Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 39

\[\boxed{\mathbf{39}\mathbf{.}}\]

\[\textbf{а)}\ {5,01}^{2} = (5 + 0,01)^{2};\]

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

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

\[f(5 + 0,01) \approx f(5) + f^{'}(5) \cdot 0,01 = 5^{2} + 2 \cdot 5 \cdot 0,01 = 25 + 0,1 = 25,1.\]

\[\textbf{б)}\ {7,98}^{2} = (8 - 0,02)^{2};\]

\[f(x) = x^{2};\ \ x_{0} = 8;\ \ \mathrm{\Delta}x = - 0,02;\]

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

\[f(8 - 0,02) \approx f(8) +\]

\[+ f^{'}(8) \cdot ( - 0,02) = 64 -\]

\[- 16 \cdot 0,02 = 64 - 0,32 =\]

\[= 63,68.\]

\[\textbf{в)}\ {2,99}^{3} = (3 - 0,01)^{3};\]

\[f(x) = x^{3};\ \ x_{0} = 3;\ \ \mathrm{\Delta}x = - 0,01\]

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

\[f(3 - 0,01) \approx f(3) +\]

\[+ f^{'}(3) \cdot ( - 0,01) = 27 +\]

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

\[= 27 - 0,27 = 26,73.\]

\[\textbf{г)}\ \sqrt{24,1} = \sqrt{(25 - 0,9)};\]

\[f(x) = \sqrt{x};\ \ x_{0} = 25;\ \ \]

\[\mathrm{\Delta}x = - 0,9:\]

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

\[f(25 - 0,9) \approx f(25) +\]

\[+ f^{'}(25) \cdot ( - 0,9) = 5 - \frac{0,9}{2 \cdot 5} =\]

\[= 5 - 0,09 = 4,91.\]

\[\textbf{д)}\ \sqrt{35,98} = \sqrt{36 - 0,02};\]

\[f(x) = \sqrt{x};\ \ x_{0} = 36;\ \ \]

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

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

\[f(36 - 0,02) \approx f(36) +\]

\[+ f^{'}(36) \cdot ( - 0,02) =\]

\[= 6 - \frac{0,02}{12} \approx 6 - 0,002 = 5,998.\]

\[\textbf{е)}\ \sqrt[3]{124} = \sqrt[3]{125 - 1}\]

\[f(x) = \sqrt[3]{x} = x^{\frac{1}{3}};\ \ x_{0} = 125;\ \ \]

\[\mathrm{\Delta}x = - 1\]

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

\[f(125 - 1) \approx f(125) +\]

\[+ f^{'}(125) \cdot ( - 1) = 5 -\]

\[- \frac{1}{75} \approx 5 - 0,013 = 4,987.\]

\[\textbf{ж)}\ \sqrt[3]{215} = \sqrt[3]{216 - 1}\]

\[f(x) = \sqrt[3]{x} = x^{\frac{1}{3}};\ \ x_{0} = 216;\ \ \]

\[\mathrm{\Delta}x = - 1\]

\[f^{'}(x) = \frac{1}{3\sqrt[3]{x^{2}}};\]

\[f(216 - 1) \approx f(216) +\]

\[+ f^{'}(216) \cdot ( - 1) = 6 -\]

\[- \frac{1}{108} \approx 6 - 0,009 = 5,991.\]

\[\textbf{з)}\ln 3 =\]

\[= \ln{(2,7 + 0,3)} \approx \ln{(e + 0,3)}\]

\[f(x) = \ln x;\ \ x_{0} = e;\ \ \mathrm{\Delta}x = 0,3\]

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

\[f(e + 0,3) \approx f(e) +\]

\[+ f^{'}(e) \cdot (0,3) = \ln e +\]

\[+ \frac{1}{e} \cdot 0,3 \approx 1 + \frac{0,3}{2,7} = 1,11.\]

\[\textbf{и)}\ {1,01}^{20} = (1 + 0,01)^{20}\]

\[f(x) = x^{20};\ \ x_{0} = 1;\ \ \mathrm{\Delta}x = 0,01\]

\[f^{'}(x) = 20x^{19};\]

\[f(1 + 0,01) \approx f(1) +\]

\[+ f^{'}(1) \cdot 0,01 = 1 + 20 \cdot 0,01 =\]

\[= 1 + 0,2 = 1,2.\]

\[к)\ {0,98}^{20} = (1 - 0,02)^{20}\]

\[f(x) = x^{20};\ \ \ x_{0} = 1;\ \ \]

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

\[f^{'}(x) = 20x^{19};\]

\[f(1 - 0,02) \approx f(1) +\]

\[+ f^{'}(1) \cdot ( - 0,02) = 1 -\]

\[- 20 \cdot 0,02 = 1 - 0,4 = 0,6.\]

\[л)\ {2,01}^{10} = (2 + 0,1)^{10}\]

\[f(x) = x^{10};\ \ \ x_{0} = 2;\ \ \mathrm{\Delta}x = 0,1\]

\[f^{'}(x) = 10x^{9};\]

\[f(2 + 0,1) \approx f(2) +\]

\[+ f^{'}(2) \cdot 0,01 = 2^{10} +\]

\[+ 10 \cdot 2^{9} \cdot 0,01 = 2^{10} + 2^{9} \cdot 0,1 =\]

\[= 2^{9}(2 + 0,1) = 512 \cdot 2,1 =\]

\[= 1075,2.\]

\[м)\ {1,99}^{10} = (2 - 0,01)^{10}\]

\[f(x) = x^{10};\ \ \ x_{0} = 2;\ \ \mathrm{\Delta}x = - 0,01\]

\[f^{'}(x) = 10x^{9};\]

\[f(2 - 0,01) \approx f(2) +\]

\[+ f^{'}(2) \cdot ( - 0,01) = 2^{10} -\]

\[- 10 \cdot 2^{9} \cdot 0,01 =\]

\[= 2^{10} - 2^{9} \cdot 0,1 = 2^{9}(2 - 0,1) =\]

\[= 512 \cdot 1,9 = 972,8.\]