Question 403290
Since the domain is {{{1<=x<=6}}}, this means that the set of possible input values is {1,2,3,4,5,6}. Say for the sake of argument that we're only dealing with integers (to make things easy for us)



Now consider that instead of plugging in 'x', you plug in x-3. So this means that instead of plugging in x=1, you'll plug x-3=1-3=-2 into the function.


So this means that the domain of f(x-3) is now



{{{1-3<=x-3<=6-3}}}



which becomes



{{{-2<=x-3<=3}}}



So the domain of f(x-3) is {{{-2<=x-3<=3}}}



Now we're going to take it a step further. Instead of plugging in x-3, we're going to double it and plug in 2(x-3). So instead of plugging in -2, we're going to plug (-2)(2)=-4 into the function.


So this means that the domain becomes



{{{2(-2)<=2(x-3)<=2(3)}}}



{{{-4<=2(x-3)<=6}}}



So the domain of f(2(x-3)) is {{{-4<=2(x-3)<=6}}} given that the domain of f(x) is {{{1<=x<=6}}}. 



Basically, the smallest number that the input 2(x-3) can be is -4 and the largest number that 2(x-3) can be is 6