Answer
Consistent, because a critical point does not have to be such that the derivative there equals zero.
Work Step by Step
The theorem states that IF $f'(c)$ is defined, then $f'(c)=0$ ...
(for a local extremum at x=c)
There is no contradiction with the theorem, as a critical point can be one where the derivative is not defined. Furthermore, there could be an extremum at that critical point (but doesn't have to be)
Here, $\quad f(x)= |x|=\left\{\begin{array}{ll}
x, & x\geq 0\\
-x & x\leq 0
\end{array}\right.$.
$f'(x)=\left\{\begin{array}{ll}
1, & x\geq 0\\
-1 & x\leq 0
\end{array}\right.$
The left- and right-derivatives at x=0 are not equal, so the derivative is not defined there, but the function is.
$f(0)=0$ and $f(x)$ is positive for all other x,
so at $x=0$, f has an absolute minimum.
