Proving the Mean Value Theorem Theorem


  The Mean Value Theorem: Let there be a function $f$ which is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $f'(c)=\dfrac{f(b)-f(a)}{b-a}$


I'd like to prove the Mean Value Theorem, assuming Rolle's Theorem has already been proved.

Proof

Let $f$ be the function whose graph is the chord between the endpoints $(a,f(a))$, and $(b,f(b))$.

Then let $g(x)=\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)$, and let $h(x)=f(x)-g(x)$, so $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Note that $h(a)=h(b)=0$, and so the hypotheses of Rolle's Theorem are met, and thus there exists a $c\in(a,b)\ni h'(c)=0$.

$$0=h'(c)=f'(c)-g'(c)=f'(c)-\dfrac{f(b)-f(a)}{b-a}$$ $$\implies f'(c)=\dfrac{f(b)-f(a)}{b-a}\tag*{$\blacksquare$}$$

Is the above proof valid (and are there any suggestions for improvement?)
Another way to prove the MVT:

Let us define a function $\phi$:[a,b]$\rightarrow IR$ by
$$\begin{matrix}
f(x)&amp;g(x)&amp;h(x) \\
f(a)&amp;g(a)&amp;h(a)\\
f(b)&amp;g(b)&amp;h(b)
\end{matrix}$$
Observe that $\phi (a)$=$\phi (b)$=0. Hence by Rolle's theorem $\exists$ c $\in$ (a,b) such that $\phi^{'} (c)$=0.
$$\begin{matrix}
f^{'}(x)&amp;g^{'}(x)&amp;h^{'}(x) \\
f(a)&amp;g(a)&amp;h(a)\\
f(b)&amp;g(b)&amp;h(b)
\end{matrix}$$
is equal to 0.Take h(x) to be a constant function,g(x)=x and expand the determinant to get the desired result.
Another way to prove the MVT:

Let us define a function $\phi$:[a,b]$\rightarrow IR$ by
$$\begin{matrix}
f(x)&amp;g(x)&amp;h(x) \\
f(a)&amp;g(a)&amp;h(a)\\
f(b)&amp;g(b)&amp;h(b)
\end{matrix}$$
Observe that $\phi (a)$=$\phi (b)$=0. Hence by Rolle's theorem $\exists$ c $\in$ (a,b) such that $\phi^{'} (c)$=0.
$$\begin{matrix}
f^{'}(x)&amp;g^{'}(x)&amp;h^{'}(x) \\
f(a)&amp;g(a)&amp;h(a)\\
f(b)&amp;g(b)&amp;h(b)
\end{matrix}$$
is equal to 0.Take h(x) to be a constant function,g(x)=x and expand the determinant to get the desired result.