# Mean Value Theorem Question?

Let $f : [a,b] \to \mathbf R$ be differentiable on $[a,b]$, let $m \in \mathbf R$ and suppose $f'(a) < m < f'(b)$. Show that there exists $c \in (a,b)$ such that $f'(c) = m$.

I tried to prove this as following:

Consider $g(x) = f(x) - mx$ and we can show that $g'(x) = f'(x) -m$ and thus $g'(a) < 0 < g'(b)$, since $g$ is continuous on $[a,b]$, by $\text{Extreme Value Theorem}$ we can have a $c \in [a,b]$ such that $g(c)$ is a local extreme value, then using $Fermat's\ Theorem$ we can have $g'(c)=0 \implies f'(c) =m$..

But I just found that I have to show that $c \in (a,b)$, which means I have to show $g$ can't obtain it's extreme value at $a$ or $b$, so I got stuck now, can someone give me hints for this question?

$g'(a)<0$ so for some (sufficiently small ) $d\in (0,b-a)$ we have $\frac {g(a+d)-g(a)}{d}<0,$ so $g(a+d)<g(a).$

Similarly, $g'(b)>0$ so for some $e\in (0,b-a)$ we have $g(b-e)<g(b)$.

So $\;\min \{g(x): x\in [a,b]\}$ is not equal to $g(a)$ nor to $g(b).$

See $g(x) = f(x) - mx$

$g'(a)<0, g'(b)>0$ => minimum of $g(x)$ in $[a,b]$ is an inner point (Minimum is exists by Weierstrass thm)

In the minimum value point $c$ $g'(c) = 0$ => $f'(c) = m$

• I don't use what derivative is continuos, but $f$ and $g$ are continuos both – kotomord Mar 3 '17 at 0:04
• @egreg Can you elaborate on that comment? (kotomord, I think there's a typo -- $g(b)$ should be $g'(b)$.) – littleO Mar 3 '17 at 0:21
• @littleO - tnx! – kotomord Mar 3 '17 at 0:22