Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 21

\[\boxed{\mathbf{21}\mathbf{.}}\]

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

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

\[f^{'}(x) = 3x^{2} - 6x + 1;\]

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

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

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

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

\[y + 1 = x\]

\[y = x - 1.\]

\[Уравнение\ касательной:\ \ \]

\[y = x - 1.\]

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

\[x_{0} = 1\]

\[f^{'}(x) = 3x^{2} - 6x + 1;\]

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

\[k = f^{'}(1) = 3 - 6 + 1 = - 2.\]

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

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

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

\[y = - 2x.\]

\[Уравнение\ касательной:\ \]

\[\ y = - 2x.\]

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

\[x_{0} = - 1\]

\[f^{'}(x) = 3x^{2} - 6x + 1;\]

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

\[= - 1 - 3 - 1 - 1 = - 6;\]

\[k = f^{'}( - 1) = 3 + 6 + 1 = 10.\]

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

\[y + 6 = 10 \cdot (x + 1)\]

\[y + 6 = 10x + 10\]

\[y = 10x + 4.\]

\[Уравнение\ касательной:\ \ \]

\[y = 10x + 4.\]

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

\[x_{0} = - 2\]

\[f^{'}(x) = 3x^{2} - 6x + 1;\]

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

\[= - 8 - 12 - 2 - 1 = - 23;\]

\[k = f^{'}( - 2) = 12 + 12 + 1 = 25.\]

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

\[y + 23 = 25 \cdot (x + 2)\]

\[y + 23 = 25x + 50\]

\[y = 25x + 27.\]

\[Уравнение\ касательной:\ \]

\[\ y = 25x + 27.\]