ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 312

\[1)\ y = 2 + 5x^{3} - 3x^{5};\]

\[y^{'} = 0 + 5 \bullet 3x^{2} - 3 \bullet 5x^{4} =\]

\[= 15x^{2} - 15x^{4};\]

\[y^{''} = 15 \bullet 2x - 15 \bullet 4x^{3} =\]

\[= 30x - 60x^{3}.\]

\[Промежуток\ возрастания:\]

\[15x^{2} - 15x^{4} \geq 0\]

\[1 - x^{2} \geq 0\]

\[(x + 1)(x - 1) \leq 0\]

\[- 1 \leq x \leq 1;\text{\ \ \ x} = 0.\]

\[Выпукла\ вниз:\]

\[30x - 60x^{3} \geq 0\]

\[2x^{3} - x \leq 0;\]

\[\left( \sqrt{2}x + 1 \right)x\left( \sqrt{2}x - 1 \right) \leq 0\]

\[x \leq - \frac{1}{\sqrt{2}};\ \ \ 0 \leq x \leq \frac{1}{\sqrt{2}}.\]

\[2)\ y = 3x^{5} - 5x^{3};\]

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

\[= 15x^{4} - 15x^{2};\]

\[y^{''} = 15 \bullet 4x^{3} - 15 \bullet 2x =\]

\[= 60x^{3} - 30x.\]

\[Промежуток\ возрастания:\]

\[15x^{4} - 15x^{2} \geq 0\]

\[x^{2} - 1 \geq 0\]

\[(x + 1)(x - 1) \geq 0\]

\[x \leq - 1;\ \ \ x \geq 1;\text{\ \ \ x} = 0.\]

\[Выпукла\ вниз:\]

\[60x^{3} - 30x \geq 0\]

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

\[\left( x\sqrt{2} + 1 \right)x\left( x\sqrt{2} - 1 \right) \geq 0\]

\[- \frac{1}{\sqrt{2}} \leq x \leq 0;\ \ \ x \geq \frac{1}{\sqrt{2}}.\]

\[Функция\ нечетная:\]

\[y( - x) = 3( - x)^{5} - 5( - x)^{3} =\]

\[= - 3x^{5} + 5x^{3} = - y(x).\]

\[3)\ y = 4x^{5} - 5x^{4};\]

\[y^{'} = 4 \bullet 5x^{4} - 5 \bullet 4x^{3} =\]

\[= 20x^{4} - 20x^{3};\]

\[y^{''} = 20 \bullet 4x^{3} - 20 \bullet 3x^{2} =\]

\[= 80x^{3} - 60x^{2}.\]

\[Промежуток\ возрастания:\]

\[20x^{4} - 20x^{3} \geq 0\]

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

\[x(x - 1) \geq 0\]

\[x \leq 0;\text{\ \ \ x} \geq 1.\]

\[Выпукла\ вниз:\]

\[80x^{3} - 60x^{2} \geq 0\]

\[4x - 3 \geq 0\]

\[x \geq \frac{3}{4}.\]

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

\[= (x - 1)^{2}(x + 1)(5x + 1) =\]

\[= (x - 1)^{2}\left( 5x^{2} + 6x + 1 \right);\]

\[= (x - 1)\left( 20x^{2} + 8x - 4 \right).\]

\[Промежуток\ возрастания:\]

\[5x^{2} + 6x + 1 \geq 0\]

\[D = 36 - 20 = 16\]

\[x_{1} = \frac{- 6 - 4}{2 \bullet 5} = - 1;\ \]

\[x_{2} = \frac{- 6 + 4}{2 \bullet 5} = - 0,2;\]

\[(x - 1)(x + 1)(x + 0,2) \geq 0\]

\[x \leq - 1;\ \ \ x \geq - 0,2;\text{\ \ \ x} = 1.\]

\[Выпукла\ вниз:\]

\[(x - 1)\left( 20x^{2} + 8x - 4 \right) \geq 0\]

\[D = 64 + 80 = 144\]

\[x_{1} = \frac{- 8 - 12}{2 \bullet 20} = - 0,5;\]

\[x_{2} = \frac{- 8 + 12}{2 \bullet 20} = 0,1;\]

\[(x + 0,5)(x - 0,1)(x - 1) \geq 0\]

\[- 0,5 \leq x \leq - 0,1;\text{\ \ \ x} \geq 1.\]