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?