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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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Hetware

Posts: 148
Registered: 4/13/13
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 13, 2013 10:01 PM
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On 10/13/2013 9:41 PM, fom wrote:
> On 10/13/2013 8:00 PM, Hetware wrote:
>> On 10/13/2013 3:32 PM, Hetware wrote:
>>

>>> "Definition 2-1.1. A function f is a set of ordered pairs, no two of
>>> which have the same first element. The set of first elements of the
>>> pairs is called the domain of the function, whereas the set of second
>>> elements of the pairs is called the range. The domain and range
>>> elements are related by a given rule"

>>
>> Alas and alack, I forgot the citation:
>>
>> Introduction to Vector and Tensor Analysis, by Robert C. Wrede.
>>

>
> Read the definition carefully.
>
> Before you can even consider the property
> of continuity, you must show that the
> relationships asserted by your description
> actually satisfy the definition of a function.
>
> For example,
>
> f(x) = x^2
>
> is a function. Its inverse, namely,
>
> f(x) = x^(1/2)
>
> is not. This is why one speaks of
> a principal square root,
>
> f(x) = surd(x)
>
> and treats (the relation)
>
> f(x) = x^(1/2)
>
> as two separate functions,
>
> f(x) = surd(x)
>
> f(x) = -surd(x)
>
> A similar situation exists when
> treating the analytic expression
> describing a circle,
>
> x^2 + y^2 = r^2
>
> where
>
> f(x) = ( r^2 - x^2 )^(1/2)
>
> is not a function and must be
> treated as
>
> f(x) = surd( r^2 - x^2 )
>
> f(x) = -surd( r^2 - x^2 )
>
>
> Now, for your formula,
>
> f(t) = (t^2-9)/(t-3)
>
> you may not form a value
> for f(t) at t=3.
>
> That is, you cannot divide by 0.
>
> So, what value completes the
> ordered pair
>
> < 3, ? >
>
> This question is motivated by
> the very definition that you
> have provided here.
>
>
> What calculus enables us to do is
> to "repair" this problem by distinguishing
> between removable discontinuity
>
> http://www.mathwords.com/r/removable_discontinuity.htm
>
> https://www.google.com/search?q=removable+discontinuity&client=firefox-a&hs=WIt&rls=org.mozilla:en-US:official&tbm=isch&tbo=u&source=univ&sa=X&ei=Z0pbUsmDPKPr2wXz_4CADA&ved=0CDcQsAQ&biw=1352&bih=634&dpr=1
>
>
>
>
> and essential discontinuity
>
> http://www.mathwords.com/e/essential_discontinuity.htm
>
> https://www.google.com/search?q=essential+discontinuity&client=firefox-a&hs=5YY&rls=org.mozilla:en-US:official&tbm=isch&tbo=u&source=univ&sa=X&ei=Z0lbUpCQDYfI2gWYzYG4DQ&ved=0CEIQsAQ&biw=1352&bih=634&dpr=1
>
>
>
>
> It seems unlikely, but I hope that helps.
>
>
>


Or I just assert that the function which is defined everywhere except
where there is a removable discontinuity is continuous, and everything
else falls in to place.

The statement "Let f(t) be a continuous function for all real numbers t"
has a concise meaning.


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