Let f(x) be defined on [-2, 2] such that f(x)=≤ft […] . and, g(x) = f…

We have,

$f(x)=\left\{\begin{array}{rr}-1, & -2 \leq x \leq 0 \\ x-1, & 0<x \leq 2\end{array}\right.$

∴ $|f(x)|=\left\{\begin{array}{cc}1, & -2 \leq x \leq 0 \\ |x-1|, & 0<x \leq 2\end{array}\right.$

$=\left\{\begin{array}{rr}1, & -2 \leq x \leq 0 \\ 1-x, & 0<x<1 \\ x-1, & 1 \leq x \leq 2\end{array}\right.$

and, |f(|x|)| = |x| - 1, for -2 ≤ x ≤ 2

$= \begin{cases}-x-1, & -2 \leq x<0 \\ x-1, & 0 \leq x \leq 2\end{cases}$

∴ $g(x)=|f(x)|+f(|x|)$

$\Rightarrow g(x)= \begin{cases}-x \quad, & -2 \leq x<0 \\ 0, & 0 \leq x<1 \\ 2(x-1), & 1 \leq x \leq 2\end{cases}$

Clearly, g(x) is continuous at x = 0, 1.

Also,

(LHD at x = 0) = $\left(\frac{d}{d x}(-x)\right)_{\text {at } x=0}=-1$

(RHD at x = 0) = $\left(\frac{d}{d x}(0)\right)_{\text {at } x=0}=0$

(LHD at x = 1) = $\left(\frac{d}{d x}(0)\right)_{\text {at } x=0}=0$

and, (RHD at x = 1) = $\left(\frac{d}{d x}\{2(x-1)\}\right)_{\text {at } x=1}=2$

∴  (LHD at x = 0) ≠ (RHD at x = 0)

and, (LHD at x = 1) ≠ (RHD at x = 1)

So, g(x) is not differentiable at x = 0, 1.

Now,

(RHD at x = -2) = $\left(\frac{d}{d x}(-x)\right)_{\text {at } x=-2}=-1$, exists finitely.

Similarly, (LHD at x = 2) = 2, exists finitely.

Hence, g(x) is differentiable at all points in the interval [-1, 1] except at x = 0 and x = 1.