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Existence of Inverse Functions

  1. Mar 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Hello. My following problem is partially about the maths concept involved but is largely to do with what the question is actually asking? It's from an online quiz and a printscreen of it has been provided as an attachment.


    2. Relevant equations
    See attachments for the questions and a graph of one of the possible answers to help illustrate my point.


    3. The attempt at a solution
    Please look at the attachment with the graph (drawn in Paint sorry) of f(x) = [tex]\sqrt{x-3}[/tex] drawn in blue and its inverse in green. The question asks if it has an inverse for x > 1 but I'm not sure whether I'm looking at the domain of x > 1 for the inverse or the original function? f(x) has a domain of [3, [tex]\infty[/tex]) so I thought perhaps I would say 'no' to it as it isn't defined for all of x > 1. I then realised that the inverse function has a domain of [0, [tex]\infty[/tex]) so perhaps I should say yes as the inverse is defined for all of x > 1

    This is my main query and it extends throughout the other 3 possible answers as well. I think I'm confidentially able to say 'yes' to (a) because both f(x) = ln(x+2) and its inverse are defined for x > 1 but that doesn't mean I fully understand what I'm being asked for.

    Thank you kindly in advance for help on what I'm actually being asked for. Help on (b) would also be much appreciated because finding the inverse gets a bit harder when the inverse isn't actually defined as a function.

    -David
     

    Attached Files:

  2. jcsd
  3. Mar 15, 2008 #2
    I wonder if it's permissible to upload pictures on something like tinypic, and then post it.
     
  4. Mar 15, 2008 #3
    Ahh okay. Here we go:

    [​IMG]

    Graph of Question B and the inverse function.

    [​IMG]
     
  5. Mar 15, 2008 #4
    I think they might mean x>1 for the original function.

    However, ln(x+2) is the only function where the inverse function is still a function and isn't ambiguous, either.
     
  6. Mar 15, 2008 #5
    That's what I was thinking. I thought that (d) would be correct too because the inverse of f(x) = |x + 1| for x > 1 is specifically defined as shown in the below graph. The circled in red parts are the relevant bits as they are the graph of f(x) for x > 1 and then its inverse.

    [​IMG]
     
  7. Mar 15, 2008 #6
    But in order for a function to be a function, it has to pass the vertical line test, which the inverse of that graph doesn't do.
     
  8. Mar 15, 2008 #7
    But the inverse of that graph does pass the vertical line test if only the inverse for the function over the domain [1, + infinity) is considered.
     
  9. Mar 15, 2008 #8
    It doesn't. The inverse of the function has two y coordinates for every x coordinate on [0,∞).

    Edit: Ah, nevermind. I get what you're saying. Quite the ambiguous question indeed. But I don't believe they are asking you to restrict the domain when plotting, just restricting the domain when seeing which has an inverse when x>1.
     
    Last edited: Mar 15, 2008
  10. Mar 15, 2008 #9
    So with that in mind you'd agree on at least (a) and (d) being correct?
     
  11. Mar 16, 2008 #10
    Does anyone else have any input on this? The quiz gets handed in in 3 hours and atm I'm ticking (a) and (d). I'm pretty positive on this but any other input would be much appreciated.
     
  12. Mar 16, 2008 #11

    tiny-tim

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    Homework Helper

    Hi David! :smile:

    "Which of the following functions has an inverse for x > 1?

    a. ln(x+2) …"

    That definitely means if the x in the definition is > 1.

    If they meant y (or whatever), they'd say so! :smile:
     
  13. Mar 16, 2008 #12
    Okay then. So to clarify, that'd mean:

    (a) Yes because the function is defined for -3 -> + infinity and has a clearly defined inverse function over the domain 1 -> + infinity

    (b) No because the function does not have a clearly defined inverse over the domain 1 to + infinity due to the absolute value effectively turning it into a hybrid function. There is also the issue of the function not being 1 to 1

    (c) No because the function is only defined for 3 -> + infinity and therefore isn't defined over all of 1 -> + infinity and therefore doesn't fit the criteria

    (d) Yes because whilst the function isn't 1 to 1 and therefore doesn't have a single inverse function defined for its entire domain, for the domain 1 -> + infinity it is 1 to 1 and has a clearly defined inverse function

    Are those answers and reasonings correct?
     
  14. Mar 16, 2008 #13

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi David! :smile:

    (a) correct.

    (b) correct - but what's a hybrid function? only the one-to-one matters.

    (c) correct, if the values are all real, not correct if they allow imaginary values!

    (d) correct.

    btw, a good guide is to differentiate - if the differential is always non-zero, then there can't be a turning-point, and and so the function must be single-valued! :smile:

    But if the differential is zero, you still have to check, since it might be an inflexion point and not a turning-point. :frown:
     
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