I observe a range chart of $f(x)$ and here you are $$\begin{array}{|r|l|l|l|l|l|l|l|l|l|l|} \hline x\,\,\,\,\,\,\, & - \infty & \,... & \,\,\,\,\,0 & \,... & \,\,\,\,\,1 & \,... & + \infty \\ \hline f(x) & + \infty & \searrow & - 2018 & \nearrow & 2018 & \searrow & - \infty \\ \hline \end{array}$$ Does $|f(x+ 2017)- 2018|= 2019$ has $2$ roots, $3$ roots, $4$ roots or $6$ roots? I can't understand it
I see that $f(x)= 4017$ has only one root and $f(x)= -1$ has $3$ distinct roots. what if $f(x)= f(\!x+ 2017\!)$
If I understand correctly, then the arrows mean "strictly increasing/decreasing" and $\lim_{x\to\pm\infty} f(x) \longleftrightarrow \mp \infty$.
About your question: We need to find all $y=x+2017$ such that $|f(y)-2018|=2019$. So $f(y)=2018+2019$ or $f(y)=-1$.
For the first equation, we have exactly one solution in $]-\infty,0]$ and no other solutions.
The second equation has exactly three solutions.
So the answer is that there are exactly $4$ such $y$ (and hence exactly $4$ such $x$).