The set of points where ,`f(x) = x|x|` is twice differentiable is

Answer

To determine the set of points where the function \( f(x) = x|x| \) is twice differentiable, we will analyze the function step by step.

### Step 1: Define the function
The function can be expressed in piecewise form based on the definition of the absolute value:
\[ f(x) = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases} \]

### Step 2: Find the first derivative
Next, we differentiate \( f(x) \) to find \( f'(x) \):
\[ f'(x) = \begin{cases} 2x & \text{if } x > 0 \\ -2x & \text{if } x < 0 \end{cases} \]
At \( x = 0 \), we need to check the left-hand derivative (LHD) and right-hand derivative (RHD):
- LHD at \( x = 0 \):
\[ \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} -h = 0 \]
- RHD at \( x = 0 \):
\[ \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0 \]

Since both derivatives at \( x = 0 \) are equal, we have:
\[ f'(0) = 0 \]

### Step 3: Find the second derivative
Now, we differentiate \( f'(x) \) to find \( f''(x) \):
\[ f''(x) = \begin{cases} 2 & \text{if } x > 0 \\ -2 & \text{if } x < 0 \end{cases} \]
At \( x = 0 \), we again check the left-hand and right-hand derivatives:
- LHD at \( x = 0 \):
\[ \lim_{h \to 0^-} \frac{f'(0 + h) - f'(0)}{h} = \lim_{h \to 0^-} \frac{-2h - 0}{h} = -2 \]
- RHD at \( x = 0 \):
\[ \lim_{h \to 0^+} \frac{f'(0 + h) - f'(0)}{h} = \lim_{h \to 0^+} \frac{2h - 0}{h} = 2 \]

Since the left-hand derivative and right-hand derivative of \( f'(x) \) at \( x = 0 \) are not equal, \( f''(0) \) does not exist.

### Conclusion
The function \( f(x) = x|x| \) is twice differentiable everywhere except at \( x = 0 \). Therefore, the set of points where \( f(x) \) is twice differentiable is:
\[ (-\infty, 0) \cup (0, \infty) \]