# About the mean value theorem

The mean value theorem is given by: If a function $g$ is continuous on the closed interval $[u,v]$, where $u<v$, and differentiable on the open interval $(u,v)$, then there exists a point $c$ in $(u,v)$ such that

$$g(v)=g(u)+g′(c)(v-u) (**)$$

My question is: For a fixed $c$ and $v$ can we find $u$ such that (**) is verified? Can we find sufficient conditions on $g$ to guarantees this property?

• @ rogerl: and for the second question – DER May 14 '14 at 18:42
• One sufficient condition would be $g$ is constant. – guest196883 May 14 '14 at 18:44
• @SamDeHority: What happen if $g$ is not constant? – DER May 14 '14 at 18:47
• @DER I doubt there are any simple conditions on $g$ that guarantee this property, but perhaps someone will prove me wrong. – rogerl May 14 '14 at 18:48
• There should be a lot of functions that satisfy this. One such class are all functions with $f'',f' > 0$ on some interval then clearly all $c$ (in the interior) will have at least one pair $u,v$ where this is the case. – Winther May 14 '14 at 19:54

No. Suppose that $g$ is increasing and that $c$ is a point at which $g'(c)=0$ (for example, let $g(x) = x^3$ and $c=0$. Then $g(v) = g(u) + 0(v-u) = g(u)$, which is impossible.
• Well, in this case $u=v$ works. What if we allow this degenerate case as well? – Andrés E. Caicedo May 14 '14 at 18:46
• @AndresCaicedo The OP assumed $u<v$. – rogerl May 14 '14 at 18:47
• @AndresCaicedo I guess I'm not sure exactly what you're asking. If we allow $u=v$, then the OP's equation is satisfied no matter what $g$ looks like or what $g'(c)$ is. – rogerl May 14 '14 at 19:08