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Need Help With Calculus Question

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    si9vr73ny4o1.jpg

    Note: The functions above are in piece wise form, I just didn't know how to put them in piece wise function here.


    2. Relevant equations



    3. The attempt at a solution

    http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:-1,-xSMB02ESMB032SMB02eSMB03+2SMB02lSMB03?p=87?p=38 [Broken] = 1

    http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:-1,4x+5SMB02lSMB03?p=83?p=38 [Broken] = 1

    1/1 = 1

    Answer: limit as x approaches f(x) = 1
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 21, 2012 #2
    I'm not sure what your question is..
    Your answer is correct
     
  4. Jan 21, 2012 #3
    Edited the question! Hopefully you will be able to see it and understand it and help me out.
     
  5. Jan 21, 2012 #4
    I understand the question in the image but I don't understand your question
    I'll try and explain piecewise functions but I may be going in the wrong direction

    your f(x) can be thought of as two seperate functions in the two seperate areas, when x is greater than or equal to -1 we treat the function as 4x+5
    when x is less than -1 we treat the function as -x^2 + 2

    In trying to find the limit, in this simple case you can just evaluate the function at x=-1. So we would use the region where x is greater than or equal to -1 and get -4+5 = 1.
    I'm assuming this is all that you're expected to be doing since it seems like this is an early calculus class that isn't going to be all rigorous.

    I attatched a graph of your function if it will help
     

    Attached Files:

  6. Jan 21, 2012 #5
    So the answer to my question is just limit as x approaches -1 f(x) = 1 ? And will it be acceptable for me to just show the two cases that I have in my original answer, where I have plugged in a -1 every where I have an x?
     
  7. Jan 21, 2012 #6

    SammyS

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    While it is true that [itex]\displaystyle \lim_{x\to-1}\,f(x)=1\,,[/itex] what you wrote isn't quite what you need to say.

    Also, what does 1/1 = 1 have to do with this question ?

    You need to show that [itex]\displaystyle \lim_{x\to-1^-}\ f(x)=\lim_{x\to-1^+}\ f(x)\,.[/itex] If that is true then that common result is the limit you're looking for.

    While it is true that [itex]\displaystyle \lim_{x\to-1^-}\ f(x)= \lim_{x\to-1}-x^2+2\,,[/itex] and [itex]\displaystyle \lim_{x\to-1^+}\ f(x)= \lim_{x\to-1}4x+5\,,[/itex] you really should state that this is what you're doing.
     
    Last edited by a moderator: May 5, 2017
  8. Jan 21, 2012 #7
    So can I just rewrite your last sentence and add something like:

    The above statements hold true therefore limit as x approaches -1 f(x) = 1. I mean would that be an acceptable answer?
     
  9. Jan 22, 2012 #8

    SammyS

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    As an instructor I would want you to indicate that you're taking a limit as x approaches -1 from the left, and taking a limit as x approaches -1 from the right.

    That may be picky on my part, and many of my students would agree.
     
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