Can you prove mean value theorem using IVT?

Question

Assuming derivative is continuous, let us first start with Rolle's theorem. The result follows easily if we don't assume that the derivative is either positive or negative in the interval. But if so, can we not show quite easily that the function must be increasing or decreasing strictly in the interval? Hence a contradiction?

Answer 1

We have our setup: $f : [a, b] \to \Bbb{R}$ is continuous, with $f(a) = f(b)$, and $f$ is differentiable on $(a, b)$. Moreover, we're assuming $f'$ is continuous on $(a, b)$.

Your argument goes, if $f'$ is not always positive and not always negative on the interval, then by the IVT, $f'$ is $0$ somewhere. This much I agree with.

However, it's not so easy to show that, if $f'$ is positive everywhere, then $f$ is increasing (nor is it easy to show the opposite, where $f$ is decreasing). How would you show this? The typical proof involves the MVT:

Suppose $x, y \in [a, b]$ with $x < y$. By the mean value theorem, there must be some $z \in (x, y)$ such that $$0 < f'(z) = \frac{f(y) - f(x)}{y - x} \implies f(y) - f(x) > 0 \implies f(y) > f(x),$$ i.e. $f$ is strictly increasing.

Without this argument, how do you propose to show that a positive derivative makes the function (strictly) increasing?

There may well be ways to dance around explicitly using the mean value theorem here, but unless you make this more explicit, I'd dismiss the argument as circular.