Finding a function given a limit and restriction

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Homework Help Overview

The discussion revolves around finding a function given specific limits and restrictions, particularly focusing on the behavior of the function at a point where it may have a hole or asymptote. Participants are exploring the implications of continuity and the definition of the function at that point.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are considering whether a function can have a hole at a certain point and how that affects its definition. There are discussions about using piecewise functions and the implications of defining a function at a point where it is otherwise undefined.

Discussion Status

There is an ongoing exploration of different function definitions and their properties. Some participants have suggested using piecewise functions, while others are questioning the assumptions about continuity and the specific requirements of the problem. No consensus has been reached, but various interpretations and approaches are being discussed.

Contextual Notes

Participants note that the problem states f(2) ≠ 4, but it does not clarify whether f(2) must be defined or what it should equal. This ambiguity is leading to varied interpretations of the function's behavior at that point.

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Homework Statement
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Relevant Equations
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Not sure how to go about this. Would relying on a hole or asymptote work?
 
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Mustard said:
Homework Statement:: Look at snippet
Relevant Equations:: Look at snippet

View attachment 268806
Not sure how to go about this. Would relying on a hole or asymptote work?
yes, the function need not be continuous.
 
ehild said:
yes, the function need not be continuous.
Hmmm... I was thinking of a function like f(x)= 4x-8/x-2 = 4(x-2)/x-2 = 4, that would make a hole at x=2.
But would also make f(2) = 4 ? Would it be safe to assume since there is a hole at (2,4) , it is undefined therefore f(2) does not equal 4 ?
 
Mustard said:
Hmmm... I was thinking of a function like f(x)= 4x-8/x-2 = 4(x-2)/x-2 = 4, that would make a hole at x=2.
But would also make f(2) = 4 ? Would it be safe to assume since there is a hole at (2,4) , it is undefined therefore f(2) does not equal 4 ?
The function has to be defined at x=2.
You can define it in a way everywhere except x=2, and define its value separately at x=2.
 
ehild said:
The function has to be defined at x=2.
You can define it in a way everywhere except x=2, and define its value separately at x=2.
Do you mean like a piece wise function ?
 
Mustard said:
I was thinking of a function like f(x)= 4x-8/x-2 = 4(x-2)/x-2 = 4
You need more parentheses.
4x-8/x-2 means ##4x - \frac 8 x - 2##
and 4(x-2)/x-2 means ##\frac{4(x - 2)}x - 2##
When a numerator or denominator of a fraction contains multiple terms, you need parentheses around the whole numerator or denominator, like so.
f(x)= (4x-8)/(x-2) = 4(x-2)/(x-2)
ehild said:
You can define it in a way everywhere except x=2, and define its value separately at x=2.
Mustard said:
Do you mean like a piece wise function ?
Yes.
 
I suppose you can take any function, and define another function as that one multiplied it by (x - 2) and also divided by (x - 2) - I defer to the mathematicians as to whether that is formally a bona fide new function but even if it is it looks to me trivial and cheating.

The problem says that f(2) ≠ 4 but it doesn't say it has to be equal something or be defined. Probably you have studied before functions which at some point become equal to 0/0 but you were able to find their limit at that point? So you could adapt one of those, I guess that is what the question is expecting.
 
epenguin said:
I suppose you can take any function, and define another function as that one multiplied it by (x - 2) and also divided by (x - 2) - I defer to the mathematicians as to whether that is formally a bona fide new function but even if it is it looks to me trivial and cheating.

Two functions f : A \to B and g: C \to D are equal if and only if A = C and B = D and for every a \in A we have f(a) = g(a). Thus f : \mathbb{R} \to \mathbb{R} and <br /> g: \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} \frac{(x-2)}{(x-2)}f(x) &amp; x \neq 2 \\<br /> f(2) &amp; x = 2 \end{cases} are the same function, but if g(2) is defined to be something other than f(2) then they are not. However, g is always equal to <br /> h: \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} f(x) &amp; x \neq 2 \\<br /> g(2) &amp; x = 2 \end{cases} which is easier to write and read.

I don't think it assists anyone to characterise an obvious or straightforward example which satisfies the conditions of the question as "trivial and cheating".

epenguin said:
The problem says that f(2) ≠ 4 but it doesn't say it has to be equal something or be defined. Probably you have studied before functions which at some point become equal to 0/0 but you were able to find their limit at that point? So you could adapt one of those, I guess that is what the question is expecting.

I would say that f(2) \neq 4 means that f(2) is defined but is not 4. If a function for which f(2) was not defined but \lim_{x \to 2} f(x) = 4 was wanted, then the question would have said exactly that; and f : \mathbb{R} \setminus \{2\} \to \mathbb{R} : x \mapsto 4 would work.

(It would be nice if the question had started "Find f : \mathbb{R} \to \mathbb{R} such that ...")
 

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