Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 20

\[\boxed{\mathbf{20}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = 4x^{3} - 3x^{2} - 2x;\ \]

\[\ x_{0} = 0;\]

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

\[- 2 \cdot (x)^{'} = 4 \cdot 3x^{2} -\]

\[- 3 \cdot 2x - 2 \cdot 1 =\]

\[= 12x^{2} - 6x - 2;\]

\[f^{'}(0) = 12 \cdot 0 - 6 \cdot 0 - 2 = - 2.\]

\[\textbf{б)}\ f(x) = - 5x^{3} + 7x^{2} + x;\ \ \]

\[x_{0} = 1\]

\[f^{'}(x) = - 5 \cdot \left( x^{3} \right)^{'} + 7 \cdot \left( x^{2} \right)^{'} +\]

\[+ x^{'} = - 5 \cdot 3x^{2} + 7 \cdot 2x + 1 =\]

\[= - 15x^{2} + 14x + 1;\]

\[f^{'}(1) = - 15 \cdot 1 + 14 \cdot 1 + 1 = 0.\]

\[\textbf{в)}\ f(x) = - x^{3} + 4x + 5;\ \]

\[\ x_{0} = - 1;\]

\[f^{'}(x) = - 1 \cdot \left( x^{3} \right)^{'} + 4 \cdot x^{'} + 5^{'} =\]

\[= - 1 \cdot 3x^{2} + 4 \cdot 1 + 0 =\]

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

\[f^{'}( - 1) = - 3 \cdot 1 + 4 = 1.\]

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

\[\ x_{0} = - 2;\]

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

\[+ 3^{'} = 4 \cdot 3x^{2} + 2x - 3 \cdot 1 + 0 =\]

\[= 12x^{2} + 2x - 3;\]

\[f^{'}( - 2) = 12 \cdot 4 + 2 \cdot ( - 2) - 3 =\]

\[= 48 - 7 = 41.\]