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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 29
Задание 29
\[\boxed{\mathbf{29.}}\]
\[\textbf{а)}\ f(x) = e^{x};\ \ x_{0} = - 2\]
\[f^{'}(x) = e^{x};\]
\[y_{0} = f( - 2) = e^{- 2} = \frac{1}{e^{2}};\]
\[k = f^{'}( - 2) = e^{- 2} = \frac{1}{e^{2}};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - \frac{1}{e^{2}} = \frac{1}{e^{2}}(x + 2)\]
\[y - \frac{1}{e^{2}} = \frac{1}{e^{2}}x + \frac{2}{e^{2}}\]
\[y = \frac{1}{e^{2}}x + \frac{3}{e^{2}}.\]
\[Уравнение\ касательной:\]
\[y = \frac{1}{e^{2}}x + \frac{3}{e^{2}}.\]
\[\textbf{б)}\ f(x) = e^{x};\ \ x_{0} = - 1\]
\[f^{'}(x) = e^{x};\]
\[y_{0} = f( - 1) = e^{- 1} = \frac{1}{e};\]
\[k = f^{'}( - 1) = e^{- 1} = \frac{1}{e};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - \frac{1}{e} = \frac{1}{e}(x + 1)\]
\[y - \frac{1}{e} = \frac{1}{e}x + \frac{1}{e}\]
\[y = \frac{1}{e}x + \frac{2}{e}.\]
\[Уравнение\ касательной:\]
\[y = \frac{1}{e}x + \frac{2}{e}.\]
\[\textbf{в)}\ f(x) = e^{x};\ \ x_{0} = 0\]
\[f^{'}(x) = e^{x};\]
\[y_{0} = f(0) = e^{0} = 1;\]
\[k = f^{'}(0) = e^{0} = 1;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 1 = 1(x - 0)\]
\[y - 1 = x\]
\[y = x + 1\]
\[Уравнение\ касательной:\]
\[y = x + 1.\]
\[\textbf{г)}\ f(x) = e^{x};\ \ x_{0} = 2\]
\[f^{'}(x) = e^{x};\]
\[y_{0} = f(2) = e^{2};\]
\[k = f^{'}(2) = e^{2};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - e^{2} = e^{2}(x - 2)\]
\[y - e^{2} = e^{2}x - 2e^{2}\]
\[y = e^{2}x - e^{2}.\]
\[Уравнение\ касательной:\]
\[y = e^{2}x - e^{2}.\]