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Derivatives and graphs

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data

    y=(2x+1)/[tex]\sqrt{x^2+1}[/tex]

    Find where are the asymptotes, where is it increasing increasing/decreasing, ect.....

    2. Relevant equations



    3. The attempt at a solution
    when I took the first derivative (im trying to find where it increases/decrease), I got
    dy/dx = [tex]\frac{2x^2-x+2}{(x^2+1)\sqrt{x^2+1}}[/tex]

    but there isn't any roots for this function... so what does that mean?
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 9, 2007 #2
    No vertical asymptotes as x^2+1 can never be 0.

    No horizontal asy as (x^2)^1/2 is one, and 2x is one.

    Slant asymptote is synthetic division (x^2+1)^(1/2) |2x+1

    If there are no 0's then that means that the equation always increases or decreases.
     
  4. Dec 10, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Grammer police: Better would be "either always increases or always decreases".


    Any graph that does not have a horizontal line segment "always increases or decreases"!
     
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