График показательной функции \(f\left( x \right) = {a^{x + b}}\) проходит через точки \(\left( {3;1} \right)\) и \(\left( {1;4} \right)\).
Следовательно: \(\left\{ {\begin{array}{*{20}{c}}{1 = {a^{3 + b}}}\\{4 = {a^{1 + b}}}\end{array}} \right.\)
Разделим первое уравнение на второе: \(\dfrac{{{a^{3 + b}}}}{{{a^{1 + b}}}} = \dfrac{1}{4}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{a^2} = \dfrac{1}{4}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,a = \dfrac{1}{2}.\)
Тогда: \({\left( {\dfrac{1}{2}} \right)^{3 + b}} = 1\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{\left( {\dfrac{1}{2}} \right)^{3 + b}} = {\left( {\dfrac{1}{2}} \right)^0}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,3 + b = 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,b = — 3.\)
Таким образом:
\(f\left( x \right) = {\left( {\dfrac{1}{2}} \right)^{x — 3}}\) и \({\left( {\dfrac{1}{2}} \right)^{x — 3}} = 64\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {\dfrac{1}{2}} \right)^{x — 3}} = {\left( {\dfrac{1}{2}} \right)^{ — 6}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x — 3 = — 6\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x = — 3.\)
Ответ: – 3.
Задача 31. На рисунке изображён график функции \(f\left( x \right) = {a^{x + b}}.\) Найдите значение x, при котором \(f\left( x \right) = 64.\)