TITLE: f(x) Changing Sign Implies f'(x) Not Changing Sign
QUESTION [8 upvotes]: Let $f(x)$ be a real-valued function with $c$ in its domain. Given
1. $f(c)=0, $
2. $f$ is not identically zero in any interval about $c$, and
3. $f$ is differentiable at $c$.   
Claim:
a. If $f(x)$ changes sign at $c$, then $f’(x)$ does not change sign at $c$.
b. If $f(x)$ does not change sign at $c$, then $f’(x)$ changes sign at $c$.  
How would I go about proving this claim?
Thoughts:
The main difficulty I have with this question is that I cannot express "changing sign" in a way involving derivatives that makes this result amenable to a proof I can come up with. 
The one precise meaning of "changing sign" I thought of is:
$f(x)$ changes sign at $c$ if there exists an open set $S$ containing $c$ such that for all $y\in S$  and $y<c$, $f(y)$ is positive (negative) and for all $y\in S$ and $y>c$, $f(y)$ is negative (positive).
It's also clear that:
If $f(x)$ changes sign at $c$, $f(c)=0$, but the converse is not true.
Suggestions would be appreciated.

REPLY [2 votes]: Here's a counterexample to (b) according to your definition.
$$f(x)=x^3sin(1/x), x \neq 0$$
$$f(0)=0$$
According to your definition, neither f nor f' changes sign at $x=0$.