Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 281

\[1)\ f(x) = x^{3} - 6x^{2} + 9\ на\ \lbrack - 1;\ 2\rbrack:\]

\[f^{'}(x) = 3x^{2} - 6 \bullet 2x + 0 =\]

\[= 3x^{2} - 12x.\]

\[Стационарные\ точки:\]

\[3x^{2} - 12x = 0\]

\[3x(x - 4) = 0\]

\[x_{1} = 0;\text{\ \ \ }x_{2} = 4.\]

\[f( - 1) = - 1 - 6 + 9 = 2;\]

\[f(0) = 0 - 6 \bullet 0 + 9 = 9;\]

\[f(2) = 8 - 24 + 9 = - 7.\]

\[Ответ:\ - 7;\ 9.\]

\[2)\ f(x) = x^{4} - 8x^{2} + 3\ на\ \lbrack - 1;\ 2\rbrack:\]

\[f^{'}(x) = 4x^{3} - 8 \bullet 2x + 0 =\]

\[= 4x^{3} - 16x.\]

\[Стационарные\ точки:\]

\[4x^{3} - 16x = 0\]

\[4x\left( x^{2} - 4 \right) = 0\]

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

\[x_{1} = - 2;\text{\ \ }x_{2} = 0;\ x_{3} = 2.\]

\[f( - 1) = 1 - 8 + 3 = - 4;\]

\[f(0) = 0 - 8 \bullet 0 + 3 = 3;\]

\[f(2) = 16 - 32 + 3 = - 13.\]

\[Ответ:\ - 13;\ 3.\]

\[3)\ f(x) = 2x^{3} + 3x^{2} - 36x\ на\ \lbrack - 2;\ 1\rbrack:\]

\[f^{'}(x) = 2 \bullet 3x^{2} + 3 \bullet 2x - 36 =\]

\[= 6x^{2} + 6x - 36.\]

\[Стационарные\ точки:\]

\[6x^{2} + 6x - 36 = 0\]

\[x^{2} + x - 6 = 0\]

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{- 1 - 5}{2} = - 3;\text{\ \ }\]

\[x_{2} = \frac{- 1 + 5}{2} = 2.\]

\[f( - 2) = - 16 + 12 + 72 = 68;\]

\[f(1) = 2 + 3 - 36 = - 31.\]

\[Ответ:\ - 31;\ 68.\]

\[4)\ f(x) = x^{3} + 9x^{2} + 15x\ на\ \lbrack - 3;\ - 2\rbrack:\]

\[f^{'}(x) = 3x^{2} + 9 \bullet 2x + 15 =\]

\[= 3x^{2} + 18x + 15.\]

\[Стационарные\ точки:\]

\[3x^{2} + 18x + 15 = 0\]

\[x^{2} + 6x + 5 = 0\]

\[D = 36 - 20 = 16\]

\[x_{1} = \frac{- 6 - 4}{2} = - 5;\text{\ \ }\]

\[x_{2} = \frac{- 6 + 4}{2} = - 1.\]

\[f( - 3) = - 27 + 81 - 45 = 9;\]

\[f( - 2) = - 8 + 36 - 30 = - 2.\]

\[Ответ:\ - 2;\ 9.\]