2. Let $n_{1}, n_{2} \in \mathbb{R}$ and $f\left(n_{1}\right)=f\left(n_{2}\right)$. Then

$6 n_{1}+5=(f \circ f)\left(n_{1}\right)=f\left(f\left(n_{1}\right)\right)=f\left(f\left(n_{2}\right)\right)=(f \circ f)\left(n_{2}\right)=6 n_{2}+5 \Rightarrow n_{1}=n_{2}$

In the given relation, we substitute $n$ with $f(n)$ and obtain

$(f \circ f)(f(n))=6 f(n)+5, f(6 n+5)=6 f(n)+5$

$n=15 \Rightarrow f(95)=11$

$n=95 \Rightarrow f(575)=71$

$n=575 \Rightarrow f(3455)=431$
2. (7p) Consider the function $f: R \rightarrow R$ with the property $(f \circ f)(n)=6n+5$. Prove that $f$ is injective on the set of real numbers and, if $f(15)=1$, calculate $f(3455)$.

Petru Vlad