Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 296

\[1)\ f(x) = \left\{ \begin{matrix} 1 - x\ \text{\ \ \ }при\ x < 1 \\ \sqrt{x - 1}\ при\ x \geq 1 \\ \end{matrix} \right.\ \ \]

\[на\ отрезке\ \lbrack - 1;\ 2\rbrack.\]

\[- 1 \leq x \leq 1:\]

\[f^{'}(x) = - 1 < 0.\]

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

\[f^{'}(x) = \frac{1}{2\sqrt{x - 1}} > 0.\]

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

\[f(1) = \sqrt{1 - 1} = 0;\]

\[f(2) = \sqrt{2 - 1} = 1.\]

\[Ответ:\ \ 0;\ 2.\]

\[2)\ f(x) = \left\{ \begin{matrix} - 2x^{2} - 12x - 17\ при\ x < - 2 \\ (x + 1)^{3}\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }при\ x \geq - 2 \\ \end{matrix} \right.\ \ \]

\[на\ отрезке\ \lbrack - 5;\ - 1\rbrack.\]

\[- 5 \leq x \leq - 2:\]

\[f^{'}(x) = - 2 \bullet 2x - 12 - 0 =\]

\[= - 4x - 12 \geq 0;\]

\[4x \leq - 12\]

\[x \leq - 3.\]

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

\[f^{'}(x) = 3(x + 1)^{2} \geq 0;\]

\[x + 1 \geq 0\]

\[x \geq - 1.\]

\[f( - 5) = - 50 + 60 - 17 = - 7;\]

\[f( - 3) = - 18 + 36 - 17 = 1;\]

\[f( - 2) = ( - 2 + 1)^{3} = - 1;\]

\[f( - 1) = ( - 1 + 1)^{3} = 0.\]

\(Ответ:\ - 7;\ 1.\)