Given that `f(x) = xg (x)//|x| g(0) = g'(0)=0 and f(x)` is continuous at x=0 then the value of f'(0)

Options

A. does not exist
B. is-1
C. is 1
D. is 0

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x g(x)}{|x|} \) and determine the value of \( f'(0) \) given the conditions \( g(0) = 0 \) and \( g'(0) = 0 \), and that \( f(x) \) is continuous at \( x = 0 \).

### Step-by-Step Solution:

1. **Understanding the function**:
The function \( f(x) \) can be rewritten for \( x > 0 \) and \( x < 0 \):
- For \( x > 0 \): \( f(x) = g(x) \)
- For \( x < 0 \): \( f(x) = -g(x) \)

2. **Finding \( f(0) \)**:
Since \( f(x) \) is continuous at \( x = 0 \), we need to ensure that \( \lim_{x \to 0} f(x) = f(0) \).
- From the definition, we have \( f(0) = \lim_{x \to 0} f(x) \).
- As \( x \) approaches 0 from the right, \( f(x) = g(x) \) and as \( x \) approaches 0 from the left, \( f(x) = -g(x) \).
- Thus, \( f(0) = g(0) = 0 \) (since \( g(0) = 0 \)).

3. **Finding the derivative \( f'(0) \)**:
To find \( f'(0) \), we use the definition of the derivative:
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]
Since \( f(0) = 0 \), this simplifies to:
\[ f'(0) = \lim_{h \to 0} \frac{f(h)}{h} \]

4. **Evaluating the limit**:
- For \( h > 0 \): \( f(h) = g(h) \)
- For \( h < 0 \): \( f(h) = -g(h) \)

Thus, we can evaluate the limit from both sides:
- As \( h \to 0^+ \):
\[ f'(0) = \lim_{h \to 0^+} \frac{g(h)}{h} \]
- As \( h \to 0^- \):
\[ f'(0) = \lim_{h \to 0^-} \frac{-g(h)}{h} = -\lim_{h \to 0^-} \frac{g(h)}{h} \]

5. **Using the derivative of \( g \)**:
Since \( g'(0) = 0 \), we know:
\[ \lim_{h \to 0} \frac{g(h)}{h} = g'(0) = 0 \]

6. **Final result**:
Therefore, both limits give us:
\[ f'(0) = 0 \]

### Conclusion:
The value of \( f'(0) \) is \( 0 \).