The function cannot be differentiable at \(x=0\) because a cusp occurs there for the specific absolute value function. This can be proven by taking the derivative of the absolute value function.

\[f'(x)= \left\{\begin{aligned} &\frac{x+h-x}{h}, && x\geq0\\ &\frac{-x-h+x}{h}, && x

As we can see, \(f'(x)\) is discontinuous at \(0\); therefore, the function \(\left|x\right|\) cannot be differentiable at \(x=0\).

The first thing we should check is if the function is continuous at \(x=2\) because that will tell us about the discontinuities at that point and whether or not it's diferentiable.

At \(x=2\) the function has a vertical asymptote; therefore, it is neither continuous or differentiable at \(x=2\).

The correct answer is \(\mbox{E.}\)

When we first see and absolute value function, we need to rewrite it into its two piecewise parts: a negative and a positive part.

\[f\left(x\right)=\sqrt{\left|x+2\right|}=\sqrt{\pm\left(x+2\right)}=\sqrt{\pm x \pm 2}\]

Choices \(\mbox{A.}\) and \(\mbox{D.}\) are incorrect.

Now we need to check if the derivative of the function is continuous at \(x=-2\)

Let \(f_{1}\left(x\right)=\sqrt{-x-2}\) and \(f_{2}\left(x\right)=\sqrt{x+2}\)

\[f'_{1}\left(x\right)=\mathop {\lim}\limits_{h \to 0}\frac{\sqrt{-x-h-2}-\sqrt{-x-2}}{h}=\]

\[\mathop {\lim}\limits_{h \to 0}\frac{\left(\sqrt{-x-h-2}-\sqrt{-x-2}\right)\left(\sqrt{-x-h-2}+\sqrt{-x-2}\right)}{h\left(\sqrt{-x-h-2}+\sqrt{-x-2}\right)}=\]

\[\mathop {\lim}\limits_{h \to 0}\frac{-x-h-2+x+2}{h\left(\sqrt{-x-h-2}+\sqrt{-x-2}\right)}=\mathop {\lim}\limits_{h \to 0}\frac{-h}{h\left(\sqrt{-x-h-2}+\sqrt{-x-2}\right)}\]

\[\mathop {\lim}\limits_{h \to 0}\frac{-1}{\sqrt{-x-h-2}+\sqrt{-x-2}}=\frac{-1}{2\sqrt{-x-2}}\]

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\[f'_{2}\left(x\right)=\mathop {\lim}\limits_{h \to 0}\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}=\]

\[\mathop {\lim}\limits_{h \to 0}\frac{\left(\sqrt{x+h+2}-\sqrt{x+2}\right)\left(\sqrt{x+h+2}+\sqrt{x+2}\right)}{h\left(\sqrt{x+h+2}+\sqrt{x+2}\right)}\]

\[\mathop {\lim}\limits_{h \to 0}\frac{x+h+2-x-2}{h\left(\sqrt{x+h+2}+\sqrt{x+2}\right)}=\mathop {\lim}\limits_{h \to 0}\frac{h}{h\left(\sqrt{x+h+2}+\sqrt{x+2}\right)}\]

\[\mathop {\lim}\limits_{h \to 0}\frac{1}{\sqrt{x+h+2}+\sqrt{x+2}}=\frac{1}{2\sqrt{x+2}}\]

\[f'(x)=\left\{\begin{aligned} &\frac{1}{2\sqrt{x+2}}, && x\>-2\\ &\frac{-1}{2\sqrt{-x-2}}, && x\]

Since both of the above functions have an vertical asymptote at \(x=-2\), the derivative function is not continuous at \(x=-2\); therefore, the original function it is not differentiable at \(x=-2\).

The correct answer is \(\mbox{C.}\)

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