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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: 10,184
Registered: 7/15/05
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 20, 2013 3:20 PM
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quasi wrote:
>Hetware wrote:
>>
>>So Quasi was correct.

>
>Of course!
>
>(just kidding)
>

>>I can define a function g(x) to be continuous,
>
>I don't think I said that.
>

>>but it is not the same function as the function f(x) used
>>to define g(x) everywhere f(x) is defined.

>
>As I've said before, the wording is key.
>
>Care needs to taken so as to be both sufficiently rigorous
>and sufficiently clear.
>
>Let's take some examples.
>
>Define f by:
>
> f(x) = (x^- 9)/(x - 3), x! = 0.


I meant:

f(x) = (x^- 9)/(x - 3)

>Then:
>
>* f(3) is undefined.
>
>* f is defined for all real numbers _except_ 3.
>
>* f is continuous on its domain.
>
>* f is not continuous at x = 3.
>
>* f has a removable discontinuity at x = 3.
>
>Now define g by:
>
> g(x) = f(x) for x != 3
>
> g(3) = 6
>
>Then:
>
>* g is continuous on R
>
>* g(x) = x + 3 for all x in R
>
>Remark:
>
>Note that I didn't _define_ g to be continuous.
>
>Rather, I defined g, then concluded that g is continuous.
>
>That assertion requires _proof_, which, although I didn't
>give it, is easily supplied:
>
>(1) Since f is continuous for on its domain, g is continuous
>when x != 3.
>
>(2) At x = 3, we have
>
> g(3) = 6 = lim (x -> 3) f(x) = lim (x -> 3) g(x)
>
>so g is continuous at x = 3.
>
>It follows that g is continuous on R.
>
>But once again, I didn't define g as continuous on R. Rather,
>I defined g and then asserted that g is continuous on R,
>initially without proof. Omitting the proof is acceptable
>provided, for the intended audience, the truth of the claim
>is obvious and the proof is easily supplied if requested.
>
>Let's try a similar example with a change of wording:
>
>Let h be a continuous function such that
>
> h(x) = f(x) for x != 3
>
>Does the above qualify as a "definition" of h?
>
>For the above to be regarded as a definition, it must be the
>case that there is one and only one function h which satisfies
>the above conditions. But you would first have to _prove_ that
>claim. Only then can you claim to have "defined" h.
>
>In other words, imposing conditions on h is not the same as
>defining h.
>
>Now consider a sample test problem ...
>
>Suppose k(x) is continuous on R and such that
>
> k(x) = (x^2 - 9)/(x - 3)
>
>for x != 3. Must k(3) = 6?
>
>The answer is yes.
>
>But the question didn't _define_ k(x). Rather, it specified
>some conditions on k(x) and then asked a question regarding
>such a function k.


quasi


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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quasi
9/28/13
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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
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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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10/6/13
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10/6/13
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10/8/13
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quasi
10/7/13
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9/29/13
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Michael F. Stemper
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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
10/7/13
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10/12/13
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Hetware
10/12/13
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10/13/13
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Richard Tobin
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fom
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magidin@math.berkeley.edu
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10/8/13
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quasi
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10/8/13
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quasi
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10/12/13
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quasi
10/13/13
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10/9/13
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fom
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fom
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9/29/13
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quasi
9/30/13
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quasi
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quasi
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10/14/13
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10/16/13
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10/19/13
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10/20/13
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10/18/13
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10/19/13
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