How to show this function is monotonic strictly increasing? Consider $f:\left[a,b\right]\rightarrow \mathbb{R}$ continuous at $\left[a,b\right]$ and differentiable at $\left(a,b\right)$.
$\forall x\:\ne \frac{a+b}{2}$, $f'\left(x\right)\:>\:0$, but at $x\:=\frac{a+b}{2}$, $f'\left(x\right)\:=\:0$.
Show that $f$ is strictly increasing at $\left[a,b\right]$. 
So far i don't find any approach how to handle with $x\:=\:\frac{a+b}{2}$.
I know that $x\:=\:\frac{a+b}{2}$ can't be minimum or maximum.
I thought about assuming there is point $c\:\ne \frac{a+b}{2}$  such that $f\left(c\right)=f\left(\frac{a+b}{2}\right)$ and get a contradict somehow but i can't figure out how..
Any ideas? thanks in advance!
 A: For sure $f$ is increasing in the sense that $f(x) \geq f(y)$ whenever $x > y$. Now assume $f(x)=f(y)$ for $x > y$. Clearly, then $f$ must be constant on $[y,x]$, contradicting the fact that the derivative vanishes at only one single point.
A: 1) $f$ is strictly increasing in $[a,c]$ and in $[c,b]$
Let $x,y \in [a,c]$ and $ x<y $ Then, from the mean value theorem, we have: $f(y)-f(x)=f'(\xi)(y-x) $ where $\xi \in (x,y) $ Since $\xi \in (a,c)$ we have $f'(\xi)>0$ and $(y-x)>0$ , so $f(y)-f(x)>0$
Proof for $[c,b]$ is identical.
2) $f$ is strictly increasing in $[a,b]$
Let $x,y \in [a,b]$ and $x<y$ 
If $x,y \in [a,c]$ or $x,y \in [c,b]$ then $f(x)<f(y) $
Let $x \in [a,c] $ and $y \in (c,b] $ then $f(x) \le f(c) < f(y) $
Let $x \in [a,c) $ and $y \in [c,b] $ then $f(x) < f(c) \le f(y) $
A: Suppose there are $\;x,y\in [a,b]\;,\;\;x,y\;$ with $\;f(x)=f(y)\;$ . Since the function is at least monotonic increasing "weakly", as $\;f'(x)\ge 0\;$ , the above means 
$$f(w)=0\;\;\;\forall\,w\in (x,y)\implies f'(w)=0\;\;\;\forall\,w\in (x,y)\implies\;\text{contradiction}$$
