ГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 47

\[\boxed{\mathbf{47}\mathbf{.}}\]

\[\frac{f(b) - f(a)}{b - a} = f^{'}(c);\ \ a < b < c.\]

\[\textbf{а)}\ f(x) = x^{3};\]

\[a = - 1;\ \ b = 2;\]

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

\[= \frac{8 - ( - 1)}{3} = 3;\]

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

\[3x^{2} = 3\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[c = 1 \in ( - 1;2).\]

\[Ответ:1.\]

\[\textbf{б)}\ f(x) = x^{3};\ \ \]

\[a = - 2;\ \ b = 1;\]

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

\[= \frac{1 - ( - 8)}{3} = 3;\]

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

\[3x^{2} = 3\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[c = - 1 \in ( - 2;1).\]

\[\textbf{в)}\ f(x) = \sqrt[3]{x};\]

\[a = 0;\ \ b = 27;\]

\[f^{'}(c) = \frac{f(27) - f(0)}{27 - 0} =\]

\[= \frac{3 - 0}{27} = \frac{1}{9};\]

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

\[\frac{1}{3\sqrt[3]{x^{2}}} = \frac{1}{9}\]

\[\sqrt[3]{x^{2}} = 3\]

\[x^{2} = 27\]

\[x = \pm 3\sqrt{3};\]

\[c = 3\sqrt{3} \in (0;27).\]

\[Ответ:3\sqrt{3}.\]

\[\textbf{г)}\ f(x) = \sqrt[3]{x}\ ;\]

\[a = - 27;\]

\[b = 0;\]

\[f^{'}(c) = \frac{f(0) - f( - 27)}{0 + 27} =\]

\[= \frac{0 + 3}{27} = \frac{1}{9};\]

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

\[\frac{1}{3\sqrt[3]{x^{2}}} = \frac{1}{9}\]

\[\sqrt[3]{x^{2}} = 3\]

\[x^{2} = 27\]

\[x = \pm 3\sqrt{3};\]

\[c = - 3\sqrt{3} \in ( - 27;0).\]

\[Ответ: - 3\sqrt{3}.\]