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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.

A proof of MVT used Rolle's theorem explicitly.

Consider a continuous function $f:[a,b]toBbb R$ that is differentiable on $(a,b)$. We are to prove MVT, i.e. to prove there exists a point $cin(a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$, assuming that Rolle's theorem is true.

Define a new function $g:[a,b]toBbb R$ by $g(x)=f(x)-fracf(b)-f(a)b-ax$. Verify yourself that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Verify also that $g(a)=g(b)$. So we can apply Rolle's theorem to this function $g$, to obtain a point $cin(a,b)$ such that $g'(c)=0$. Note that the derivative of $g$ is $g'(x)=f'(x)-fracf(b)-f(a)b-a$. The point $c$ also satisfies $f'(c)-fracf(b)-f(a)b-a=0$.

You don't need any linear transformations in $mathbbR^2$; you can do this by subtracting a linear polynomial from your function (which is, itself, a shear map). Simply take your continuous function $f : [a, b] to mathbbR$ that is differentiable on $(a, b)$, and let $g(x) = f(x) - fracf(b) - f(a)b - a (x - a) - f(a)$. 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
$$f'(c) = fracf(b)- f(a)b - a.$$

Since none of the existing answers have explicitly pointed out your error:

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. [...]

No you cannot do this! You can apply Rolle's theorem to any differentiable function, but rotating the graph of a function in the cartesian plane may not result in a graph of another function, much less a differentiable function. This is the reason edm mentioned a shear in the comments, because a shear with an invariant vertical line is indeed the kind of transformation that would preserve differentiable functions.

Not really.
Rolle's theorum is a special case of Legrange's MVT. Not the other way around. So just by assuming Rolle's theorum, MVT is not automatically proved. We do of course use RT in the MVT proof. But just because Rolle's theorum is true, MVT need'nt be.

If you are wondering why we use this special case to prove the MVT, it is because Rolle's theorum appeared first.