Vertical, Horizontal and Oblique Asymptotes Explaining

Click For Summary
SUMMARY

This discussion focuses on identifying vertical, horizontal, and oblique asymptotes for the functions f(x) = (e^(-x) x^5 + 2)/(x^5 − x^4 + x^3 − x^2 + x + 1) and k(x) = sin(1/x) arctan(x). The first function has a vertical asymptote due to the denominator equating to zero, confirmed using the Intermediate Value Theorem. Horizontal asymptotes are determined by analyzing the degrees of the numerator and denominator, with the conclusion that there is no horizontal asymptote for f(x) because the degrees are equal, while oblique asymptotes do not exist since the degree of the numerator is not greater than that of the denominator. The second function, k(x), remains unexplored in detail.

PREREQUISITES
  • Understanding of asymptotes in calculus, including vertical, horizontal, and oblique types.
  • Familiarity with the Intermediate Value Theorem.
  • Knowledge of limits as x approaches infinity.
  • Basic proficiency in polynomial functions and their properties.
NEXT STEPS
  • Study the application of the Intermediate Value Theorem in determining roots of functions.
  • Learn about limits and their role in finding horizontal asymptotes.
  • Explore the concept of oblique asymptotes and how to derive them using polynomial long division.
  • Investigate the behavior of trigonometric functions like sin(1/x) and their asymptotic properties.
USEFUL FOR

Students and educators in calculus, particularly those focusing on asymptotic analysis of functions, as well as anyone seeking to deepen their understanding of polynomial behavior and limits.

tmlrlz
Messages
28
Reaction score
0

Homework Statement



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)

Homework Equations





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)
 
Physics news on Phys.org
tmlrlz said:

Homework Statement



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)

Homework Equations



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 denominator 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 synthetic 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)
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 \displaystyle f(x)=e^{-x} x^5 + \frac{2}{x^5}-x^4 + x^3 − x^2 + x + 1


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.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K