6.3.Polynomials are continuous and have continuous derivatives. So, exists and (since it is continuous) is bounded over [a, b]. By Theorem 6.6, the function is of bounded variation.

Suppose are the solutions of . Then, over each interval the function is monotonic. So, the total variation of over is. By the additive property (Theorem 6.11) we have

.

6.5. It suffices to show that . That would tell us that 1 is in A and it would also give us that 1 is the supremum of A (for, obviously, A is a subset of the interval [0,1], and nothing in A could be larger than 1.]

Suppose . Denote by B the set of all numbers x in [0,1] for which . Under our assumption (that ), 1 is in B and so B is not empty. Since B is bounded (it is a subset of [0,1]), there must be a number , such that .

Case 1: . That makes . Consider the interval . Nothing there is in B (else, ). Take a number in . We have: even though . Contradiction. So, B must be empty. So .

Case 2: . Similar.

(a) Since f is not a constant on any subinterval , is strictly increasing (Th.6.19). So, is one-to-one. Since it is a one-to-one continuous function over a compact set, the inverse function is also continuous over [0,L].

(b) We need to show that is real valued, continuous and strictly monotonic. The first is obvious, the second claim is done in part (a) above. So we need only the last claim: (both inequalities strict). But is equivalent to (since is strictly increasing) and that in turn means .