1. ## Discontinuity

For the absolute value of x-1 (or abs(x-1) on a calculator),
find the points of continuity and the points of discontinuity of the function. Identify each type of discontinuity.

From what I knew of absolute value functions, there are no breaks in the graph.

Is this just a really easy problem or is there something I'm missing?

Thanks!

2. Originally Posted by Truthbetold
For the absolute value of x-1 (or abs(x-1) on a calculator),
find the points of continuity and the points of discontinuity of the function. Identify each type of discontinuity.

From what I knew of absolute value functions, there are no breaks in the graph.

Is this just a really easy problem or is there something I'm missing?

Thanks!
|x - 1| is continuous everywhere

beware, just because it is an absolute value function does not mean it's continuous. for instance, $\displaystyle \left| \frac 1x \right|$ is not continuous at x = 0. the continuity of |x -1| had much to do with the continuity of x - 1

3. You are correct, this line is continuous

From what my teacher tells me, as long as you can draw the line without having to lift up your pencil it's continuous.

Ahh you beat me to it!

Anywho, I think you only run into continuity issues if you have weird piecewise defined functions, or if you have holes or asymptotes. Asymptotes are nonremovable because there's nothing you can do to fix them, but if you have a piecewise defined function you can usually put a coefficient on one or both of the functions to make it continuous, which makes it removable

4. Originally Posted by vesperka
You are correct, this line is continuous

From what my teacher tells me, as long as you can draw the line without having to lift up your pencil it's continuous.

Ahh you beat me to it!

Anywho, I think you only run into continuity issues if you have weird piecewise defined functions, or if you have holes or asymptotes. Asymptotes are nonremovable because there's nothing you can do to fix them, but if you have a piecewise defined function you can usually put a coefficient on one or both of the functions to make it continuous, which makes it removable
What about $\displaystyle y = \sqrt{x}$? Is this continuous at x = 0?

-Dan

5. Yes? I'm sorry, I think I'm missing your point here (not trying to sound rude). I'd say that function is continuous at x=0, and for every value below 0 I would say it's neither continuous or discontinuous, but rather nonexistent.

6. Originally Posted by vesperka
Yes? I'm sorry, I think I'm missing your point here (not trying to sound rude). I'd say that function is continuous at x=0, and for every value below 0 I would say it's neither continuous or discontinuous, but rather nonexistent.
Whoops! Sorry about that. The derivative is not continuous at 0. My bad.

-Dan