Minimum value of $f$ occurs at $x\implies f'(x)=0$. Let $f$ be a continuous and differentiable function on the interval $[a,\,b]$, and suppose that $f$ attains its minimum at point $c\in(a,\,b)$. Show that $f'(c)=0$. 
I think an appropriate theorem to apply here would be the Mean Value Theorem which states if a function is continuous and differentiable on $(a,\,b)$ then we can find a $c\in(a,\,b)$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}.$$ 
However $(f(b)-f(a))/(b-a)=0\iff f(a)=f(b)$ which we cannot assume so I am certain this is incorrect. I also haven't used the fact that $f(c)$ is the minimum value of $f$ or in other words $f(x)>f(c)$ for every $x\in[a,\,b]$ such that $x\not=c$. However i'm not really sure how to connect these bits of information. Any hints?
 A: No, the $c$ given by the MVT most likely isn't the $c$ that's fixed in the statement of your 
theorem. 
Instead, all you need is the definition of the derivative: $$f'(c)=\lim\limits_{h\rightarrow0}{f(c+h)-f(c)\over h}.$$ If you assume $f'(c)<0$, you can show, using the definition of limit, that there is an $h>0$  with $c+h<b$ such that $f(c+h)-f(c)<0$. This contradicts the hypothesis that $x=c$ gives a minimum. Similarly, you can argue that $f'(c)>0$ can't hold (take small negative $h$).
Note you only need that $f'(c)$ exists, here.
A: Here is another approach:
Once $f'$ is differentiable, you have that for any fixed $z\in\mathbb{R}$ $$f(c+tz)-f(c)=f'(c)tz+o(t),\ \mbox{small}\ t,\ t\neq 0,$$
thus $$\frac{f(c+tz)-f(c)}{t}=f'(c)z+\frac{o(t)}{t}\, \mbox{small}\ t,\ t\neq 0 .$$
Because $c$ is the minimum, you can conlude that $$0\le f'(c)z+\frac{o(t)}{t},\ \mbox{small}\ t,\ t\neq 0.$$
Now, let $t\to 0$ and vary $z$ to conclude.
A: Let $f$ be a continuous and differentiable function on the interval $[a,\,b]$, and suppose that $f$ attains its minimum at point $c\in(a,\,b)$. Show that $f'(c)=0$. 

Since $f(c)=m$ is the minimum and $f$ is differentiable on $[a,\,b]$ consider $$\lim_{x\to c^{-}} \dfrac{f(x)-f(c)}{x-c} = \lim_{x\to c^{-}}\dfrac{f(x)-m}{x-c} \le0 \text{ since numerator +ve and denominator -ve.}$$ $$\lim_{x\to c^{+}} \dfrac{f(x)-f(c)}{x-c} = \lim_{x\to c^{+}} \dfrac{f(x)-m}{x-c} \ge0 \text{ since numerator and denominator +ve.}$$
Therefore the limit only exists if $f'(c) = 0$.
