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

\[\boxed{\mathbf{5}\mathbf{.}}\]

\[\mathbf{На\ отрезке\ }\lbrack - 2;\ 2\rbrack.\]

\[\textbf{а)}\ y = \left\{ \begin{matrix} - x;\ \ \ \ \ \ \ \ \ \ \ \ \ \ x < - 1 \\ x + 2;\ - 1 \leq x < 1 \\ - 3x + 6;\ \ \ \ \ \ \ x \geq 1 \\ \end{matrix} \right.\ \]

\[f( - 2) = 2;\]

\[f( - 1) = 1;\]

\[f(1) = 3;\]

\[f(2) = 0;\]

\[\max_{\lbrack - 2;2\rbrack}{f(x)} = 3;\]

\[\min_{\lbrack - 2;2\rbrack}{f(x)} = 0.\]

\[\textbf{б)}\ y = \left\{ \begin{matrix} - x + 2;\ \ \ \ \ x < - 1 \\ - 3x;\ \ - 1 \leq x < 1 \\ x - 4;\ \ \ \ \ \ \ \ \ \ \ \ x \geq 1 \\ \end{matrix} \right.\ \]

\[f( - 2) = 4;\]

\[f( - 1) = 3;\]

\[f(1) = - 3;\]

\[f(2) = - 2;\]

\[\max_{\lbrack - 2;2\rbrack}{f(x)} = 4;\]

\[\min_{\lbrack - 2;2\rbrack}{f(x)} = - 3.\]

\[\textbf{в)}\ y = |x - 1| + |2x + 1|\]

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

\[f( - 2) = 6;\]

\[f\left( - \frac{1}{2} \right) = \frac{3}{2};\]

\[f(1) = 3;\]

\[f(2) = 6;\]

\[\max_{\lbrack - 2;2\rbrack}{f(x)} = 6;\]

\[\min_{\lbrack - 2;2\rbrack}{f(x)} = 1,5.\]

\[\textbf{г)}\ y = |x - 1| - |2x + 1|\]

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

\[f( - 2) = 0;\]

\[f\left( - \frac{1}{2} \right) = \frac{3}{2};\]

\[f(1) = - 3;\]

\[f(2) = - 4;\]

\[\max_{\lbrack - 2;2\rbrack}{f(x)} = 1,5;\]

\[\min_{\lbrack - 2;2\rbrack}{f(x)} = - 4.\]