anonymous
  • anonymous
URGENT!!!: What is the range of f^-1[g(x)]???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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sammixboo
  • sammixboo
Maybe @triciaal or @jim_thompson5910 could help :)
jim_thompson5910
  • jim_thompson5910
There seems to be missing information. Can you post a screenshot?
anonymous
  • anonymous
ok one sec

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anonymous
  • anonymous
anonymous
  • anonymous
here!!
jim_thompson5910
  • jim_thompson5910
so this is all part of #6 and \[\Large f(x) = \frac{2}{x+3}\] \[\Large g(x) = 2^{x-1}\] right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
im stuck on part b). i have the domain i just cant figure out the range
jim_thompson5910
  • jim_thompson5910
I agree that \[\Large f^{-1}(x) = \frac{2-3x}{x}\] \[\Large f^{-1}(g(x)) = \frac{2-3(2^{x-1})}{2^{x-1}}\]
jim_thompson5910
  • jim_thompson5910
you have the correct domain as well
jim_thompson5910
  • jim_thompson5910
as for the range, let me think
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
\[\Large f^{-1}(x) = \frac{2-3x}{x}\] \[\Large f^{-1}(x) = \frac{2}{x} - \frac{3x}{x}\] \[\Large f^{-1}(x) = \frac{2}{x} - 3\] \[\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3\]
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
does that make sense?
anonymous
  • anonymous
wait so whats the range?
jim_thompson5910
  • jim_thompson5910
if you plug in x = 0, what do you get?
anonymous
  • anonymous
idont know :(((((
anonymous
  • anonymous
@triciaal
jim_thompson5910
  • jim_thompson5910
\[\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3\] \[\Large f^{-1}(g(0)) = \frac{2}{2^{0-1}} - 3\] \[\Large f^{-1}(g(0)) = ???\]
triciaal
  • triciaal
consider when x is negative when x = 0 and when x is positive
triciaal
  • triciaal
you said you had the actual domain the range is the set of y values
anonymous
  • anonymous
can i pls just have the answer pls. this homework q is the last one and im rly tired plz
triciaal
  • triciaal
maybe -infinity to -3 not sure
anonymous
  • anonymous
ok thanks
jim_thompson5910
  • jim_thompson5910
if you graph the entire thing, you'll get this
jim_thompson5910
  • jim_thompson5910
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)
anonymous
  • anonymous
omg thank you guys so much
jim_thompson5910
  • jim_thompson5910
no problem
anonymous
  • anonymous
i can finally sleep!!!!!
triciaal
  • triciaal
I see my error when x = -infinity y is positive so the solution is like Jim has it
anonymous
  • anonymous
guys!!!
anonymous
  • anonymous
is anyone here still able to help me with one more q?
jim_thompson5910
  • jim_thompson5910
go ahead

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