I want to prove that a function $f(x)$ has its unique minimum at $f(c)$. By that I mean, $f(c) < f(x)$ for all $x \neq c$. The strict inequality is obviously vital.
My question is, does this follow automatically if $f'(x) = 0$ if and only if $x = c$?
I've heard the words "sufficient, necessary" be thrown around when it comes to derivatives. What do they mean?
edit:
I forgot to mention: $f$ is bounded below by zero. It is quadratic in $\beta$.
On
Although I'm not entirely sure I understand what you are asking, I'm going to have to say no: being bounded and having a global local minima is not sufficient to show that the derivative will only be 0 when $x=c$.
For example take the dirty function that I just whipped up in Desmos to show a counterexample
$$ f(x) = \dfrac{x^5-x^4+x^3+x}{20e^{|x|}}+4 $$
This function is bounded both above and below and the local minima you find at $c=-4.756..$ is also equivilent to the global minima, which holds your condition that $f(x) > f(c) \forall x\in \mathbb{R}$. However there is also a local maxima at $c'=5.134..$. By definition the derivative will be zero at both of these points, which runs counter to your original claim that $f'(x)=0$ only when $x=c$
Take $f(x)=x-x^3/3$ at $ [-1,1].$
for all $x\in ]-1, 1],$
$$f(-1)<f(x)$$
but $$f'(1)=0$$