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Draw the graph of arctan((x-1)/(x+1))

  1. Jan 29, 2016 #1
    I am so far able to find the domain, intercepts, concavity and increase/decrease. But I am stuck at finding the asymptotes for the graph.

    I think there is no vertical asymptote or is there one at x=-1?
    I think the horizontal asymptote is y=pi/4
     
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  3. Jan 30, 2016 #2

    Mark44

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    ##\frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}##
    What happens to ##arctan(1 - \frac{2}{x + 1})## as x gets very large or very negative?
     
  4. Jan 31, 2016 #3
    For the vertical asymptote, ask yourself this: can you take the arctan of an undefined value? When will ##\frac{x-1}{x+1}## be undefined?
     
  5. Jan 31, 2016 #4
    I think the limit of arctan((x-1)/(x+1)) as x approaches -1 from the right is pi/2 and as x approaches -1 from the left is -pi/2, which means there is no vertical asymptote here but a jump discontinuity here right?
     
  6. Jan 31, 2016 #5
    when x=-1 arctan is undefined but as x approaches x=-1 (x-1)/(x+1) approaches negative infinity and positive infinity from left and right. Let (x-1)/(x+1) = t, then arctan(t) approaches pi/2 and -pi/2 as t approaches the two infinity is this correct? then there is no vertical asymptote?
     
  7. Jan 31, 2016 #6
    is the horizontal asymptote correct? is there any other horizontal asymptote that I am missing? I got a graph like this, but I'm not sure if it is correct
     

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  8. Jan 31, 2016 #7
    I don't believe that is so. It is undefined at x=-1, but I don't believe that as x approaches -1 that the function approaches infinity with either sign.
    No, x=-1 is the vertical asymptote.
     
  9. Jan 31, 2016 #8
    The graph is good.
     
  10. Jan 31, 2016 #9

    Samy_A

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    That's not correct.
     
  11. Jan 31, 2016 #10
    Ok, I am probably thinking of an incorrect definition of asymptote. The function does get infinitely close to but never reaches π/4 as x approaches -1. Does this not make it an asymptote?
     
  12. Jan 31, 2016 #11
    x approaches infinity and function goes to pi/4 thats the definition of a horizontal asymptote, correct?
     
  13. Jan 31, 2016 #12
    Yes that it definitely an asymptote. I'm just not sure about x=-1.
     
  14. Jan 31, 2016 #13

    Samy_A

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    No. A function f has a vertical asymptote at x=a if ##\displaystyle\lim_{x\rightarrow a^+}f(x)=\pm \infty## or ##\displaystyle\lim_{x\rightarrow a^-}f(x)=\pm \infty##.
    That's definitely not the case with an arctan.
     
  15. Jan 31, 2016 #14
    Ok, my bad.
     
  16. Jan 31, 2016 #15
    I'm not sure if the graph looks like that. Because the calculator gives a different image.
     

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  17. Jan 31, 2016 #16
    That's the graph my calculator gives.
     
  18. Feb 1, 2016 #17

    Samy_A

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    You misunderstood me. I was answering @Isaac0427 's post concerning vertical asymptotes. The definition of a vertical asymptote makes it evident that your function doesn't have vertical asymptotes. At x=-1 it has a jump discontinuity, exactly as you wrote.
     
    Last edited: Feb 1, 2016
  19. Feb 2, 2016 #18

    Svein

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    upload_2016-2-2_21-48-13.png
    I used Atan2, which takes the x and y values separately.
     
  20. Feb 2, 2016 #19
    The trickiest part is figuring out what happens as x → -1 from above and from below. It appears from your graph that you've already done that.

    But " The function does get infinitely close to but never reaches π/4 as x approaches -1." is not true, because of a mistake in this statement. It's also important to consider the cases x < -1 and x > -1 separately when x → -1.
     
    Last edited: Feb 2, 2016
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