Finding a function given a limit and restriction

[Mustard]

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?

 

[ehild]

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.

 

[Mustard]

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 ?

 

[ehild]

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.

 

[Mustard]

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 ?

 

[Mark44]

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)

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 ?

Yes.

 

[epenguin]

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.

 

[pasmith]

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 $$g: \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} \frac{(x-2)}{(x-2)}f(x) & x \neq 2 \\<br /> f(2) & x = 2 \end{cases}$$ 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 $$h: \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} f(x) & x \neq 2 \\<br /> g(2) & x = 2 \end{cases}$$ 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".

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 ...")