Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 16

\[\boxed{\mathbf{16}\mathbf{.}}\]

\[\textbf{а)}\ y = |x - a|;\ \ \lbrack - 2;3\rbrack\]

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

\[x = - 2:\]

\[y = | - 2 - a|\]

\[| - 2 - a| = 4,5\]

\[- 2 - a = 4,5\]

\[a = 4,5 + 2\]

\[a = 6,5.\]

\[2 + a = 4,5\]

\[a = 4,5 - 2\]

\[a = 2,5.\]

\[x = 3:\]

\[y = |3 - a|\]

\[|3 - a| = 4,5\]

\[3 - a = 4,5\]

\[a = - 1,5.\]

\[- 3 + a = 4,5\]

\[a = 7,5.\]

\[Наибольшее\ значение\ при\]

\[\ a = - 1,5;\ \ a = 2,5.\]

\[\textbf{б)}\ y = |x - a|;\ \ \lbrack - 2;3\rbrack\]

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

\[x = - 2:\]

\[y = | - 2 - a|\]

\[| - 2 - a| = 3,5\]

\[- 2 - a = 3,5\]

\[a = - 5,5.\]

\[2 + a = 3,5\]

\[a = 1,5.\]

\[x = 3:\]

\[|3 - a| = 3,5\]

\[3 - a = 3,5\]

\[a = - 0,5.\]

\[- 3 + a = 3,5\]

\[a = 6,5.\]

\[Наибольшее\ значение\ при\]

\[\ a = - 0,5;\ \ a = 1,5.\]

\[\textbf{в)}\ y = |x - a|;\ \ \lbrack - 2;3\rbrack\]

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

\[x = - 2;\ \ x = 3:\]

\[| - 2 - a| = |3 - a|\]

\[1) - 2 - a = 3 - a\]

\[0a = 5\]

\[нет\ корней.\]

\[2)\ 2 + a = - 3 + a\]

\[0a = 5\]

\[нет\ корней.\]

\[3)\ 2 + a = 3 - a\]

\[2a = 1\]

\[a = 0,5.\]

\[4) - 2 - a = - 3 + a\]

\[2a = 1\]

\[a = 0,5.\]

\[Ответ:при\ a = 0,5.\]