If ` f : R to R` , `g : R to R ` be two given functions then h(x) = 2 min {f (x) - g(x), 0} equals

Options

A. ` f(x) + g (x) - |g (x)-f(x)|`
B. ` f(x) + g (x) + |g (x)-f(x)|`
C. ` f(x) - g (x) + |g (x)-f(x)|`
D. ` f(x) - g (x) - |g (x)-f(x)|`

Answer

To solve the problem, we need to analyze the function \( h(x) = 2 \min \{ f(x) - g(x), 0 \} \).

### Step-by-Step Solution:

1. **Understanding the Minimum Function**:
The function \( h(x) \) is defined as \( 2 \) times the minimum of \( f(x) - g(x) \) and \( 0 \). This means we will compare the value of \( f(x) - g(x) \) with \( 0 \).

2. **Case Analysis**:
We will consider two cases based on the relationship between \( f(x) \) and \( g(x) \).

**Case 1**: \( f(x) \geq g(x) \)
- If \( f(x) \geq g(x) \), then \( f(x) - g(x) \geq 0 \).
- Therefore, \( \min \{ f(x) - g(x), 0 \} = 0 \).
- Hence, \( h(x) = 2 \cdot 0 = 0 \).

**Case 2**: \( f(x) < g(x) \)
- If \( f(x) < g(x) \), then \( f(x) - g(x) < 0 \).
- Therefore, \( \min \{ f(x) - g(x), 0 \} = f(x) - g(x) \).
- Hence, \( h(x) = 2(f(x) - g(x)) \).

3. **Combining the Results**:
We can summarize the results from both cases:
- If \( f(x) \geq g(x) \), then \( h(x) = 0 \).
- If \( f(x) < g(x) \), then \( h(x) = 2(f(x) - g(x)) \).

4. **Final Expression**:
Thus, we can express \( h(x) \) as:
\[ h(x) = \begin{cases} 0 & \text{if } f(x) \geq g(x) \\ 2(f(x) - g(x)) & \text{if } f(x) < g(x) \end{cases} \]