Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 960

\[\boxed{\mathbf{960}\mathbf{.}}\]

\[1)\ y = \frac{x^{3}}{3} + 3x^{2}\]

\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]

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

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

\[= x^{2} + 6x.\]

\[\textbf{в)}\ Стационарные\ точки:\]

\[x^{2} + 6x = 0\]

\[(x + 6) \bullet x = 0\]

\[x_{1} = - 6;\ x_{2} = 0.\]

\[\textbf{г)}\ f( - 6) = \frac{( - 6)^{3}}{3} + 3 \bullet ( - 6)^{2} =\]

\[= - \frac{216}{3} + 3 \bullet 36 = - 72 + 108 =\]

\[= 36;\]

\[f(0) = \frac{0^{3}}{3} + 3 \bullet 0^{2} = 0.\]

\[\textbf{д)}\ Возрастает\ \]

\[на\ ( - \infty;\ - 6) \cup (0;\ + \infty)\ и\ \]

\[убывает\ на\ ( - 6;\ 0);\]

\[x = 0 - точка\ минимума;\ \ \]

\[x = - 6 - точка\ максимума.\]

\[\textbf{е)}\ \]

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

\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]

\[\textbf{б)}\ y^{'}(x) = - \frac{1}{4} \bullet \left( x^{4} \right)^{'} + \left( x^{2} \right)^{'};\]

\[y^{'}(x) = - \frac{1}{4} \bullet 4x^{3} + 2x =\]

\[= - x^{3} + 2x;\]

\[\textbf{в)}\ Стационарные\ точки:\]

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

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

\[\left( \sqrt{2} + x \right) \bullet x \bullet \left( \sqrt{2} - x \right) = 0\]

\[x_{1} = - \sqrt{2},\ \ \ x_{2} = 0,\ \ \ x_{3} = \sqrt{2}.\]

\[\textbf{г)}\ f\left( \pm \sqrt{2} \right) =\]

\[= - \frac{\left( \pm \sqrt{2} \right)^{4}}{4} + \left( \pm \sqrt{2} \right)^{2} =\]

\[= - \frac{4}{4} + 2 = - 1 + 2 = 1;\]

\[f(0) = - \frac{0^{4}}{4} + 0^{2} = 0;\]

\[\textbf{д)}\ Возрастает\ \]

\[на\ \left( - \infty;\ - \sqrt{2} \right) \cup \left( 0;\ \sqrt{2} \right);\ \]

\[убывает\ \]

\[на\ \left( - \sqrt{2};\ 0 \right) \cup \left( \sqrt{2};\ + \infty \right);\]

\[x = 0 - точка\ минимума;\ \]

\[x = \pm \sqrt{2} - точки\ максимума.\]

\[\textbf{е)}\ \]