For the domain we are looking for what #x# cannot be we can do that by breaking down the functions and seeing if any of them yield a result where x is undefined
#u=x+1#
With this function x is defined for all #RR# on the number line i.e. all numbers.
#s=3^u#
With this function u is defined for all #RR# as u can be negative, positive or 0 without a problem. So through transitivity we know that x is also defined for all #RR# or defined for all numbers
Lastly
#f(s)=-2(s)+2#
With this function s is defined for all #RR# as u can be negative, positive or 0 without a problem. So through transitivity we know that x is also defined for all #RR# or defined for all numbers
So we know that x is also defined for all #RR# or defined for all numbers
#{x in RR}#
For the range we have to look at what the y values will be for the function
#u=x+1#
With this function we that there is no value on the number line that won't be u. I.e. u is defined for all #RR#.
#s=3^u#
With this function we can see that if we place in all the positive numbers #s=3^(3)=27# we get out another positive number.
Whilst if we place in a negative number #s=3^-1=1/3# we get a positive number so y can't be negative and will also never be but will approach 0 at #-oo#
#{s in RR| s>0}#
Lastly
#f(s)=-2(s)+2#
We see that there is no value #f(s)# can equal any value if we disregard what #s# and #u# actually state.
But when we look carefully and we consider what #s# can actually be i.e. only greater than 0. We know that this will effect our final range, as what we see is that every #s# value is moved up 2 and stretched by -2 when it is placed on the y axis.
So all of the values in s become negative #{f(s) in RR| f(s)<0}#
Then we know that every value is moved up two
#{f(s) in RR| f(s)<2}#
so as #f(x)=f(s)# we can say the range is every y value lower than 2
or
#{f(x) in RR| f(x)<2}#
graph{-2(3^(x+1))+2 [-10, 10, -5, 5]}
