I just called them and the dude figured out my question in less than 2 minutes from me reading him the problem...but I still don't understand

 

Got it.

 

...and? :bs:

 

this might help

 

Thanks, but don't forget the open hole at (-1, 11)

 

I don't understand whats happening:

-1<=x<=3 then y=x^2+2
x<2 then y=11
x > 3 then y = 11

You have a conflict with the definition: x can't take on two values (i.e. be x^2+2 @x<=3 and also be 11@x<2). Are you sure its not x<-2 ?

But by either definition I don't think its discontinuous anywhere. A discontinuity (I thought) was when you had no y-values for a range of x. It appears you have either more than 1 value of y for an x OR only a 1:1 map (a defined, continuous function).

 

My bad I merged two problems

11 if x < -1

 

came here to post that

 

For this one you have to know how the range is defined. Is this function defined for all real numbers? If so, then it is discontinuous when x >= -1. If y=0 when x>-1 then its discontinuous (there's a jump) @ x=-1.

Similarly, this would be discontinuous outside the range if you are dealing with real numbers. So whenever x<-1.

In all cases, if x was limited by the definition in the right way(say x could only take on values >-1) then the function in problem #2 could never be discontinuous.

Notice that @x=3, y=11 which is the same when x>3, this means there is no 'jump' or discontinuity at that point. i.e. the limit as you approach from the right is the same as the left.

 

and heres the red one.