Let $$f(x) = \begin{cases} \max(|x|, x^2), & |x| \leq 2 \\ 8-2|x|, & 2 < |x| \leq 4 \end{cases}$$. Let $$S$$ be the set of points in the interval $$(-4, 4)$$ at which $$f$$ is not differentiable. Then $$S$$:

We begin with the definition of the function

$$ f(x)= \begin{cases} \max\!\bigl(|x|,\;x^{2}\bigr), & |x|\le 2\\[4pt] 8-2|x|, & 2<|x|\le 4 \end{cases} $$

We have to find all points in the open interval $$(-4,\,4)$$ where $$f$$ is not differentiable. Whenever a piece-wise definition changes its analytic form, or whenever the chosen expression itself is non-smooth, a lack of differentiability can occur. Therefore we examine

1. the internal switch between $$|x|$$ and $$x^{2}$$ when $$|x|\le 2$$,

2. the boundary points $$|x|=2$$ where the first case is replaced by $$8-2|x|$$.

First we focus on the part $$|x|\le 2$$, where

$$ f(x)=\max\!\bigl(|x|,\;x^{2}\bigr). $$

To understand which of the two quantities $$|x|$$ and $$x^{2}$$ is larger, we solve the equality

$$ |x| = x^{2}. $$

For $$x\ge 0$$ we have $$|x|=x$$, so

$$ x^{2}=x \;\;\Longrightarrow\;\; x^{2}-x=0 \;\;\Longrightarrow\;\; x(x-1)=0, $$

giving $$x=0$$ or $$x=1$$. For $$x\le 0$$ we have $$|x|=-x$$, hence

$$ x^{2}=-x \;\;\Longrightarrow\;\; x^{2}+x=0 \;\;\Longrightarrow\;\; x(x+1)=0, $$

yielding $$x=0$$ or $$x=-1$$. Thus the two expressions intersect at

$$ x=-1,\;0,\;1. $$

Now test which term dominates in each sub-interval:

• If $$0<|x|<1$$, observe that $$x^{2}<|x|$$, so $$f(x)=|x|$$ there.

• If $$1<|x|\le 2$$, the inequality reverses to $$x^{2}>|x|$$, hence $$f(x)=x^{2}$$ there.

• Exactly at $$|x|=1$$ and at $$x=0$$ the two alternatives are equal, i.e. $$f(1)=f(-1)=f(0)=1$$ or $$0$$ respectively.

Because $$|x|$$ itself contains a corner at $$x=0$$, we inspect the derivative around that point. For small positive $$h$$ we have

$$ \frac{f(0+h)-f(0)}{h} \;=\; \frac{|h|-0}{h}=+1, $$

while for small negative $$h$$ we get

$$ \frac{f(0+h)-f(0)}{h} \;=\; \frac{|h|-0}{h}=-1. $$

The two one-sided limits differ, so $$f$$ is not differentiable at $$x=0$$.

Next consider the points $$x=\pm1$$ where the maximum operator switches from $$|x|$$ to $$x^{2}$$. We compute the derivatives of each candidate expression:

For $$x>0$$,

$$ \frac{d}{dx}|x| = 1, \qquad \frac{d}{dx}x^{2}=2x. $$

Approaching $$x=1$$ from the left (values slightly less than 1) we are in the region $$f(x)=|x|$$, so the left-hand derivative equals $$1$$. Approaching from the right we use $$f(x)=x^{2}$$, so the right-hand derivative equals $$2(1)=2$$. Because $$1\neq2$$, $$f$$ is not differentiable at $$x=1$$.

By symmetry the same reasoning applies at $$x=-1$$. For $$x<0$$ we have $$\dfrac{d}{dx}|x|=-1$$, whereas $$\dfrac{d}{dx}x^{2}=2x=-2$$ at $$x=-1$$. The values $$-1$$ and $$-2$$ are unequal, hence $$f$$ is not differentiable at $$x=-1$$.

We have thus far obtained the points $$x=-1,\,0,\,1$$ inside the interval $$|x|\le2$$.

Now we turn to the boundary $$|x|=2$$, i.e. $$x=2$$ and $$x=-2$$, where the analytic form of $$f$$ switches from the “maximum” description to the linear expression $$8-2|x|$$.

Inside $$|x|\le2$$ and close to $$x=2$$ we are in the sub-region $$1<|x|\le2$$, so $$f(x)=x^{2}$$ there. Its derivative is

$$ \frac{d}{dx}x^{2}=2x, $$

which at $$x=2$$ equals $$2\cdot2=4$$. Outside that boundary, for $$x>2$$, we have $$f(x)=8-2|x|=8-2x$$ (because $$x$$ is positive). The derivative of $$8-2x$$ is $$-2$$. Since $$4\neq-2$$, the derivatives do not match, hence $$f$$ is not differentiable at $$x=2$$.

For $$x=-2$$ the inside expression is still $$x^{2}$$, whose derivative equals $$2x=2(-2)=-4$$ at that point. For $$x<-2$$ we write $$|x|=-x$$, so

$$ f(x)=8-2(-x)=8+2x, $$

whose derivative is $$2$$. Again $$-4\neq2$$, so $$f$$ is not differentiable at $$x=-2$$.

Collecting all points found, the set where $$f$$ fails to be differentiable inside $$(-4,4)$$ is

$$ S=\{-2,\,-1,\,0,\,1,\,2\}. $$

Looking at the given alternatives, this coincides with Option A.

Hence, the correct answer is Option A.