11.8. Find all continuous functions $f: \mathbf{R} \rightarrow \mathbf{R}$, which satisfy the relation

$$
3 \cdot f(2 x+1)=f(x)+5 x, \quad \forall x \in \mathbf{R}
$$
Solution. First, we will look for solutions in the class of first-degree functions, of the form $f(x)=a x+b$.

For such functions, we obtain the equivalent relations:

$$
\begin{gathered}
3 \cdot(a \cdot(2 x+1)+b)=a x+b+5 x, \quad \forall x \in \mathbf{R}, \quad \Leftrightarrow \\
6 a x+3 a+3 b=(a+5) x+b, \quad \forall x \in \mathbf{R}, \quad \Leftrightarrow \\
\left\{\begin{array} { c } 
{ 6 a = a + 5 , } \\
{ 3 a + 3 b = b }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
a=1 \\
b=-\frac{3}{2}
\end{array}\right.\right.
\end{gathered}
$$

It is easy to verify that the function $f_{1}, f_{1}(x)=x-\frac{3}{2}$, satisfies the conditions of the problem.

We will show that $f_{1}$ is the only such function.

Let $f$ be an arbitrary continuous function that satisfies the conditions of the problem and denote the function $g, g(x)=$ $f(x)-\left(x-\frac{3}{2}\right)$. The function $g$ is continuous. We will show that $g(x)=0, \forall x \in \mathbf{R}$.

Substituting $f(x)=g(x)+x+\frac{3}{2}$ into the given relation and for any $x \in \mathbf{R}$, we have the following equivalent relations:

$$
\begin{gathered}
3 \cdot\left(g(2 x+1)+2 x+1-\frac{3}{2}\right)=g(x)+x-\frac{3}{2}+5 x \Leftrightarrow \\
3 \cdot g(2 x+1)+6 x-\frac{3}{2}=g(x)+6 x-\frac{3}{2} \Leftrightarrow 3 \cdot g(2 x+1)=g(x)
\end{gathered}
$$

In the last relation, we substitute $x \rightarrow \frac{x-1}{2}$ and obtain equivalent relations:

$$
\begin{aligned}
3 \cdot g\left(2 \cdot \frac{x-1}{2}+1\right) & =g\left(\frac{x-1}{2}\right) \Leftrightarrow 3 \cdot g(x)=g\left(\frac{x-1}{2}\right) \Leftrightarrow \\
g(x) & =\frac{1}{3} \cdot g\left(\frac{x-1}{2}\right)
\end{aligned}
$$

We will show by induction that

$$
g(x)=\frac{1}{3^{n}} \cdot g\left(\frac{x-2^{n}+1}{2^{n}}\right), \quad \forall x \in \mathbf{R}, \quad \forall n \in \mathbf{N}^{*}
$$

For $n=1$ we have obtained (1).

Assume that for some $m \geq 1$ we have

$$
g(x)=\frac{1}{3^{m}} \cdot g\left(\frac{x-2^{m}+1}{2^{m}}\right), \quad \forall x \in \mathbf{R}
$$

Then, from (1) and (3), in which we will substitute $x \rightarrow \frac{x-1}{2}$, we obtain:

$$
g(x)=\frac{1}{3} \cdot g\left(\frac{x-1}{2}\right)=\frac{1}{3} \cdot \frac{1}{3^{m}} \cdot g\left(\frac{\frac{x-1}{2}-2^{m}+1}{2^{m}}\right)=\frac{1}{3^{m+1}} \cdot g\left(\frac{x-2^{m+1}+1}{2^{m+1}}\right)
$$

Thus, relation (2) is proved.

Since $g$ is a continuous function, then for any $x \in \mathbf{R}$ we have

$$
\lim _{n \rightarrow \infty} g\left(\frac{x-2^{n}+1}{2^{n}}\right)=g\left(\lim _{n \rightarrow \infty}\left(-1+\frac{x+1}{2^{n}}\right)\right)=g(-1)
$$

Thus, passing to the limit in (2), we obtain

$$
g(x)=\lim _{n \rightarrow \infty} \frac{1}{3^{n}} \cdot g\left(\frac{x-2^{n}+1}{2^{n}}\right)=\lim _{n \rightarrow \infty} \frac{1}{3^{n}} \cdot \lim _{n \rightarrow \infty} g\left(\frac{x-2^{n}+1}{2^{n}}\right)=0 \cdot g(-1)=0, \quad \forall x \in \mathbf{R}
$$

Therefore, the function $f, f(x)=x-\frac{3}{2}$, is the only function that satisfies the conditions of the problem.