Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 986

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\[1)\ f^{'}(x) = x\]

\[f^{'}(x) = \frac{1}{2} \bullet 2x + 0 =\]

\[= \frac{1}{2} \bullet \left( x^{2} \right)^{'} + (C)^{'} = \left( \frac{x^{2}}{2} + C \right)^{'}.\]

\[M( - 1;\ 3):\]

\[3 = \frac{( - 1)^{2}}{2} + C\]

\[C = 3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2}.\]

\[Ответ:\ \ f(x) = \frac{x^{2} + 5}{2}.\]

\(2)\ f^{'}(x) = \sqrt{x}\)

\[f^{'}(x) = \frac{2}{3} \bullet \frac{3}{2}x^{\frac{1}{2}} + 0 =\]

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

\[= \left( \frac{2x\sqrt{x}}{3} + C \right)^{'}.\]

\[M(9;\ 10):\]

\[10 = \frac{2 \bullet 9 \bullet \sqrt{9}}{3} + C\]

\[C = 10 - \frac{18 \bullet 3}{3} = 10 - 18 = - 8.\]

\[Ответ:\ \ f(x) = \frac{2x\sqrt{x}}{3} - 8.\]