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Is this a valid proof of the mean value theorem?

Suppose $f$ is a continuous differentiable function on the interval $[a,b]$ then there exists some $c\in(a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.

Proof:

Let $B=\{e_1,e_2\}$ be the standard basis for $\mathbb{R}^2$. Then let $\theta = \tan^{-1}\left(\frac{f(b)-f(a)}{b-a}\right)$, that is, the angle that the secant line forms with the $x$-axis. Then consider the the linear transformation, $T$, defined by rotation matrix by $\theta$ on the standard basis:

$$ T = \left(\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right) $$

Then $T(B)$ is a basis one of whose component vectors is parallel with the secant line. Thus we may apply Rolle's theorem and the result follows. $\blacksquare$

My thinking here is that by rotating the basis vectors we can just consider it as $f(a)=f(b)$ and Rolle's theorem gives it to us right away then. I guess the intuition behind the proof is that you can just turn your head until it's a curve that satisfies the conditions of Rolle's theorem, but is that valid can I actually do this?

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  • $\begingroup$ You haven't written a proper proof, but the MVT can indeed be shown easily using Rolle's theorem. $\endgroup$ – Olivier Oct 1 '17 at 1:25
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    $\begingroup$ Normally this would be done by a shear rather than by a rotation. $\endgroup$ – Michael Hardy Oct 1 '17 at 1:28
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It's not valid that you can do this; when you rotate the graph of a function, it may no longer be a function. I think it's a good exercise to come up with an example where this happens.

However, you can prove the mean value theorem from Rolle's theorem: define a new function to be equal to $f$ minus the line through $(a,f(a))$ and $(b,f(b))$ and then apply Rolle's Theorem.

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If the function is concave downward everywhere and has a vertical tangent at both endpoints, then this method will not work. Draw a picture and you'll see why. In fact, if the tangent is steep enough but not vertical, this won't work because the resulting graph will fail the vertical line test.

The linear transformation whose matrix is $$ \begin{bmatrix} 1 & 0 \\ -m & 1 \end{bmatrix}, $$ where $m$ is the slope of the secant line, will work. And that is just what is usually done, even if it's not phrased in the language of linear algebra and matrices.

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