ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1498

\[\boxed{\mathbf{1498}\mathbf{.}}\]

\[y = f(a) + f^{'}(a)(x - a).\]

\[1)\ f(x) = x \bullet \ln{2x};\text{\ \ }x_{0} = 0,5;\]

\[f^{'}(x) = (x)^{'} \bullet \ln{2x} + x \bullet \left( \ln{2x} \right)^{'} =\]

\[= 1 \bullet \ln{2x} + x \bullet \frac{2}{2x} = \ln{2x} + 1;\]

\[f^{'}(0,5) = \ln(2 \bullet 0,5) + 1 =\]

\[= \ln 1 + 1 = 0 + 1 = 1;\]

\[f(0,5) = 0,5 \bullet \ln(2 \bullet 0,5) =\]

\[= 0,5 \bullet \ln 1 = 0,5 \bullet 0 = 0;\]

\[y = 0 + 1(x - 0,5) = x - 0,5.\]

\[Ответ:\ \ y = x - 0,5.\]

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

\[u = - x;\ f(u) = 2^{u}:\]

\[f^{'}(x) = ( - x)^{'} \bullet \left( 2^{u} \right)^{'} =\]

\[= - 1 \bullet 2^{u} \bullet \ln 2 = - 2^{- x} \bullet \ln 2;\]

\[f^{'}(1) = - 2^{- 1} \bullet \ln 2 = - \frac{1}{2}\ln 2;\]

\[f(1) = 2^{- 1} = \frac{1}{2};\]

\[y = \frac{1}{2} - \frac{1}{2}\ln 2(x - 1) =\]

\[= \frac{1}{2} - \frac{1}{2}x\ln 2 + \frac{1}{2}\ln 2 =\]

\[= \frac{1}{2}\left( 1 + \ln 2 - x\ln 2 \right).\]

\[Ответ:\ \ y = \frac{1}{2}\left( 1 + \ln 2 - x\ln 2 \right).\]