Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 325

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

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

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

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

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

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

\[f( - 2) = - 8 - 24 + 9 = - 23;\]

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

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

\[Ответ:\ - 23;\ 9.\]

\[2)\ f(x) = x^{3} + 6x^{2} + 9x;\ \lbrack - 4;\ 0\rbrack:\]

\[f^{'}(x) = 3x^{2} + 6 \bullet 2x + 9 =\]

\[= 3x^{2} + 12x + 9.\]

\[3x^{2} + 12x + 9 = 0\]

\[x^{2} + 4x + 3 = 0\]

\[D = 16 - 12 = 4\]

\[x_{1} = \frac{- 4 - 2}{2} = - 3;\]

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

\[f( - 4) = - 64 + 96 - 36 = - 4;\]

\[f( - 3) = - 27 + 54 - 27 = 0;\]

\[f( - 1) = - 1 + 6 - 9 = - 4;\]

\[f(0) = 0 + 0 + 0 = 0.\]

\[Ответ:\ - 4;\ 0.\]

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

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

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

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

\[4x\left( x^{2} - 1 \right) = 0\]

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

\[f( - 4) = 256 - 32 + 3 = 227;\]

\[f( \pm 1) = 1 - 2 + 3 = 2;\]

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

\[f(3) = 81 - 18 + 3 = 66.\]

\[Ответ:\ \ 2;\ 227.\]

\[4)\ f(x) = x^{4} - 8x^{2} + 5;\ \lbrack - 3;\ 2\rbrack:\]

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

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

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

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

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

\[f( - 3) = 81 - 72 + 5 = 14;\]

\[f( \pm 2) = 16 - 32 + 5 = - 11;\]

\[f(0) = 0 - 0 + 5 = 5.\]

\[Ответ:\ - 11;\ 14.\]