Can I drop one hypothesis in the First Derivative Test? The First Derivative Test says:

Let $I\subseteq\mathbb{R}$ be an open interval, $c$ is an interior
   point of $I$, $f:I\rightarrow\mathbb{R}$ and is continuous at $c$.
If $c$ is a critical
   point of $f$, and there exists a $\delta>0$ such that $\forall
 x\in(c-\delta,c),f'(x)>0$ and $\forall x\in(c,c+\delta),f'(x)<0$, then
   $c$ is a relative maximum of $f$. (The opposite part of theorem is
   omitted.)

In order to determine whether a point $c$ is a relative extremum of a function $f$ by the First Derivative Test, one should first examine the point $c$ is a critical point of $f$. I think the requirement of $c$'s being a critical point is redundant, can we just drop this without getting something wrong? That is to say, are there some counterexamples that $c$ is not a critical point and $f'$ changes signs from left to right but $c$ is not a relative extremum?
Maybe I should ask if there a function $f$ such that $f'(c)=10$ and $\forall
 x\in(c-\delta,c),f'(x)>0$ and $\forall x\in(c,c+\delta),f'(x)<0$? (Maybe such a function originally not exist.)And secondly ask that if some of, or maybe all of, this type's function doesn't have a relative extremum at $c$.
 A: We need the additional hypothesis. Consider $f(x)=\frac{1}{x^2}$. The derivate is positive for negative $x$ and negative for positive $x$, but $0$ is not a local maximum , but a pole.
If $f$ is continous at the point $x_0$, then your condition is sufficient to show that the function has a local maximum at $x_0$.
But keep in mind that the places of such sign changes are hard to detect without calculating the roots of the first derivate. Therefore it makes sense to begin with the roots of the first derivate. 
There is a good chance that they are extrema.
A: The definition of a critical point $c$ of a function $f$ which I've seen in many texts anyway is a value $c$ such that either $f'(c)=0$ or $f'(c)$ does not exist [the last needed to take care of things like $f(x)=x^{2/3}$ at $c=0.$]
So if in fact $c$ were not a critical point [according to the above definition] then $f'(c)$ would exist and be nonzero, and one could show that the two conditions you have cited could not both occur anyway near $c.$ So in a way you're thus correct, but IMO only in a kind of vacuous sense.
