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Let $f(a)=c$ and $f(b)=d$. Now we start at $x=a$. At this point, if we increased $f$ by a constant rate $\frac{d-c}{b-a}$ for the interval $b-a$, we'd arrive at $f(b)=d$ at $x=b$..

Now if we start with a derivative higher than $\frac{d-c}{b-a}$ at $x=a$ and the derivative remains higher for the whole interval, then we'll obviously find that $f(b)>d$. So if we want $ f(b)=d $ , the derivative must transition to values lower than $\frac{d-c}{b-a}$ at some point in the interval. Since the derivative is continuous, this transition can't be abrupt. So the derivative has to equal $\frac{d-c}{b-a}$ at least at one point.

Similarly, if $f'(x)<\frac{d-c}{b-a}$ at $ x=a$, then the derivative must transition to values greater than $\frac{d-c}{b-a}$ at some point in the interval, for $f(b)$ to be equal to$ d$. So again, the derivative has to equal $\frac{d-c}{b-a}$ at least at one point in the interval.

The Mean Value Theorem for integrals is similar:

Let $A=c(b-a)$, where $ c, b$ and $a$ are constants. Let A also be the area of a continuous function over the interval $(a,b).$ If $f(a)>c$, then it can't remain greater than $ c$ for the whole interval, because then we'll end up with an area higher than $A$. So $f$ has to transition to values lower than $c$ at some point. Since $f$ is continuous, so it must be equal to c at some point.

Is my proof correct?

Robert Shore
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Ryder Rude
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    Have you proven that $f’(x)$ has the intermediate value property, that is, that if $f’(c)\gt L$ anda $f’(d)\lt L$, then there is a point $x_0$ between $c$ and $d$ where $f’(c)=L$? Because you are using that property, which is known as Darboux’s Theorem, and is non-trivial. It certainly is not part of the hypotheses of the MVT. – Arturo Magidin Mar 07 '19 at 01:55
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    In fact you explicitly say “since the derivative is continuous”... On what grounds do you make that assertion? The hypotheses of the MVT are: the function $f$ itself is continuous on $[a,b]$, and differentiable on $(a,b)$; there is no assumption that $f’(x)$ is continuous. – Arturo Magidin Mar 07 '19 at 01:57
  • @ArturoMagidin I just assumed that $f'(x)$ is continuous over the whole interval. Continuous functions can't change abrupty so they have to pass through L. – Ryder Rude Mar 07 '19 at 02:00
  • @ArturoMagidin I just checked that that's indeed not an assumption of MVT. Thanks for correcting me. But I also have a proof of the integral MVT in the end. That doesn't use that assumption. – Ryder Rude Mar 07 '19 at 02:04
  • That’s the standard proof of the Mean Value Theorem for integrals, as I recall. If $m$ and $M$ are the min and max values for $f$, then $m\leq \frac{1}{b-a}\int_a^b f(t),dt \leq M$, and hence by the IVT there is a point where $f(c) = \frac{1}{b-a}\int_a^b f(t),dt$. That’s essentially what you are doing, only with more words. – Arturo Magidin Mar 07 '19 at 02:09
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    In fact, it’s possible to have functions that are differentiable but with derivatives that are not continuous, so you cannot assume that $f’(x)$ will be continuous. See for example https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be – Arturo Magidin Mar 07 '19 at 02:11

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