[Пусть\ u = x^{2} - 1;\ \ y(u) = e^{u}:\]

\[y^{'}(x) = \left( x^{2} - 1 \right)^{'} \bullet \left( e^{u} \right)^{'} =\]

\[= 2x \bullet e^{u} = 2x \bullet e^{x^{2} - 1}.\]

\[Стационарные\ точки:\]

\[2x = 0\ \]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]

\[4)\ y = 2^{x^{2} + x}\]

\[Пусть\ u = x^{2} + x;\ \ y(u) = 2^{u}:\]

\[y^{'}(x) = \left( x^{2} + x \right)^{'} \bullet \left( 2^{u} \right)^{'};\]

\[y^{'}(x) = (2x + 1) \bullet 2^{u} \bullet \ln 2 =\]

\[= (2x + 1) \bullet 2^{x^{2} + x} \bullet \ln 2.\]

\[Стационарные\ точки:\]

\[2x + 1 = 0\]

\[2x = - 1\]

\[x = - 0,5.\]

\[Ответ:\ \ x = - 0,5.\]