Find all the functions $f(x),$ continuous on the whole real axis, such that for every real $x$ \[f(3x-2)\leq f(x)\leq f(2x-1).\]

[i]Proposed by A. Golovanov[/i]
1. Given the functional inequality:
   \[
   f(3x-2) \leq f(x) \leq f(2x-1)
   \]
   for every real \( x \), we need to find all continuous functions \( f(x) \) that satisfy this inequality.

2. We start by considering the inequality \( f(3x-2) \leq f(x) \leq f(2x-1) \). To explore the behavior of \( f \), we will use induction to show that:
   \[
   f\left(\frac{x}{2^n}\right) \leq f(x) \leq f\left(\frac{x}{3^n}\right)
   \]
   for all \( x \in \mathbb{R} \) and \( n \in \mathbb{N} \).

3. **Base Case**: For \( n = 1 \), we have:
   \[
   f\left(\frac{x}{2}\right) \leq f(x) \leq f\left(\frac{x}{3}\right)
   \]
   This follows directly from the given inequality by substituting \( x \) with \( \frac{x}{2} \) and \( \frac{x}{3} \).

4. **Inductive Step**: Assume that for some \( n \in \mathbb{N} \), the inequality holds:
   \[
   f\left(\frac{x}{2^n}\right) \leq f(x) \leq f\left(\frac{x}{3^n}\right)
   \]
   We need to show that it holds for \( n+1 \):
   \[
   f\left(\frac{x}{2^{n+1}}\right) \leq f(x) \leq f\left(\frac{x}{3^{n+1}}\right)
   \]

5. Using the given inequality, we substitute \( x \) with \( \frac{x}{2^n} \) and \( \frac{x}{3^n} \):
   \[
   f\left(3\left(\frac{x}{2^n}\right) - 2\right) \leq f\left(\frac{x}{2^n}\right) \leq f\left(2\left(\frac{x}{2^n}\right) - 1\right)
   \]
   Simplifying, we get:
   \[
   f\left(\frac{3x - 2 \cdot 2^n}{2^n}\right) \leq f\left(\frac{x}{2^n}\right) \leq f\left(\frac{2x - 2^n}{2^n}\right)
   \]
   Similarly, for \( \frac{x}{3^n} \):
   \[
   f\left(3\left(\frac{x}{3^n}\right) - 2\right) \leq f\left(\frac{x}{3^n}\right) \leq f\left(2\left(\frac{x}{3^n}\right) - 1\right)
   \]
   Simplifying, we get:
   \[
   f\left(\frac{3x - 2 \cdot 3^n}{3^n}\right) \leq f\left(\frac{x}{3^n}\right) \leq f\left(\frac{2x - 3^n}{3^n}\right)
   \]

6. By the induction hypothesis, we have:
   \[
   f\left(\frac{x}{2^{n+1}}\right) \leq f\left(\frac{x}{2^n}\right) \leq f(x) \leq f\left(\frac{x}{3^n}\right) \leq f\left(\frac{x}{3^{n+1}}\right)
   \]
   Thus, the inequality holds for \( n+1 \).

7. By induction, we have:
   \[
   f\left(\frac{x}{2^n}\right) \leq f(x) \leq f\left(\frac{x}{3^n}\right)
   \]
   for all \( x \in \mathbb{R} \) and \( n \in \mathbb{N} \).

8. Taking the limit as \( n \to \infty \) and using the continuity of \( f \), we get:
   \[
   \lim_{n \to \infty} f\left(\frac{x}{2^n}\right) = f(0) \quad \text{and} \quad \lim_{n \to \infty} f\left(\frac{x}{3^n}\right) = f(0)
   \]
   Therefore:
   \[
   f(0) \leq f(x) \leq f(0)
   \]
   which implies:
   \[
   f(x) = f(0)
   \]
   for all \( x \in \mathbb{R} \).

9. Hence, \( f \) is a constant function.

\(\blacksquare\)

The final answer is \( \boxed{ f(x) = c } \) for some constant \( c \).