1. Given functions

$$f(x)= \begin{cases} -1, & -2\le x<0 \\ x^2-1, & 0\le x\le 2 \end{cases}$$

and

$$g(x)=|f(x)|+f(|x|).$$

We need to study continuity/differentiability of $g$ in the interval $(-2,2)$.

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2. Find $f(|x|)$

Since $|x|\in [0,2)$ for $x\in(-2,2)$, we always use the second branch of $f$:

$$f(|x|)=|x|^2-1=x^2-1.$$

So,

$$g(x)=|f(x)|+x^2-1.$$

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3. Find $|f(x)|$ piecewise

Case 1: $-2<x<0$

Here,

$$f(x)=-1 \implies |f(x)|=1.$$

Hence

$$g(x)=1+x^2-1=x^2.$$

Case 2: $0\le x<2$

Here,

$$f(x)=x^2-1.$$

Thus

$$|f(x)|=|x^2-1|.$$

Now split according to sign of $x^2-1$:

• For $0\le x<1$, $x^2-1<0$, so $|x^2-1|=1-x^2.$ Therefore, $g(x)=(1-x^2)+(x^2-1)=0.$

• For $1\le x<2$, $x^2-1\ge 0$, so $|x^2-1|=x^2-1.$ Therefore, $g(x)=(x^2-1)+(x^2-1)=2x^2-2.$

So overall,

$$g(x)= \begin{cases} x^2, & -2<x<0,\\ 0, & 0\le x<1,\\ 2x^2-2, & 1\le x<2. \end{cases}$$

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4. Check continuity

Potential trouble points are where formula changes: $x=0$ and $x=1$.

At $x=0$

Left limit:

$$\lim_{x\to 0^-} g(x)=\lim_{x\to 0^-} x^2=0.$$

Right limit:

$$\lim_{x\to 0^+} g(x)=\lim_{x\to 0^+} 0=0.$$

And

$$g(0)=0.$$

So $g$ is continuous at $x=0$.

At $x=1$

Left limit:

$$\lim_{x\to 1^-} g(x)=0.$$

Right limit:

$$\lim_{x\to 1^+} g(x)=2(1)^2-2=0.$$

And

$$g(1)=2(1)^2-2=0.$$

So $g$ is continuous at $x=1$.

Hence $g$ is continuous throughout $(-2,2)$.

So option A is false.

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5. Check differentiability

For $-2<x<0$

$$g(x)=x^2 \implies g'(x)=2x.$$

Differentiable.

For $0<x<1$

$$g(x)=0 \implies g'(x)=0.$$

Differentiable.

For $1<x<2$

$$g(x)=2x^2-2 \implies g'(x)=4x.$$

Differentiable.

Now check the junctions.

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6. At $x=0$

Left derivative:

$$g'_-(0)=\lim_{x\to 0^-} \frac{x^2-0}{x}=\lim_{x\to 0^-} x=0.$$

Right derivative:

$$g'_+(0)=\lim_{x\to 0^+} \frac{0-0}{x}=0.$$

Thus,

$$g'_-(0)=g'_+(0)=0,$$

so $g$ is differentiable at $x=0$.

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7. At $x=1$

Left derivative:

$$g'_-(1)=0$$

since $g(x)=0$ for $0<x<1$.

Right derivative:

$$g'_+(1)=\left.\frac{d}{dx}(2x^2-2)\right|_{x=1}=4.$$

Since

$$g'_-(1)\ne g'_+(1),$$

$g$ is not differentiable at $x=1$.

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8. Conclusion

In $(-2,2)$, $g$ is continuous everywhere and not differentiable at exactly one point, namely $x=1$.

Therefore the correct option is:

$$\boxed{\text{D: not differentiable at one point}}$$

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9. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They agree.