Continuity of Functions Exercises

Answer

Answer. We tweak the definition of continuous to involve the left-sided limit at x = b. f must be continuous on (a, b) and 

• f(b) must exist.

•  must exist.

• f(b) must equal .

Answer

f must be continuous on both [a, b) and on (a, b]. Equivalently, f must be continuous on (a, b) and

• f(a) and f(b) must exist

•  and  must exist

• f(a) must equal  and f(b) must equal 

To think of this with more words and fewer expressions, the value of f at each endpoint must be what we'd expect if we let x approach the endpoint from within the interval.

Answer

• [a, b]: No, because f(a) and f(b) are undefined.

• [a, b): No, because f(a) is undefined.

• (a, b]: No, because f(b) is undefined.

• (a, b) Yes, because we don't need the consider the endpoints.

Answer

• [a, b]: No, because f(b) is undefined.

• [a, b): Yes, because  and f is continuous on (a,b).

• (a, b]: No, because f(b) is undefined.

• (a, b): Yes, because we have a nice, smooth curve between these points.

Answer

Nope on all accounts, because f is not continuous on the open interval (a, b) to begin with.

Answer

If we draw the function, we get something like this:

This makes it easier to see what's going on. Since f(0) disagrees with both  and , f will be discontinuous on any interval containing 0, and continuous on any other interval. Here's the breakdown:

• [-1, 0] - No
• [-1, 0) - Yes
• (-1, 0] - No
• (-1, 0) - Yes
• [0, 1] - No
• [0, 1) - No
• (0, 1] - Yes
• (0, 1) - Yes

Answer

Answer. If we graph the function, we see this:

Although f is discontinuous at 1 when we look at the whole graph, f(1) agrees with its right-sided limit (that is, ). This means if x = 1 is the left endpoint of an interval, f can be continuous on that interval.

• [0, 1] - No, because .
• [0, 1) - Yes, because x = 1 is not included in this interval.
• (0, 1] - No, because .
• (0, 1) - Yes, because x = 1 is not included in this interval.
• [1, 5] - Yes, because f is continuous on (1, 5] and 
• [1, 5) - Yes, because f is continuous on (1, 5) and 
• (1, 5] - Yes, because f is continuous at all points greater than 1.
• (1, 5) - Yes, because f is continuous at all points greater than 1.