Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1475

\[\boxed{\mathbf{1475}\mathbf{.}}\]

\[1)\ y = - x^{4} + 4x^{2} - 5;\]

\[y( - x) = - ( - x)^{4} + 4( - x)^{2} - 5 =\]

\[= - x^{4} + 4x^{2} - 5 = y(x) - четная;\]

\[y^{'}(x) = - \left( x^{4} \right)^{'} + 4\left( x^{2} \right)^{'} - (5)^{'} =\]

\[= - 4x^{3} + 4 \bullet 2x - 0 =\]

\[= 4\left( 2x - x^{3} \right);\]

\[Точки\ экстремума:\]

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

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

\[x_{1} = 0\ \ \ и\ \ x_{2} = \sqrt{2};\]

\[y_{1} = - 0^{4} + 4 \bullet 0^{2} - 5 = - 5;\]

\[y_{2} = - \left( \sqrt{2} \right)^{4} + 4 \bullet \left( \sqrt{2} \right)^{2} - 5 =\]

\[= - 4 + 8 - 5 = - 1.\]

\[2)\ y = x^{3} - 4x\]

\[y( - x) = ( - x)^{3} - 4 \bullet ( - x) =\]

\[= - x^{3} + 4x = - y(x) - нечетная;\]

\[y^{'}(x) = \left( x^{3} \right)^{'} - (4x)^{'} = 3x^{2} - 4.\]

\[Точки\ экстремума:\]

\[3x^{2} - 4 = 0\]

\[3x^{2} = 4\]

\[x^{2} = \frac{4}{3}\]

\[x = \frac{2}{\sqrt{3}};\]

\[y = \left( \frac{2}{\sqrt{3}} \right)^{3} - 4 \bullet \frac{2}{\sqrt{3}} = \frac{8}{3\sqrt{3}} - \frac{8}{\sqrt{3}} =\]

\[= \frac{8 - 24}{3\sqrt{3}} = - \frac{16}{3\sqrt{3}}.\]