ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 323

\[1)\ y = - x^{4} + 8x^{2} - 16;\]

\[y^{'} = - 4x^{3} + 8 \bullet 2x - 0 =\]

\[= 16x - 4x^{3};\]

\[y^{''} = 16 - 4 \bullet 3x^{2} = 16 - 12x^{2}.\]

\[Промежуток\ возрастания:\]

\[16x - 4x^{3} \geq 0\]

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

\[(x + 2)x(x - 2) \leq 0\]

\[x \leq - 2;\ \ \ 0 \leq x \leq 2.\]

\[Выпукла\ вниз:\]

\[16 - 12x^{2} \geq 0\]

\[\left( 2\sqrt{3}x + 4 \right)\left( 2\sqrt{3}x - 4 \right) \leq 0\]

\[- \frac{4}{2\sqrt{3}} \leq x \leq \frac{4}{2\sqrt{3}}.\]

\[Функция\ четная:\]

\[y( - x) = - ( - x)^{4} + 8( - x)^{2} - 16 =\]

\[= - x^{4} + 8x^{2} - 16 = y(x).\]

\[2)\ y = \frac{1}{4}x^{4} - \frac{1}{24}x^{6};\]

\[y^{'} = \frac{1}{4} \bullet 4x^{3} - \frac{1}{24} \bullet 6x^{5} = x^{3} - \frac{1}{4}x^{5};\]

\[y^{''} = 3x^{2} - \frac{1}{4} \bullet 5x^{4} = 3x^{2} - \frac{5}{4}x^{4}.\]

\[Промежуток\ возрастания:\]

\[x^{3} - \frac{1}{4}x^{5} \geq 0\]

\[\frac{1}{4}x^{3}\left( 4 - x^{2} \right) \geq 0\]

\[(x + 2)x(x - 2) \leq 0\]

\[x \leq - 2;\ \ \ 0 \leq x \leq 2.\]

\[Выпукла\ вниз:\]

\[3x^{2} - \frac{5}{4}x^{4} \geq 0\]

\[\frac{1}{4}x^{2}\left( 12 - 5x^{2} \right) \geq 0\]

\[\left( x\sqrt{5} + 2\sqrt{3} \right)\left( x\sqrt{5} - 2\sqrt{3} \right) \leq 0\]

\[- \frac{2\sqrt{3}}{\sqrt{5}} \leq x \leq \frac{2\sqrt{3}}{\sqrt{5}}.\]

\[Функция\ четная:\]

\[y( - x) = \frac{1}{4}( - x)^{4} - \frac{1}{24}( - x)^{6} =\]

\[= \frac{1}{4}x^{4} - \frac{1}{24}x^{6} = y(x).\]

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

\[y^{'} = \frac{1}{3} \bullet 3x^{2} + 3 \bullet 2x = x^{2} + 6x;\]

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

\[Промежуток\ возрастания:\]

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

\[x(x + 6) \geq 0\]

\[x \leq - 6;\text{\ \ \ x} \geq 0.\]

\[Выпукла\ вниз:\]

\[2x + 6 \geq 0\]

\[2x \geq - 6\]

\[x \geq - 3.\]

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

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

\[y^{''} = - 3x^{2} + 2.\]

\[Промежуток\ возрастания:\]

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

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

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

\[x \leq - \sqrt{2};\ \ \ 0 \leq x \leq \sqrt{2}.\]

\[Выпукла\ вниз:\]

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

\[\left( x\sqrt{3} + \sqrt{2} \right)\left( x\sqrt{3} - \sqrt{2} \right) \leq 0\]

\[- \frac{\sqrt{2}}{\sqrt{3}} \leq x \leq \frac{\sqrt{2}}{\sqrt{3}}.\]

\[Функция\ четная:\]

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

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