Number of Non-Differentiable Points for Piecewise Composite Function

To find the number of points in the interval $x \in (-\pi, \pi)$ at which the function $f(x)$ is not differentiable, we decompose $f(x)$ into two components: $f(x) = g(x) + h(x)$ where $g(x) = \max \left\{ 6x, 2 + 3x^2 \right\}$ and $h(x) = |x - 1| \cos \left| x^2 - \frac{1}{4} \right|$

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Step 1: Differentiability of $g(x)$

To find where $g(x) = \max \left\{ 6x, 2 + 3x^2 \right\}$ changes its definition, we solve for the points of intersection: $2 + 3x^2 = 6x \implies 3x^2 - 6x + 2 = 0$

Using the quadratic formula: $x = \frac{6 \pm \sqrt{36 - 24}}{6} = 1 \pm \frac{1}{\sqrt{3}}$

Let these two roots be: $x_1 = 1 - \frac{1}{\sqrt{3}} \quad \text{and} \quad x_2 = 1 + \frac{1}{\sqrt{3}}$

Since $3x^2 - 6x + 2 > 0$ for $x \in (-\infty, x_1) \cup (x_2, \infty)$ and $3x^2 - 6x + 2 < 0$ for $x \in (x_1, x_2)$, the function $g(x)$ is defined piecewise as: $g(x) = \begin{cases} 2 + 3x^2, & x \le x_1 \text{ or } x \ge x_2 \\ 6x, & x_1 < x < x_2 \end{cases}$

Now, we evaluate the left-hand and right-hand derivatives at $x_1$ and $x_2$:

• At $x = x_1$: $g'(x_1^-) = 6x_1 = 6\left(1 - \frac{1}{\sqrt{3}}\right) = 6 - 2\sqrt{3}$ $g'(x_1^+) = 6$ Since $g'(x_1^-) \neq g'(x_1^+)$, $g(x)$ is not differentiable at $x = x_1$.

• At $x = x_2$: $g'(x_2^-) = 6$ $g'(x_2^+) = 6x_2 = 6\left(1 + \frac{1}{\sqrt{3}}\right) = 6 + 2\sqrt{3}$ Since $g'(x_2^-) \neq g'(x_2^+)$, $g(x)$ is not differentiable at $x = x_2$.

At all other points $x \in (-\pi, \pi) \setminus \{x_1, x_2\}$, $g(x)$ is smooth and differentiable.

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Step 2: Differentiability of $h(x)$

Using the property that $\cos(-t) = \cos(t)$ for all $t \in \mathbb{R}$, we have: $\cos \left| x^2 - \frac{1}{4} \right| = \cos \left( x^2 - \frac{1}{4} \right)$

Let $k(x) = \cos \left( x^2 - \frac{1}{4} \right)$, which is a composition of smooth functions and is thus infinitely differentiable everywhere on $\mathbb{R}$.

Therefore, $h(x) = |x - 1| k(x)$:

• At $x = 1$: $k(1) = \cos \left( 1 - \frac{1}{4} \right) = \cos \left( \frac{3}{4} \right) \neq 0$ The left-hand and right-hand derivatives of $h(x)$ at $x = 1$ are: $h'(1^-) = \lim_{x \to 1^-} \frac{-(x-1)k(x) - 0}{x - 1} = -k(1) = -\cos\left(\frac{3}{4}\right)$ $h'(1^+) = \lim_{x \to 1^+} \frac{(x-1)k(x) - 0}{x - 1} = k(1) = \cos\left(\frac{3}{4}\right)$ Since $h'(1^-) \neq h'(1^+)$, $h(x)$ is not differentiable at $x = 1$.

• For $x \neq 1$: Both $|x - 1|$ and $k(x)$ are differentiable, so $h(x)$ is differentiable everywhere on $(-\pi, \pi) \setminus \{1\}$.

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Step 3: Differentiability of $f(x) = g(x) + h(x)$

Using the principle that the sum of a differentiable function and a non-differentiable function at a point is non-differentiable:

1. At $x = x_1 = 1 - \frac{1}{\sqrt{3}}$: $g(x)$ is non-differentiable, while $h(x)$ is differentiable. Thus, $f(x)$ is not differentiable.
2. At $x = x_2 = 1 + \frac{1}{\sqrt{3}}$: $g(x)$ is non-differentiable, while $h(x)$ is differentiable. Thus, $f(x)$ is not differentiable.
3. At $x = 1$: $g(x)$ is differentiable (since $1 \neq x_1, x_2$), while $h(x)$ is non-differentiable. Thus, $f(x)$ is not differentiable.
4. For all other $x \in (-\pi, \pi) \setminus \{x_1, x_2, 1\}$: Both $g(x)$ and $h(x)$ are differentiable, so $f(x)$ is differentiable.

Checking the interval constraints: $-\pi < 1 - \frac{1}{\sqrt{3}} \approx 0.423 < 1 < 1 + \frac{1}{\sqrt{3}} \approx 1.577 < \pi$

All three non-differentiable points lie within $(-\pi, \pi)$.

Thus, the number of points at which $f(x)$ is not differentiable is 3.