# AP Calculus AB : An intuitive understanding of continuity

## Example Questions

### Example Question #1 : An Intuitive Understanding Of Continuity

Is the following piecewise function continuous for all ? If not, state where it is discontinuous.

No. The function is not continuous at .

No. The function is not continuous at both  and .

No. There are sharp turns at  and .

Yes. The function is continuous at all .

No. The function is not continuous at .

Yes. The function is continuous at all .

Explanation:

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that  and  have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have  and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there are sharp turns at  and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

### Example Question #2 : An Intuitive Understanding Of Continuity

when  and

when

At the funciton described above is:

differentiable but not continuous

undefined

continuous but not differentiable

both continuous and diffentiable

neither differentiable or continuous

both continuous and diffentiable

Explanation:

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

### Example Question #3 : An Intuitive Understanding Of Continuity

Which of the following functions contains a removeable discontinuity?

Explanation:

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as  approaches  exists, but the value of  does not.

For example, the function contains a removeable discontinuity at . Notice that we could simplify as follows:

, where .

Thus, we could say that .

As we can see, the limit of  exists at , even though  is undefined.

What this means is that  will look just like the parabola with the equation EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

, and

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

and are continuous over all the real values of ; they have no discontinuities of any kind.