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]}
