The functions $g:\mathbb{R}\to\mathbb{R}, g(x)=x+1$ and $h:\mathbb{R}\to\mathbb{R}, h(x)=2x+1$ are bijective. With the substitution $y=x+1$ results $f(y)\le f(2y-1), \forall y\in\mathbb{R}$. With the substitution $y=3x+1$ results $f(y)\ge f(2y-1), \forall y\in\mathbb{R}$. Hence, $f(y)=f(2y-1), \forall y\in\mathbb{R}\quad(1)$. Consider the sequence $a_1=3, a_{n+1}=2a_n-1, \forall n\in\mathbb{N}$. Using $(1)$ results: $f(a_n)=f(a_1)=f(3)=2, \forall n\in\mathbb{N}$. The sequence $(a_n)_{n\in\mathbb{N}}$ is strictly increasing, hence contains an infinity of distinct numbers. Conclusion: exist an infinity number of values $a_n$ for which $f(a_n)=2$.