Solve
$4 x = x + [x]$ .

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Solution

We have,
$\begin{matrix} 4 x = x + [x] & [\begin{matrix} f (x) = x - [x] R f = [0, 1] \end{matrix}] \Rightarrow 3 x = [x] \Rightarrow 2 x + x - [x] = 0 \Rightarrow \begin{matrix} 2 x + 0 = 0 o r \end{matrix}, \begin{matrix} x = 0 o r \end{matrix} \Rightarrow 2 x + 1 = 0, x = - \frac{1}{2}, ∴ x \in [0, - \frac{1}{2}] . \end{matrix}$

Hence, this is the answer.

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