Thread: [SOLVED] Little help with "continuity"

I have trouble grasping the concept of "continuity". I understand it when the teacher explains it to me but when faced with problems I don't seem to be able to do them. Will someone help me with this?

Loosely, a continuous function is one that has no "holes" (i.e., the function is undefined at a point), no "jumps" (i.e., the limit as x approaches a value c from the left is different from the limit as x approaches the same value c from the right), and no vertical asymptotes.

If you say that a function is continuous on an interval, then it has all of these properties on that interval. However, if the interval is open, then the function may have a vertical asymptote at one of the open endpoints and it will still be continuous.

Okay so how exactly must I approach this type of problem?

Find the numbers a and b so that the function f (x) is continuous.

-x^2 + 2 if x < 0
ax + b if 0 less than or equal to x < 1
(x - 1)^2 if 1 less than or equal to x

The function is defined piecewise. At the x-values of 0 and 1, the limit of f(x) from the left must be the same as the limit of f(x) from the right.

In terms of your question, this means that

when and

when .

Thus,

Wow it seems a lot simpler than I thought. My thanks to you sir.