first derivative test and its converse According to the first derivative test, if $f′≥0$ on an open interval to the left of c, $f′(c)=0$, and $f′≤0$ on an open interval to the right of c, then $f(c)$ is a local maximum. If the inequalities are both reversed, then $f(c)$ is a local minimum. 
Then for a local maximum, shouldn't $f′(x)≥0$ for $x<c$, and $f′(x)≤0$ for $x>c$, if $x$ is sufficiently close to c? Why isn't this always true? 
update: my teacher told me that we can't just naively take the above conditions to be true all the time and wanted me to try find a counterexample where "$f′(x)≥0$ for $x<c$, and $f′(x)≤0$ for $x>c$, if $x$ is sufficiently close to c" would be false. 

Your question is not clear, but you want to ask for a differentiable function $f$ on reals that has a local maximum at $c$ but fails to satisfy ( $f'(c) = 0$ and $f'(x) \ge 0 \ge f'(y)$ as $x \to c^-$ and $y \to c^+$ ).
You should know by Rolle's theorem that the first condition will be satisfied, so you want to find an $f$ that fails the second condition. Well, construct $f$ to oscillate faster and faster near $c$ but still have a maximum at $c$ and still be differentiable everywhere.
Hint 1:

 

Hint 2:

 Combine $\cos(1/x)$ with $x^2$ in an appropriate way (for $x \ne 0$), and separately patch the hole (for $x = 0$).

