Continuity of Functions Exercises

Answer

Yes. For starters, f is continuous on the closed interval [-2, 2]. We have a = -2 and b = 2, so

Since f(a) = -11 < 0 < 13 = f(b), the IVT says that there is some c with -2 < c < 2 and f(c) = 0.

Answer

No. With a = 0 and b = 1, f(a) = 1 and f(b) = e. The value 0.1 is not between 1 and e, so we can't use the IVT here.

Answer

Yes. f is continuous on the closed interval . With  and b = π, we have

 and f(b) = sin(π) = 0. Since f(b) = 0 < 0.4 < 1 = f(a), the IVT says there is some c in  with f(c) = 0.4.

Answer

No. The function tan(x) is not continuous on [0, π] (it has a vertical asymptote on this interval), so we can't use the IVT.

Answer

No. The function f is indeed continuous on the closed interval [-1, 1], but with a = -1 and b = 1, f(a) = f(b) = 1. The value 1 is not strictly in between 1 and 1 (if we wrote down 1 < 1 < 1 we would be lying), so the IVT won't fly here.

Answer

There are infinitely many right answers to this. Here's one of them:

We could change the above problem to make f(0.5) equal anything we want.

Be Careful: We can never use the IVT to conclude that a function f fails to hit a value M.

Another thing to be aware of with the IVT is that it doesn't tell us where a function hits a value M, or how many times it does so. On its best day, the IVT can only guarantee we hit the value M at least once.

Answer

No. f could look like this, for example:

This function hits every value from 0 to 1 on the interval [0, 1], but f is discontinuous on this interval.