1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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.

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: I have a question about HA, VA and minimum points
  1. Minimum Point (Replies: 7)

  2. Ok I have a question (Replies: 2)