18 IVT e(-x) = In x Use the Intermediate Value Theorem to verify there is a solution on the interval (1,2). QUESTION 19 IVT (Intermediate Value Theorem) not valid when interval has a discontinuity Consider the function g(x) given as follows g(x) = 4/(2 x) { g(x) = 4/(2 - x) } Note that g(1) > 0 and g(3) < 0. If the IVT applied in this case on the interval [1, 3], we could conclude that g(x) would have a zero in the interval (1,3). Verify this is not the case (i.e. show g(x) = 0 has no solutions). Then explain why this does not violate the IVT.