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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 'ln' 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.

    Thanks,
    Roni
     
    Last edited: Oct 2, 2009
  2. jcsd
  3. Oct 2, 2009 #2
    I) You are correct. However, note that if you happen to define h(0)=0, which is what you get after canceling the x in the denominator, then there isn't a hole. Though either way, the function does not have an asymptote, as you said, and as stated, the function is not defined at x=0.

    II) What does 'lan' mean? To find the horizontal asymptotes, if there are any, you take the limit of h(x) as x goes to infinity, and then you take the limit as x goes to -infinity. If you get a number, then y=that number is a horizontal asymptote. It is always possible to get infinity (or -infinity) for the limit, which in that case, there is no horizontal asymptote. That's why you have to check both cases.

    III) There is no critical point at x=0 because the function isn't even defined there. So x=-1 is your only critical point. How did you determine the maximum? Show your methods, and check again, because you should get a maximum from that.

    Also, there's a concern that you're tutoring calculus I if you couldn't figure this out on your own or with the use of a book. You should either have the book you're tutoring for, or check one out of the library. As a tutor, you should be in a position to explain and to not just being able to solve the problems by getting help of your own.
     
  4. Oct 2, 2009 #3
    A few things, for III, right, its a min point..

    for II, I did to infinite and negative infinite but only with L'Hopital's rule. I don't think they are supposed to know how to use L'Hopital's rule. Anyways, this exercise allows them to use the calculator. I thought there was a simpler way to do it.

    And right, as a tutor I should know how to solve all this and know how to explain it. I know that if I understand something I can explain it.
    So, now I am solving final exams and that was my only problem ( though not a real problem because I could solve it with a more advanced way).
    And you are right, I should open a book for the next time I face a problem :)

    Thank you for your help.
     
  5. Oct 2, 2009 #4
    There is no reason to use l'Hopital's rule here. You can just cancel out the x in the denominator when taking the limit, so that you are taking the limit of xex as x goes to plus/minus infinity. This limit is infinity as x goes to positive infinity, and the limit is zero as x goes to negative infinity. You need to know why though.

    There are also methods to find the minimum of such a function without looking at the graph on a calculator, so you need to review in a text the sections that cover the first and second derivative tests.
     
  6. Oct 2, 2009 #5
    Yes, I found the minimum, I did something wrong and I got a max point but I found my mistake.

    And when we are trying to find lim of x*e^x when it goes to negative infinite, we get negative infinite over infinite, which is undefined.

    What method did you use for the limit?
     
  7. Oct 2, 2009 #6
    As x goes to negative infinity, the function ex decreases (approaches zero) much faster than x goes to negative infinity. So that's why the limit is zero in that case. I guess I misspoke as you technically need to apply l'Hopital's rule here, but the reasoning above is fine.
     
  8. Oct 2, 2009 #7
    oh right, how could I forget this :\ ... lol e^x increases faster than x ....

    guess I need to keep reviewing...
     
  9. Oct 2, 2009 #8
    How can you tell that ex goes to 0 fast enough for that limit to be 0?
     
  10. Oct 2, 2009 #9
    Look at the http://www.wolframalpha.com/input/?i=graph+y=x+and+y=exp(x)" of y=ex and y=x. You can see as x gets more and more negative, y=ex gets closer to zero much faster than x gets more negative. The exponential function is said to dominate the limit, because it has more of an effect than the x term.
     
    Last edited by a moderator: Apr 24, 2017
  11. Oct 2, 2009 #10
    I meant to say without a calculator because the OP said he wanted to do it without a calculator.
     
  12. Oct 2, 2009 #11
    I didn't say you need a calculator. I included the graph as a reference. You should be able to graph those two functions. Also, just understand that the exponential function often dominates limits and grows much faster than x does (or decreases to zero much faster than x decreases). To prove this limit exactly, you need l'Hopital's rule as mentioned above.
     
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