ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 806

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

\[f^{'}(x) = \left( x^{2} \right)^{'} - (2x - 1)^{'} =\]

\[= 2x - 2\]

\[f^{'}(0) = 2 \bullet 0 - 2 = - 2\]

\[f^{'}(2) = 2 \bullet 2 - 2 = 4 - 2 = 2.\]

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

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

\[f^{'}(0) = 3 \bullet 0^{2} - 2 = - 2\]

\[f^{'}(2) = 3 \bullet 2^{2} - 2 = 3 \bullet 4 - 2 =\]

\[= 12 - 2 = 10.\]

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

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

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

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

\[f^{'}(2) = - 3 \bullet 2^{2} + 2 \bullet 2 =\]

\[= - 3 \bullet 4 + 4 = - 12 + 4 = - 8.\]

\[4)\ f(x) = x^{2} + x + 1\]

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

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

\[f^{'}(2) = 2 \bullet 2 + 1 = 4 + 1 = 5.\]