In `f (x)= [{:(cos x ^(2),, x lt 0), ( sin x ^(3) -|x ^(3)-1|,, x ge 0):}` then find the number of points where `g (x) =f (|x|)` is non-differentiable.

Answer

To solve the problem, we need to analyze the function \( g(x) = f(|x|) \) based on the given piecewise function \( f(x) \).

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} \cos(x^2) & \text{if } x < 0 \\ \sin(x^3) - |x^3 - 1| & \text{if } x \geq 0 \end{cases} \]

### Step 2: Rewrite \( g(x) \)
Since \( g(x) = f(|x|) \), we need to consider the absolute value of \( x \):
- For \( x < 0 \), \( |x| = -x \) so \( g(x) = f(-x) = \cos((-x)^2) = \cos(x^2) \).
- For \( x \geq 0 \), \( |x| = x \) so \( g(x) = f(x) = \sin(x^3) - |x^3 - 1| \).

Thus, we can rewrite \( g(x) \) as:
\[ g(x) = \begin{cases} \cos(x^2) & \text{if } x < 0 \\ \sin(x^3) - |x^3 - 1| & \text{if } x \geq 0 \end{cases} \]

### Step 3: Analyze \( g(x) \) for differentiability
We need to check the differentiability of \( g(x) \) at the points where the definition of \( g(x) \) changes, which are \( x = 0 \) and \( x = 1 \).

#### Check at \( x = 0 \):
For \( x < 0 \):
\[ g(x) = \cos(x^2) \]
For \( x \geq 0 \):
\[ g(x) = \sin(x^3) - |x^3 - 1| \]

To check differentiability at \( x = 0 \), we need to find the left-hand and right-hand derivatives.

**Left-hand derivative at \( x = 0 \)**:
\[ g'(0^-) = \lim_{h \to 0^-} \frac{g(h) - g(0)}{h} = \lim_{h \to 0^-} \frac{\cos(h^2) - g(0)}{h} \]
Since \( g(0) = \sin(0) - |0 - 1| = 0 - 1 = -1 \),
\[ g'(0^-) = \lim_{h \to 0^-} \frac{\cos(h^2) + 1}{h} \]
As \( h \to 0 \), \( \cos(h^2) \to 1 \), so:
\[ g'(0^-) = \lim_{h \to 0^-} \frac{2}{h} \to -\infty \text{ (not defined)} \]

**Right-hand derivative at \( x = 0 \)**:
\[ g'(0^+) = \lim_{h \to 0^+} \frac{g(h) - g(0)}{h} = \lim_{h \to 0^+} \frac{\sin(h^3) - |h^3 - 1| + 1}{h} \]
For small \( h \), \( |h^3 - 1| = 1 - h^3 \), thus:
\[ g'(0^+) = \lim_{h \to 0^+} \frac{\sin(h^3) + h^3}{h} \]
As \( h \to 0 \), \( \sin(h^3) \to 0 \):
\[ g'(0^+) = \lim_{h \to 0^+} \frac{h^3}{h} = 0 \]

Since \( g'(0^-) \) and \( g'(0^+) \) are not equal, \( g(x) \) is non-differentiable at \( x = 0 \).

#### Check at \( x = 1 \):
For \( x < 1 \):
\[ g(x) = \sin(x^3) - |x^3 - 1| = \sin(x^3) - (1 - x^3) = \sin(x^3) + x^3 - 1 \]
For \( x \geq 1 \):
\[ g(x) = \sin(x^3) - (x^3 - 1) = \sin(x^3) - x^3 + 1 \]

**Left-hand derivative at \( x = 1 \)**:
\[ g'(1^-) = \lim_{h \to 1^-} \frac{g(h) - g(1)}{h - 1} \]
Calculating \( g(1) \):
\[ g(1) = \sin(1^3) + 1 - 1 = \sin(1) \]
Thus:
\[ g'(1^-) = \lim_{h \to 1^-} \frac{\sin(h^3) + h^3 - 1 - \sin(1)}{h - 1} \]

**Right-hand derivative at \( x = 1 \)**:
\[ g'(1^+) = \lim_{h \to 1^+} \frac{g(h) - g(1)}{h - 1} \]
Calculating:
\[ g'(1^+) = \lim_{h \to 1^+} \frac{\sin(h^3) - h^3 + 1 - \sin(1)}{h - 1} \]

Since both derivatives yield different results, \( g(x) \) is also non-differentiable at \( x = 1 \).

### Conclusion
The points where \( g(x) \) is non-differentiable are \( x = 0 \) and \( x = 1 \). Therefore, the number of points where \( g(x) \) is non-differentiable is **2**.