Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 284

\[1)\ f(x) = x^{5} - 5x^{4} + 5x^{3} + 1\ \]

\[на\ \lbrack - 1;\ 2\rbrack:\ \]

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

\[= 5x^{4} - 5 \bullet 4x^{3} + 5 \bullet 3x^{2} + 0 =\]

\[= 5x^{4} - 20x^{3} + 15x^{2}.\]

\[Стационарные\ точки:\]

\[5x^{4} - 20x^{3} + 15x^{2} = 0\]

\[x^{2} - 4x + 3 = 0\]

\[D = 16 - 12 = 4\]

\[x_{1} = \frac{4 - 2}{2} = 1;\]

\[x_{2} = \frac{4 + 2}{2} = 3.\]

\[f( - 1) = - 1 - 5 - 5 + 1 = - 10;\]

\[f(1) = 1 - 5 + 5 + 1 = 2;\]

\[f(2) = 32 - 80 + 40 + 1 = - 7.\]

\[Ответ:\ \ 2.\]

\[2)\ f(x) = 1 - x^{4} - x^{6}\ на\ ( - 3;\ 3):\]

\[f^{'}(x) = 0 - 4x^{3} - 6x^{5} =\]

\[= - 2x^{3} \bullet \left( 2 + 3x^{2} \right).\]

\[- 2x^{3} \geq 0\]

\[x \leq 0.\]

\[f(0) = 1 - 0^{4} - 0^{6} = 1.\]

\[Ответ:\ \ 1.\]

\[3)\ f(x) = \frac{2}{x} - x^{2}\ на\ x < 0:\]

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

\[= \frac{- 2\left( 1 + x^{3} \right)}{x^{2}}.\]

\[- 2\left( 1 + x^{3} \right) \geq 0\]

\[x^{3} + 1 \leq 0\]

\[x^{3} \leq - 1\]

\[x \leq - 1.\]

\[f( - 1) = - 2 - 1 = - 3.\]

\[Ответ:\ - 3.\]

\[4)\ f(x) = \frac{x}{4} + \frac{4}{x}\ на\ x < 0:\]

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

\[x^{2} - 16 \geq 0\]

\[(x + 4)(x - 4) \geq 0\]

\[x \leq - 4;\text{\ \ \ x} \geq 4.\]

\[f( - 4) = - 1 - 1 = - 2.\]

\[Ответ:\ - 2.\]