1. Not finding help here? Sign up for a free 30min 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!

Vertical, Horizontal and Oblique Asymptotes Explaining

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Explain why each of the following functions have or do not have vertical, horizontal,
    and/or oblique asymptotes:
    (a) f(x) = e-x x5 + 2/ x5 − x4 + x3 − x2 + x + 1
    (b) k(x) = sin(1/x) arctan(x)

    2. Relevant equations



    3. The attempt at a solution
    For the first one, to explain the vertical asymptote, i knew that it will occur where the denominator is equal to zero, however I did not know exactly where that was so i used the Intermediate Value Theorem to prove that the function g(x) = x5 − x4 + x3 − x2 + x + 1
    has at least one root which will prove that the original function f(x) has at least one VA. The problem i run into is with the Horizontal and Oblique Asymptote. Before I had always been thought that if the highest power of x in the denominater is of the same degree as the numerator then there is a horizontal asymptote at the ratio of the leading coefficients of the highest power of x. If the degree in the numerator is less than that of the denominator then there is a horizontal asymptote at y = 0. As for oblique asymptotes, i was taught that if the degree of the numerator is EXACTLY one greater than the degree of the denominator then there is an oblique asymptote which you can find using synthtic or long division. However in this case it is different because if you look at the powers of x, the highest power of x in numerator and denominator is x5 so you just look at the coefficients because you multiply both top and bottom by 1/x5 but then you end up with e-x. Can an exponential function be an oblique asymptote? How do i explain that there is no horizontal asymptote, because the degree of the numerator and denominator is the same which would contradict what i was taught. How do i prove this is an oblique asymptote because for that the degree has to be greater in the numerator which it isn't. Any help is greatly appreciated.
    (As for the second one I have no clue how to do that one and thought that if i got help on the first one, then i would build on that and ask for help on the second one afterwards)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 24, 2012 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    If the numerator of f(x) is e-x x5 + 2 and the denominator is x5 − x4 + x3 − x2 + x + 1, you really should enclose each in parentheses.
    f(x)=(e-x x5 + 2)/(x5 − x4 + x3 − x2 + x + 1)​

    What you have written is equivalent to [itex]\displaystyle f(x)=e^{-x} x^5 + \frac{2}{x^5}-x^4 + x^3 − x^2 + x + 1[/itex]


    Those rules you list for asymptotes are for rational functions -- functions in which the numerators & denominators are strictly polynomials. Some of those ideas are also aplicable here but degree is meaningless except for the denominator of f(x).

    To find find horizontal asymptotes, you need to take limits as x→-∞ and as x→+∞.

    Vertical asymptotes may still occur when the denominator is zero.

    These are pretty interesting functions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Vertical, Horizontal and Oblique Asymptotes Explaining
Loading...