TITLE: Application of mean value theorem
QUESTION [6 upvotes]: $f$ is differentiable in $[a,b]$, and satisfies $f(a)<f(b)$, $f'(a)=f'(b)=0$.
Show that there exists $c$ in $[a,b]$
\begin{equation}
\frac{f(c)-f(a)}{c-a}=f'(c)
\end{equation}
How is the mean value theorem 
\begin{equation}
\frac{f(b)-f(a)}{b-a}=f'(c)
\end{equation}
applicable here?

REPLY [5 votes]: Consider the auxiliary function
$$g(x):={f(x)-f(a)\over x-a} \quad(a<x\leq b), \quad g(a):=0\ .$$
Then $g$ is continuous on $[a,b]$ and differentiable in $(a,b)$. One computes
$$g'(x)={(x-a) f'(x)-\bigl(f(x)-f(a)\bigr)\over (x-a)^2}\qquad(0<x\leq b)\ .$$
If $g(b)=0$, then this is a consequence of Rolle's Theorem. 
If $g(b)>0$, then $g'(b)<0$; therefore $g(x)>g(b)>0$ immediately to the left of $b$. This allows us to conclude that $g$ assumes its maximum on $[0,b]$ in an interior point $c$. From $g'(c)=0$ it then follows that
$${f(c)-f(a)\over c-a}=f'(c)\ .$$
The case for $g(b)<0$ is similar.