Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 800

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\[Уравнение\ функции\ имеет\ вид:\ \ \]

\[y = ax^{2} + c.\]

\[\textbf{а)}\ (0\ 1)\ и\ (1\ 2):\]

\[1 = a \bullet 0^{2} + c = 0 + c = c\]

\[2 = a \bullet 1^{2} + 1 = a + 1\ \]

\[a = 2 - 1 = 1.\]

\[y(x) = x^{2} + 1\]

\[y^{'}(x) = \left( x^{2} \right)^{'} + (1)^{'} =\]

\[= 2x^{2 - 1} + 0 = 2x.\]

\[Ответ:\ \ y(x) = x^{2} + 1\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }y^{'}(x) = 2x.\]

\[\textbf{б)}\ (0\ 1)\ и\ \left( \frac{3}{2}\ - \frac{3}{2} \right):\]

\[1 = a \bullet 0^{2} + c = 0 + c = c\]

\[- \frac{3}{2} = a \bullet \left( \frac{3}{2} \right)^{2} + 1 = \frac{9}{4}a + 1\]

\[a = \left( - \frac{3}{2} - 1 \right) \bullet \frac{4}{9}\]

\[a = - \frac{5}{2} \bullet \frac{4}{9} = - \frac{10}{9}.\]

\[y(x) = 1 - \frac{10}{9}x^{2}\]

\[y^{'}(x) = (1)^{'} - \frac{10}{9} \bullet \left( x^{2} \right)^{'} =\]

\[= 0 - \frac{10}{9} \bullet 2x^{2 - 1} = - \frac{20}{9}x.\]

\[Ответ:\ \ y(x) = 1 - \frac{10}{9}x^{2}\text{\ \ }\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }y^{'}(x) = - \frac{20}{9}\text{x.}\]