Proving the derivative is $0$ at the extremum and all derivatives are $0$. The pictures below show the proof that Apostol uses in his book.


I can't understand why Apostol introduces the function $Q(x)$ and proves the theorem by contradiction using the sign preserving property.
What we want to prove:$$\lim_{h\to 0}\frac{f(c+h)-f(c)}h=0$$
The way I proved it was:
Since $f'(c)$ exists, this means that the function $f$ is continuous at point $c$. Using the definition of continuity at a point $c$ this means that $$\lim_{x\to c}f(x)=f(c)$$
Let $x=c+h$ so the above limit becomes $$\lim_{c+h \to c} f(c+h)=f(c)$$ which is equivalent to $$\lim_{h \to 0} f(c+h)-f(c)=0$$  This is all assuming $c+h$ lies in open interval $I$. So since the above limit appoahes $0$ as $h$ approaches $0$ the limit to be proved is proved. $\square$
Is that right ??
 A: There are two problems with your approach.
1) Calculation of the limit $\lim_{h \to 0}\{f(c + h) - f(c)\}/h$
2) not using the fact that $f$ has a relative min/max at $c$.
The first problem seems to be result of some misunderstanding of limit concept (please don't mind, many people do have problems with it in the beginning). If the numerator of a fraction tends to $0$ then it does not necessarily mean that the fraction tends to $0$. We also need to analyze the behavior of denominator (after all a fraction is not just numerator). Had it been the case the denominator tends to some non-zero value then your conclusion that fraction tends to $0$ would have been OK. But when both numerator and denominator of a fraction tend to $0$, then the fraction itself could tend to any value ($\pm \infty$ also) or it may oscillate.
Regarding the proof from Apostol, you need to see how he defines a continuous function $Q(x)$ such that $Q(c) = f'(c)$. He wants to do this only to use the fact that $Q(x)$ will have same sign as $Q(c) = f'(c)$ in a certain neighborhood of $c$. This is a technique and it is not really necessary. Even without creating a new function $Q(x)$ you can proceed as follows:
Suppose $f'(c) > 0$ i.e. $$\lim_{h \to 0}\frac{f(c + h) - f(c)}{h} > 0$$ Now if a limit is positive then it means that the function itself is positive for values of $h$ under consideration (informal common sense, a negative thing can not tend to a positive limit, but this can be justified formally. See update below.) Thus $$\frac{f(c + h) - f(c)}{h} > 0$$ for all values of $h$ satisfying $0 < |h| < \delta$ for some $\delta > 0$.
This means that $f(c + h) - f(c)$ has same sign as that of $h$. In other words if $h > 0$ then $f(c + h) > f(c)$ and if $h < 0$ then $f(c + h) < f(c)$. This means that $f$ has neither a relative minimum nor a relative maximum at $c$. This is exactly what Apostol does through $Q(x)$.
The case when $f'(c) < 0$ can be handled similarly. And thus the only option left is $f'(c) = 0$.
Update: If $\lim_{x \to a}f(x) = A > 0$ then show that the function $f(x)$ is positive in a certain neighborhood of $a$ (except possibly at $a$).
Clearly since $A > 0$ we have $\epsilon = A/2 > 0$. Now there exists a $\delta > 0$ such that $|f(x) - A| < \epsilon$ whenever $0 < |x - a| < \delta$. Thus we have $$A - \epsilon < f(x) < A + \epsilon$$ whenever $0 < |x - a| < \delta$. This clearly means that $$0 < \frac{A}{2} = A - \epsilon < f(x)$$ and thus $f(x)$ is positive for $0 < |x - a| < \delta$.
A: No, that is not right ; if your proof was correct, the derivatives of all functions would be zero! Didn't you notice that you never used anywhere the fact that $f$ had a maximum at $c$? What if $f(x) = x$ and the open interval was the real numbers?
Then your proof shows that for $c = 0$ for example,
$$
\lim_{h \to 0} f(c+h) - f(c) = \lim_{h \to 0} f(h) - f(0) = \lim_{h \to 0} h = 0,
$$
but that certainly does not imply that 
$$
0 = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h}{h} = 1.
$$
He introduces the function $Q$ because he needs to. Read the proof more carefully! Feel free to ask more questions about the proof though.
Hope that helps,
