URGENT!!!: What is the range of f^-1[g(x)]???

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URGENT!!!: What is the range of f^-1[g(x)]???

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Maybe @triciaal or @jim_thompson5910 could help :)
There seems to be missing information. Can you post a screenshot?
ok one sec

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Other answers:

here!!
so this is all part of #6 and \[\Large f(x) = \frac{2}{x+3}\] \[\Large g(x) = 2^{x-1}\] right?
yes
im stuck on part b). i have the domain i just cant figure out the range
I agree that \[\Large f^{-1}(x) = \frac{2-3x}{x}\] \[\Large f^{-1}(g(x)) = \frac{2-3(2^{x-1})}{2^{x-1}}\]
you have the correct domain as well
as for the range, let me think
ok
\[\Large f^{-1}(x) = \frac{2-3x}{x}\] \[\Large f^{-1}(x) = \frac{2}{x} - \frac{3x}{x}\] \[\Large f^{-1}(x) = \frac{2}{x} - 3\] \[\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3\]
as x gets larger and larger, the term \(\LARGE \frac{2}{2^{x-1}}\) gets smaller and smaller effectively getting to small that it becomes 0
so as x gets larger and larger \[\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3\] gets closer and closer to -3. It will never actually get to -3 itself though
does that make sense?
wait so whats the range?
if you plug in x = 0, what do you get?
idont know :(((((
\[\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3\] \[\Large f^{-1}(g(0)) = \frac{2}{2^{0-1}} - 3\] \[\Large f^{-1}(g(0)) = ???\]
consider when x is negative when x = 0 and when x is positive
you said you had the actual domain the range is the set of y values
can i pls just have the answer pls. this homework q is the last one and im rly tired plz
maybe -infinity to -3 not sure
ok thanks
if you graph the entire thing, you'll get this
it goes up forever but doesn't do the same in the other direction. Instead it gets closer and closer to -3 (but never actually gets to -3) so the range is y > -3 which in interval notation is (-3, infinity)
omg thank you guys so much
no problem
i can finally sleep!!!!!
I see my error when x = -infinity y is positive so the solution is like Jim has it
guys!!!
is anyone here still able to help me with one more q?
go ahead

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