Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 19

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\[\textbf{а)}\ f(x) = x^{2};\ \ x_{0} = 0\]

\[f^{'}(x) = 2x;\]

\[y_{0} = f(0) = 0;\]

\[k = f^{'}(0) = 2 \cdot 0 = 0.\]

\[y - 0 = 0(x - 0)\]

\[y = 0.\]

\[Уравнение\ касательной:\ \ y = 0.\]

\[\textbf{б)}\ f(x) = x^{2};\ \ x_{0} = 1\]

\[f^{'}(x) = 2x;\]

\[y_{0} = f(1) = 1;\]

\[k = f^{'}(1) = 2 \cdot 1 = 2.\]

\[y - 1 = 2(x - 1)\]

\[y - 1 = 2x - 2\]

\[y = 2x - 1.\]

\[Уравнение\ касательной:\ \ \]

\[y = 2x - 1.\]

\[\textbf{в)}\ f(x) = x^{2};\ \ x_{0} = 2\]

\[f^{'}(x) = 2x;\]

\[y_{0} = f(2) = 4;\]

\[k = f^{'}(2) = 2 \cdot 2 = 4.\]

\[y - 4 = 4(x - 2)\]

\[y - 4 = 4x - 8\]

\[y = 4x - 4.\]

\[Уравнение\ касательной:\ \]

\[\ y = 4x - 4.\]

\[\textbf{г)}\ f(x) = x^{2};\ \ x_{0} = - 1\]

\[f^{'}(x) = 2x;\]

\[y_{0} = f( - 1) = 1;\]

\[k = f^{'}( - 1) = - 2.\]

\[y - 1 = - 2(x + 1)\]

\[y - 1 = - 2x - 2\]

\[y = - 2x - 1.\]

\[Уравнение\ касательной:\]

\[\ \ y = - 2x - 1.\]