# Thread: [SOLVED] Little help with &quot;continuity&quot;

1. ## [SOLVED] Little help with &quot;continuity&quot;

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?

2. 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.

3. 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

4. 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

$-x^2 + 2 = ax + b$ when $x = 0$ and

$ax + b = (x-1)^2$ when $x = 1$.

Thus,

$-0^2 + 2 = a(0) + b$

$b = 2$

$a(1) + 2 = (1-1)^2$

$a + 2 = 0$

$a = -2$

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