Let $$f(x) = \begin{cases} -1, & -2 \le x \lt 0 \\ x^2 - 1, & 0 \le x \le 2 \end{cases}$$ and $$g(x) = |\eta(x)| + f(|x|)$$. Then, in the interval $$(-2, 2)$$, $$g$$ is:

We have the function

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

and the second function

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

First we examine the term $$f(|x|).$$ Since $$|x|\ge 0$$ for every real $$x,$$ the argument $$|x|$$ always falls in the interval $$[0,2].$$ For this non-negative branch the definition of $$f$$ is

$$f(t)=t^{2}-1\quad\text{when }0\le t\le 2.$$

Substituting $$t=|x|,$$ we obtain

$$f(|x|)=|x|^{2}-1$$

for every $$x$$ with $$-2\lt x\lt 2.$$

Therefore

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

Continuity of $$g$$ in $$(-2,2)$$

For $$x\neq 0$$ the expression $$|x|+|x|^{2}-1$$ is a polynomial in either $$x$$ (when $$x\gt 0$$) or $$-x$$ (when $$x\lt 0$$); polynomials are continuous, so $$g$$ is continuous at every non-zero point.

We now check continuity at $$x=0.$$ We compute the left-hand and right-hand limits:

For $$h\to 0^{+}$$ (that is, $$x=h\gt 0$$)

$$\displaystyle\lim_{h\to 0^{+}}g(h)=\lim_{h\to 0^{+}}\bigl(h+h^{2}-1\bigr)=-1.$$

For $$h\to 0^{-}$$ (that is, $$x=h\lt 0$$)

$$\displaystyle\lim_{h\to 0^{-}}g(h)=\lim_{h\to 0^{-}}\bigl(-h+h^{2}-1\bigr)=-1.$$

The two one-sided limits are equal, so

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

Because $$g(0)=|0|+|0|^{2}-1=-1,$$ the limit equals the function value, proving that $$g$$ is continuous at $$x=0.$$ Hence $$g$$ is continuous at every point of the open interval $$(-2,2).$$

Differentiability of $$g$$ in $$(-2,2)$$

We next differentiate the two explicit expressions for $$g(x).$$

For $$x\gt 0$$ we have $$g(x)=x+x^{2}-1,$$ so

$$g'(x)=1+2x\quad\text{when }x\gt 0.$$

For $$x\lt 0$$ we have $$g(x)=-x+x^{2}-1,$$ so

$$g'(x)=-1+2x\quad\text{when }x\lt 0.$$

Thus a derivative exists at every non-zero point.

The only point that still needs attention is $$x=0.$$ We use the limit definition of the derivative.

The right-hand derivative at zero is

$$\begin{aligned} g'_{+}(0)&=\lim_{h\to 0^{+}}\frac{g(h)-g(0)}{h} =\lim_{h\to 0^{+}}\frac{(h+h^{2}-1)-(-1)}{h} \\ &=\lim_{h\to 0^{+}}\frac{h+h^{2}}{h} =\lim_{h\to 0^{+}}(1+h)=1. \end{aligned}$$

The left-hand derivative at zero is

$$\begin{aligned} g'_{-}(0)&=\lim_{h\to 0^{-}}\frac{g(h)-g(0)}{h} =\lim_{h\to 0^{-}}\frac{(-h+h^{2}-1)-(-1)}{h} \\ &=\lim_{h\to 0^{-}}\frac{-h+h^{2}}{h} =\lim_{h\to 0^{-}}(-1+h)=-1. \end{aligned}$$

Because $$g'_{+}(0)=1$$ and $$g'_{-}(0)=-1$$ are unequal, the derivative does not exist at $$x=0.$$

Combining our findings:

• $$g$$ is continuous everywhere in \((-2,2).$$
• $$g$$ is differentiable at every point of \((-2,2)$$ except $$x=0.$$

Therefore $$g$$ fails to be differentiable at exactly one point in the open interval \((-2,2).$$

Hence, the correct answer is Option D.