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МатематикаГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 231
Задание 231
\[1)\ y = x^{4},\ \ \ y = x^{6} + 2x^{2}:\]
\[x^{4} = x^{6} + 2x^{2}\]
\[x^{6} - x^{4} + 2x^{2} = 0\]
\[x^{2} \bullet \left( x^{4} - x^{2} + 2 \right) = 0\]
\[x^{2} = 0\]
\[x = 0.\]
\[Уравнение\ касательной:\]
\[y(0) = 0^{4} = 0;\]
\[y^{'}(0) = 4x^{3} = 0;\]
\[y^{'}(0) = 6x^{5} + 2 \bullet 2x = 0;\]
\[y = 0 + 0(x - 0) = 0;\]
\[Ответ:\ \ y = 0.\]
\[2)\ y = x^{4},\ \ \ y = x^{3} - 3x^{2}:\]
\[x^{4} = x^{3} - 3x^{2}\]
\[x^{4} - x^{3} + 3x^{2} = 0\]
\[x^{2} \bullet \left( x^{2} - x + 3 \right) = 0\]
\[x^{2} = 0\]
\[x = 0.\]
\[Уравнение\ касательной:\]
\[y(0) = 0^{4} = 0;\]
\[y^{'}(0) = 4x^{3} = 0;\]
\[y^{'}(0) = 3x^{2} - 3 \bullet 2x = 0;\]
\[y = 0 + 0(x - 0) = 0.\]
\[Ответ:\ \ y = 0.\]
\[3)\ y = (x + 2)^{2},\ \ \ y = 2 - x^{2}:\]
\[(x + 2)^{2} = 2 - x^{2}\]
\[x^{2} + 4x + 4 = 2 - x^{2}\]
\[2x^{2} + 4x + 2 = 0\]
\[x^{2} + 2x + 1 = 0\]
\[(x + 1)^{2} = 0\]
\[x = - 1.\]
\[Уравнение\ касательной:\]
\[y( - 1) = 2 - 1 = 1;\]
\[y^{'}( - 1) = 2(x + 2) = 2;\]
\[y^{'}( - 1) = - 2x = 2;\]
\[y = 1 + 2(x + 1) = 2x + 3.\]
\[Ответ:\ \ y = 2x + 3.\]
\[4)\ y = x(2 + x),\ \ \ y = x(2 - x):\]
\[x(2 + x) = x(2 - x)\]
\[2x + x^{2} = 2x - x^{2}\]
\[2x^{2} = 0\]
\[x = 0.\]
\[Уравнение\ касательной:\]
\[y(0) = 0 \bullet 2 = 0;\]
\[y^{'}(0) = 2 + 2x = 2;\]
\[y^{'}(0) = 2 - 2x = 2;\]
\[y = 0 + 2(x - 0) = 2x.\]
\[Ответ:\ \ y = 2x.\]
\[5)\ y = \sqrt{x + 1};\ y = x^{2} + \frac{1}{2}x + 1:\]
\[\sqrt{x + 1} = x^{2} + \frac{1}{2}x + 1\]
\[x + 1 =\]
\[= x^{4} + 2x^{2}\left( \frac{1}{2}x + 1 \right) + \left( \frac{1}{2}x + 1 \right)^{2}\]
\[x + 1 =\]
\[= x^{4} + x^{3} + 2x^{2} + \frac{1}{4}x^{2} + x + 1\]
\[x^{4} + x^{3} + \frac{9}{4}x^{2} = 0\]
\[4x^{4} + 4x^{3} + 9x^{2} = 0\]
\[x^{2}\left( 4x^{2} + 4x + 9 \right) = 0\]
\[x^{2} = 0\]
\[x = 0.\]
\[Уравнение\ касательной:\]
\[y(0) = \sqrt{1} = 1;\]
\[y^{'}(0) = \frac{1}{2\sqrt{x + 1}} = \frac{1}{2};\]
\[y^{'}(0) = 2x + \frac{1}{2} = \frac{1}{2};\]
\[y = 1 + \frac{1}{2}(x - 0) = \frac{1}{2}x + 1.\]
\[Ответ:\ \ y = \frac{1}{2}x + 1.\]
\[6)\ y = \sqrt{x + 1};\ y = 2 - \sqrt{1 - x}:\]
\[\sqrt{x + 1} = 2 - \sqrt{1 - x}\]
\[\sqrt{x + 1} + \sqrt{1 - x} = 4\]
\[x + 1 + 2\sqrt{(x + 1)(1 - x)} + 1 - x = 4\]
\[2\sqrt{1 - x^{2}} = 2\]
\[1 - x^{2} = 1\]
\[x^{2} = 0\]
\[x = 0.\]
\[Уравнение\ касательной:\]
\[y(0) = \sqrt{1} = 1;\]
\[y^{'}(0) = \frac{1}{2\sqrt{x + 1}} = \frac{1}{2};\]
\[y^{'}(0) = - \left( - \frac{1}{2\sqrt{x + 1}} \right) = \frac{1}{2};\]
\[y = 1 + \frac{1}{2}(x - 0) = \frac{1}{2}x + 1.\]
\[Ответ:\ \ y = \frac{1}{2}x + 1.\]
