11-5. The function $f(x)$ is defined for all positive values of $x$. It turns out that $f\left(\frac{4 y+1}{y+1}\right)=\frac{1}{y}$ for any $y>0$. Find $f(3)$.
Answer: 0.5.

Solution: We need to substitute for $y$ such a number that $\frac{4 y+1}{y+1}=3$, which means $y=2$. We get $f(3)=\frac{1}{2}$.

Comment: Note that such a function exists. The proof of this is almost identical to the solution: $\frac{4 y+1}{y+1}=a, y=\frac{a-1}{4-a}$, from which for $a \neq 4$ we get $f(a)=\frac{4-a}{a-1}$. A question arises - what are $f(1)$ and $f(4)$? It turns out they can be any numbers! After all, the condition $f\left(\frac{4 y+1}{y+1}\right)=\frac{1}{y}$ says nothing about the value of $f(1)$ or the value of $f(4)$: the expression $\frac{4 y+1}{y+1}$ is never equal to 4, and $\frac{4 y+1}{y+1}=1$ when $y=0$, which we cannot substitute according to the condition.