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МатематикаРешебник по алгебре и начала математического анализа 11 класс Колягин Задание 309
Задание 309
\[1)\ y = x^{4} - 2x^{2} + 2;\]
\[y^{'} = 4x^{3} - 2 \bullet 2x + 0 = 4x^{3} - 4x;\]
\[y^{''} = 4 \bullet 3x^{2} - 4 = 12x^{2} - 4.\]
\[Промежуток\ возрастания:\]
\[4x^{3} - 4x \geq 0\]
\[4x\left( x^{2} - 1 \right) \geq 0\]
\[(x + 1)x(x - 1) \geq 0\]
\[- 1 \leq x \leq 0;\text{\ \ \ x} \geq 1.\]
\[Выпукла\ вниз:\]
\[12x^{2} - 4 \geq 0\]
\[3x^{2} - 1 \geq 0\]
\[\left( x\sqrt{3} + 1 \right)\left( x\sqrt{3} - 1 \right) \geq 0\]
\[x \leq - \frac{1}{\sqrt{3}};\ \ \ x \geq \frac{1}{\sqrt{3}}.\]
\[Функция\ четная:\]
\[y( - x) = ( - x)^{4} - 2( - x)^{2} + 2 =\]
\[= x^{4} - 2x^{2} + 2 = y(x).\]
\[2)\ y = \frac{1}{9}x^{3}(x + 4) = \frac{1}{9}x^{4} + \frac{4}{9}x^{3};\]
\[y^{'} = \frac{1}{9} \bullet 4x^{3} + \frac{4}{9} \bullet 3x^{2} = \frac{4}{9}x^{3} + \frac{4}{3}x^{2};\]
\[y^{''} = \frac{4}{9} \bullet 3x^{2} + \frac{4}{3} \bullet 2x = \frac{4}{3}x^{2} + \frac{8}{3}x.\]
\[Промежуток\ возрастания:\]
\[\frac{4}{9}x^{3} + \frac{4}{3}x^{2} \geq 0\]
\[x^{3} + 3x^{2} \geq 0\]
\[x + 3 \geq 0\]
\[x \geq - 3;\text{\ \ \ x} = 0.\]
\[Выпукла\ вниз:\]
\[\frac{4}{3}x^{2} + \frac{8}{3}x \geq 0\]
\[x^{2} + 2x \geq 0\]
\[x(x + 2) \geq 0\]
\[x \leq - 2;\text{\ \ \ x} \geq 0.\]
\[3)\ y = \frac{1}{5}x^{3}(8 - 3x) = \frac{8}{5}x^{3} - \frac{3}{5}x^{4};\]
\[y^{'} = \frac{8}{5} \bullet 3x^{2} - \frac{3}{5} \bullet 4x^{3} =\]
\[= \frac{24}{5}x^{2} - \frac{12}{5}x^{3};\]
\[y^{''} = \frac{24}{5} \bullet 2x - \frac{12}{5} \bullet 3x^{2} =\]
\[= \frac{48}{5}x - \frac{36}{5}x^{2}.\]
\[Промежуток\ возрастания:\]
\[\frac{24}{5}x^{2} - \frac{12}{5}x^{3} \geq 0\]
\[2x^{2} - x^{3} \geq 0\]
\[x - 2 \leq 0\]
\[x \leq 2;\text{\ \ \ x} = 0.\]
\[Выпукла\ вниз:\]
\[\frac{48}{5}x - \frac{36}{5}x^{2} \geq 0\]
\[4x - 3x^{2} \geq 0\]
\[x(3x - 4) \leq 0\]
\[0 \leq x \leq 1\frac{1}{3}.\]
\[4)\ y = 6x^{4} - 4x^{6};\]
\[y^{'} = 6 \bullet 4x^{3} - 4 \bullet 6x^{5} =\]
\[= 24x^{3} - 24x^{5};\]
\[y^{''} = 24 \bullet 3x^{2} - 24 \bullet 5x^{4} =\]
\[= 72x^{2} - 120x^{4}.\]
\[Промежуток\ возрастания:\]
\[24x^{3} - 24x^{5} \geq 0\]
\[x - x^{3} \geq 0\]
\[x\left( x^{2} - 1 \right) \leq 0\]
\[(x + 1)x(x - 1) \leq 0\]
\[x \leq - 1;\ \ \ 0 \leq x \leq 1.\]
\[Выпукла\ вниз:\]
\[72x^{2} - 120x^{4} \geq 0\]
\[3 - 5x^{2} \geq 0\]
\[\left( x\sqrt{5} - \sqrt{3} \right)\left( x\sqrt{5} + \sqrt{3} \right) \leq 0\]
\[- \frac{\sqrt{3}}{\sqrt{5}} \leq x \leq \frac{\sqrt{3}}{\sqrt{5}}.\]
\[Функция\ четная:\]
\[y( - x) = 6( - x)^{4} - 4( - x)^{6} =\]
\[= 6x^{4} - 4x^{6} = y(x).\]
