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I have a question about HA, VA and minimum points

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data
    For the function h(x)=[tex]\frac{x^2*e^x}{x}[/tex] , which of the following are true about the graph of y=h(x)?

    I. The graph has a vertical asymptote at x=0
    II. The graph has a horizontal asymptote at y=0
    III. The graph has a minimum point

    2. Relevant equations

    h(x)=[tex]\frac{x^2*e^x}{x}[/tex]

    3. The attempt at a solution

    'I' is wrong because we can cancel X in the denominator and it becomes a hole (or I am wrong?), so its not an asymptote.

    For 'II', I tried different ways.
    My first way was putting 'lan' on each side, and trying to find the limit of that function when x goes to infinite.

    For the third one, when I take the derivative, I get X=-1 when the derivative is equal to 0 and a critical point at X=0...
    But I am getting a max point and not a min point.

    Plus, according to my answers, 'II' and 'III' are correct.

    The thing is that I haven't took cal I since the 10th grade and now I am tutoring cal I at my college, so, I need to refresh the very basic things (even though it doesn't look very basic)
    Also, this question allows using a calculator, but I want to know how to solve it without.

    P.S: I posted it in precal, but I think its more cal than precal.

    Thanks,
    Roni
     
  2. jcsd
  3. Oct 2, 2009 #2

    rock.freak667

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    Homework Helper

    You are correct for I being wrong. I also believe II is wrong as xex=0 can be solved to give a real value for x.


    EDIT: plotted the wrong graph :redface: ignore my post please. :)
     
    Last edited: Oct 3, 2009
  4. Oct 2, 2009 #3
    2 and 3 are indeed correct. X=-1 is a minimum. You can prove 2 by taking the limit as x goes to -infinity, and using lhospital's rule
     
  5. Oct 2, 2009 #4
    This problem has already been discussed and solve https://www.physicsforums.com/showthread.php?t=342328", as the original poster made a double post.

    I'm also not really for sure what rock.freak667 meant by solving xex=0 for number II, because it's asking for a horizontal asymptote.
     
    Last edited by a moderator: Apr 24, 2017
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