Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 82

\[\boxed{\mathbf{82}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = x + \frac{4}{x};\ \ \ (0; + \infty)\]

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

\[f^{''}(x) = 0 - \left( 4x^{- 2} \right)^{'} =\]

\[= - 4 \cdot ( - 2)x^{- 3} = \frac{8}{x^{3}};\]

\[1 - \frac{4}{x^{2}} = 0\]

\[\frac{4}{x^{2}} = 1\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[Подставим\ x = 2:\]

\[f^{''}(2) = \frac{8}{2^{3}} = \frac{8}{8} = 1 > 0 \rightarrow x =\]

\[= 2 - точка\ минимума\ \]

\[функции\ на\ \]

\[данном\ промежутке.\]

\[f_{наим} = f(2) = 2 + \frac{4}{2} = 2 + 2 = 4.\]

\[\textbf{б)}\ f(x) = x + \frac{4}{x};\ \ \ ( - \infty;0)\]

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

\[f^{''}(x) = 0 - \left( 4x^{- 2} \right)^{'} =\]

\[= - 4 \cdot ( - 2)x^{- 3} = \frac{8}{x^{3}};\]

\[1 - \frac{4}{x^{2}} = 0\]

\[\frac{4}{x^{2}} = 1\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[Подставим\ x = - 2:\]

\[f^{''}( - 2) = \frac{8}{( - 2)^{3}} = - \frac{8}{8} =\]

\[= - 1 < 0;\]

\[x = - 2 \rightarrow точка\ максимума\ \]

\[f(x)\ на\ данном\ промежутке.\]

\[f_{наиб} = f( - 2) = - 2 + \frac{4}{- 2} =\]

\[= - 2 - 2 = - 4.\]

\[\textbf{в)}\ f(x) = 8x^{2} - \frac{1}{4x};\ \ \ ( - \infty;0)\]

\[f^{'}(x) = 16x + \frac{1}{4x^{2}};\]

\[f^{''}(x) = 16 + \left( \frac{1}{4}x^{- 2} \right)^{'} = 16 +\]

\[+ \frac{1}{4} \cdot ( - 2)x^{- 3} = 16 - \frac{1}{2x^{3}};\]

\[16x + \frac{1}{4x^{2}} = 0\]

\[64x^{3} + 1 = 0\]

\[64x^{3} = - 1\]

\[x^{3} = - \frac{1}{64}\]

\[x = - \frac{1}{4} \in ( - \infty;0).\]

\[Подставим\ x = - \frac{1}{4}:\]

\[f^{''}\left( - \frac{1}{4} \right) = 16 - \frac{1}{2 \cdot \left( - \frac{1}{4} \right)^{3}} =\]

\[= 16 + \frac{1}{2 \cdot \frac{1}{64}} = 16 + 1\ :\frac{1}{32} =\]

\[= 16 + 32 = 48 > 0\]

\[x = - \frac{1}{4} \rightarrow точка\ минимума\ \]

\[функции\ f(x)\ на\ данном\ \]

\[промежутке.\]

\[f_{наим} = f\left( - \frac{1}{4} \right) = 8 \cdot \frac{1}{16} +\]

\[+ \frac{1}{4 \cdot \frac{1}{4}} = \frac{1}{2} + 1 = 1,5.\]

\[\textbf{г)}\ f(x) = 8x^{2} + \frac{1}{4x};\ \ \ (0; + \infty)\]

\[f^{'}(x) = 16x - \frac{1}{4x^{2}};\]

\[f^{''}(x) = 16 - \left( \frac{1}{4}x^{- 2} \right)^{'} = 16 -\]

\[- \frac{1}{4} \cdot ( - 2)x^{- 3} = 16 + \frac{1}{2x^{3}};\]

\[16x - \frac{1}{4x^{2}} = 0\]

\[64x^{3} - 1 = 0\]

\[64x^{3} = 1\]

\[x^{3} = \frac{1}{64}\]

\[x = \frac{1}{4} \in (0; + \infty).\]

\[Подставим\ x = \frac{1}{4}:\]

\[f^{''}\left( \frac{1}{4} \right) = 16 + \frac{1}{2 \cdot \left( \frac{1}{4} \right)^{3}} = 16 +\]

\[+ \frac{1}{2 \cdot \frac{1}{64}} = 16 + 1\ :\frac{1}{32} =\]

\[= 16 + 32 = 48 > 0\]

\[x = \frac{1}{4} \rightarrow точка\ минимума\]

\[\ функции\ f(x)\ на\ данном\]

\[\ промежутке.\]

\[f_{наим} = f\left( \frac{1}{4} \right) = 8 \cdot \frac{1}{16} + \frac{1}{4 \cdot \frac{1}{4}} =\]

\[= 1,5.\]