If `f(x) = x^(3)`sgn (x), then


A. f is differentiable at x = 0
B. f is continuous but not differentiable at x = 0
C. `f'(0^(-)) = 1`
D. None of these

Options

A. f is differentiable at x = 0
B. f is continuous but not differentiable at x = 0
C. `f'(0^(-)) = 1`
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) = x^3 \cdot \text{sgn}(x) \) and determine its continuity and differentiability at \( x = 0 \).

### Step 1: Define the function
The signum function, \( \text{sgn}(x) \), is defined as:
- \( \text{sgn}(x) = -1 \) for \( x < 0 \)
- \( \text{sgn}(x) = 0 \) for \( x = 0 \)
- \( \text{sgn}(x) = 1 \) for \( x > 0 \)

Thus, we can express \( f(x) \) in piecewise form:
\[ f(x) = \begin{cases} -x^3 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ x^3 & \text{if } x > 0 \end{cases} \]

### Step 2: Check continuity at \( x = 0 \)
To check if \( f(x) \) is continuous at \( x = 0 \), we need to evaluate:
\[ \lim_{x \to 0^-} f(x) \quad \text{and} \quad \lim_{x \to 0^+} f(x) \]

**Left-hand limit:**
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^3) = -0^3 = 0 \]

**Right-hand limit:**
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^3) = 0^3 = 0 \]

Since both limits equal \( f(0) = 0 \), we conclude that:
\[ \lim_{x \to 0} f(x) = f(0) \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 3: Check differentiability at \( x = 0 \)
To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the left-hand derivative and the right-hand derivative.

**Left-hand derivative:**
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^3 - 0}{h} = \lim_{h \to 0^-} -h^2 = 0 \]

**Right-hand derivative:**
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^3 - 0}{h} = \lim_{h \to 0^+} h^2 = 0 \]

Since both derivatives exist and are equal:
\[ f'(0^-) = f'(0^+) = 0 \]
Thus, \( f(x) \) is differentiable at \( x = 0 \).

### Conclusion
Based on our analysis:
- \( f(x) \) is continuous at \( x = 0 \).
- \( f(x) \) is differentiable at \( x = 0 \).
- Both derivatives at \( x = 0 \) are equal to \( 0 \).

The correct answer is:
**A. f is differentiable at x = 0.**