ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 191

\[f^{'}(x) > 0.\]

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

\[f^{'}(x) = 4x^{3} - 4 \bullet 2x > 0;\]

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

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

\[- \sqrt{2} < x < 0;\]

\[x > \sqrt{2}.\]

\[Ответ:\ \ \left( - \sqrt{2};\ 0 \right) \cup \left( \sqrt{2};\ + \infty \right).\]

\[2)\ f(x) = 3x^{4} - 4x^{3} - 12x^{2} + 3;\]

\[f^{'}(x) = 3 \bullet 4x^{3} - 4 \bullet 3x^{2} - 12 \bullet 2x > 0;\]

\[12x^{3} - 12x^{2} - 24x > 0\]

\[12x\left( x^{2} - x - 2 \right) > 0\]

\[D = 1 + 8 = 9\]

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

\[x_{2} = \frac{1 + 3}{2} = 2;\]

\[(x + 1)x(x - 2) > 0\]

\[- 1 < x < 0;\ \ \ x > 2.\]

\[Ответ:\ \ ( - 1;\ 0) \cup (2;\ + \infty).\]

\[3)\ f(x) = \left( x + \frac{1}{x} \right)^{2};\]

\[f^{'}(x) = \left( 1 - \frac{1}{x^{2}} \right) \bullet 2\left( x + \frac{1}{x} \right) > 0;\]

\[\frac{x^{2} - 1}{x^{2}} \bullet \frac{x^{2} + 1}{x} > 0\]

\[\frac{(x + 1)(x - 1)}{x^{3}} > 0\]

\[- 1 < x < 0;\ \ \ x > 1.\]

\[Ответ:\ \ ( - 1;\ 0) \cup (1;\ + \infty).\]

\[4)\ f(x) = \frac{x^{3} + 16}{x};\]

\[f^{'}(x) = \frac{3x^{2} \bullet x - \left( x^{3} + 16 \right)}{x^{2}} > 0;\]

\[\frac{2x^{3} - 16}{x^{2}} > 0\]

\[x^{3} - 8 > 0\]

\[x^{3} > 8\]

\[x > 2.\]

\[Ответ:\ \ (2;\ + \infty).\]

\[5)\ f(x) = (x + 2)^{2}\sqrt{x};\]

\[f^{'}(x) = 2(x + 2)\sqrt{x} + (x + 2)^{2} \bullet \frac{1}{2\sqrt{x}} > 0;\]

\[(x + 2) \bullet \frac{4x + (x + 2)}{2\sqrt{x}} > 0\]

\[\frac{(x + 2)(5x + 2)}{2\sqrt{x}} > 0\]

\[x > 0.\]

\[Ответ:\ \ (0;\ + \infty).\]

\[6)\ f(x) = (x - 3)\sqrt{x};\]

\[f^{'}(x) = \sqrt{x} + (x - 3) \bullet \frac{1}{2\sqrt{x}} > 0;\]

\[\frac{2x + x - 3}{2\sqrt{x}} > 0\]

\[3x - 3 > 0\]

\[3x > 3\]

\[x > 1.\]

\[Ответ:\ \ (1;\ + \infty).\]