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

- anonymous

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

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- schrodinger

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- sammixboo

Maybe @triciaal or @jim_thompson5910 could help :)

- jim_thompson5910

There seems to be missing information. Can you post a screenshot?

- anonymous

ok one sec

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## More answers

- anonymous

##### 1 Attachment

- anonymous

here!!

- 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

yes

- anonymous

im stuck on part b). i have the domain i just cant figure out the range

- 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

you have the correct domain as well

- jim_thompson5910

as for the range, let me think

- anonymous

ok

- 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

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

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

does that make sense?

- anonymous

wait so whats the range?

- jim_thompson5910

if you plug in x = 0, what do you get?

- anonymous

idont know :(((((

- anonymous

@triciaal

- 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

consider when x is negative
when x = 0 and when x is positive

- triciaal

you said you had the actual domain
the range is the set of y values

- anonymous

can i pls just have the answer pls. this homework q is the last one and im rly tired plz

- triciaal

maybe -infinity to -3 not sure

- anonymous

ok thanks

- jim_thompson5910

if you graph the entire thing, you'll get this

##### 1 Attachment

- 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

omg thank you guys so much

- jim_thompson5910

no problem

- anonymous

i can finally sleep!!!!!

- triciaal

I see my error when x = -infinity y is positive so the solution is like Jim has it

- anonymous

guys!!!

- anonymous

is anyone here still able to help me with one more q?

- jim_thompson5910

go ahead

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