Решебник по алгебре 11 класс Никольский Параграф 1. Функции и их графики Задание 49

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Год:2020-2021-2022
Тип:учебник

Задание 49

\[\boxed{\mathbf{49.}}\]

\[\textbf{а)}\ y = x^{2} - 4;\ \ X = R;\]

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

\[x^{2} = 4\]

\[x = \pm 2.\]

\[y > 0\ при\]

\[\ x \in ( - \infty; - 2) \cup (2; + \infty);\]

\[y < 0\ при\ x \in ( - 2;2).\]

\[\textbf{б)}\ y = x^{2} - 4x\]

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

\[x(x - 4) = 0\]

\[x = 0;\ \ \ x = 4.\]

\[y > 0\ при\ x \in ( - \infty;0) \cup (4; + \infty);\]

\[y < 0\ при\ x \in (0;4).\]

\[\textbf{в)}\ y = x^{2} - 5x + 4\]

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

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = 4\]

\[x_{1} = 1;\ \ \ x_{2} = 4.\]

\[y > 0\ при\ x \in ( - \infty;1) \cup (4; + \infty);\]

\[y < 0\ при\ x \in (1;4).\]

\[\textbf{г)}\ y = 9 - x^{2}\]

\[9 - x^{2} = 0\]

\[x^{2} = 9\]

\[x = \pm 3.\]

\[y < 0\ при\ x \in ( - \infty; - 3) \cup (3; + \infty);\]

\[y > 0\ при\ x \in ( - 3;3).\]

\[\textbf{д)}\ y = - x^{2} + 2x\]

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

\[- x(x - 2) = 0\]

\[x = 0;\ \ \ x = 2.\]

\[y < 0\ при\ x \in ( - \infty;0) \cup (2; + \infty);\]

\[y > 0\ при\ x \in (0;2).\]

\[\textbf{е)}\ y = - 2x^{2} - 3x + 5\]

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

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

\[D = 9 + 40 = 49\]

\[x_{1} = \frac{- 3 + 7}{4} = 1;\ \ \]

\[x_{2} = \frac{- 3 - 7}{4} = - 2,5;\]

\[y < 0\ при\ \]

\[x \in ( - \infty; - 2,5) \cup (1; + \infty);\]

\[y > 0\ при\ x \in ( - 2,5;1).\]

\[\textbf{ж)}\ y = \frac{4}{x + 3} + 1;\ \ x \neq - 3;\]

\[\frac{4}{x + 3} + 1 = 0\ \ \ \ | \cdot (x + 3)\]

\[4 + x + 3 = 0\]

\[x + 7 = 0\]

\[x = - 7.\]

\[y > 0\ при\]

\[\ x \in ( - \infty; - 7) \cup ( - 3; + \infty);\]

\[y < 0\ при\ x \in ( - 7; - 3).\]

\[\textbf{з)}\ y = \frac{- 2}{x - 2} - 1;\ \ \ x \neq 2;\]

\[\frac{- 2}{x - 2} - 1 = 0\ \ \ \ \ \ | \cdot (x - 2)\]

\[- 2 - (x - 2) = 0\]

\[- 2 - x + 2 = 0\]

\[x = 0.\]

\[y < 0\ при\ x \in ( - \infty;0) \cup (2; + \infty);\]

\[y > 0\ при\ x \in (0;2).\]

\[\textbf{и)}\ y = - |x - 2| + 2 = 0\]

\[- |x - 2| + 2 = 0\]

\[- |x - 2| = - 2\]

\[|x - 2| = 2\]

\[x - 2 = 2\ \ \ \ \ \ x - 2 = - 2\]

\[x = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0\]

\[y < 0\ при\ x \in ( - \infty;0) \cup (4; + \infty);\]

\[y > 0\ при\ x \in (0;4).\]

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