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Were you able to figure out the domain and range of the transformed graph?
can someone help me i realy need help
No I can't find out the domain or range because none of the points I've plotted correspond to it
What is the left most point on the given graph?
yes, so the left half of the domain of the original function starts with x = -1 the question is: which x value, when plugged into (1/3)*(x-1), gives a result of -1? ie what is the solution to (1/3)*(x-1) = -1 ?
this means that the left most part of the domain in the transformed function is x = -2
if you plugged in something smaller than -2 (say -3), then (1/3)*(x-1) produces a number that is smaller than -1 (which is outside the domain of f(x))
the goal is to stay in the domain of f(x) because we can't use x values that aren't defined for f(x)
does that make sense?
Not really. Why would x-values need to be defined for the original function when it's being transformed?
notice how we have f[ `(1/3)*(x-1)` ] basically f[ T ] where T = (1/3)*(x-1)
is it possible to have f[ T ] when T is say, T = 5 ?
I guess in a way, you have to think backwards
The lowest you can go is T = -1 So you solve for x in T = (1/3)*(x-1) and like you said, you'll get x = -2 So f[ (1/3)*(x-1) ] has the lower part of the domain be x = -2
My final answer is "a."
Hopefully I don't have to graph anything
You don't. They just want the domain and range of the transformed function.