Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 308

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

\[y^{'} = 3x^{2} - 3 \bullet 2x + 0 = 3x^{2} - 6x;\]

\[y^{''} = 3 \bullet 2x - 6 = 6x - 6.\]

\[Промежуток\ возрастания:\]

\[3x^{2} - 6x \geq 0\]

\[3x(x - 2) \geq 0\]

\[x \leq 0;\text{\ \ \ x} \geq 2.\]

\[Выпукла\ вниз:\]

\[6x - 6 \geq 0\]

\[6x \geq 6\]

\[x \geq 1.\]

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

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

\[y^{''} = 0 - 3 \bullet 2x = - 2x.\]

\[Промежуток\ возрастания:\]

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

\[3(x + 1)(x - 1) \leq 0\]

\[- 1 \leq x \leq 1.\]

\[Выпукла\ вниз:\]

\[- 2x \geq 0\]

\[x \leq 0.\]

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

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

\[= - 3x^{2} + 8x - 4;\]

\[y^{''} = - 3 \bullet 2x + 8 - 0 = - 6x + 8.\]

\[Промежуток\ возрастания:\]

\[- 3x^{2} + 8x - 4 \geq 0\]

\[3x^{2} - 8x + 4 \leq 0\]

\[D = 64 - 48 = 16,\]

\[x_{1} = \frac{8 - 4}{2 \bullet 3} = \frac{2}{3};\text{\ \ }\]

\[x_{2} = \frac{8 + 4}{2 \bullet 3} = 2;\]

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

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

\[Выпукла\ вниз:\]

\[- 6x + 8 \geq 0\]

\[6x \leq 8\]

\[x \leq 1\frac{1}{3}.\]

\[4)\ y = x^{3} + 6x^{2} + 9x;\]

\[y^{'} = 3x^{2} + 6 \bullet 2x + 9 =\]

\[= 3x^{2} + 12x + 9;\]

\[y^{''} = 3 \bullet 2x + 12 + 0 = 6x + 12.\]

\[Промежуток\ возрастания:\]

\[3x^{2} + 12x + 9 \geq 0\]

\[x^{2} + 4x + 3 \geq 0\]

\[D = 16 - 12 = 4\]

\[x_{1} = \frac{- 4 - 2}{2} = - 3;\text{\ \ }\]

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

\[(x + 3)(x + 1) \geq 0\]

\[x \leq - 3;\text{\ \ \ x} \geq - 1.\]

\[Выпукла\ вниз:\]

\[6x + 12 \geq 0\]

\[6x \geq - 12\]

\[x \geq - 2.\]