Let

f is continuous on each interval.

Answer

Since we are asked about the same function on so many intervals, first we'll figure out all the values at which *f* is discontinuous.

- Where is
*f* undefined?*f* is undefined at *x* = 1 because the function definition forgot to say what to do when *x* = 1. *f* is also undefined at 2, since is undefined at 2.

- Where does the limit of
*f* not exist?We already know *f* is undefined at *x* = 1 and *x* = 2, therefore we don't need to worry about those values. Since *x* = 0 is a spot where the piecewise definition changes, we'd better check out the limit there.The left-hand limit is The right-hand limit is

Since the one-sided limits disagree,does not exist, and *f* is discontinuous at *x* = 0.

- There are no spots where
*f* exists, the limit exists, and they disagree. We've already taken care of all the possible trouble spots.

To summarize, *f* is discontinuous at *x* = 0, *x* = 1, and *x* = 2.Now we can answer the real questions.

- (-10,0) - Yes,
*f* is continuous on this interval since none of 0, 1, and 2 are in the interval. - (0,1) - Yes,
*f* is continuous on this interval since none of 0, 1, and 2 are in the interval. - (0,2) - No,
*f* is not continuous on this interval since 1 is in the interval. - (1,10) - No,
*f* is not continuous on this interval since 2 is in the interval. - (1,2) - Yes,
*f* is continuous on this interval since none of 0, 1, and 2 are in the interval. )