Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 14

\[\boxed{\mathbf{14}\mathbf{.}}\]

\[\textbf{а)}\ y = \frac{x^{4}}{4} - 2x^{2};\ \ \lbrack - 2;2\rbrack\]

\[y^{'} = \frac{1}{4} \cdot 4x^{3} - 2 \cdot 2x = x^{3} - 4x;\]

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

\[x\left( x^{2} - 4 \right) = 0\]

\[x = 0;\ \ x = \pm 2.\]

\[- 2;0;2 \in \lbrack - 2;2\rbrack.\]

\[f( - 2) = \frac{16}{4} - 8 = 4 - 8 = - 4;\]

\[f(0) = 0;\]

\[f(2) = - 4.\]

\[\max{f(x)} = 0;\]

\[\min{f(x)} = - 4.\]

\[\textbf{б)}\ y = \frac{x^{4}}{4} - 2x^{2};\ \ ( - 2;2)\]

\[y^{'} = \frac{1}{4} \cdot 4x^{3} - 2 \cdot 2x = x^{3} - 4x;\]

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

\[x\left( x^{2} - 4 \right) = 0\]

\[x = 0;\ \ x = \pm 2.\]

\[0 \in ( - 2;2);\]

\[f(0) = 0;\]

\[f( - 2) = - 4 \rightarrow ( - 2;0\rbrack -\]

\[функция\ возрастает;\]

\[f(2) = - 4 \rightarrow \lbrack 0; - 2) -\]

\[функция\ убывает.\]

\[\max{f(x)} = 0;\]

\[\min{f(x)} = нет.\]

\[\textbf{в)}\ y = \frac{x^{4}}{4} - 2x^{2};\ \ ( - 2;2\rbrack\]

\[y^{'} = \frac{1}{4} \cdot 4x^{3} - 2 \cdot 2x = x^{3} - 4x;\]

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

\[x\left( x^{2} - 4 \right) = 0\]

\[x = 0;\ \ x = \pm 2.\]

\[0;2 \in ( - 2;2\rbrack.\]

\[f(0) = 0;\]

\[f(2) = - 4.\]

\[\max{f(x)} = 0;\]

\[\min{f(x)} = - 4.\]

\[\textbf{г)}\ y = \frac{x^{4}}{4} - 2x^{2};\ \ \lbrack - 2;2)\]

\[y^{'} = \frac{1}{4} \cdot 4x^{3} - 2 \cdot 2x = x^{3} - 4x;\]

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

\[x\left( x^{2} - 4 \right) = 0\]

\[x = 0;\ \ x = \pm 2.\]

\[- 2;0 \in \lbrack - 2;2).\]

\[f( - 2) = \frac{16}{4} - 8 = 4 - 8 = - 4;\]

\[f(0) = 0.\]

\[\max{f(x)} = 0;\]

\[\min{f(x)} = - 4.\]