find the x values (if any) at which f is not continuous. Which of the discontinuities are removable? f(x)=x-3/x^2-9
I got an answer, then my teacher said it was wrong.... She said you created a hole in the graph at x=3 which is a removable discontinuity. A non-removable discontinuity happens when the denominator =0 (after you factored and reduced). HELP?
@ParthKohli @Calcmathlete @Hero
well the question asks where are the removable discontinuities... You said at x=3, which is correct. Is there another part to the question?
Also, do you understand why it's removable?
I know that x=3 creates a hole, so that is removable discontinuity. Idk the rest..
No the only question is the one written above^^
It's not that it's because it creates a hole. There's another reason why. Can you factor x^2-9 to be in the form, (x - )(x + ) ?
(x-3)(x+3). then it leaves you with just 1/x+3
So the reason why x=3 is a removable discontinuity is because when you factor out the denom. part of it is divided out and "removed". :D Does that help make better sense so far?
Ok yes it does!!
So can you help me with figuring out the rest of the prob?
Awesome. Now that you've simplified the question a bit, we have a new problem.. what can x= where it will again, be zero?
-3
so is -3 a non removable disconuntity?
and that is non-removable because we can't factor or anything like that anymore.
Correctamundo!
:D ok!
is that all I have to do then for this prob?
I believe so, it's just asking where it's discontinuous.
ok! what about the x values where f is not continuous
or do I already have that?
That is just what we found :D
ok thanks so much
It was basically asking you to find removable and non-removable ones.
aka, all kinds. :D
ok cool
be sure to keep this in mind when integrating! you'll end up having to take limits at those points eventually.
OK!
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