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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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quasi

Posts: 10,188
Registered: 7/15/05
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Sep 28, 2013 3:02 PM
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Hetware wrote:
>quasi wrote:
>> Hetware wrote:
>>>
>>> I'm reading a 1953 edition of Thomas's Calculus and
>>> Analytic Geometry.
>>>
>>> In it he states that given:
>>>
>>> F(t) = (t^2-9)/(t-3)
>>>
>>> F(t) = (t-3)(t+3)/(t-3) = t+3 when t!=3.
>>>
>>> But F(t) is not defined at t=3 because it evaluates to 0/0.
>>>
>>> If someone were to ask me if (t^2-9)/(t-3) is defined when t=3,
>>> I would say it is

>>
>> Then you would be wrong.
>>

>>> because it can be simplified to t+3.
>>
>> To get that result, you had to cancel the common factor t-3 in
>> numerator and denominator. But that cancellation depends on the
>> simplification
>>
>> (t - 3)/(t - 3) = 1
>>
>> which is only valid if t != 3.
>>
>> In a first level algebra course (Elementary Algebra) where
>> function concepts are not yet in play, the simplification
>>
>> (t^2 - 9)/(t - 3)
>>
>> = (t + 3)(t - 3))/(t - 3)
>>
>> = t + 3
>>
>> is allowed, without worrying about exceptional values of t for
>> which the simplification fails.
>>
>> But at the next level of algebra, algebraic expressions are
>> often being regarded as functions, so more care is taken to
>> identify those exceptional values.
>>

>>> Am I (and/or Thomas) engaging in meaningless hair-splitting
>>> regarding the question of F(3) being defined?

>>
>> For functions, identifying the precise domain is key.
>>
>> Thomas is correct.
>>
>> Those hairs _need_ to be split.
>>
>> All modern precalculus and calculus texts are careful (in the
>> context of deciding whether functions are equal) to identify
>> exceptional values where simplifications fail.

>
>The reason I have some misgiving about saying that F(3) is
>undefined is


F is being regarded as a _function_ not just an expression,
so the definition of F is taken literally. You don't have the
right to simplify the definition before evaluating the
function unless you can guarantee that the results would
always be the same, before and after.

Thus if F(t) = (t^2 - 9)/(t - 3), then by direct substitution

F(3) = (3^2 - 9)/(3 - 3) = 0/0

which is undefined.

If we let G(t) = t + 3, then G(3) = 3 + 3 = 6.

Thus, F and G are not equal as functions.

They _are_ equal for all values t _except_ t = 3.

Thus the graphs are not the same.

The graph of the equation y = G(x) is a straight line.

The graph of the equation y = F(x) is the same straight
line, but with a missing point at (3,6), usually symbolized
by placing a small open circle around (3,6) to make clear
that there is a missing point there (a "hole").

With regard to continuity, the function G is continuous,
whereas the function F has a discontinuity at t = 3.

However since the limit of F, as t approaches 3 _exists_,
we say that F has a _removable_ discontinuity at t = 3.

You need to drop your preconceptions on this so as to make
progress in your study of Calculus. All modern Calculus texts
are consistent with regard to this issue.

quasi


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9/28/13
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Hetware
9/28/13
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Michael F. Stemper
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quasi
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