#### Answer

$f (x)$ is not differentiable at $x = 2$

#### Work Step by Step

Given $$f(x)=|x^2-4| $$
Since
$$ f^{\prime}(x) =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$
Then
\begin{align*}
f^{\prime}(2)&=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\left|(2+h)^{2}-4\right|-0}{h}\\
&=\lim _{h \rightarrow 0} \frac{\left|h^{2}+4 h\right|}{h}\\
&=\lim _{h \rightarrow 0}|h+4| \cdot \frac{|h|}{h}\\
&= \lim _{h \rightarrow 0}|h+4| \lim _{h \rightarrow 0} \frac{|h|}{h}\\
&=4\lim _{h \rightarrow 0} \frac{|h|}{h}
\end{align*}
We consider the one-sided limits:
$$\lim _{h \rightarrow 0^+} \frac{|h|}{h}=1,\ \ \ \ \lim _{h \rightarrow 0^-} \frac{|h|}{h}=-1 $$
Since the limits are not equal, then $f (x)$ is not differentiable at $x = 2$