Vertical Asymptote: Is f Defined at x=1?

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Homework Help Overview

The discussion revolves around the concept of vertical asymptotes in relation to the definition of a function at specific points, particularly focusing on whether a function can be defined at a point where a vertical asymptote exists.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between vertical asymptotes and function definitions, questioning whether a function can be defined at a point where an asymptote is present. Various examples are proposed to illustrate these points.

Discussion Status

Participants have engaged in a back-and-forth regarding the validity of examples and the implications of function definitions at points of vertical asymptotes. Some guidance has been offered, but multiple interpretations of the problem are still being explored.

Contextual Notes

There is an emphasis on the need for clarity regarding the conditions under which a function can be defined at a point associated with a vertical asymptote. Participants are also referencing homework constraints and examples from calculus materials.

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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.
 
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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.
 

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