Understanding the Problem

We are given a piecewise function f(x) defined as:

$f\left(x\right)=\{\begin{array}{cc}\max \left\{\right|x|,x^2\}& \left|x\right|\le 2\\ 8-2\left|x\right|& 2<\left|x\right|\le 4\end{array}$

We need to find the set S of all points in the interval (-4, 4) where this function is not differentiable.

Key Concepts for Differentiability

A function is not differentiable at a point x = a if any of the following conditions occur:

1. The function is not continuous at x = a.
2. The function has a sharp corner or cusp (the left-hand derivative does not equal the right-hand derivative).
3. The function is not defined at x = a (not applicable here as f(x) is defined for all |x| ≤ 4).

Since our function is piecewise and involves absolute values and a maximum function, we must check for continuity and sharp corners at the boundaries between pieces and within the pieces themselves.

Step 1: Analyze the Function on |x| ≤ 2

For |x| ≤ 2, f(x) = max{ |x|, x² }.

To understand this part, we need to find out where |x| = x² and which function is larger on either side of those points.

Solve |x| = x²:

$\left|x\right|=x^2$

For x ≥ 0, this becomes x = x² => x² - x = 0 => x(x - 1) = 0. So, x = 0 or x = 1.

For x < 0, this becomes -x = x² => x² + x = 0 => x(x + 1) = 0. So, x = 0 or x = -1.

The critical points within this interval are therefore at x = -1, 0, 1.

Now, let's determine which function is the maximum in the intervals created by these points:

• For x < -1 (e.g., x = -2): |x| = 2, x² = 4. Since 4 > 2, max{ |x|, x² } = x².
• For -1 < x < 0 (e.g., x = -0.5): |x| = 0.5, x² = 0.25. Since 0.5 > 0.25, max{ |x|, x² } = |x|.
• For 0 < x < 1 (e.g., x = 0.5): |x| = 0.5, x² = 0.25. Since 0.5 > 0.25, max{ |x|, x² } = |x|.
• For x > 1 (e.g., x = 2): |x| = 2, x² = 4. Since 4 > 2, max{ |x|, x² } = x².

So, we can rewrite f(x) for |x| ≤ 2 in a simpler, piecewise manner:

$f\left(x\right)=\{\begin{array}{cc}{x}^2& -2\le x\le -1\\ -x& -1<x<0\\ x& 0<x<1\\ x^2& 1\le x\le 2\end{array}$

Potential non-differentiability points in [-2, 2]: The boundaries where the definition changes: x = -2, -1, 0, 1, 2. We must check the derivatives from the left and right at these points.

Step 2: Analyze the Function on 2 < |x| ≤ 4

For 2 < |x| ≤ 4, f(x) = 8 - 2|x|.

This is a V-shaped function (because of the |x|) with its vertex at (0, 8). Its slope changes at x = 0, but note that this piece is only defined for |x| > 2. Therefore, the only relevant point for a corner in this piece is at the boundary x = ±2, where it meets the first piece. We must check if the function is continuous and differentiable at x = ±2.

Let's write this piece explicitly for positive and negative x:

$f\left(x\right)=\{\begin{array}{cc}8-2x& 2<x\le 4\\ 8+2x& -4\le x<-2\end{array}$

Potential non-differentiability points in this region: The function 8 - 2|x| itself is not differentiable at x=0, but x=0 is not in this piece (2 < |0| is false). Therefore, the only candidate points from this piece are the boundaries x = 2 and x = -2.

Step 3: Check for Continuity at Critical Points

First, we must ensure the function is continuous at the points we want to test. A function cannot be differentiable if it is not continuous.

Let's check the values of the pieces at their meeting points: x = -2, -1, 0, 1, 2.

At x = -2:
From first piece (x²): f(-2) = (-2)² = 4.
From second piece (8 + 2x): f(-2) = 8 + 2*(-2) = 4.
Continuous.

At x = -1:
From left (x²): f(-1) = (-1)² = 1.
From right (-x): f(-1) = -(-1) = 1.
Continuous.

At x = 0:
From left (-x): f(0) = -0 = 0.
From right (x): f(0) = 0.
Continuous.

At x = 1:
From left (x): f(1) = 1.
From right (x²): f(1) = (1)² = 1.
Continuous.

At x = 2:
From first piece (x²): f(2) = (2)² = 4.
From second piece (8 - 2x): f(2) = 8 - 2*2 = 4.
Continuous.

The function is continuous at all critical points x = -2, -1, 0, 1, 2. Now we check differentiability.

Step 4: Check for Differentiability at Critical Points

We calculate the left-hand derivative (LHD) and right-hand derivative (RHD) at each point. If LHD ≠ RHD, the function is not differentiable there.

At x = -2:
For x < -2, we use f(x) = 8 + 2x. Derivative: f'(x) = 2.
LHD = 2.
For x > -2 (approaching from the right, in [-2, -1]), we use f(x) = x². Derivative: f'(x) = 2x.
At x = -2, RHD = 2*(-2) = -4