Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 825

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

\[f^{'}(x) = \left( x^{4} \right)^{'} - 4 \bullet \left( x^{2} \right)^{'} + (1)^{'} =\]

\[= 4x^{3} - 4 \bullet 2x + 0 =\]

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

\[4x^{3} - 8x > 0\]

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

\[\left( x - \sqrt{2} \right) \bullet 4x \bullet \left( x + \sqrt{2} \right) > 0\]

\[- \sqrt{2} < x < 0\ или\ x > \sqrt{2}.\]

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

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

\[f^{'}(x) =\]

\[= 3 \bullet \left( x^{4} \right)^{'} - 4 \bullet \left( x^{3} \right)^{'} - 12 \bullet \left( x^{2} \right)^{'} + (3)^{'} =\]

\[= 3 \bullet 4x^{3} - 4 \bullet 3x^{2} - 12 \bullet 2x + 0 =\]

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

\[= 12x \bullet \left( x^{2} - x - 2 \right)\]

\[x^{2} - x - 2 = 0\]

\[D = 1 + 8 = 9\]

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

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

\[(x + 1) \bullet 12x \bullet (x - 2) > 0\]

\[- 1 < x < 0\ или\text{\ x} > 2.\]

\[x \in ( - 1\ 0) \cup (2\ + \infty).\]

\[3)\ f(x) = (x + 2)^{2} \bullet \sqrt{x}\]

\[f^{'}(x) =\]

\[= {(x + 2)^{2}}^{'} \bullet \sqrt{x} + (x + 2)^{2} \bullet \left( \sqrt{x} \right)^{'} =\]

\[= 2\sqrt{x} \bullet (x + 2) + \frac{(x + 2)^{2}}{2\sqrt{x}} =\]

\[= \frac{4x(x + 2) + (x + 2)^{2}}{2\sqrt{x}} =\]

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

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

\[(x + 2)(5x + 2) > 0\]

\[x < - 2\ или\ x > - 0,4\]

\[Выражение\ имеет\ смысл\ при:\]

\[x > 0.\]

\[x \in (0\ + \infty).\]

\[4)\ f(x) = (x - 3) \bullet \sqrt{x}\]

\[f^{'}(x) =\]

\[= (x - 3)^{'} \bullet \sqrt{x} + (x - 3) \bullet \left( \sqrt{x} \right)^{'} =\]

\[= 1 \bullet \sqrt{x} + \frac{x - 3}{2\sqrt{x}} =\]

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

\[3x - 3 > 0\]

\[3(x - 1) > 0\]

\[x - 1 > 0\]

\[x > 1.\]

\[Выражение\ имеет\ смысл\ при:\]

\[x > 0.\]

\[Ответ:\ \ x \in (1\ + \infty).\]