Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 38

\[\boxed{\mathbf{38.}}\]

\[f\left( x_{0} + \mathrm{\Delta}x \right) \approx f\left( x_{0} \right) + f^{'}\left( x_{0} \right)\mathrm{\Delta}x\]

\[\textbf{а)}\ f(x) = x^{2};\ \ x_{0} = 5;\ \ \mathrm{\Delta}x = 0,01\]

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

\[f^{'}\left( x_{0} \right) = f^{'}(5) = 10;\]

\[f\left( x_{0} \right) = f(5) = 25;\]

\[f\left( x_{0} + \mathrm{\Delta}x \right) \approx 25 + 10 \cdot 0,01 =\]

\[= 25 + 0,1 = 25,1.\]

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

\[\mathrm{\Delta}x = - 0,01\]

\[f^{'}(x) = 3x^{2};\]

\[f^{'}\left( x_{0} \right) = 27;\]

\[f\left( x_{0} \right) = f(3) = 27;\]

\[f\left( x_{0} + \mathrm{\Delta}x \right) \approx 27 +\]

\[+ 27 \cdot ( - 0,01) = 27 - 0,27 =\]

\[= 26,73.\]

\[\textbf{в)}\ f(x) = \sqrt{x} = x^{\frac{1}{2}};\ \ x_{0} = 16;\ \]

\[\ \mathrm{\Delta}x = 0,02\]

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

\[f^{'}\left( x_{0} \right) = f^{'}(16) = \frac{1}{2 \cdot 4} =\]

\[= \frac{1}{8} = 0,125;\ \]

\[f\left( x_{0} \right) = f(16) = \sqrt{16} = 4;\]

\[f\left( x_{0} + \mathrm{\Delta}x \right) \approx 4 + 0,125 \cdot 0,02 =\]

\[= 4 + 0,0025 = 4,0025.\]

\[\textbf{г)}\ f(x) = \ln x;\ \ x_{0} = e;\ \ \]

\[\mathrm{\Delta}x = 0,01\]

\[f^{'}(x) = \frac{1}{x};\]

\[f^{'}\left( x_{0} \right) = f^{'}(e) = \frac{1}{e};\]

\[f\left( x_{0} \right) = f(e) = \ln e = 1;\]

\[f\left( x_{0} + \mathrm{\Delta}x \right) \approx 1 + \frac{1}{e} \cdot 0,01 \approx 1 +\]

\[+ \frac{0,01}{2,7} = 1 + 0,0037 = 1,0037.\]

\[\textbf{д)}\ f(x) = 2^{x};\ \ x_{0} = 2;\ \]

\[\ \mathrm{\Delta}x = - 0,02\]

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

\[f(2 - 0,02) \approx f(2) +\]

\[+ f^{'}(2) \cdot ( - 0,02) = 2^{2} +\]

\[+ 2^{2}\ln 2 \cdot ( - 0,02) =\]

\[= 4 - 0,08\ln 2 =\]

\[= 4 - 0,08 \cdot 0,7 = 3,94.\]