Страница 244 номер 825, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(f(x) = x^{4} - 4x^{2} + 1\)
\(f'(x) = 4x^{3} - 8x, 4x^{3} - 8x > 0, 4x(x^{2} - 2) > 0,\)

Методом интервалов: \(x_{1} = 0, x_{2{,}3} = \pm \sqrt{2};\)

\(x \in \left(-\sqrt{2};\ 0\right) \cup \left(\sqrt{2};\ +\infty\right);\)

2) \(f(x) = 3x^{4} - 4x^{3} - 12x^{2} + 3\)
\(f'(x) = 12x^{3} - 12x^{2} - 24x\)
\(12x(x^{2} - x - 2) > 0\)
\(x(x^{2} - x - 2) = 0\)
\(x_{1} = 0\)
\(x_{2{,}3} = \dfrac{1 \pm \sqrt{1 + 8}}{2}, x_{2} = 2, x_{3} = -1,\)
\(x \in (-1;\ 0) \cup (2;\ +\infty);\)

3) \(f(x) = (x + 2)^{2}\sqrt{x}\)
\(f'(x) = ((x + 2)^{2})'\sqrt{x} + (x + 2)^{2}\left(\sqrt{x}\right)' =\)
\(= 2(x + 2)\sqrt{x} + \dfrac{(x + 2)^{2}}{2\sqrt{x}} =\)
\(= (x + 2)\left(\dfrac{4x + x + 2}{2\sqrt{x}}\right) = \dfrac{(x + 2)(5x + 2)}{2\sqrt{x}},\)
\(f'(x) > 0, \sqrt{x} > 0, x > 0; (x + 2)(5x + 2) > 0,\)
\(x_{1} = -2, x_{2} = -\frac{2}{5}. x \in (0;\ +\infty)\), т.к. \(x > 0\);

4) \(f(x) = (x - 3)\sqrt{x}\)
\(f'(x) = (x - 3)'\sqrt{x} + (x - 3)\left(\sqrt{x}\right)' = \sqrt{x} + \dfrac{x - 3}{2\sqrt{x}} = \dfrac{3x - 3}{2\sqrt{x}},\)
\(f'(x) > 0, 2\sqrt{x} > 0, x > 0; 3x - 3 > 0, x > 1,\)
\(x \in (1;\ +\infty).\)