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anonymous

  • one year ago

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

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  1. sammixboo
    • one year ago
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    Maybe @triciaal or @jim_thompson5910 could help :)

  2. jim_thompson5910
    • one year ago
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    There seems to be missing information. Can you post a screenshot?

  3. anonymous
    • one year ago
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    ok one sec

  4. anonymous
    • one year ago
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  5. anonymous
    • one year ago
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    here!!

  6. jim_thompson5910
    • one year ago
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    so this is all part of #6 and \[\Large f(x) = \frac{2}{x+3}\] \[\Large g(x) = 2^{x-1}\] right?

  7. anonymous
    • one year ago
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    yes

  8. anonymous
    • one year ago
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    im stuck on part b). i have the domain i just cant figure out the range

  9. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    you have the correct domain as well

  11. jim_thompson5910
    • one year ago
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    as for the range, let me think

  12. anonymous
    • one year ago
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    ok

  13. jim_thompson5910
    • one year ago
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    \[\Large f^{-1}(x) = \frac{2-3x}{x}\] \[\Large f^{-1}(x) = \frac{2}{x} - \frac{3x}{x}\] \[\Large f^{-1}(x) = \frac{2}{x} - 3\] \[\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3\]

  14. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    does that make sense?

  17. anonymous
    • one year ago
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    wait so whats the range?

  18. jim_thompson5910
    • one year ago
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    if you plug in x = 0, what do you get?

  19. anonymous
    • one year ago
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    idont know :(((((

  20. anonymous
    • one year ago
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    @triciaal

  21. jim_thompson5910
    • one year ago
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    \[\Large f^{-1}(g(x)) = \frac{2}{2^{x-1}} - 3\] \[\Large f^{-1}(g(0)) = \frac{2}{2^{0-1}} - 3\] \[\Large f^{-1}(g(0)) = ???\]

  22. triciaal
    • one year ago
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    consider when x is negative when x = 0 and when x is positive

  23. triciaal
    • one year ago
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    you said you had the actual domain the range is the set of y values

  24. anonymous
    • one year ago
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    can i pls just have the answer pls. this homework q is the last one and im rly tired plz

  25. triciaal
    • one year ago
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    maybe -infinity to -3 not sure

  26. anonymous
    • one year ago
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    ok thanks

  27. jim_thompson5910
    • one year ago
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    if you graph the entire thing, you'll get this

  28. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    omg thank you guys so much

  30. jim_thompson5910
    • one year ago
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    no problem

  31. anonymous
    • one year ago
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    i can finally sleep!!!!!

  32. triciaal
    • one year ago
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    I see my error when x = -infinity y is positive so the solution is like Jim has it

  33. anonymous
    • one year ago
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    guys!!!

  34. anonymous
    • one year ago
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    is anyone here still able to help me with one more q?

  35. jim_thompson5910
    • one year ago
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    go ahead

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