Can we use the IVT to conclude that f(x) = x2 passes through 1 on the interval (-1,1)?
No. The function f is indeed continuous on the closed interval [-1,1]. We have a = -1 and b = 1, therefore
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!), we can't use the IVT here.