Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 897

\[\boxed{\mathbf{897}\mathbf{.}}\]

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

\[g(x) = - x^{2} + 6x - 10:\]

\[g^{'}(x) = - \left( x^{2} \right)^{'} + (6x - 10)^{'} =\]

\[= - 2x + 6\]

\[f^{'}(a) = \left( x^{2} \right)^{'} - (4x - 3)^{'} =\]

\[= 2x - 4 = 2a - 4\]

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

\[y = a^{2} - 4a + 3 + (2a - 4) \bullet (x - a) =\]

\[= a^{2} - 4a + 3 + 2xa - 2a^{2} - 4x + 4a =\]

\[= - a^{2} + 2ax - 4x + 3 =\]

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

\[g^{'}(b) = - \left( x^{2} \right) + (6x - 10)^{'} =\]

\[= - 2x + 6 = 6 - 2b\]

\[g(b) = 6b - b^{2} - 10\]

\[y = 6b - b^{2} - 10 + (6 - 2b) \bullet (x - b) =\]

\[= 6b - b^{2} - 10 + 6x - 6b - 2bx + 2b^{2} =\]

\[= b^{2} - 2bx + 6x - 10 =\]

\[= (6 - 2b)x + \left( b^{2} - 10 \right).\]

\[Касательные\ совпадают:\]

\[\left\{ \begin{matrix} 2a - 4 = 6 - 2b\ \\ 3 - a^{2} = b^{2} - 10 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 2a + 2b = 10\ \ \ \ \ \ \ \\ b^{2} + a^{2} - 13 = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a = 5 - b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b^{2} + a^{2} - 13 = 0 \\ \end{matrix} \right.\ \]

\[b^{2} + (5 - b)^{2} - 13 = 0\]

\[b^{2} + 25 - 10b + b^{2} - 13 = 0\]

\[2b^{2} - 10b + 12 = 0\]

\[b^{2} - 5x + 6 = 0\]

\[D = 25 - 24 = 1\]

\[b_{1} = \frac{5 - 1}{2} = 2\text{\ \ }\]

\[b_{2} = \frac{5 + 1}{2} = 3\]

\[a_{1} = 5 - 2 = 3\]

\[a_{2} = 5 - 3 = 2.\]

\[Общие\ касательные:\]

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

\[= (6 - 4)x + (3 - 9) = 2x - 6\]

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

\[= (4 - 4)x + (3 - 4) = - 1.\]

\[Ответ:\ \ y = 2x - 6\ \ и\ \ y = - 1.\]