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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 118
Задание 118
\[\boxed{\mathbf{118}\mathbf{.}}\]
\[\textbf{а)}\ y = \frac{x^{2} - 3x + 1}{x^{2} + 1}\]
\[y\left( x^{2} + 1 \right) = x^{2} - 3x + 1\]
\[x^{2}(y - 1) + 3x + (y - 1) = 0\]
\[D = 9 - 4(y - 1)^{2} \geq 0\]
\[(y - 1)^{2} \leq \frac{9}{4}\]
\[|y - 1| \leq \frac{3}{2}\]
\[- \frac{1}{2} \leq y \leq \frac{5}{2}.\]
\[\textbf{б)}\ y = \frac{x^{2} - 2x + 1}{x^{2} + 3};\]
\[y\left( x^{2} + 3 \right) = x^{2} - 2x + 1\]
\[x^{2}(y - 1) + 2x + (3y - 1) = 0\]
\[D = 4 - 4(y - 1)(3y - 1) =\]
\[= 4y(4 - 3y) \geq 0\]
\[- 12y\left( y - \frac{4}{3} \right) \leq 0\]
\[0 \leq y \leq \frac{4}{3}.\]
\[\textbf{в)}\ y = \frac{x^{2} + 3x + 1}{x^{2} + 1};\]
\[y\left( x^{2} + 1 \right) = x^{2} + 3x + 1\]
\[x^{2}(y - 1) - 3x + (y - 1) = 0\]
\[D = 9 - 4(y - 1)^{2} \geq 0\]
\[(y - 1)^{2} \leq \frac{9}{4}\]
\[|y - 1| \leq \frac{3}{2}\]
\[- \frac{1}{2} \leq y \leq \frac{5}{2}.\]
\[\textbf{г)}\ y = \frac{x^{2} + 2x + 1}{x^{2} + 3};\]
\[y\left( x^{2} + 3 \right) = x^{2} + 2x + 1\]
\[x^{2}(y - 1) - 2x + (3y - 1) = 0\]
\[D = 4 - 4(y - 1)(3y - 1) =\]
\[= 4y(4 - 3y) \geq 0\]
\[- 12y\left( y - \frac{4}{3} \right) \leq 0\]
\[0 \leq y \leq \frac{4}{3}.\]
