ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 181

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

\[f^{'}(x) = - 3x^{- 4} - 2x^{- 3};\]

\[f^{'}(3) = - \frac{3}{3^{4}} - \frac{2}{3^{3}} =\]

\[= - \frac{1}{3^{3}} - \frac{2}{3^{3}} = - \frac{1}{9};\]

\[f^{'}(1) = - \frac{3}{1^{4}} - \frac{2}{1^{3}} = - 3 - 2 = - 5.\]

\[Ответ:\ - \frac{1}{9};\ - 5.\]

\[2)\ f(x) = \sqrt{x} + \frac{1}{x} + 1 =\]

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

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

\[f^{'}(3) = \frac{1}{2\sqrt{3}} - \frac{1}{3^{2}} = \frac{\sqrt{3}}{6} - \frac{1}{9};\]

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

\[Ответ:\ \ \frac{\sqrt{3}}{6} - \frac{1}{9};\ - \frac{1}{2}.\]

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

\[f^{'}(x) = 3 \bullet \left( - \frac{1}{3}x^{- \frac{4}{3}} \right) - 2 \bullet \left( - 3x^{- 4} \right);\]

\[f^{'}(3) = 3 \bullet \left( - \frac{1}{3 \bullet 3\sqrt[3]{3}} \right) + \frac{6}{3^{4}} =\]

\[= \frac{2}{27} - \frac{1}{3\sqrt[3]{3}};\]

\[f^{'}(1) = 3 \bullet \left( - \frac{1}{3 \bullet 1 \bullet \sqrt[3]{1}} \right) + \frac{6}{1^{4}} =\]

\[= 6 - 1 = 5.\]

\[Ответ:\ \ \frac{2}{27} - \frac{1}{3\sqrt[3]{3}};\ 5.\]

\[4)\ f(x) = x^{\frac{3}{2}} - x^{- \frac{3}{2}};\]

\[f^{'}(x) = \frac{3}{2}x^{\frac{1}{2}} + \frac{3}{2}x^{- \frac{5}{2}};\]

\[f^{'}(3) = \frac{3\sqrt{3}}{2} + \frac{3}{2 \bullet 3^{2} \bullet \sqrt{3}} =\]

\[= \frac{27\sqrt{3} + \sqrt{3}}{2 \bullet 9} = \frac{14\sqrt{3}}{9}\ ;\]

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

\[= 1,5 + 1,5 = 3.\]

\[Ответ:\ \ \frac{14\sqrt{3}}{9};\ 3.\]

\[5)\ f(x) = (x - 1)^{2}(x - 3);\]

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

\[f^{'}(3) = 2 \bullet 2 \bullet 0 + 2^{2} = 0 + 4 = 4;\]

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

\[= 0 + 0 = 0.\]

\[Ответ:\ \ 4;\ 0.\]

\[6)\ f(x) = (x - 3)^{3}(x - 1);\]

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

\[f^{'}(3) = 3 \bullet 0^{2} \bullet 2 + 0^{3} = 0 + 0 = 0;\]

\[f^{'}(1) = 3 \bullet ( - 2)^{2} \bullet 0 + ( - 2)^{3} = - 8.\]

\[Ответ:\ \ 0;\ - 8.\]

\[7)\ f(x) = \left( x^{2} - 1 \right)(x + 3);\]

\[f^{'}(x) = 2x(x + 3) + \left( x^{2} - 1 \right);\]

\[f^{'}(3) = 6 \bullet 6 + 8 = 36 + 8 = 44;\]

\[f^{'}(1) = 2 \bullet 4 + 0 = 8 + 0 = 8.\]

\[Ответ:\ \ 44;\ 8.\]

\[8)\ f(x) = \left( x^{2} - 9 \right)(x + 1);\]

\[f^{'}(x) = 2x(x + 1) + \left( x^{2} - 9 \right);\]

\[f^{'}(3) = 6 \bullet 4 + 0 = 24 + 0 = 24;\]

\[f^{'}(1) = 2 \bullet 2 + ( - 8) =\]

\[= 4 - 8 = - 4.\]

\[Ответ:\ \ 24;\ - 4.\]