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Homework Help: Why this function has an inverse function

  1. Feb 1, 2010 #1
    Ok, i need to explain why this function has an inverse function. fr^-1

    [math]fr(x) = 1.5/sinx[/math]

    and state its image set & domain.

    Is that (0,1/2Pi]

    Thanks :)
     
  2. jcsd
  3. Feb 1, 2010 #2

    Mark44

    Staff: Mentor

    Re: Triangle

    You realize that 1.5/sin(x) = 1.5csc(x), right?

    On what interval(s) is 1.5 csc(x) one-to-one? There are a bunch, so pick one. If you haven't already graphed y = 1.5 csc(x), you should do that now, since it will give you a good start on determining domain and range for the inverse.
     
  4. Feb 1, 2010 #3
    Re: Triangle

    here we are plotted:

    so how do we get the image set and domain
     

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    • iii.jpg
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  5. Feb 1, 2010 #4
    Re: Triangle

    how does this explain that is has an inverse function
     
  6. Feb 1, 2010 #5

    Mark44

    Staff: Mentor

    Re: Triangle

    You should increase the size of the interval to, say, -pi to pi. You're getting only a small part of the graph of y = 1.5 cscx.

    When you find an interval for which the graph is one-to-one, then the domain for your inverse will be the range of y = 1.5 csc x, and the range for the inverse will be the domain for y = 1.5 csc x.
     
  7. Feb 1, 2010 #6

    Mark44

    Staff: Mentor

    Re: Triangle

    If you have a function that is strictly increasing or strictly decreasing, it is one-to-one. Every one-to-one function has an inverse.

    What you're trying to do is to find an interval for which y = 1.5 csc(x) is one-to-one, and on which there is an inverse for this function.
     
  8. Feb 1, 2010 #7
    Re: Triangle

    ok Thanks - so how do i go about do that? basically from my attachment, i need to explain why fr has fr^-1, stating the domain and image set a hand would be great with it please
     
  9. Feb 1, 2010 #8

    Mark44

    Staff: Mentor

    Re: Triangle

    I'll say it again:
    If you have a function that is strictly increasing or strictly decreasing, it is one-to-one. Every one-to-one function has an inverse.

    What you're trying to do is to find an interval for which y = 1.5 csc(x) is one-to-one, and on which there is an inverse for this function.
     
  10. Feb 1, 2010 #9
    Re: Triangle

    i hear what you're saying bud, but any advice how to take that further? I dont have a clue :(
     
  11. Feb 1, 2010 #10

    Mark44

    Staff: Mentor

    Re: Triangle

    You have a graph of y = 1.5 csc x. See post #5 and what I said about using a larger interval. Can you look at a graph and tell whether a function is one-to-one, either overall or over some part of its domain?
     
  12. Feb 2, 2010 #11

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Triangle

    You were told in your problem that the domain of fr is [itex](0, \pi/2][/itex]. Do you see why 0 is not included? For x arbitrarily close to 0, fr(x) is arbitrarily large. What is [itex]fr(\pi/2)[/itex]? What is the range of fr?

    Taking the inverse of a function "swaps" domain and range.
     
  13. Feb 2, 2010 #12
    Re: Triangle

    the range is the function is 0 < x < 1/2pi

    Thanks for helping
     
  14. Feb 2, 2010 #13

    Mark44

    Staff: Mentor

    Re: Triangle

    The principal domain of fr(x) = 1.5*csc(x) is [-pi/2, 0) U (0, pi/2]. The range of this function is (-infinity, -1.5) U (1.5, infinity). In the original post of this thread, you guessed that the domain of fr was (0, pi/2], but the domain is actually quite a bit larger.
     
  15. Feb 3, 2010 #14
    Re: Triangle

    so do we make it?

    for the inverse the image set (range)

    (0,1/2pi]

    for the domain

    (1.5, infinity)
     
    Last edited: Feb 3, 2010
  16. Feb 3, 2010 #15

    Mark44

    Staff: Mentor

    Re: Triangle

    Yes, but those domain/range sets are only half of the story. The inverse function is defined at, for example, -2.

    Did you ever get a better graph of y = 1.5csc(x)? The one you had in post #3 was too limited.
     
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