# Proving Mean Value Theorem with Rolle's Theorem?

How to prove the Mean Value Theorem using Rolle's Theorem? I am getting the impression that it is possible by adding a linear function to a function where Rolle's theorem applies to prove the MVT. However, I can't quite turn this idea into a rigorous mathematical argument.

• Use the function defined by $\phi(x)=f(x)-f(a)-{f(b)-f(a)\over b-a}(x-a)$ (this gives the difference of $f$ and the function whose graph is the line segment joining the points $\bigl(a,f(a)\bigr)$ and $\bigl(b, f(b)\bigr)$. – David Mitra Dec 15 '12 at 1:33
• @David I think I understand. If I subtract the line from the function, then the point c between a and b such that $f'(c) = (f(b) - f(a))/(b - a)$ becomes $f'(c) = 0$. However, how can I now mathematically use that function to formally prove the MVT? – hesson Dec 15 '12 at 1:45
• My analysis professor's one-sentence proof was "tilt your head". – MJD Dec 15 '12 at 3:00

For $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, the standard proofs I've seen use the function that gives the difference of $f$ and the function whose graph is the line segment joining the points $\bigl(a,f(a)\bigr)$ and $(b,f(b)\bigr)$; $$\phi(x)=f(x)-f(a)-{f(b)-f(a)\over b-a}(x-a).$$
From the continuity and differentiablity of $f$ (and standard theorems such as the difference of continuous functions is continuous) it follows that $\phi$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Since $\phi(a)=\phi(b)=0$, Rolle's theorem applies to $\phi$ on $[a,b]$. Writing down the result obtained from Rolle's Theorem gives all that is desired.