A Continuity question.

[woohoo1]

This is the problem:

If the following function has a removable discontinuity at a, find a function that agrees with f if x not equal to a and is continuos on the real numbers.

f(x) = (3 - sqrt(x))/(9 -x), a = 9

So the function has a removable discontinuity at a = 9, which I found by taking the limit as x approaches 9.

Now, most of the other problems like this I had, I just redefined the function at the value where there was a discontinuity to be equal the limit. This then made the new piecewise defined function continuos on the set of reals. But the original function in this problem is not defined on the set of reals so do I just assign the negative values of x any arbitrary value in the new piecewise defined function, and then just define f(9) as 1/6 (the value of the limit as x approaches 9)?

I hope it is clear what I'm asking. I'll try and clarify if I need to.

 

[Gene]

I could be wrong, but can't you just reduce it to
(3-sqrt(x)/9-x =
(3-sqrt(x)/(3-sqrt(x)*(3+sqrt(x)) =
1/(3+sqrt(x))
----------------
Gene

 

[woohoo1]

Yeah, I got that far, but that still isn't continuous on the set of reals is it? because sqrt(x) doesn't give a real number as a result.

so would I just make the new function:

g(x) = 1/3 if x < 0



1/3 because the function needs to be continuous of the set of reals and lim as x->0 f(x) = 1/3

 

[woohoo1]

sorry I pressed enter by accident:

g(x) = 1/3 if x < 0
1/(3+sqrt(x)) if 0<x<9 and x > 9
1/6 if x = 9

??????????

 

[Gene]

Hmmmm. It is continuous for all f(x) that are defined (0 to infinity) including x=9. They seemed only bothered by that discontinuity at x=9 which has been removed. Whether you have to have continuity in areas where f(x) is un-defined is questionable (to me at least.) Your g(x) doesn't agree with f(x) which was a requirement as I read it.
Any one else have a comment????
----------------
Gene