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МатематикаРешебник по алгебре и начала математического анализа 10 класс Алимов Задание 809
Задание 809
\[\boxed{\mathbf{809}\mathbf{.}}\]
\[1)\ f(x) = x^{3} - 2x\]
\[f^{'}(x) = \left( x^{3} \right)^{'} - (2x)^{'} = 3x^{2} - 2\]
\[3x^{2} - 2 = 0\]
\[3x^{2} = 2\]
\[x^{2} = \frac{2}{3}\]
\[x = \pm \sqrt{\frac{2}{3}} = \pm \sqrt{\frac{6}{9}} = \pm \frac{\sqrt{6}}{3}.\]
\[Ответ:\ \ \pm \frac{\sqrt{6}}{3}.\]
\[2)\ f(x) = - x^{2} + 3x + 1\]
\[f^{'}(x) = - \left( x^{2} \right)^{'} + (3x + 1)^{'} =\]
\[= - 2x + 3\]
\[- 2x + 3 = 0\]
\[2x = 3\]
\[x = \frac{3}{2} = 1,5.\]
\[Ответ:\ \ 1,5.\]
\[3)\ f(x) = 2x^{3} + 3x^{2} - 12 - 3\]
\[f^{'}(x) =\]
\[= 2 \bullet \left( x^{3} \right)^{'} + 3 \bullet \left( x^{2} \right)^{'} - (12 + 3)^{'} =\]
\[= 2 \bullet 3x^{2} + 3 \bullet 2x - 12 =\]
\[= 6x^{2} + 6x - 12\]
\[6x^{2} + 6x - 12 = 0\]
\[x^{2} + x - 2 = 0\]
\[D = 1 + 8 = 9\]
\[x_{1} = \frac{- 1 - 3}{2} = - 2\ \]
\[x_{2} = \frac{- 1 + 3}{2} = 1.\]
\[Ответ:\ \ - 2\ \ 1.\]
\[4)\ f(x) = x^{3} + 2x^{2} - 7x + 1\]
\[f^{'}(x) = \left( x^{3} \right)^{'} + 2 \bullet \left( x^{2} \right)^{'} - (7x - 1)^{'} =\]
\[= 3 \bullet x^{2} + 2 \bullet 2x - 7 =\]
\[= 3x^{2} + 4x - 7\]
\[3x^{2} + 4x - 7 = 0\]
\[D = 16 + 84 = 100\]
\[x_{1} = \frac{- 4 - 10}{2 \bullet 3} = \frac{- 14}{6} = - \frac{7}{3}\]
\[x_{2} = \frac{- 4 + 10}{2 \bullet 3} = 1.\]
\[Ответ:\ \ - \frac{7}{3}\ \ 1.\]
\[5)\ f(x) = 3x^{4} - 4x^{3} - 12x^{2}\]
\[f^{'}(x) =\]
\[= 3 \bullet \left( x^{4} \right)^{'} - 4 \bullet \left( x^{3} \right)^{'} - 12 \bullet \left( x^{2} \right)^{'} =\]
\[= 3 \bullet 4x^{3} - 4 \bullet 3x^{2} - 12 \bullet 2x =\]
\[= 12x^{3} - 12x^{2} - 24x\]
\[12x^{3} - 12x^{2} - 24x = 0\]
\[12x \bullet \left( x^{2} - x - 2 \right) = 0\]
\[12x = 0\]
\[x = 0\]
\[x^{2} - x - 2 = 0\]
\[D = 1 + 8 = 9\]
\[x_{1} = \frac{1 - 3}{2} = - 1\]
\[x_{2} = \frac{1 + 3}{2} = 2.\]
\[Ответ:\ \ - 1\ \ 0\ \ 2.\]
\[6)\ f(x) = x^{4} + 4x^{3} - 8x^{2} - 5\]
\[f^{'}(x) =\]
\[= \left( x^{4} \right)^{'} + 4 \bullet \left( x^{3} \right)^{'} - 8 \bullet \left( x^{2} \right)^{'} - (5)^{'} =\]
\[= 4x^{3} + 4 \bullet 3x^{2} - 8 \bullet 2x - 0 =\]
\[= 4x^{3} + 12x^{2} - 16x\]
\[4x^{3} + 12x^{2} - 16x = 0\]
\[4x \bullet \left( x^{2} + 3x - 4 \right) = 0\]
\[4x = 0\]
\[x = 0\]
\[x^{2} + 3x - 4 = 0\]
\[D = 9 + 16 = 25\]
\[x_{1} = \frac{- 3 - 5}{2} = - 4\]
\[x_{2} = \frac{- 3 + 5}{2} = 1.\]
\[Ответ:\ \ - 4\ \ 0\ \ 1.\]
