ГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 88

\[\boxed{\mathbf{88}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = (x - b)^{2};\ \ \lbrack - 1;1\rbrack\]

\[Функция\ определена\ для\ всех\]

\[\text{\ x}\ из\ данного\ интервала.\]

\[f^{'}(x) = 2(x - b) \cdot 1 = 2x - 2b;\]

\[2x - 2b = 0\]

\[2x = 2b\]

\[x = b \rightarrow критическая\ точка\ на\]

\[\ отрезке\ \lbrack - 1;1\rbrack.\]

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

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

\[f(b) = (b - b)^{2} = 0.\]

\[b < - 1:\]

\[\min{f(x)} = f( - 1) = (1 + b)^{2}.\]

\[- 1 \leq b \leq 1:\]

\[\min{f(x)} = f(b) = 0.\]

\[b > 1:\]

\[\min{f(x)} = f(1) = (1 - b)^{2}.\]

\[\textbf{б)}\ f(x) = (x - b)^{2};\ \ \lbrack - 1;1\rbrack\]

\[Функция\ определена\ для\ всех\]

\[\text{\ x}\ из\ данного\ интервала.\]

\[f^{'}(x) = 2(x - b) \cdot 1 = 2x - 2b;\]

\[2x - 2b = 0\]

\[2x = 2b\]

\[x = b \rightarrow критическая\ точка\ на\]

\[\ отрезке\ \lbrack - 1;1\rbrack.\]

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

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

\[f(b) = (b - b)^{2} = 0.\]

\[b \leq 0:\]

\[\max{f(x)} = f(1) = (1 - b)^{2}.\]

\[b > 0:\]

\[\max{f(x)} = f( - 1) = (1 + b)^{2}.\]