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МатематикаРешебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 28
Задание 28
\[\boxed{\mathbf{28.}}\]
\[\textbf{а)}\ f(x) = 2^{x};\ \ \ x_{0} = - 1\]
\[f^{'}(x) = 2^{x}\ln 2;\]
\[y_{0} = f( - 1) = 2^{- 1} = \frac{1}{2};\]
\[k = f^{'}( - 1) = 2^{- 1}\ln 2 = \frac{\ln 2}{2};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - \frac{1}{2} = \frac{\ln 2}{2}(x + 1)\]
\[y - \frac{1}{2} = \frac{\ln 2}{2}x + \frac{\ln 2}{2}\]
\[y = \frac{\ln 2}{2}x + \frac{\ln 2}{2} + \frac{1}{2}\]
\[y = \frac{\ln 2}{2}x + \frac{\ln 2 + 1}{2}.\]
\[Уравнение\ касательной:\ \]
\[\ y = \frac{\ln 2}{2}x + \frac{\ln 2 + 1}{2}.\]
\[\textbf{б)}\ f(x) = 2^{x};\ \ \ x_{0} = 0\]
\[f^{'}(x) = 2^{x}\ln 2;\]
\[y_{0} = f(0) = 2^{0} = 1;\]
\[k = f^{'}(0) = 2^{0}\ln 2 = \ln 2;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 1 = \ln 2(x + 0)\]
\[y - 1 = \ln 2x\]
\[y = x\ln 2 + 1.\]
\[Уравнение\ касательной:\ \ \]
\[y = x\ln 2 + 1.\]
\[\textbf{в)}\ f(x) = 2^{x};\ \ \ x_{0} = 2\]
\[f^{'}(x) = 2^{x}\ln 2;\]
\[y_{0} = f(2) = 2^{2} = 4;\]
\[k = f^{'}(2) = 2^{2}\ln 2 = 4\ln 2;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 4 = 4\ln 2(x - 2)\]
\[y - 4 = 4\ln 2x - 8\ln 2\]
\[y = 4x\ln 2 - 8\ln 2 + 4.\]
\[Уравнение\ касательной:\ \ \]
\[y = 4x\ln 2 - 8\ln 2 + 4.\]
\[\textbf{г)}\ f(x) = 2^{x};\ \ \ x_{0} = 3\]
\[f^{'}(x) = 2^{x}\ln 2;\]
\[y_{0} = f(3) = 2^{3} = 8;\]
\[k = f^{'}(3) = 2^{3}\ln 2 = 8\ln 2;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 8 = 8\ln 2(x - 3)\]
\[y - 8 = 8\ln 2x - 24\ln 2\]
\[y = 8x\ln 2 - 24\ln 2 + 8.\]
\[Уравнение\ касательной:\]
\[\ \ y = 8x\ln 2 - 24\ln 2 + 8.\]
