If Rolle's Theorem is assumed to be true, doesn't that prove the MVT? If we assume that Rolle's Theorem is true is it practical to say that it also proves the MVT?
My reasoning is that even though Rolle's Theorem is the special case for when $f(a)=f(b)$ and the secant line between $(a,f(a))$ and $(b,f(b))$ is horizontal, doesn't that mean you can 'rotate' and 'stretch' the function $f$ and thus 'stretch' and 'rotate' the straight line between $(a,f(a))$ and $(b,f(b))$ as well as the line tangent to $(c,f(c))$ to maintain its symmetry to the line secant between $(a,f(a))$ and $(b,f(b))$ to thus prove that for the straight line intersecting points $(a,f(a))$ and $(b,f(b))$, there must be a point $c$ on $[a,b]$ for which $f'(c)$ is parallel to the aforementioned line between $(a,f(a))$ and $(b, f(b))$?
I feel that this could be better explained with something like linear algebra.  I do not know it well enough to use it.
EDIT: My question was just as unclear as I thought it was.  I am going to try and explain it differently:
For a function $f$, if it is differentiable/continuous over $[a,b] $/$(a,b)$ etc. then there is a straight line between $(a, f(a))$ and $(b,f(b))$.  You can rotate $f$ around the origin until this straight line is parallel to the x-axis.  By Rolle's Theorem, there must exist a $c$ on the transformed $(a,b)$ such that $f'(c)=0$ and thus there is at least one line tangent to $f$ on $(a,b)$ that is parallel to the secant of $(a,f(a))$ and $(b,f(b))$.  Then, you can 'undo' the original rotation of $f$ and know that the lines proven to exist by Rolle's Theorem exist with the same symmetries with respect to $f$, meaning the MVT must hold even when the secant line is not horizontal.

You don't need any linear transformations in $\mathbb{R}^2$; you can do this by subtracting a linear polynomial from your function (which is, itself, a shear map). Simply take your continuous function $f : [a, b] \to \mathbb{R}$ that is differentiable on $(a, b)$, and let $g(x) = f(x) - \frac{f(b) - f(a)}{b - a} (x - a) - f(a)$. In particular $f(a) = f(b) = 0$, and when you apply Rolle's theorem to find $c \in (a, b)$ such that $g'(c) = 0$, you'll find that
$$f'(c) = \frac{f(b)- f(a)}{b - a}.$$
