ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 886

\[\boxed{\mathbf{886}\mathbf{.}}\]

\[f'x) = 0\]

\[1)\ f(x) = ax^{2} - \frac{1}{x^{2}}\]

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

\[= a \bullet 2x - ( - 2) \bullet x^{- 3} =\]

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

\[Не\ имеет\ корней:\]

\[ax + \frac{1}{x^{3}} = 0\]

\[ax^{4} + 1 = 0\]

\[ax^{4} = - 1\]

\[x^{4} = - \frac{1}{a}.\]

\[Ответ:\ \ a \geq 0.\]

\[2)\ f(x) = ax + \frac{1}{x}\]

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

\[Не\ имеет\ корней:\]

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

\[ax^{2} - 1 = 0\]

\[ax^{2} = 1\]

\[x^{2} = \frac{1}{a}.\]

\[Ответ:\ \ a \leq 0.\]

\[3)\ f(x) = ax^{3} + 3x^{2} + 6x\]

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

\[= a \bullet \left( x^{3} \right)^{'} + 3 \bullet \left( x^{2} \right)^{'} + (6x)^{'} =\]

\[= a \bullet 3x^{2} + 3 \bullet 2x + 6 =\]

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

\[Не\ имеет\ корней:\]

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

\[D = 2^{2} - 4 \bullet a \bullet 2 = 4 - 8a =\]

\[= 4 \bullet (1 - 2a) < 0\]

\[1 - 2a < 0\]

\[2a > 1\]

\[a > 0,5.\]

\[Ответ:\ \ a > 0,5.\]

\[4)\ f(x) = x^{3} + 6x^{2} + ax\]

\[f^{'}(x) = \left( x^{3} \right)^{'} + 6 \bullet \left( x^{2} \right) + a \bullet (x)^{'} =\]

\[= 3x^{2} + 6 \bullet 2x + a =\]

\[= 3x^{2} + 12x + a\]

\[Не\ имеет\ корней:\]

\[3x^{2} + 12x + a = 0\]

\[D = 12^{2} - 4 \bullet 3 \bullet a = 144 - 12a =\]

\[= 12 \bullet (12 - a) < 0\]

\[12 - a < 0\]

\[a > 12.\]

\[Ответ:\ \ a > 12.\]