Help understanding proof: If $c$ is a maximum\minimum point then $f'(c)=0$ if $f$ is differentiable at $c$. Goal is to prove the following statement: Let $I$ be an interval and $c$ be an interior point of I. If $c$ is a maximum or a minimum point for $f$ and $f$ is differentiable at $c$, then $f'(c)=0$.
The proof of this statement given in my book is based upon the use of an an o-function, and since I have no idea what the english term is, I will define such a function below.
Note: From what I've understood an o-function is not normally used in this context.
Definition: An o-function denoted $\varphi$ satisfies following properties:
(i) $\exists\hspace{2pt}r>0$ such that $\varphi:(-r,r)\to\mathbb{R}$.
(ii) $\varphi(0)=0$.
(iii) $\varphi$ is continuous at $0$.  
Proof by contraposition:
Assume that $f'(c)\neq0$, then show that $c$ is neither a maximum or a minimum point.  
Since $f$ is differentiable at $c$, there exists an o-function $\varphi$ such that:
\begin{equation}
f(c+h)=f(c)+(f'(c)+\varphi(h))h, \hspace{8pt}\forall\hspace{1pt}h\in(-r,r).
\end{equation}
Assume w.l.o.g. that $f'(c)<0$. Since $\varphi$ is an o-function there exists a $\delta\in(0,r)$ such that $f'(c)+\varphi(h)<0$ for $\lvert h\rvert<\delta$, and from this it follows that:
\begin{equation}
\begin{split}
f(c+h)&<f(c)\quad\textrm{for}\quad 0<h<\delta\qquad(*)\\
f(c+h)&>f(c)\quad\textrm{for}\quad -\delta<h<0\qquad(**).
\end{split}
\end{equation}
The first inequality (*) shows that $f$ does not attain its maximum value at $c$.
The second inequality (**) shows that $f$ does not attain its minimum value at $c$.
What I dont understand:
1) What ensures that $\delta$ exists, and why does it exist in the interval $(0,r)$ and not $(-r,0)$?
2) Why is $f'(c)+\varphi(h)<0$ for $\lvert h\rvert<\delta$?
3) How are the equations (*) and (**) derived?
Any help is very appreciated and thanks in advance!
 A: 1) The function $g(h) = f'(c) + \phi(h)$ is continuous at $0$ (because $\phi$ is continuous at $0$) and $g(0) = f'(c) + \phi(0) = f'(c) < 0$. This way, it exists an interval around $0$ where $g$ is negative, where the sign is the same, this is called the sign conservation theorem (in spanish at least, I actually can't find this theorem name in english, sorry) and it's a basic theorem using the definition of continuous function. You can find a proof (in spanish, I'm sorry, but at least it has few words, if you don't understand something feel free to ask) here. Edit: I've also found a proof in english here. You can see this easily, if you make the graph of your function and you know a point is above $y=0$ then the theorem is saying that you can draw two vertical lines around this point where the function remains above $y=0$. This $\delta$ is defining the interval around the $0$ where the function keeps the sign, and this is usually done with a positive $\delta$ so you can define the interval as $(-\delta, \delta)$, that's why it's on $(0,r)$.
2) Because of the property in 1).
3) $f(c+h) = f(c) + g(h)h$, now we know in the interval $(-\delta, \delta)$ that $g(h) < 0$, so when $h < 0$ the product $g(h)h > 0$, and when $h > 0$ we have the product $g(h)h < 0$, and that's where the equations are derived.
A: $(1).$ If  $f'(c)$ satisfies $0< f'(c)+r$ for every $r>0$ then $f'(c)\ge 0.$ Because if $f'(c)<0$ then when $r=-f(c)> 0$ we would have $0< f'(c)+r=f'(c)+(-f'(c))=0.$
(2). Suppose $f$ has a  local maximum at $c$.  There exists $\delta>0$ such that $(c-\delta,c+\delta)\subset I$ and such that $f(c)\ge f(x)$ whenever $x\in (c-\delta,c+\delta).$
Take any $r>0.$ Since $f'(c)$ exists, there exists $\delta'_r>0$ such that $(c-\delta'_r,c+\delta'_r)\subset I$ and such that $\left|\frac {f(c)-f(x)}{c-x}-f'(c)\right|<r$ whenever $c\ne x\in (c-\delta'_r,c+\delta'_r).$
Now let $\delta''_r=\min (\delta,\delta'_r).$ When $x\in (c-\delta''_r,c)$ we have $0\le \frac {f(c)-f(x)}{c-x}$ so $$\left|\frac {f(c)-f(x)}{c-x}-f'(c)\right|<r\implies$$ $$ \implies 0\le \frac {f(c)-f(x)}{c-x}<f'(c)+r\implies$$ $$\implies 0<f'(c)+r.$$ The last line above is satisfied by every $r>0$ so by $(1)$ we have $$f'(c)\ge 0.$$ A similar argument for $x\in (c,c+\delta''_r)$ shows that $$f'(c)\le 0$$ where we use the fact that if $f'(c)-r<0$ for every $r>0$ then $f'(c)\le 0.$
$(3).$ If $f$ has a  local minimum at $c$ then $(-f)$ has a local maximum at $c,$ so by $(2)$ we have $0=(-f)'(c)$. And we have $(-f)'(c)=-f'(c).$
