First we note that when , and therefore that

We want to compare and to show that does not surpass on .

To do this, we may use a proof by contradiction. It can be easily shown that there are for which . Suppose that for some ,

By the Intermediate Value Theorem, there must be a number between and such that

.

If this is so, then we must have

By the Mean Value Theorem, there must be a number between and such that the derivative of at , , vanishes.

However, solving for , we see that

implies

provided that , contradicting the vanishing of where was supposed to lie between and .