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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: Sep 28, 2013 2:29 PM
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On 9/28/2013 2:35 PM, 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.
>
> quasi
>


The reason I have some misgiving about saying that F(3) is undefined is
that it's possible to arrive at such an indeterminate form in the
process of simplifying or solving for variables, but we tend to accept
results which are defined for the same argument values.

For example, Mathematica tells me that (t^2 - 9)/(t - 3) /. t -> 3 is
an indeterminate form, but Solve[(t^2 - 9)/(t - 3) == 6] returns t->3.




Date Subject Author
9/28/13
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Hetware
9/28/13
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Michael F. Stemper
9/28/13
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Hetware
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quasi
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Hetware
9/28/13
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quasi
9/28/13
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Peter Percival
9/29/13
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quasi
9/28/13
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Hetware
9/28/13
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Richard Tobin
9/28/13
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Hetware
9/28/13
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tommyrjensen@gmail.com
9/29/13
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Hetware
10/6/13
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Hetware
10/6/13
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Peter Percival
10/6/13
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Hetware
10/6/13
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quasi
10/8/13
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quasi
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9/29/13
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Michael F. Stemper
9/29/13
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Hetware
9/29/13
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quasi
9/29/13
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Hetware
9/29/13
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magidin@math.berkeley.edu
10/6/13
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Hetware
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magidin@math.berkeley.edu
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Hetware
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10/12/13
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10/12/13
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fom
10/13/13
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10/13/13
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Richard Tobin
10/13/13
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Hetware
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10/13/13
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fom
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fom
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fom
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fom
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Virgil
10/18/13
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10/19/13
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10/19/13
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10/19/13
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