Can we use the IVT to conclude that f(x) = sin(x) passes through y = 0 on ?
Yes. The function f is continuous on the closed interval . With and , and f(b) = 1. Since f(a) = -1 < 0 < 1 = f(b),
the IVT says there is some c in between a and b (that is, ), with f(c) = 0. In this case, c = 0.
Can we use the IVT to conclude that passes through y = 1 on ?
No. We can't use the IVT in this case because the function f is discontinuous at x = 0.
Can we use the IVT to conclude that passes through y = 1 on (0, 1)?
No. This question doesn't even make sense. In order to use the IVT we need to know the function values at the endpoints of the interval, but f(0) is undefined.
Can we use the IVT to conclude that f(x) = x2 passes through y = 0 on (-1, 1)?
No. Since f(a) = (-1) 2 = 1 and f(b) = (1) 2 = 1, 0 is not between f(a) and f(b). We cannot use the IVT in this situation. The function f does in fact pass throughy = 0, but we know that by observation, not by using the IVT.
Reminder: It doesn't matter whether we have f(a) < M < f(b) or f(b) < M < f(a), as long as M is between f(a) and f(b).