Key Concepts and Formulas

• Differentiability at a point: A function $f(x)$ is differentiable at a point $x=a$ if the limit $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists. This is equivalent to checking if the left-hand derivative (LHD) equals the right-hand derivative (RHD) at that point.
  • LHD at $x=a$: $\lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h} = f'(a^-)$
  • RHD at $x=a$: $\lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h} = f'(a^+)$
• Continuity: A function must be continuous at a point to be differentiable at that point. If a function is not continuous at a point, it cannot be differentiable there.
• Differentiability of standard functions:
  • Polynomials are differentiable everywhere.
  • $|x|$ is not differentiable at $x=0$.
  • $x^2$ is differentiable everywhere.
• Piecewise functions: Differentiability needs to be checked at the points where the definition of the function changes. We also need to check for continuity at these points.

Step-by-Step Solution

First, let's analyze the given function $f(x)$. The function is defined in pieces based on the value of $|x|$. The interval of interest is $(-4, 4)$.

The function is given by: f\left( x \right) = \left\{ {\matrix{ {\max \left\{ {\left| x \right|,{x^2}} \right\}} & {\left| x \right| \le 2} \cr {8 - 2\left| x \right|} & {2 < \left| x \right| \le 4} \cr } } \right.

Let's break down the definition of $f(x)$ into different intervals for $x$ in $(-4, 4)$.

Part 1: Analyzing $f(x)$ for $|x| \le 2$

This condition means $-2 \le x \le 2$. In this interval, $f(x) = \max\{|x|, x^2\}$. We need to determine when $|x| \ge x^2$ and when $x^2 \ge |x|$.

• Case 1: $|x| \ge x^2$. This occurs when $x^2 - |x| \le 0$. Let $y = |x|$. Then $y^2 - y \le 0 \implies y(y-1) \le 0$. This inequality holds for $0 \le y \le 1$. Substituting back $y = |x|$, we get $0 \le |x| \le 1$. This means $-1 \le x \le 1$. So, for $x \in [-1, 1]$, $f(x) = |x|$.

• Case 2: $x^2 \ge |x|$. This occurs when $x^2 - |x| \ge 0$. This inequality holds for $|x| \le 0$ or $|x| \ge 1$. Since $|x| \ge 0$, $|x| \le 0$ implies $|x|=0$, which means $x=0$. So, we have $x=0$ or $|x| \ge 1$. Considering the interval $|x| \le 2$, this means $1 \le |x| \le 2$. This corresponds to $x \in [-2, -1] \cup [1, 2]$. So, for $x \in [-2, -1] \cup [1, 2]$, $f(x) = x^2$.

Combining these, for $|x| \le 2$ (i.e., $-2 \le x \le 2$): f(x) = \left\{ {\matrix{ {x^2} & { - 2 \le x \le - 1} \cr {\left| x \right|} & { - 1 < x < 1} \cr {{x^2}} & {1 \le x \le 2} \cr } } \right.

Part 2: Analyzing $f(x)$ for $2 < |x| \le 4$

This condition means $2 < x \le 4$ or $-4 \le x < -2$. In this interval, $f(x) = 8 - 2|x|$.

• If $2 < x \le 4$, then $|x| = x$, so $f(x) = 8 - 2x$.
• If $-4 \le x < -2$, then $|x| = -x$, so $f(x) = 8 - 2(-x) = 8 + 2x$.

Putting it all together, the piecewise definition of $f(x)$ on $(-4, 4)$ is: f\left( x \right) = \left\{ {\matrix{ {8 + 2x,} & { - 4 \le x \le - 2} \cr {{x^2},} & { - 2 \le x \le - 1} \cr {\left| x \right|,} & { - 1 < x < 1} \cr {{x^2},} & {1 \le x \le 2} \cr {8 - 2x,} & {2 < x \le 4} \cr } } \right.

Now, we need to find the points in $(-4, 4)$ where $f(x)$ is not differentiable. The potential points of non-differentiability are the points where the definition of the function changes, which are $x = -2, -1, 1, 2$. We also need to consider points where the constituent functions are not differentiable, such as $x=0$ for $|x|$.

Step 1: Check for continuity at the points where the definition changes.

• At $x = -2$:

  • Left limit: $\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} (8 + 2x) = 8 + 2(-2) = 8 - 4 = 4$.
  • Right limit: $\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} x^2 = (-2)^2 = 4$.
  • Function value: $f(-2) = (-2)^2 = 4$. Since the left limit, right limit, and function value are equal, $f(x)$ is continuous at $x = -2$.

• At $x = -1$:

  • Left limit: $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} x^2 = (-1)^2 = 1$.
  • Right limit: $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} |x| = |-1| = 1$.
  • Function value: $f(-1) = (-1)^2 = 1$. Since the left limit, right limit, and function value are equal, $f(x)$ is continuous at $x = -1$.

• At $x = 1$:

  • Left limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} |x| = |1| = 1$.
  • Right limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1$.
  • Function value: $f(1) = 1^2 = 1$. Since the left limit, right limit, and function value are equal, $f(x)$ is continuous at $x = 1$.

• At $x = 2$:

  • Left limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4$.
  • Right limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (8 - 2x) = 8 - 2(2) = 8 - 4 = 4$.
  • Function value: $f(2) = 2^2 = 4$. Since the left limit, right limit, and function value are equal, $f(x)$ is continuous at $x = 2$.

The function is continuous at all these points. Now we check differentiability.

Step 2: Check for differentiability at the points where the definition changes and at $x=0$.

We need to compare the left-hand derivative (LHD) and the right-hand derivative (RHD) at each point.

• At $x = -2$:

  • For $x < -2$, $f(x) = 8 + 2x$. The derivative is $f'(x) = 2$. So, LHD at $x=-2$ is $2$.
  • For $x > -2$ (and $x \le -1$), $f(x) = x^2$. The derivative is $f'(x) = 2x$. So, RHD at $x=-2$ is $2(-2) = -4$. Since LHD ($2$) $\ne$ RHD ($-4$), $f(x)$ is not differentiable at $x = -2$.

• At $x = -1$:

  • For $x < -1$ (and $x \ge -2$), $f(x) = x^2$. The derivative is $f'(x) = 2x$. So, LHD at $x=-1$ is $2(-1) = -2$.
  • For $x > -1$ (and $x < 1$), $f(x) = |x|$.
    • For $-1 < x < 0$, $f(x) = -x$, so $f'(x) = -1$.
    • For $0 < x < 1$, $f(x) = x$, so $f'(x) = 1$. The RHD at $x=-1$ is the derivative of $|x|$ as $x$ approaches $-1$ from the right. Since $-1 < x < 0$, $f(x) = -x$, so the derivative is $-1$. Thus, RHD at $x=-1$ is $-1$. Since LHD ($-2$) $\ne$ RHD ($-1$), $f(x)$ is not differentiable at $x = -1$.

• At $x = 0$:

  • The function in the neighborhood of $x=0$ is $f(x) = |x|$.
  • The function $|x|$ is known to be not differentiable at $x=0$.
  • Let's verify:
    • LHD at $x=0$: $\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$.
    • RHD at $x=0$: $\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$. Since LHD ($-1$) $\ne$ RHD ($1$), $f(x)$ is not differentiable at $x = 0$.

• At $x = 1$:

  • For $x < 1$ (and $x > -1$), $f(x) = |x|$.
    • For $0 < x < 1$, $f(x) = x$, so $f'(x) = 1$. The LHD at $x=1$ is the derivative of $|x|$ as $x$ approaches $1$ from the left. Since $0 < x < 1$, $f(x) = x$, so the derivative is $1$. Thus, LHD at $x=1$ is $1$.
  • For $x > 1$ (and $x \le 2$), $f(x) = x^2$. The derivative is $f'(x) = 2x$. So, RHD at $x=1$ is $2(1) = 2$. Since LHD ($1$) $\ne$ RHD ($2$), $f(x)$ is not differentiable at $x = 1$.

• At $x = 2$:

  • For $x < 2$ (and $x \ge 1$), $f(x) = x^2$. The derivative is $f'(x) = 2x$. So, LHD at $x=2$ is $2(2) = 4$.
  • For $x > 2$, $f(x) = 8 - 2x$. The derivative is $f'(x) = -2$. So, RHD at $x=2$ is $-2$. Since LHD ($4$) $\ne$ RHD ($-2$), $f(x)$ is not differentiable at $x = 2$.

Step 3: Identify the set S.

The set S is the set of points in the interval (– 4, 4) at which f is not differentiable. From our analysis, the points are $x = -2, -1, 0, 1, 2$.

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

Common Mistakes & Tips

• Forgetting to check continuity first: If a function is not continuous at a point, it cannot be differentiable there. While we checked continuity here, it's a good first step.
• Incorrectly defining $|x|$ in piecewise form: Remember $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$. This is crucial when dealing with intervals involving negative numbers.
• Assuming differentiability of standard functions within intervals: While $x^2$ is differentiable everywhere and $|x|$ is differentiable everywhere except $x=0$, you still need to check differentiability at the junction points of the piecewise definition. The derivative of $|x|$ from the left and right at $x=0$ are $-1$ and $1$ respectively.

Summary

We first correctly expanded the piecewise definition of $f(x)$ by analyzing the $\max\{|x|, x^2\}$ and $8 - 2|x|$ conditions over the given intervals. We identified the critical points where differentiability might fail: the points where the function definition changes ($x=-2, -1, 1, 2$) and points where the constituent functions are not differentiable ($x=0$ for $|x|$). We then checked for continuity at these points, which was confirmed. Finally, we calculated the left-hand and right-hand derivatives at each critical point and found that they were unequal at $x = -2, -1, 0, 1, 2$. Thus, these points constitute the set S of non-differentiable points.

The final answer is $\boxed{\left\{ { - 2, - 1,0,1,2} \right\}}$. This corresponds to option (B).