Given y=[x-1]+2 (Assuming [x] is the greatest intger function):

The greatest integer function returns the greatest integer that is less than or equal to the given input.

For values of x>0 [x] is just the integer part of the number (also called the truncate function or integer function.) e.g. [1.6]=1

For values of x<0 [x] is the integer part minus 1. e.g. [-1.3]=-2

Take the graph of [x], shift it 1 unit to the right and two units up.

The graph of [x] in red, [x-1] in blue, and [x-1]+2 in green:

In tabular form:

`-5<=x<-4 ` y=-4

`-4<=x<-3` y=-3

`-3<=x<-2` y=-2

etc...

This is graph of function `f(x)=[x-2] +2` that has zero for `-1 <= x <= 0`

the set of definition and the range of this function is `RR`

Note , if we divide the set of definition in open intervall `I_n=(n; n+1)` with `n` integer, `AA x in I_n` `f'(x)=0`

This means all around set of definition `f(x) ` derivative zero. even tought function isn't costant.

(This is the clear test against of assertion: "a funcion with derivative zero all its define set zero is costant")