Problema
(Indicado a partir do 1º ano do E. M.)
A função [tex]f:[0,1]\rightarrow \mathbb{R}[/tex] é tal que
[tex]\qquad f(x) = \begin{cases}
\dfrac{1}{3}f(2x), ~~~~~~0\leq x\leq \dfrac{1}{2}\\ \\
\dfrac{1}{3}+\dfrac{2}{3}f(2x-1), ~~\dfrac{1}{2}\leq x\leq 1
\end{cases}[/tex]
Quanto vale [tex]f\left(\dfrac{1}{7}\right)[/tex]?
Adaptado de PUC – RIO, 2022.
Solução
Como [tex]0\leq \dfrac{1}{7}\leq \dfrac{1}{2}[/tex], então
[tex]\qquad f\left(\dfrac{1}{7}\right) = \dfrac{1}{3}f\left(2\cdot \dfrac{1}{7}\right) =\dfrac{1}{3}f\left(\dfrac{2}{7}\right) [/tex]
[tex]\qquad f\left(\dfrac{1}{7}\right)=\dfrac{1}{3}f\left(\dfrac{2}{7}\right). ~~(I)[/tex]
Como [tex]0\leq \dfrac{2}{7}\leq \dfrac{1}{2}[/tex], então
[tex]\qquad f\left(\dfrac{2}{7}\right) = \dfrac{1}{3}f\left(2\cdot \dfrac{2}{7}\right) =\dfrac{1}{3}f\left(\dfrac{4}{7}\right)[/tex]
[tex]\qquad f\left(\dfrac{2}{7}\right) =\dfrac{1}{3}f\left(\dfrac{4}{7}\right).~~(II)[/tex]
Como [tex]\dfrac{1}{2}\leq \dfrac{4}{7}\leq 1[/tex], então
[tex]\qquad f\left(\dfrac{4}{7}\right) = \dfrac{1}{3}+\dfrac{2}{3}f\left(2\cdot \dfrac{4}{7}-1\right) = \dfrac{1}{3}+\dfrac{2}{3}f\left(\dfrac{1}{7}\right)[/tex]
[tex]\qquad f\left(\dfrac{4}{7}\right)= \dfrac{1}{3}+\dfrac{2}{3}f\left(\dfrac{1}{7}\right). ~~(III)[/tex]
Substituindo [tex](III)[/tex] em [tex](II)[/tex], temos
[tex]\qquad f\left(\dfrac{2}{7}\right)=\dfrac{1}{3}f\left(\dfrac{4}{7}\right) = \dfrac{1}{3}\left[\dfrac{1}{3}+\dfrac{2}{3}f\left(\dfrac{1}{7}\right)\right][/tex]
[tex]\qquad f\left(\dfrac{2}{7}\right)= \dfrac{1}{9} + \dfrac{2}{9}f\left(\dfrac{1}{7}\right)[/tex]
Substituindo a última expressão acima em [tex](I)[/tex], obtemos
[tex]\qquad f\left(\dfrac{1}{7}\right)=\dfrac{1}{3}\left[\dfrac{1}{9} + \dfrac{2}{9}f\left(\dfrac{1}{7}\right)\right][/tex]
[tex]\qquad f\left(\dfrac{1}{7}\right)=\dfrac{1}{27} + \dfrac{2}{27}f\left(\dfrac{1}{7}\right)[/tex]
[tex]\qquad f\left(\dfrac{1}{7}\right)-\dfrac{2}{27}f\left(\dfrac{1}{7}\right)=\dfrac{1}{27}[/tex]
[tex]\qquad \dfrac{25}{27}f\left(\dfrac{1}{7}\right)=\dfrac{1}{27}[/tex]
[tex]\qquad f\left(\dfrac{1}{7}\right)=\dfrac{1}{27}\cdot \dfrac{27}{25}[/tex]
[tex]\qquad \boxed{f\left(\dfrac{1}{7}\right)=\dfrac{1}{25}}.[/tex]
Solução elaborada pelos Moderadores do Blog.