ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 887

\[\boxed{\mathbf{887}\mathbf{.}}\]

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

\[1)\ f(x) = ax^{7} + x^{3} - 1\]

\[f^{'}(x) = a \bullet \left( x^{7} \right)^{'} + \left( x^{3} \right)^{'} - (1)^{'} =\]

\[= 7ax^{6} + 3x^{2} - 0 =\]

\[= x^{2} \bullet \left( 7ax^{4} + 3 \right)\]

\[7ax^{4} + 3 < 0\]

\[7ax^{4} < - 3\]

\[ax^{4} < - \frac{3}{7}\]

\[x^{4} < - \frac{3}{7a}.\]

\[Ответ:\ \ a \geq 0.\]

\[2)\ f(x) = x^{5} + ax^{3} + 3\]

\[f^{'}(x) = \left( x^{5} \right)^{'} + a \bullet \left( x^{3} \right)^{'} + (3)^{'} =\]

\[= 5x^{4} + 3ax^{2} + 0 =\]

\[= 5x^{4} + 3ax^{2} = x^{2} \bullet \left( 5x^{2} + 3a \right)\]

\[5x^{2} + 3a < 0\]

\[5x^{2} < - 3a\]

\[x^{2} < - \frac{3a}{5}.\]

\[Ответ:\ \ a \geq 0.\]

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

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

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

\[= 1 \bullet \sqrt{x} + (x + a) \bullet \frac{1}{2\sqrt{x}} =\]

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

\[Имеет\ смысл\ при:\ \]

\[x < 0.\]

\[3x + a < 0\]

\[3x < - a\]

\[x < - \frac{a}{3}.\]

\[Ответ:\ \ a \geq 0.\]

\[4)\ f(x) = x + \frac{a}{x}\]

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

\[1 - \frac{a}{x^{2}} < 0\]

\[x^{2} - a < 0\]

\[x^{2} < a.\]

\[Ответ:\ \ a \leq 0.\]