Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 24

\[\boxed{\mathbf{24}\mathbf{.}}\]

\[\textbf{а)}\ y = \sqrt{x^{2} + 6x + 9} =\]

\[= \sqrt{(x + 3)^{2}} = |x + 3|\]

\[\textbf{б)}\ y = \sqrt{4x^{2} - 4x + 1} =\]

\[= \sqrt{(2x - 1)^{2}} = |2x - 1|\]

\[\textbf{в)}\ y = \sqrt{- x^{2} + x + 6}\]

\[y^{2} = - x^{2} + x + 6\]

\[y^{2} + x^{2} - x = 6\]

\[y^{2} + x^{2} - x + \frac{1}{4} - \frac{1}{4} = 6\]

\[y^{2} + \left( x - \frac{1}{2} \right)^{2} = 6 + \frac{1}{4}\]

\[\left( x - \frac{1}{2} \right)^{2} + y^{2} = \frac{25}{4}\]

\[окружность\ O\left( \frac{1}{2};0 \right);\ \ R = \frac{5}{2}.\]

\[\textbf{г)}\ y = \sqrt{- x^{2} + 2x + 8}\]

\[y^{2} = - x^{2} + 2x + 8\]

\[y^{2} + x^{2} - 2x = 8\]

\[y^{2} + x^{2} - 2x + 1 - 1 = 8\]

\[y^{2} + (x - 1)^{2} = 8 + 1\]

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

\[окружность\ O(1;0);\ \ R = 3.\]