Minima of a continuous function A function $f:\mathbb{R}\mapsto \mathbb{R}$ is continuous and smooth in $[a,b]$. How can I prove that the minimum of the function will be either at the end points or where the derivative goes to zero. 
The extreme value theorem states that there is at least one minimum and one maximum. However, I was looking for a proof that the global minimum is either at the end points or the derivative. 
EDIT
As pointed out in the comments, it would be more appropriate to say if $f(c)$ is a minimum, then how do we prove that $c$ is either the end points or $f'(c)=0$ ?
 A: Suppose that $f$ attains its minimum at $x=c\in(a,b)$. then
$$\forall x\in(a,c) \;\;\frac{f(x)-f(c)}{x-c}\leq 0$$
$$\implies \lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}=f'(c)\leq 0.$$
You can finish for $x\in (c,b)$.
A: Proof by contradiction to show that the global minimum is either at the end points or at an interior point where the derivative is zero.
Suppose it is not at the end points (say, at $x_0 \in(a,b)$) and the derivative is non-zero. Then for all $\epsilon > 0$, there exist $\delta > 0$ such that
$$\left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)\right|<\epsilon$$
for $|x-x_0|<\delta$.
In other words
$$f'(x)-\epsilon <\frac{f(x)-f(x_0)}{x-x_0}<f'(x_0)+\epsilon$$
If $f'(x)<0$, we can choose small enough $\epsilon$ such that
$$\frac{f(x)-f(x_0)}{x-x_0}<0$$
or in other words for $x_0+\delta>x>x_0$
$$f(x)<f(x_0)$$
Similarly, for $f'(x)>0$, we can choose small enough $\epsilon$ such that
$$\frac{f(x)-f(x_0)}{x-x_0}>0$$
or in other words for $x_0-\delta<x<x_0$
$$f(x)<f(x_0)$$
Contradiction.
A: A minimum occurs at a non-endpoint $c$ if and only if for every $x\in[a,b]$ you have $f(c)\le f(x).$ That is equivalent to $f(x)-f(c)\ge0.$
If $x>c$, then $x-c>0$ so $\dfrac{f(x) - f(c)}{x-c} \ge0.$ Therefore $\displaystyle\lim_{x\,\downarrow\,0} \frac{f(x)-f(c)}{x-c}\ge0.$
If $x<c$, then $x-c<0$ so $\dfrac{f(x) - f(c)}{x-c} \le 0.$ Therefore $\displaystyle\lim_{x\,\uparrow\,0} \frac{f(x)-f(c)}{x-c}\le0.$
Therefore $\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c} \ge0$ and $\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c} \le0.$
Therefore $\displaystyle0\le\lim_{x\,\to\,0} \frac{f(x)-f(c)}{x-c}=0.$
