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?
Best Answer
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.