Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 41

\[\boxed{\mathbf{41}\mathbf{.}}\]

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

\[x_{0} = 1;\]

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

\[f^{'}(x) = \left( x^{n} \right)^{'} = nx^{n - 1};\]

\[f^{'}\left( x_{0} \right) = f^{'}(1) = n.\]

\[Получим\ формулу:\]

\[(1 + \mathrm{\Delta}x)^{n} \approx 1 + n\mathrm{\Delta}x.\]

\[\textbf{а)}\ (1,001)^{100} = (1 + 0,001)^{100}\]

\[(1 + 0,001)^{100} =\]

\[= 1 + 100 \cdot 0,001 = 1,1.\]

\[\textbf{б)}\ (0,998)^{100}\]

\[(1 - 0,002)^{100} = 1 +\]

\[+ 100 \cdot ( - 0,002) =\]

\[= 1 - 0,2 = 0,8.\]

\[\textbf{в)}\ (1,003)^{25}\]

\[(1 + 0,003)^{25} = 1 +\]

\[+ 25 \cdot (0,003) = 1 + 0,075 =\]

\[= 1,075.\]

\[\textbf{г)}\ (0,9997)^{25}\]

\[(1 - 0,0003)^{25} = 1 -\]

\[- 25 \cdot (0,0003) = 1 -\]

\[- 0,0075 = 0,9925.\]

\[\textbf{д)}\ \left( 1 - \frac{1}{1001} \right)^{10} = 1 -\]

\[- 10 \cdot \left( \frac{1}{1001} \right) \approx 1 - 0,01 = 0,99.\]

\[\textbf{е)}\ \left( 1 + \frac{2}{998} \right)^{15} = 1 +\]

\[+ 15 \cdot \frac{2}{998} \approx 1 + 0,03 = 1,03.\]

\[\textbf{ж)}\ \left( 1 - \frac{3}{1003} \right)^{20} = 1 -\]

\[- 20 \cdot \frac{3}{1003} \approx 1 - 0,06 = 0,94.\]

\[\textbf{з)}\ \left( 1 + \frac{3}{9997} \right)^{35} = 1 +\]

\[+ 35 \cdot \frac{3}{9997} \approx 1 + 0,01 = 1,01.\]