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I'm reading the Wikipedia proof for the MVT, and it uses Rolle's theorem. In fact, many other websites that prove MVT do the same. When I first read the statement of the mean value theorem, I thought it must obviously be true because the alternative, that $f'(x) > \frac{f(b)-f(a)}{b-a} \ \forall x \in (a,b)$ was absurd (and same for $f'(x)$ strictly less than the right hand side), because the rate of increase of the function being higher than the average rate of increase at all points contradicts the very definition of the average rate of increase.

By this, I mean that if the function $f$ was increasing at the average rate $f'(x) = \frac{f(b)-f(a)}{b-a} \ \forall x \in (a,b)$ then it would exactly go straight from $(a, f(a))$ to $(b, f(b))$ (this is the definition of average), and so if it's always increasing strictly faster, then surely it increases "too much" to be able to get down to $(b, f(b))$ in time, so to speak.

Yet, I do not see sites formalizing this to give a proof by contradiction of the mean value theorem in a line or two, so I imagine this must be going wrong somewhere. Could someone tell me where?

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    $\begingroup$ How does it contradict the "very definition" of the average rate of increase? What is that definition, exactly? $\endgroup$ Commented Feb 16, 2022 at 16:16
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    $\begingroup$ The alternative is "$\ne$", not "$>$". $\endgroup$ Commented Feb 16, 2022 at 16:18
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    $\begingroup$ Your reasoning is questionable. In fact, $(f(b)-f(a))/(b-a)$ is called "the average rate of increase" only because of the fundamental theorem of calculus, which is proved using the MVT. $\endgroup$
    – GEdgar
    Commented Feb 16, 2022 at 16:18
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    $\begingroup$ @TanishqKumar The thing you just said is an invocation of the intermediate value theorem - which, as the answer to this question better than mine points out, does not immediately apply to $f'(x)$. $\endgroup$ Commented Feb 16, 2022 at 16:21
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    $\begingroup$ For your argument to work would require the other meaning of "average" -- it would require that the "average rate of change" is some sort of arithmetic mean of the instantaneous rates of change. Where have you calculated such an arithmetic mean of instantaneous rates of change. (For your contradiction to work, the "average $X$" has to actually inspect all the relevant $X$s. The difference quotient of the endpoints inspects none of the instantaneous rates of change.) $\endgroup$ Commented Feb 17, 2022 at 6:44

2 Answers 2

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If $f'$ is continuous then you can use your argument and the intermediate value theorem to prove MVT.

As a first, somewhat minor criticism of this approach, MVT doesn't assume that $f'$ is continuous. Consequently, to prove the "full" MVT in this fashion, you have to show that a derivative, even if it isn't continuous, has the intermediate value property. This is true, actually; the result is called Darboux's theorem. Of course, you could prove a weaker MVT that relies on $f'$ being continuous without Darboux's theorem, and from here you could do a lot of practical analysis.

More importantly though, there is some setup required to even invoke the intermediate value property, and this setup is a little bit more difficult to formally prove than you seem to think it is. Specifically, you need to prove that for $m=\frac{f(b)-f(a)}{b-a}$, if there is no point where $f'=m$ then there must be a point where $f'<m$ and another point where $f'>m$.

From the point of view of someone who has prior exposure to calculus, the obvious way to do that is to assume one inequality or the other, integrate both sides, and get a contradiction. But in order to do that, you need the fundamental theorem of calculus (FTC). The issue is that generally one develops the theory in the other direction, with MVT preceding FTC, and even being used to prove FTC. I won't claim it's impossible to work in the other order, but there is definitely more involved in proving FTC than there is in proving Rolle's theorem.

Now as was pointed out in the comments, there is a different way to go. That is to immediately look at $g(x)=f(x)-m(x-a)$ and derive a contradiction from the assumption that $g'$ has a definite sign. From there you can use the intermediate value theorem to prove "weak MVT", while Darboux's theorem gets you "full MVT". But this route is basically the same idea as proving and then applying Rolle's theorem. You're just skipping directly to the more general scenario of MVT rather than identifying Rolle's theorem as a special case along the way. I would argue that this way of developing the theory is worse because the picture for MVT is not as easy to see as the picture for Rolle's theorem.

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    $\begingroup$ I agree. In fact, if we examine the proofs of the fundamental theorem of calculus in the textbooks, we find that every one of them uses MVT. $\endgroup$
    – GEdgar
    Commented Feb 16, 2022 at 16:22
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    $\begingroup$ @TanishqKumar The argument goes like this: suppose $f'<m$ everywhere, then $\int_a^b f'(x) dx = f(b)-f(a)<m(b-a)=f(b)-f(a)$. On the other side, suppose $f'>m$ everywhere, then $\int_a^b f'(x) dx = f(b)-f(a)>m(b-a)=f(b)-f(a)$. Obviously neither of those can be true, so if $f'$ is never equal to $m$ then there must be a point on each side. But those integration steps used FTC. If you can't use FTC in this argument, how do you prove that it is impossible that $f'<m$ everywhere? $\endgroup$
    – Ian
    Commented Feb 16, 2022 at 16:35
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    $\begingroup$ @TanishqKumar The usual ways to prove that "differentiation and integration are opposites" uses the mean value theorem to connect the independent definitions of differentiation and integration. $\endgroup$ Commented Feb 16, 2022 at 16:37
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    $\begingroup$ Another thing to note is that one of the two common proofs of Darboux theorem is based on proof technique used in Rolle's and the other uses mean value theorem. So @TanishqKumar approach can't avoid Rolle's. $\endgroup$
    – Paramanand Singh
    Commented Feb 16, 2022 at 23:47
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    $\begingroup$ There is a different way. If $g'>0$ at every point of some interval then $g$ is strictly increasing at every point and then we can use Heine Borel to show strictly increasing over whole interval. $\endgroup$
    – Paramanand Singh
    Commented Feb 17, 2022 at 0:10
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Suppose we actually put in the definitions of both terms. Then

$\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ being higher than $\frac{f(b) - f(a)}{b-a}$ at all points $x \in [a,b]$ contradicts the very definition of $\frac{f(b) - f(a)}{b-a}$

does not make any sense.

What's going on with the very plausible-sounding phrasing with "rate of increase" and "average rate of increase" is that those are just intuition for the definitions of $f'(x)$ and of the secant slope. They're very good intuition: in this case, they're suggesting a true statement about the relationship here that's not at all obvious from the definitions!

But in an actual proof, we need to use the definitions instad.

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    $\begingroup$ +1 This is the best answer to OP's assertion that something is "obviously absurd" and "contradicts the very definition of ...". For sure, the names which we have given the mathematical quantities $f'(c)$ and $(f(b)-f(a))/(b-a)$ were chosen so that MVT feels very true and "obvious" (that's why they were chosen). But that does not suffice as a rigorous proof using their exact mathematical definitions, as you strikingly show. $\endgroup$ Commented Feb 17, 2022 at 18:11

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