Vertical Asymptote: Is f Defined at x=1?

  • Thread starter Thread starter dkotschessaa
  • Start date Start date
  • Tags Tags
    Asymptote Vertical
Click For Summary
SUMMARY

The discussion centers on the concept of vertical asymptotes in relation to function definitions, specifically addressing the statement: "If the line x=1 is a vertical asymptote of y = f(x), then f is not defined at 1." The consensus is that this statement is false, as demonstrated by the function f(x) = 1/(x-1), which is undefined at x=1, while other functions can be defined at that point despite having vertical asymptotes elsewhere. The key takeaway is that a function can be defined at a point where a vertical asymptote exists, depending on the function's definition.

PREREQUISITES
  • Understanding of vertical asymptotes in calculus
  • Familiarity with function definitions and piecewise functions
  • Basic knowledge of limits and continuity
  • Experience with algebraic manipulation of functions
NEXT STEPS
  • Study the properties of vertical asymptotes in rational functions
  • Learn about piecewise functions and their definitions
  • Explore limit concepts in calculus, particularly at points of discontinuity
  • Investigate examples of functions with defined values at asymptotes
USEFUL FOR

Students studying calculus, educators teaching asymptotic behavior, and anyone interested in the nuances of function definitions and their implications in mathematical analysis.

dkotschessaa
Messages
1,063
Reaction score
763

Homework Statement



True False

If the line x=1 is a vertical asymptote of y = f(x), then f is not defined at 1.

Homework Equations



none

The Attempt at a Solution



I originally believed this was true, but on finding it was false it sought a counter example:

if for example f(x) = 1/x if x != 0
5 if x = 0

Then the function is defined, but the asymptote still is at x=1, correct?

This is very basic - I just want to make sure I understand it thoroughly. Thanks.
 
Physics news on Phys.org
dkotschessaa said:

Homework Statement



True False

If the line x=1 is a vertical asymptote of y = f(x), then f is not defined at 1.

Homework Equations



none

The Attempt at a Solution



I originally believed this was true, but on finding it was false it sought a counter example:

if for example f(x) = 1/x if x != 0
5 if x = 0

Then the function is defined, but the asymptote still is at x=1, correct?

This is very basic - I just want to make sure I understand it thoroughly. Thanks.

The vertical asymptote for that example is at x=0.

So yes the function is defined at x=1 since if we plug x=1 into the equation, we get 1. The asymptote isn't at x=1 though.
 
Thanks Mentallic!

So you're right. Not a great example. So how about [1/(x-1)] - 1 with f(1) = 5 (or some number)

Point being I guess, that a function can still be defined where there is an asymptote.
 
p.s. Posting a limit problem over in the calc forum, if you're feeling especially helpful today. This question was actually from my calc book.
 
dkotschessaa said:
Thanks Mentallic!

So you're right. Not a great example. So how about [1/(x-1)] - 1 with f(1) = 5 (or some number)
If you define the function to be defined at x=1, then that's what it's going to be. But the function f(x)=1/(x-1) alone is not defined at x=1.

dkotschessaa said:
Point being I guess, that a function can still be defined where there is an asymptote.
As you've done, yes, but the question was implying there are conditions such as the obvious - you can't define it to be defined at that x value :wink:
 
Well, I was just trying to come up with any example that would serve as a situation where 1) - there is an asymptote at some x and
2) the function is defined at x

I'm sure there are other examples.

Thanks again!
 
dkotschessaa said:
Well, I was just trying to come up with any example that would serve as a situation where 1) - there is an asymptote at some x and
2) the function is defined at x

I'm sure there are other examples.

Thanks again!

Well yes, under a certain set of conditions. The answer to the problem is no however.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
8
Views
4K
Replies
21
Views
4K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K