## anonymous one year ago URGENT!!!: What is the range of f^-1[g(x)]???

1. sammixboo

Maybe @triciaal or @jim_thompson5910 could help :)

2. jim_thompson5910

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

3. anonymous

ok one sec

4. anonymous

5. anonymous

here!!

6. jim_thompson5910

so this is all part of #6 and $\Large f(x) = \frac{2}{x+3}$ $\Large g(x) = 2^{x-1}$ right?

7. anonymous

yes

8. anonymous

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

9. jim_thompson5910

I agree that $\Large f^{-1}(x) = \frac{2-3x}{x}$ $\Large f^{-1}(g(x)) = \frac{2-3(2^{x-1})}{2^{x-1}}$

10. jim_thompson5910

you have the correct domain as well

11. jim_thompson5910

as for the range, let me think

12. anonymous

ok

13. jim_thompson5910

$\Large f^{-1}(x) = \frac{2-3x}{x}$ $\Large f^{-1}(x) = \frac{2}{x} - \frac{3x}{x}$ $\Large f^{-1}(x) = \frac{2}{x} - 3$ $\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3$

14. 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

15. 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

16. jim_thompson5910

does that make sense?

17. anonymous

wait so whats the range?

18. jim_thompson5910

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

19. anonymous

idont know :(((((

20. anonymous

@triciaal

21. jim_thompson5910

$\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3$ $\Large f^{-1}(g(0)) = \frac{2}{2^{0-1}} - 3$ $\Large f^{-1}(g(0)) = ???$

22. triciaal

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

23. triciaal

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

24. anonymous

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

25. triciaal

maybe -infinity to -3 not sure

26. anonymous

ok thanks

27. jim_thompson5910

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

28. 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)

29. anonymous

omg thank you guys so much

30. jim_thompson5910

no problem

31. anonymous

i can finally sleep!!!!!

32. triciaal

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

33. anonymous

guys!!!

34. anonymous

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

35. jim_thompson5910