Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 20

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

\[\textbf{а)}\ f(x) = x^{2} + 2x - 3;\ \ x_{0} = 0\]

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

\[y_{0} = f(0) = - 3;\]

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

\[y - y_{0} = k\left( x - x_{0} \right)\]

\[y + 3 = 2 \cdot (x - 0)\]

\[y + 3 = 2x\]

\[y = 2x - 3.\]

\[Уравнение\ касательной:\ \]

\[\ y = 2x - 3.\]

\[\textbf{б)}\ f(x) = x^{2} + 2x - 3;\ \ x_{0} = 1\]

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

\[y_{0} = f(1) = 1 + 2 - 3 = 0;\]

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

\[y - y_{0} = k\left( x - x_{0} \right)\]

\[y + 0 = 4 \cdot (x - 1)\]

\[y = 4x - 4.\]

\[Уравнение\ касательной:\ \]

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

\[\textbf{в)}\ f(x) = x^{2} + 2x - 3;\ \ x_{0} = - 1\]

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

\[y_{0} = f( - 1) = 1 - 2 - 3 = - 4;\]

\[k = f^{'}( - 1) = - 2 + 2 = 0.\]

\[y - y_{0} = k\left( x - x_{0} \right)\]

\[y + 4 = 0 \cdot (x + 1)\]

\[y + 4 = 0\]

\[y = - 4.\]

\[Уравнение\ касательной:\ \ y = - 4.\]

\[\textbf{г)}\ f(x) = x^{2} + 2x - 3;\ \ x_{0} = - 2\]

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

\[y_{0} = f( - 2) = 4 - 4 - 3 = - 3;\]

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

\[y - y_{0} = k\left( x - x_{0} \right)\]

\[y + 3 = - 2 \cdot (x + 2)\]

\[y + 3 = - 2x - 4\]

\[y = - 2x - 7.\]

\[Уравнение\ касательной:\ \]

\[\ y = - 2x - 7.\]