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Homework Help: Injectivity & Surjectivity

  1. Jan 9, 2010 #1
    Find whether the following function is injective and/or surjective:
    f(x)= (x^2-2x+2)/(2x^2+2x+2)


    Basically, I just need to know if it is surjective. I don't want any proof because I should work it out myself seen as it is homework haha. I have a done it a few different ways and came out with the answer that it is neither, however, the way the question is worded seems like it should be one or the other or both.
    Any help would be greatly appreciated.
    Thank you
     
    Last edited: Jan 9, 2010
  2. jcsd
  3. Jan 9, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Dollydaggerxo! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Show us how you got your answer, and then we'll see what went wrong. :wink:
     
  4. Jan 9, 2010 #3
    oh i didnt see that haha thanks :)

    and well, I constructed the graph of it, and since there is a horizontal line that intersects it more than once, I have concluded that it is not injective.

    for the surjectvity, I have it isnt surjective because if you were to construct a horizontal line at say, -1, it wouldnt intersect it. Therefore it is not surjective for R -> R but perhaps for R -> R+o
    also for the surjectivity, I found that there is no real number, x, such that the function = -1
    (i apologise for the rubbish symbols!)
     
  5. Jan 9, 2010 #4

    tiny-tim

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    ok but messy

    in particular, for the surjectivity, it would be neater to complete the squares, and then consider whether certain things can be negative. :wink:
     
  6. Jan 9, 2010 #5
    oh yes i could do that....
    neither could be negative, which means its not surjective because a positive/positive cannot be negative and therefore cannot be mapped onto every real number?

    sorry for all the questions, I appreciate your help.

    so is it neither? was that right or not?
    Thank you
     
  7. Jan 9, 2010 #6

    tiny-tim

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    Yup! :biggrin:
    Yes, that was right. :smile:
     
  8. Jan 9, 2010 #7
    Okay, thanks for your help, greatly appreciated! :)
     
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