Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 295

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

\[на\ отрезке\ \lbrack - 1;\ 1\rbrack:\]

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

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

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

\[= \frac{4x^{5} + 4x^{3} - 2x^{5} - 2x}{\left( x^{2} + 1 \right)^{2}} =\]

\[= \frac{2x^{5} + 4x^{3} - 2x}{\left( x^{2} + 1 \right)^{2}}.\]

\[Промежуток\ возрастания:\]

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

\[2x\left( x^{4} + 2x^{2} - 1 \right) \geq 0\]

\[D = 4 + 4 = 8\]

\[x^{2} = \frac{- 2 \pm \sqrt{8}}{2} = \frac{- 2 \pm 2\sqrt{2}}{2} =\]

\[= - 1 \pm \sqrt{2};\]

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

\[- \sqrt{\sqrt{2} - 1} \leq x \leq 0\]

\[x \geq \sqrt{\sqrt{2} - 1}.\]

\[f( \pm 1) = \frac{1 + 1}{1 + 1} = \frac{2}{2} = 1;\]

\[f(0) = \frac{0 + 1}{0 + 1} = \frac{1}{1} = 1;\]

\[f\left( \pm \sqrt{\sqrt{2} - 1} \right) =\]

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

\[f\left( \pm \sqrt{\sqrt{2} - 1} \right) = \frac{4\sqrt{2} - 4}{2} =\]

\[= 2\sqrt{2} - 2.\]

\[Ответ:\ \ 1;\ 2\sqrt{2} - 2.\]

\[2)\ f(x) = \left| x^{2} + 2x - 3 \right| + \frac{3}{2}\ln x\ \]

\[на\ отрезке\ \left\lbrack \frac{1}{2};\ 2 \right\rbrack.\]

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

\[D = 4 + 12 = 16\]

\[x_{1} = \frac{- 2 - 4}{2} = - 3;\]

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

\[(x + 3)(x - 1) \geq 0\]

\[x \leq - 3;\text{\ \ \ x} \geq 1.\]

\[0,5 \leq x \leq 1:\]

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

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

\[= \frac{3 - 4x - 4x^{2}}{2x} \geq 0;\]

\[D = 16 + 48 = 64\]

\[x_{1} = \frac{- 4 - 8}{2 \bullet 4} = - 1,5;\]

\[x_{2} = \frac{- 4 + 8}{2 \bullet 4} = 0,5.\]

\[\frac{(x + 1,5)(x - 0,5)}{x} \leq 0\]

\[x \leq - 1,5;\ \ \ 0 < x \leq 0,5.\]

\[1 \leq x \leq 2:\]

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

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

\[= \frac{4x^{2} + 4x + 3}{2x} \geq 0;\]

\[D = 16 - 48 = - 32 < 0\]

\[x \geq 0.\]

\[f\left( \frac{1}{2} \right) = \left| \frac{1}{4} + 1 - 3 \right| + \frac{3}{2}\ln\frac{1}{2} =\]

\[= \frac{7}{4} - \frac{3}{2}\ln 2;\]

\[f(1) = |1 + 2 - 3| + \frac{3}{2}\ln 1 = 0;\]

\[f(2) = |4 + 4 - 3| + \frac{3}{2}\ln 2 =\]

\[= 5 + \frac{3}{2}\ln 2.\]

\[Ответ:\ \ 0;\ 5 + \frac{3}{2}\ln 2.\]