>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.
* 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
* g is continuous on R
* g(x) = x + 3 for all x in R
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.