Mapping closed intervals under a given function $f$

Question

Currently I am reading a paper which deals with the Converse of Sarkovskii's Theorem. Here is the part where I have got a problem understanding it:

(Source: A. Ochoa, Sarkovskii's Theorem and its Converse, 2005)

The yellow marked part is where I stuck. I do not understand how the closed interval $[1,2]$ is mapped under $f$, which is defined in the paper. In particular I do not understand why $1$ is mapped to $2k+1$ instead of $k+1$.

Can anyone help?

Answer 1

Note that $f([1,k+2]) = [a,b]$ where $a = min\{f(x) : x \in [1,k+2]\}$ and $b = max\{f(x):x \in [1,k+2] \}$

So the max rise when $x = 2 \in [1,k+2]$, $f(2) = 2k +1$

And min when $x = k+2 \in [1,k+2]$ $f(k+2) = k$

Thus $f([1,k+2]) = [k,2k+1]$